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diff --git a/.gitattributes b/.gitattributes new file mode 100644 index 0000000..6833f05 --- /dev/null +++ b/.gitattributes @@ -0,0 +1,3 @@ +* text=auto +*.txt text +*.md text diff --git a/36670-pdf.pdf b/36670-pdf.pdf Binary files differnew file mode 100644 index 0000000..1c85c2a --- /dev/null +++ b/36670-pdf.pdf diff --git a/36670-pdf.zip b/36670-pdf.zip Binary files differnew file mode 100644 index 0000000..9c22cb1 --- /dev/null +++ b/36670-pdf.zip diff --git a/36670-t.zip b/36670-t.zip Binary files differnew file mode 100644 index 0000000..f974ea9 --- /dev/null +++ b/36670-t.zip diff --git a/36670-t/36670-t.tex b/36670-t/36670-t.tex new file mode 100644 index 0000000..2b26f09 --- /dev/null +++ b/36670-t/36670-t.tex @@ -0,0 +1,14166 @@ +% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % +% % +% The Project Gutenberg EBook of The First Steps in Algebra, by % +% G. A. (George Albert) Wentworth % +% % +% This eBook is for the use of anyone anywhere at no cost and with % +% almost no restrictions whatsoever. You may copy it, give it away or % +% re-use it under the terms of the Project Gutenberg License included % +% with this eBook or online at www.gutenberg.net % +% % +% % +% Title: The First Steps in Algebra % +% % +% Author: G. A. (George Albert) Wentworth % +% % +% Release Date: July 9, 2011 [EBook #36670] % +% Most recently updated: June 11, 2021 % +% % +% Language: English % +% % +% Character set encoding: UTF-8 % +% % +% *** START OF THIS PROJECT GUTENBERG EBOOK THE FIRST STEPS IN ALGEBRA ***% +% % +% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % + +\def\ebook{36670} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%% %% +%% Packages and substitutions: %% +%% %% +%% book: Required. %% +%% inputenc: Standard DP encoding. Required. %% +%% %% +%% fix-cm: For larger title page fonts. Optional. %% +%% ifthen: Logical conditionals. Required. %% +%% %% +%% amsmath: AMS mathematics enhancements. Required. %% +%% amssymb: Additional mathematical symbols. Required. %% +%% %% +%% alltt: Fixed-width font environment. Required. %% +%% array: Enhanced tabular features. Required. %% +%% %% +%% indentfirst: Indent first word of each sectional unit. Optional. %% +%% textcase: Apply \MakeUppercase (et al.) only to text, not math. %% +%% Required. %% +%% %% +%% calc: Length calculations. Required. %% +%% soul: Spaced text. Optional. %% +%% %% +%% fancyhdr: Enhanced running headers and footers. Required. %% +%% %% +%% graphicx: Standard interface for graphics inclusion. Required. %% +%% wrapfig: Illustrations surrounded by text. Required. %% +%% %% +%% geometry: Enhanced page layout package. Required. %% +%% hyperref: Hypertext embellishments for pdf output. Required. %% +%% %% +%% %% +%% Producer's Comments: %% +%% %% +%% Changes are noted in this file in two ways. %% +%% 1. \DPtypo{}{} for typographical corrections, showing original %% +%% and replacement text side-by-side. %% +%% 2. [** TN: Note]s for lengthier or stylistic comments. %% +%% %% +%% Compilation Flags: %% +%% %% +%% The following behavior may be controlled by boolean flags. %% +%% %% +%% ForPrinting (false by default): %% +%% Compile a screen-optimized PDF file. Set to true for print- %% +%% optimized file (large text block, two-sided layout, black %% +%% hyperlinks). %% +%% %% +%% Both print and screen layout are relatively loose, and contain %% +%% hard-coded page breaks (\PrintBreak, \ScreenBreak, \newpage). %% +%% %% +%% %% +%% PDF pages: 269 (if ForPrinting set to false) %% +%% PDF page size: 4.75 x 7" (non-standard) %% +%% PDF bookmarks: created, point to ToC entries %% +%% PDF document info: filled in %% +%% Images: 3 pdf diagrams %% +%% %% +%% Summary of log file: %% +%% * Large numbers of visually-harmless over-full hboxes and vboxes %% +%% from DPalign* and DPgather* environments. %% +%% %% +%% %% +%% Compile History: %% +%% %% +%% July, 2011: adhere (Andrew D. Hwang) %% +%% texlive2007, GNU/Linux %% +%% %% +%% Command block: %% +%% %% +%% pdflatex x2 %% +%% %% +%% %% +%% July 2011: pglatex. %% +%% Compile this project with: %% +%% pdflatex 36670-t.tex ..... TWO times %% +%% %% +%% pdfTeXk, Version 3.141592-1.40.3 (Web2C 7.5.6) %% +%% %% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\listfiles +\documentclass[12pt]{book}[2005/09/16] + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%% PACKAGES %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\usepackage[utf8]{inputenc}[2006/05/05] + +\usepackage{ifthen}[2001/05/26] %% Logical conditionals + +\usepackage{amsmath}[2000/07/18] %% Displayed equations +\usepackage{amssymb}[2002/01/22] %% and additional symbols + +\usepackage{alltt}[1997/06/16] %% boilerplate, credits, license + +\usepackage{array}[2005/08/23] %% extended array/tabular features + +\usepackage{multicol} + +\usepackage{graphicx}[1999/02/16]%% For diagrams + +\usepackage{indentfirst}[1995/11/23] +\usepackage{textcase}[2004/10/07] + +\usepackage{calc}[2005/08/06] + +% for running heads +\usepackage{fancyhdr} + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%% Interlude: Set up PRINTING (default) or SCREEN VIEWING %%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + +% ForPrinting=true (default) false +% Asymmetric margins Symmetric margins +% Black hyperlinks Blue hyperlinks +% Start Preface, ToC, etc. recto No blank verso pages +% +% Chapter-like ``Sections'' start both recto and verso in the scanned +% book. This behavior has been retained. +\newboolean{ForPrinting} + +%% UNCOMMENT the next line for a PRINT-OPTIMIZED VERSION of the text %% +%\setboolean{ForPrinting}{true} + +%% Initialize values to ForPrinting=false +\newcommand{\Margins}{hmarginratio=1:1} % Symmetric margins +\newcommand{\HLinkColor}{blue} % Hyperlink color +\newcommand{\PDFPageLayout}{SinglePage} +\newcommand{\TransNote}{Transcriber's Note} +\newcommand{\TransNoteCommon}{% + Minor typographical corrections and presentational changes have + been made without comment. + \bigskip +} + +\newcommand{\TransNoteText}{% + \TransNoteCommon + + This PDF file is optimized for screen viewing, but may easily be + recompiled for printing. 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\Preface calls \pagestyle{fancy} + \frontmatter + \BookMark{-1}{Front Matter.} +} + +\newcommand{\MainMatter}{% + \FlushRunningHeads + \mainmatter + \BookMark{-1}{Main Matter.} +} + +\newcommand{\BackMatter}{% + \FlushRunningHeads + \pagenumbering{Roman} + \backmatter + \BookMark{-1}{Back Matter.} + \BookMark{0}{PG License.} + \SetRunningHeads[License.]{License.} +} + +\newcommand{\Tag}[1]{\tag*{#1}} + +% DPalign, DPgather +\makeatletter +\providecommand\shortintertext\intertext +\newcount\DP@lign@no +\newtoks\DP@lignb@dy +\newif\ifDP@cr +\newif\ifbr@ce +\def\f@@zl@bar{\null} +\def\addto@DPbody#1{\global\DP@lignb@dy\@xp{\the\DP@lignb@dy#1}} +\def\parseb@dy#1{\ifx\f@@zl@bar#1\f@@zl@bar + \addto@DPbody{{}}\let\@next\parseb@dy + \else\ifx\end#1 + \let\@next\process@DPb@dy + \ifDP@cr\else\addto@DPbody{\DPh@@kr&\DP@rint}\@xp\addto@DPbody\@xp{\@xp{\the\DP@lign@no}&}\fi + \addto@DPbody{\end} + \else\ifx\intertext#1 + \def\@next{\eat@command0}% + \else\ifx\shortintertext#1 + \def\@next{\eat@command1}% + \else\ifDP@cr\addto@DPbody{&\DP@lint}\@xp\addto@DPbody\@xp{\@xp{\the\DP@lign@no}&\DPh@@kl} + \DP@crfalse\fi + \ifx\begin#1\def\begin@stack{b} + \let\@next\eat@environment + \else\ifx\lintertext#1 + \let\@next\linter@text + \else\ifx\rintertext#1 + \let\@next\rinter@text + \else\ifx\\#1 + \addto@DPbody{\DPh@@kr&\DP@rint}\@xp\addto@DPbody\@xp{\@xp{\the\DP@lign@no}&\\}\DP@crtrue + \global\advance\DP@lign@no\@ne + \let\@next\parse@cr + \else\check@braces#1!Q!Q!Q!\ifbr@ce\addto@DPbody{{#1}}\else + \addto@DPbody{#1}\fi + \let\@next\parseb@dy + \fi\fi\fi\fi\fi\fi\fi\fi\@next} +\def\process@DPb@dy{\let\lintertext\@gobble\let\rintertext\@gobble + \@xp\start@align\@xp\tw@\@xp\st@rredtrue\@xp\m@ne\the\DP@lignb@dy} +\def\linter@text#1{\@xp\DPlint\@xp{\the\DP@lign@no}{#1}\parseb@dy} +\def\rinter@text#1{\@xp\DPrint\@xp{\the\DP@lign@no}{#1}\parseb@dy} +\def\DPlint#1#2{\@xp\def\csname DP@lint:#1\endcsname{\text{#2}}} +\def\DPrint#1#2{\@xp\def\csname DP@rint:#1\endcsname{\text{#2}}} +\def\DP@lint#1{\ifbalancedlrint\@xp\ifx\csname +DP@lint:#1\endcsname\relax\phantom + {\csname DP@rint:#1\endcsname}\else\csname DP@lint:#1\endcsname\fi + \else\csname DP@lint:#1\endcsname\fi} +\def\DP@rint#1{\ifbalancedlrint\@xp\ifx\csname +DP@rint:#1\endcsname\relax\phantom + {\csname DP@lint:#1\endcsname}\else\csname DP@rint:#1\endcsname\fi + \else\csname DP@rint:#1\endcsname\fi} +\def\eat@command#1#2{\ifcase#1\addto@DPbody{\intertext{#2}}\or + \addto@DPbody{\shortintertext{#2}}\fi\DP@crtrue + \global\advance\DP@lign@no\@ne\parseb@dy} +\def\parse@cr{\new@ifnextchar*{\parse@crst}{\parse@crst{}}} +\def\parse@crst#1{\addto@DPbody{#1}\new@ifnextchar[{\parse@crb}{\parseb@dy}} +\def\parse@crb[#1]{\addto@DPbody{[#1]}\parseb@dy} +\def\check@braces#1#2!Q!Q!Q!{\def\dp@lignt@stm@cro{#2}\ifx + \empty\dp@lignt@stm@cro\br@cefalse\else\br@cetrue\fi} +\def\eat@environment#1{\addto@DPbody{\begin{#1}}\begingroup + \def\@currenvir{#1}\let\@next\digest@env\@next} +\def\digest@env#1\end#2{% + \edef\begin@stack{\push@begins#1\begin\end \@xp\@gobble\begin@stack}% + \ifx\@empty\begin@stack + \@checkend{#2} + \endgroup\let\@next\parseb@dy\fi + \addto@DPbody{#1\end{#2}} + \@next} +\def\lintertext{lint}\def\rintertext{rint} +\newif\ifbalancedlrint +\let\DPh@@kl\empty\let\DPh@@kr\empty +\def\DPg@therl{&\omit\hfil$\displaystyle} +\def\DPg@therr{$\hfil} + +\newenvironment{DPalign*}[1][a]{% + \setlength{\abovedisplayskip}{8pt plus 4pt minus 6pt} + \setlength{\belowdisplayskip}{8pt plus 4pt minus 6pt} + \if m#1\balancedlrintfalse\else\balancedlrinttrue\fi + \global\DP@lign@no\z@\DP@crfalse + \DP@lignb@dy{&\DP@lint0&}\parseb@dy +}{% + \endalign +} +\newenvironment{DPgather*}[1][a]{% + \setlength{\abovedisplayskip}{8pt plus 4pt minus 6pt} + \setlength{\belowdisplayskip}{8pt plus 4pt minus 6pt} + \if m#1\balancedlrintfalse\else\balancedlrinttrue\fi + \global\DP@lign@no\z@\DP@crfalse + \let\DPh@@kl\DPg@therl + \let\DPh@@kr\DPg@therr + \DP@lignb@dy{&\DP@lint0&\DPh@@kl}\parseb@dy +}{% + \endalign +} +\makeatother +%%%%%%%%%%%%%%%%%%%%%%%% START OF DOCUMENT %%%%%%%%%%%%%%%%%%%%%%%%%% + +\begin{document} +%%%% PG BOILERPLATE %%%% +\PGBoilerPlate + +\begin{center} +\begin{minipage}{\textwidth} +\small +\begin{PGtext} +The Project Gutenberg EBook of The First Steps in Algebra, by +G. A. (George Albert) Wentworth + +This eBook is for the use of anyone anywhere at no cost and with +almost no restrictions whatsoever. You may copy it, give it away or +re-use it under the terms of the Project Gutenberg License included +with this eBook or online at www.gutenberg.net + + +Title: The First Steps in Algebra + +Author: G. A. (George Albert) Wentworth + +Release Date: July 9, 2011 [EBook #36670] +Most recently updated: June 11, 2021 + +Language: English + +Character set encoding: UTF-8 + +*** START OF THIS PROJECT GUTENBERG EBOOK THE FIRST STEPS IN ALGEBRA *** +\end{PGtext} +\end{minipage} +\end{center} +\clearpage +%%%% Credits and transcriber's note %%%% +\begin{center} +\begin{minipage}{\textwidth} +\begin{PGtext} +Produced by Andrew D. Hwang, Peter Vachuska, Chuck Greif +and the Online Distributed Proofreading Team at +http://www.pgdp.net. +\end{PGtext} +\end{minipage} +\end{center} +\vfill + +\begin{minipage}{0.85\textwidth} +\small +\BookMark{0}{Transcriber's Note} +\subsection*{\centering\normalfont\scshape +\normalsize\MakeLowercase{\TransNote}} + +\raggedright +\TransNoteText +\end{minipage} +%%%%%%%%%%%%%%%%%%%%%%%%%%% FRONT MATTER %%%%%%%%%%%%%%%%%%%%%%%%%% +\FrontMatter +%% -----File: 001.png---Folio i------- +\begin{center} +\Large THE +\vfill + +\textbf{\LARGE FIRST STEPS IN ALGEBRA.} +\vfill + +\normalsize BY \\[12pt] +\Large G. A. WENTWORTH, A.M. \\[12pt] +\footnotesize AUTHOR OF A SERIES OF TEXT-BOOKS IN MATHEMATICS. +\vfill\vfill + +\large BOSTON, U.S.A.: \\ +PUBLISHED BY GINN \& COMPANY. \\ +1904. +\end{center} +%% -----File: 002.png---Folio ii------- +\newpage +\null\vfill +\begin{center} +\footnotesize +Entered according to Act of Congress, in the year 1894, by \\[4pt] +G. A. WENTWORTH, \\[4pt] +in the Office of the Librarian of Congress, at Washington. \\[4pt] +\tb +\textsc{All Rights Reserved.} +\vfill + +\textsc{Typography by J. S. Cushing \& Co., Boston, U.S.A.} \\ +\tb[2.5in] +\textsc{Presswork by Ginn \& Co., Boston, U.S.A.} +\end{center} +%% -----File: 003.png---Folio iii------- + + +\Preface + +\First{This} book is written for pupils in the upper grades of +grammar schools and the lower grades of high schools. +The introduction of the simple elements of Algebra into +these grades will, it is thought, so stimulate the mental +activity of the pupils, that they will make considerable +progress in Algebra without detriment to their progress +in Arithmetic, even if no more time is allowed for the +two studies than is usually given to Arithmetic alone. + +The great danger in preparing an Algebra for very +young pupils is that the author, in endeavoring to smooth +the path of the learner, will sacrifice much of the educational +value of the study. To avoid this real and serious +danger, and at the same time to gain the required simplicity, +great care has been given to the explanations of +the fundamental operations and rules, the arrangement +of topics, the model solutions of examples, and the making +of easy examples for the pupils to solve. + +Nearly all the examples throughout the book are new, +and made expressly for beginners. + +The first chapter clears the way for quite a full treatment +of simple integral equations with one unknown number. +In the first two chapters only \emph{positive} numbers are +%% -----File: 004.png---Folio iv------- +involved, and the learner is led to see the practical advantages +of Algebra in its most interesting applications before +he faces the difficulties of negative numbers. + +The third chapter contains a simple explanation of negative +numbers. The recognition of the facts that the real +nature of subtraction is counting backwards, and that the +real nature of multiplication is forming the product from +the multiplicand precisely as the multiplier is formed from +unity, makes an easy road to the laws of addition and subtraction +of algebraic numbers, and to the law of signs in +multiplication and division. All the principles and rules +of this chapter are illustrated and enforced by numerous +examples involving \emph{simple} algebraic expressions only. + +The ordinary processes with \emph{compound} expressions, including +simple cases of resolution into factors, and the +treatment of fractions, naturally follow the third chapter. +The immediate succession of topics that require similar +work is of the highest importance to the beginner, and it +is hoped that the half-dozen chapters on algebraic expressions +will prove interesting, and give sufficient readiness +in the use of symbols. + +A chapter on fractional equations with one unknown +number, a chapter on simultaneous equations with two +unknown numbers, and a chapter on quadratics follow in +order. Only one method of elimination is given in simultaneous +equations and one method of completing the +square in quadratics. Moreover, the solution of the examples +in quadratics requires the square roots of only small +numbers such as every pupil knows who has learned the +%% -----File: 005.png---Folio v------- +multiplication table. In each of these three chapters a +considerable number of problems is given to \emph{state} and solve. +By this means the learner is led to exercise his reasoning +faculty, and to realize that the methods of Algebra require +a strictly logical process. These problems, however, are +divided into classes, and a model solution of an example +of each class is given as a guide to the solution of other +examples of that class. + +The course may end with the chapter on quadratics, but +the simple questions of arithmetical progression and of +geometrical progression are so interesting in themselves, +and show so clearly the power of Algebra, that it will +be a great loss not to take the short chapters on these +series. + +The last chapter is on square and cube roots. It is +expected that pupils who use this book will learn how to +extract the square and cube roots by the simple formulas +of Algebra, and be spared the necessity of committing to +memory the long and tedious rules given in Arithmetic, +rules that are generally forgotten in less time than they +are learned. + +Any corrections or suggestions will be thankfully received +by the author. + +A teachers' edition is in press, containing solutions of +examples, and such suggestions as experience with beginners +has shown to be valuable. + +\Signature{G\Add{.} A. WENTWORTH.} +{\textsc{Exeter}, NH, April, 1894} +%% -----File: 006.png---Folio vi------- +\TableofContents +\iffalse +CONTENTS. + +Chapter Page + +I. Introduction.............. 1 + +II. Simple Equations............. 19 + +III. Positive and Negative Numbers....... 33 + +IV. Addition and Subtraction......... 46 + +V. Multiplication and Division........ 53 + +VI. Special Rules in Multiplication and Division . 64 + +VII. Factors............... 71 + +VIII. Common Factors and Multiples....... 84 + +IX. Fractions................ 89 + +X. Fractional Equations........... 103 + +XI. Simultaneous Equations of the First Degree . 122 + +XII. Quadratic Equations........... 132 + +XIII. Arithmetical Progression......... 142 + +XIV. Geometrical Progression.......... 148 + +XV. Square and Cube Roots.......... 152 + + Answers................ 165 +\fi +%% -----File: 007.png---Folio 1------- +\MainMatter +% FIRST STEPS IN ALGEBRA. +% [** TN: Chapter macro prints preceding line] + +\Chapter{I.}{Introduction.} + +\begin{Remark}[\First{Note}\Add{.}] +The principal definitions are put at the beginning of the +book for convenient reference. They are not to be committed to +memory. It is a good plan to have definitions and explanations +read aloud in the class, and to encourage pupils to make comments +upon them, and ask questions about them. +\end{Remark} + +\Paragraph{1. Algebra.} Algebra, like Arithmetic, treats of numbers. + +\Paragraph{2. Units.} In counting separate objects or in measuring +magnitudes, the \emph{standards} by which we count or measure +are called \Defn{units}. + +\begin{Remark} +Thus, in counting the boys in a school, the unit is a boy; in selling +eggs by the dozen, the unit is a dozen eggs; in selling bricks by +the thousand, the unit is a thousand bricks; in measuring short distances, +the unit is an inch, a foot, or a yard; in measuring long +distances, the unit is a rod or a mile. +\end{Remark} + +\Paragraph{3. Numbers.} \emph{Repetitions of the unit} are expressed by +numbers. + +\Paragraph{4. Quantities.} A number of specified units of any kind +is called a quantity; as, $4$~pounds, $5$~oranges. + +\Paragraph{5. Number-Symbols in Arithmetic.} Arithmetic employs +the arbitrary symbols, $1$,~$2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$,~$0$, called +\Defn{figures}, to represent numbers. +%% -----File: 008.png---Folio 2------- + +\Paragraph{6. Number-Symbols in Algebra.} Algebra employs \emph{the +letters of the alphabet} in addition to the figures of Arithmetic +to represent numbers. Letters are used as \emph{general} +symbols of numbers to which \emph{any particular values} may be +assigned. + + +\Section{PRINCIPAL SIGNS OF OPERATIONS.} + +\Paragraph{7.} The signs of the fundamental operations are the same +in Algebra as in Arithmetic. + +\Paragraph{8. The Sign of Addition,~$+$.} The sign~$+$ is read \emph{plus}. + +\begin{Remark} +Thus, $4 + 3$, read $4$~plus~$3$, indicates that the number~$3$ is to be +added to the number~$4$, $a + b$, read $a$~plus~$b$, indicates that the number~$b$ +is to be added to the number~$a$. +\end{Remark} + +\Paragraph{9. The Sign of Subtraction,~$-$\Add{.}} The sign~$-$ is read \emph{minus}. + +\begin{Remark} +Thus, $4 - 3$, read $4$~minus~$3$, indicates that the number~$3$ is to be +subtracted from the number~$4$, $a - b$, read $a$~minus~$b$, indicates that +the number~$b$ is to be subtracted from the number~$a$. +\end{Remark} + +\Paragraph{10. The Sign of Multiplication,~$×$\Add{.}} The sign~$×$ is read +\emph{times}. + +\begin{Remark} +Thus, $4 × 3$, read $4$~times~$3$, indicates that the number~$3$ is to be +multiplied by~$4$, $a × b$, read $a$~times~$b$, indicates that the number~$b$ +is to be multiplied by the number~$a$. +\end{Remark} + +A dot is sometimes used for the sign of multiplication. +Thus $2 · 3 · 4 · 5$ means the same as $2 × 3 × 4 × 5$. Either +sign is read \emph{multiplied by} when followed by the multiplier. +\$$a × b$, or \$$a · b$, is read $a$~dollars multiplied by~$b$. + +\Paragraph{11. The Sign of Division,~$÷$.} The sign~$÷$ is read \emph{divided by}. + +\begin{Remark} +Thus, $4 ÷ 2$, read $4$~divided by~$2$, indicates that the number~$4$ is +to be divided by~$2$, $a ÷ b$, read $a$~divided by~$b$, indicates that the +number~$a$ is to be divided by the number~$b$. +\end{Remark} +%% -----File: 009.png---Folio 3------- + +Division is also indicated by writing the dividend above +the divisor with a horizontal line between them. + +\begin{Remark} +Thus, $\dfrac{4}{2}$ means the same as~$4 ÷ 2$; $\dfrac{a}{b}$~means the same as~$a ÷ b$. +\end{Remark} + + +\Section{OTHER SIGNS USED IN ALGEBRA.} + +\Paragraph{12. The Sign of Equality,~$=$.} The sign~$=$ is read \emph{is equal +to}, when placed between two numbers and indicates that +these two numbers are equal. + +\begin{Remark} +Thus, $8 + 4 = 12$ means that $8 + 4$ and~$12$ stand for \emph{equal} numbers; +$x + y = 20$ means that $x + y$ and~$20$ stand for equal numbers. +\end{Remark} + +\Paragraph{13. The Sign of Inequality, $>$~or~$<$.} The sign $>$~or~$<$ is +read \emph{is greater than} and \emph{is less than} respectively, and when +placed between two numbers indicates that these two numbers +are unequal, and that the number toward which the +sign opens is the greater. + +\begin{Remark} +Thus, $9 + 6 > 12$ means that $9 + 6$ is greater than~$12$; and +$9 + 6 < 16$ means that $9 + 6$ is less than~$16$. +\end{Remark} + +%[** TN: [sic] no punctuation after \therefore] +\Paragraph{14. The Sign of Deduction,~$\therefore$}\quad The sign~$\therefore$ is read \emph{hence} +or \emph{therefore}. + +\Paragraph{15. The Sign of Continuation,~$\dots$.} The sign~$\dots$ is read +\emph{and so on}. + +\Paragraph{16. The Signs of Aggregation.} The signs of aggregation +are the bar~$|$, the vinculum~$\overline{\phantom{a+x}}$, the parenthesis~$(\ )$, the +bracket~$[\ ]$, and the brace~$\{\ \}$. + +\begin{Remark} +Thus, each of the expressions $\VSum{a}{b}$, $\Vinc{a + b}$, $(a + b)$, $[a + b]$, $\{a + b\}$, +signifies that $a + b$ is to be treated as a single number. +\end{Remark} +%% -----File: 010.png---Folio 4------- + + +\Section{FACTORS. COEFFICIENTS. POWERS.} + +\Paragraph{17. Factors.} When a number consists of the product of +two or more numbers, each of these numbers is called a +\Defn{factor} of the product. + +The sign~$×$ is generally omitted between a figure and a +letter, or between letters; thus, instead of $63 × a × b$, we +write~$63ab$; instead of $a × b × c$, we write~$abc$. + +The expression~$abc$ must not be confounded with $a + b + c$. +$abc$~is a product; $a + b + c$ is a sum. +\begin{DPalign*} +\lintertext{\indent If} +a = 2,\quad b &= 3,\quad c = 4, \\ +\lintertext{then} +abc &= 2 × 3 × 4 = 24; \\ +\lintertext{but} +a + b + c &= 2 + 3 + 4 = 9. +\end{DPalign*} + +\begin{Remark}[Note.] +When a sign of operation is omitted in the notation of +Arithmetic, it is always the \emph{sign of addition}; but when a sign of +operation is omitted in the notation of Algebra, it is always the +\emph{sign of multiplication}. Thus, $456$~means $400 + 50 + 6$, but $4ab$ +means $4 × a × b$. +\end{Remark} + +\Paragraph{18.} Factors expressed by letters are called \Defn{literal} factors; +factors expressed by figures are called \Defn{numerical} factors. + +\Paragraph{19.} If one factor of a product is equal to~$0$, the product +is equal to~$0$, whatever the values of the other factors. +Such a factor is called a \Defn{zero factor}. + +\Paragraph{20. Coefficients.} A known factor of a product which is +prefixed to another factor, to show the number of times that +factor is taken, is called a \Defn{coefficient}. + +\begin{Remark} +Thus, in~$7c$, $7$~is the coefficient of~$c$; in~$7ax$, $7$~is the +coefficient of~$ax$, +or, if $a$~is known, $7a$~is the coefficient of~$x$. +\end{Remark} +%% -----File: 011.png---Folio 5------- + +By coefficient, we generally mean the \textbf{numerical coefficient +with its sign}. If no numerical coefficient is written, $1$~is +understood. Thus, $ax$~means the same as~$1ax$. + +\Paragraph{21. Powers and Roots.} A product consisting of two or +more \textbf{equal factors} is called a \Defn{power} of that factor, and one +of the equal factors is called a \Defn{root} of the number. + +\begin{Remark} +Thus, $9 = 3 × 3$; that is, $9$~is a power of~$3$, and $3$~is a root of~$9$. +\end{Remark} + +\Paragraph{22. Indices or Exponents.} An index or exponent is a +number-symbol written at the right of, and a little above, +a number. + +If the index is a \emph{whole number}, it shows the number +of times the given number is taken as a factor. + +\begin{Remark} +Thus, $a^{1}$, or simply~$a$, denotes that $a$~is taken \emph{once} as a +factor; $a^{2}$~denotes +that $a$~is taken \emph{twice} as a factor; $a^{3}$~denotes that $a$~is taken +\emph{three times} as a factor; and $a^{4}$~denotes that $a$~is taken \emph{four times} as a +factor; and so on. These are read: the first power of~$a$; the second +power of~$a$; the third power of~$a$; the fourth power of~$a$; and so on. + +$a^{3}$~is written instead of~$aaa$. + +$a^{4}$~is written instead of~$aaaa$. +\end{Remark} + +\Paragraph{23.} The meaning of coefficient and exponent must be +carefully distinguished. Thus, +\begin{DPalign*} +4a &= a + a + a + a; \\ +a^{4} &= a× a× a× a. \displaybreak[1] \\ +\lintertext{\indent If $a = 3$,} +4a &= 3 + 3 + 3 + 3 = 12. \\ +a^{4} &= 3 × 3 × 3 × 3 = 81. +\end{DPalign*} + +\begin{Remark} +The second power of a number is generally called the \emph{square} of +that number; thus, $a^{2}$~is called the \emph{square} of~$a$, because if $a$~denotes +the number of units of length in the side of a square, $a^{2}$~denotes the +number of units of surface in the square. The third power of a number +is generally called the \emph{cube} of that number; thus, $a^{3}$~is called the +\emph{cube} of~$a$, because if $a$~denotes the number of units of length in the +edge of a cube, $a^{3}$~denotes the number of units of volume in the +cube. +\end{Remark} +%% -----File: 012.png---Folio 6------- + + +\Section{ALGEBRAIC EXPRESSIONS.} + +\Paragraph{24. An Algebraic Expression.} An algebraic expression is +a number written with algebraic symbols. An algebraic +expression may consist of one symbol, or of several symbols +connected by signs. + +\begin{Remark} +Thus, $a$, $3abc$, $5a + 2b - 3c$, are algebraic expressions. +\end{Remark} + +\Paragraph{25. Terms.} A \Defn{term} is an algebraic expression, the parts +of which are not separated by the sign $+$~or~$-$. + +\begin{Remark} +Thus, $a$, $5xy$, $2ab × 4cd$, $\dfrac{3ab}{4cd}$ are algebraic expressions of one +term each. A term may be separated into parts by the sign $×$~or~$÷$. +\end{Remark} + +\Paragraph{26. Simple Expressions.} An algebraic expression of \emph{one +term} is called a \Defn{simple expression} or \Defn{monomial}. + +\begin{Remark} +Thus, $5xy$, $7a × 2b$, $7a ÷ 2b$, are simple expressions. +\end{Remark} + +\Paragraph{27. Compound Expressions.} An algebraic expression of +\emph{two or more terms} is called a \Defn{compound expression} or \Defn{polynomial}. + +\begin{Remark} +Thus, $5xy + 7a$, $2x - y - 3z$, $4a - 3b + 2c - 3d$ are compound +expressions. +\end{Remark} + +\Paragraph{28.} A polynomial of two terms is called a \Defn{binomial}; of +three terms, a \Defn{trinomial}. + +\begin{Remark} +Thus, $3a - b$ is a binomial; and $3a - b + c$ is a trinomial. +\end{Remark} + +\Paragraph{29. Positive and Negative Terms.} The terms of a compound +expression preceded by the sign~$+$ are called \Defn{positive +terms}, and the terms preceded by the sign~$-$ are called +\Defn{negative terms}. The sign~$+$ before the first term is omitted. + +\Paragraph{30.} A positive and a negative term of the same numerical +value cancel each other when combined. +%% -----File: 013.png---Folio 7------- + +\Paragraph{31. Like Terms.} Terms which have the same combination +of \emph{letters} are called \Defn{like} or \Defn{similar} terms; terms which +do not have the same combination of letters are called +\Defn{unlike} or \Defn{dissimilar} terms. + +\begin{Remark} +Thus, $5a^{2}bc$, $-7a^{2}bc$, $a^{2}bc$, are like terms; but $5a^{2}bc$, $5ab^{2}c$, +$5abc^{2}$, are unlike terms. +\end{Remark} + +\Paragraph{32. Degree of a Term.} A term that is the product of +three letters is said to be of the \emph{third degree}; a term of +four letters is of the \emph{fourth degree}; and so on. + +\begin{Remark} +Thus, $5abc$~is of the third degree; $2a^{2}b^{2}c^{2}$, that is, $2aabbcc$, is of +the sixth degree. +\end{Remark} + +\Paragraph{33. Degree of a Compound Expression.} The degree of a +compound expression is the degree of that term of the +expression which is of the \emph{highest degree}. + +\begin{Remark} +Thus, $a^{2}x^{2} + bx + c$ is of the fourth degree, since $a^{2}x^{2}$~is of the +fourth degree. +\end{Remark} + +\Paragraph{34. Dominant Letter.} It often happens that there is one +letter in an expression of more importance than the rest, +and this is, therefore, called the \Defn{dominant letter}. In such +cases the degree of the expression is generally called by +the degree of the \emph{dominant letter}. + +\begin{Remark} +Thus, $a^{2}x^{2} + bx + c$ is of the \emph{second degree in~$x$}. +\end{Remark} + +\Paragraph{35. Arrangement of a Compound Expression.} A compound +expression is said to be \emph{arranged} according to the powers +of some letter when the exponents of that letter, reckoning +from left to right, either descend or ascend in \emph{the order of +magnitude}. + +\begin{Remark} +Thus, $3ax^{3} - 4bx^{2} - 6ax + 8b$ is arranged according to the descending +powers of~$x$, and $8b - 6ax - 4bx^{2} + 3ax^{3}$ is arranged +according to the ascending powers of~$x$. +\end{Remark} +%% -----File: 014.png---Folio 8------- + + +\Section{PARENTHESES.} + +\Paragraph{36.} If a compound expression is to be treated as a whole, +it is enclosed in a parenthesis. + +\begin{Remark} +Thus, $2 × (10 + 5)$ means that we are to add $5$~to~$10$ and multiply +the result by~$2$; if we were to omit the parenthesis and write +$2 × 10 + 5$, the meaning would be that we were to multiply $10$~by~$2$ +and add~$5$ to the result. +\end{Remark} + +Like the parenthesis, we use with the same meaning any +other sign of aggregation. + +\begin{Remark} +Thus, $(5 + 2)$, $[5 + 2]$, $\{5 + 2\}$, $\Vinc{5 + 2}$, $\VSum{5}{2}$, all mean that the +expression $5 + 2$ is to be treated as the single symbol~$7$. +\end{Remark} + +\Paragraph{37. Parentheses preceded by~$+$.} If a man has $10$~dollars +and afterwards collects $3$~dollars and then $2$~dollars, it +makes no difference whether he adds the $3$~dollars to his +$10$~dollars, and then the $2$~dollars, or puts the $3$~and~$2$ +dollars together and adds their sum to his $10$~dollars. + +The first process is represented by $10 + 3 + 2$. + +The second process is represented by $10 + (3 + 2)$. + +\begin{DPgather*} +\lintertext{\indent Hence,} +10 + (3 + 2) = 10 + 3 + 2. +\Tag{(1)} +\end{DPgather*} + +If a man has $10$~dollars and afterwards collects $3$~dollars +and then pays a bill of $2$~dollars, it makes no difference +whether he adds the $3$~dollars collected to his $10$~dollars +and pays out of this sum his bill of $2$~dollars, or pays the +$2$~dollars from the $3$~dollars collected and adds the remainder +to his $10$~dollars. + +The first process is represented by $10 + 3 - 2$. + +The second process is represented by $10 + (3 - 2)$. + +\begin{DPgather*} +\lintertext{\indent Hence,} +10 + (3 - 2) = 10 + 3 - 2. +\Tag{(2)} +\end{DPgather*} +%% -----File: 015.png---Folio 9------- + +From (1)~and~(2) it follows that + +\begin{Theorem} +If an expression within a parenthesis is preceded by the +sign~$+$, the parenthesis can be removed without making any +change in the signs of the expression. +\end{Theorem} + +\begin{Theorem}[\textsc{Conversely.}] Any part of an expression can lie enclosed +within a parenthesis and the sign~$+$ prefixed, without making +any change in the signs of the terms thus enclosed. +\end{Theorem} + +\Paragraph{38. Parentheses preceded by~$-$.} If a man has $10$~dollars +and has to pay two bills, one of $3$~dollars and one of $2$~dollars, +it makes no difference whether he takes $3$~dollars +and $2$~dollars in succession, or takes the $3$~and~$2$ dollars at +one time, from his $10$~dollars. + +The first process is represented by $10 - 3 - 2$. + +The second process is represented by $10 - (3 + 2)$. + +\begin{DPgather*} +\lintertext{\indent Hence,} +10 - (3 + 2) = 10 - 3 - 2. +\Tag{(3)} +\end{DPgather*} + +If a man has $10$~dollars consisting of $2$~five-dollar bills, +and has a debt of $3$~dollars to pay, he can pay his debt by +giving a five-dollar bill and receiving $2$~dollars. + +This process is represented by $10 - 5 + 2$. + +Since the debt paid is $3$~dollars, that is, $(5 - 2)$~dollars, +the number of dollars he has left can evidently be +expressed by +\begin{DPalign*} +10 &- (5 - 2). \\ +\lintertext{\indent Hence,} +10 &- (5 - 2) = 10 - 5 + 2. +\Tag{(4)} +\end{DPalign*} + +From (3)~and~(4) it follows that + +\begin{Theorem} +If an expression within a parenthesis is preceded by the +sign~$-$, the parenthesis can be removed, provided the sign +before each term within the parenthesis is changed, the +sign~$+$ to~$-$, and the sign~$-$ to~$+$. +\end{Theorem} +%% -----File: 016.png---Folio 10------- + +\begin{Theorem}[\textsc{Conversely.}] Any part of an expression can be enclosed +within a parenthesis and the sign~$-$ prefixed, provided the +sign of each term enclosed is changed, the sign~$+$ to~$-$, and +the sign~$-$ to~$+$. +\end{Theorem} + + +\Exercise{1.} + +Remove the parentheses, and combine: +\begin{multicols}{2} +\Item{1.} $9 + (3 + 2)$. + +\Item{2.} $9 + (3 - 2)$. + +\Item{3.} $7 + (5 + 1)$. + +\Item{4.} $7 + (5 - 1)$. + +\Item{5.} $6 + (4 + 3)$. + +\Item{6.} $6 + (4 - 3)$. + +\Item{7.} $3 + (8 - 2)$. + +\Item{8.} $9 - (8 - 6)$. + +\Item{9.} $10 - (9 - 5)$. + +\Item{10.} $9 - (6 + 1)$. + +\Item{11.} $8 - (3 + 2)$. + +\Item{12.} $7 - (3 - 2)$. + +\Item{13.} $9 - (4 + 3)$. + +\Item{14.} $9 - (4 - 3)$. + +\Item{15.} $7 - (5 - 2)$. + +\Item{16.} $7 - (7 - 3)$. + +\Item{17.} $(8 - 6) - 1$. + +\Item{18.} $(3 - 2) - (1 - 1)$. + +\Item{19.} $(7 - 3) - (3 - 2)$. + +\Item{20.} $(8 - 2) - (5 - 3)$. + +\Item{21.} $15 - (10 - 3 - 2)$. +\end{multicols} + +\Paragraph{39. Multiplying a Compound Expression.} The expression +$4(5 + 3)$ means that we are to take the sum of the numbers +$5$~and~$3$ four times. The process can be represented by +placing five dots in a line, and a little to the right three +more dots in the same line, and then placing a second, +third, and fourth line of dots underneath the first line and +exactly similar to it. +\[ +\begin{array}{*{9}{>{\ }r}} +\DOT & \DOT & \DOT & \DOT & \DOT & \quad & \DOT & \DOT & \DOT \\[-4pt] +\DOT & \DOT & \DOT & \DOT & \DOT & & \DOT & \DOT & \DOT \\[-4pt] +\DOT & \DOT & \DOT & \DOT & \DOT & & \DOT & \DOT & \DOT \\[-4pt] +\DOT & \DOT & \DOT & \DOT & \DOT & & \DOT & \DOT & \DOT \\[-4pt] +\end{array} +\] + +There are $(5 + 3)$ dots in each line, and $4$~lines. The +total number of dots, therefore, is $4 × (5 + 3)$. + +We see that in the left-hand group there are $4 × 5$ dots, +and in the right-hand group $4 × 3$ dots. The sum of these +%% -----File: 017.png---Folio 11------- +two numbers $(4 × 5) + (4 × 3)$ must be equal to the total +number; that is, +\begin{align*} +4(5 + 3) &= (4 × 5) + (4 × 3) \\ + &= 20 + 12. +\end{align*} + +Again, the expression $4(8 - 3)$ means that we are to +take the difference of the numbers $8$~and~$3$ four times. +The process can be represented by placing eight dots in a +line and crossing the last three, and then placing a second, +third, and fourth line of dots underneath the first line and +exactly similar to it. +\[ +%[** TN: Added gap between dot groups.] +\begin{array}{*{9}{>{\ }r}} +\DOT & \DOT & \DOT & \DOT & \DOT & \quad & \NOT & \NOT & \NOT \\[-4pt] +\DOT & \DOT & \DOT & \DOT & \DOT & & \NOT & \NOT & \NOT \\[-4pt] +\DOT & \DOT & \DOT & \DOT & \DOT & & \NOT & \NOT & \NOT \\[-4pt] +\DOT & \DOT & \DOT & \DOT & \DOT & & \NOT & \NOT & \NOT \\[-4pt] +\end{array} +\] + +The whole number of dots not crossed in each line is +evidently $(8 - 3)$, and the whole number of lines is~$4$. +Therefore the total number of dots not crossed is +\[ +4 × (8 - 3). +\] + +The total number of dots (crossed and not crossed) is +$(4 × 8)$, and the total number of dots crossed is~$(4 × 3)$. +Therefore the total number of dots not crossed is +\begin{DPalign*} +(4 × 8) &- (4 × 3); \\ +\lintertext{that is,} +4(8 - 3) &= (4 × 8) - (4 × 3) \\ + &= 32 - 12. \displaybreak[1] \\ +\intertext{\indent If $a$, $b$, and~$c$ stand for any three numbers, we have} +a (b + c) &= ab + ac, \\ +\lintertext{and} +a(b - c) &= ab - ac. +\EqText{Therefore,} +\end{DPalign*} + +\Dictum{To multiply a compound expression by a simple one}, +\begin{Theorem} +Multiply each term by the multiplier, and write the successive +products with the same signs as those of the original +terms. +\end{Theorem} +%% -----File: 018.png---Folio 12------- + +\Exercise{2.} + +Multiply and remove parentheses: +\begin{multicols}{3} +\Item{1.} $7(8 + 5)$. + +\Item{2.} $7(8 - 5)$. + +\Item{3.} $6(7 + 3)$. + +\Item{4.} $6(7 - 3)$. + +\Item{5.} $8(7 + 5)$. + +\Item{6.} $8(7 - 5)$. + +\Item{7.} $9(6 - 2)$. + +\Item{8.} $4(a + b)$. + +\Item{9.} $4(a - b)$. + +\Item{10.} $2(a^{2} + b^{2})$. + +\Item{11.} $2(a^{2} - b^{2})$. + +\Item{12.} $3(ab + c)$. + +\Item{13.} $3(ab - c)$. + +\Item{14.} $3(c - ab)$. + +\Item{15.} $a(b + c)$. + +\Item{16.} $a(b - c)$. + +\Item{17.} $3a(b + c)$. + +\Item{18.} $3a(b - c)$. + +\Item{19.} $5a(b^{2} + c)$. + +\Item{20.} $5a(b^{2} - c^{2})$. + +\Item{21.} $5a^{2}(b^{2} - c)$. +\end{multicols} + +\Paragraph{40.} The numerical value of an algebraic expression is the +number obtained by putting for the letters involved the +numbers for which these letters stand, and then performing +the operations required by the signs. + +\Item{1.} If $b = 4$, find the value of~$3b^{2}$. +\begin{Soln} +\begin{DPgather*} +\lintertext{\indent Here} +3b^{2} = 3 × 4^{2} = 3 × 16 = 48. +\end{DPgather*} +\end{Soln} + +\Item{2.} If $a = 7$, $b = 2$, $c = 3$, find the value of~$5ab^{2}c^{3}$. +\begin{Soln} +\begin{DPgather*} +\lintertext{\indent Here} +5ab^{2}c^{3} = 5 × 7 × 2^{2} × 3^{3} = 3780. +\end{DPgather*} +\end{Soln} + +\Exercise{3.} + +If $a = 7$, $b = 5$, $c = 3$, find the value of +\begin{multicols}{3} +\Item{1.} $9a$. + +\Item{2.} $8ab$. + +\Item{3.} $4b^{2}c$. + +\Item{4.} $2a^{2}$. + +\Item{5.} $3c^{3}$. + +\Item{6.} $2b^{4}$. + +\Item{7.} $5ac$. + +\Item{8.} $abc$. + +\Item{9.} $abc^{2}$. + +\Item{10.} $\frac{1}{3}abc$. + +\Item{11.} $\frac{1}{5}ab^{2}c$. + +\Item{12.} $\frac{1}{7}a^{2}bc$. +\end{multicols} + +If $a = 5$, $b = 2$, $c = 0$, $x = 1$, $y = 3$, find the value of +\begin{multicols}{3} +\Item{13.} $4acy^{2}$. + +\Item{14.} $3ax^{5}y^{2}$. + +\Item{15.} $2ab^{2}y$. + +\Item{16.} $2a^{2}b^{2}c^{2}y^{2}$. + +\Item{17.} $2a^{2}b^{2}x^{2}y^{2}$. + +\Item{18.} $2abx^{3}y^{3}$. + +\Item{19.} $3abcxy$. + +\Item{20.} $3abx^{3}y^{2}$. + +\Item{21.} $3ab^{2}xy^{2}$. +\end{multicols} +%% -----File: 019.png---Folio 13------- + +\Paragraph{41. The Numerical Value of a Compound Expression.} + +If $a$~stands for~$10$, $b$~for~$4$, and $c$~for~$3$, find the value of +the expression $5ab - 10c^{2} - 5b^{2}$. + +Find the value of each term, and combine the results. +\begin{Soln} +\begin{gather*} +\begin{alignedat}{3} + 5ab &\text{ stands for } &5 × 10 &× 4 &&= 200; \\ +10c^{2} &\text{ stands for } & 10 &× 3^{2} &&= \Z90; \\ +5b^{2} &\text{ stands for } & 5 &× 4^{2} &&= \Z80. +\end{alignedat} \\ +\begin{aligned} +&\therefore 5ab - 10c^{2} - 5b^{2} \\ +&= 200 - 90 - 80 \\ +&= 30. +\end{aligned} +\end{gather*} +\end{Soln} + +\Paragraph{42.} In finding the value of a compound expression the +operations indicated \emph{for each term} must be performed \emph{before} +the operation indicated by the sign prefixed to the term. + +When there is no sign expressed between single symbols +or between \emph{simple} and \emph{compound expressions}, it must be +remembered that the sign understood is the \emph{sign of multiplication}. +Thus $2(a - b)$ has the same meaning as $2 × (a - b)$. + +\Exercise{4.} + +If $a = 5$, $b = 4$, $c = 3$, find the value of +\begin{multicols}{2} +\Item{1.} $9a - 2bc$. + +\Item{2.} $ab + 2c$. + +\Item{3.} $abc + bc$. + +\Item{4.} $5ac + 2a$. + +\Item{5.} $2abc - 2ac^{2}$. + +\Item{6.} $ab + bc - ac$. + +\Item{7.} $ac - (b + c)$. + +\Item{8.} $a^{2} + (b^{2} + c^{2})$\Add{.} + +\Item{9.} $2a + (2b + 2c)$. + +\Item{10.} $a^{2} - b^{2} - c^{2}$. + +\Item{11.} $3(a - b + c)$. + +\Item{12.} $6ab - (bc + 8)$. + +\Item{13.} $7bc - c^{2} + a$. + +\Item{14.} $5ac - b^{2} + 3b$. + +\Item{15.} $4b^{2}c - 5c^{2} - 2b$. + +\Item{16.} $2a + (b + c)$. + +\Item{17.} $b + 2(a - c)$. + +\Item{18.} $c + 2(a - b)$. + +\Item{19.} $2a - (b + c)$. + +\Item{20.} $2b - (a - c)$. + +\Item{21.} $2c - (a - b)$. + +\Item{22.} $2c - 5(a - b)$. + +\Item{23.} $2b - 3(a - c)$. + +\Item{24.} $2c - b(a - b)$. +\end{multicols} +%% -----File: 020.png---Folio 14------- + + +\Section{ALGEBRAIC NOTATION.} + +\Exercise{5.} + +\Item{1.} Read $a + b$; $a - b$; $ab$; $a÷b$. + +\Item{2.} Write six increased by four. \Ans{$6 + 4$.} + +\Item{3.} Write $a$ increased by~$b$. + +\Item{4.} Write six diminished by four. \Ans{$6 - 4$.} + +\Item{5.} Write $a$ diminished by~$b$. + +\Item{6.} By how much does twenty-five exceed sixteen? +\Ans{$25 - 16$.} + +\Item{7.} By how much does $x$ exceed~$y$? + +\Item{8.} Write four times three; the fourth power of three. +\Ans{$4 × 3$; $3^{4}$.} + +\Item{9.} Write four times~$x$; the fourth power of~$x$. + +\Item{10.} If one part of twenty-five is fifteen, what is the +other part? +\Ans{$25 - 15$.} + +\Item{11.} If one part of~$35$ is~$x$, what is the other part? + +\Item{12.} If one part of~$x$ is~$a$, what is the other part? + +\Item{13.} How much does ten lack of being twelve? +\Ans{$12 - 10$.} + +\Item{14.} How much does $x$ lack of being fourteen? + +\Item{15.} How much does $x$ lack of being~$a$? + +\Item{16.} If a man walks four miles an hour, how many miles +will he walk in three hours? +\Ans{$3 × 4$.} + +\Item{17.} If a man walks $y$~miles an hour, how many miles +will he walk in $x$~hours? + +\Item{18.} If a man walks $y$~miles an hour, how many hours +will it take him to walk $x$~miles? +%% -----File: 021.png---Folio 15------- + +\Exercise{6.} + +\Item{1.} If the dividend is twenty and the quotient five, +what is the divisor? +\Ans{$\frac{20}{5}$.} + +\Item{2.} If the dividend is~$a$ and the quotient~$b$, what is the +divisor? + +\Item{3.} If John is twenty years old to-day, how old was he +four years ago? How old will he be five years hence? +\Ans{$20 - 4$; $20 + 5$.} + +\Item{4.} If James is $x$~years old to-day, how old was he three +years ago? How old will he be seven years hence? + +\Item{5.} Write four times the expression seven minus five. +\Ans{$4(7 - 5)$.} + +\Item{6.} Write seven times the expression $2x$~minus~$y$. + +\Item{7.} Write the next integral number above four. +\Ans{$4 + 1$.} + +\Item{8.} If $x$~is an integral number, write the next integral +number above it; the next integral number below it. + +\Item{9.} What number is less than $20$ by~$d$? + +\Item{10.} If the difference of two numbers is five, and the +smaller number is fifteen, what is the greater number? +\Ans{$15 + 5$.} + +\Item{11.} If the difference of two numbers is eight, and the +smaller number is~$x$, what is the greater number? + +\Item{12.} If the sum of two numbers is~$30$, and one of them +is~$20$, what is the other? +\Ans{$30 - 20$.} + +\Item{13.} If the sum of two numbers is~$x$, and one of them is~$10$, +what is the other? + +\Item{14.} If $100$ contains~$x$ ten times, what is the value of~$x$? +%% -----File: 022.png---Folio 16------- + +\Exercise{7.} + +\Item{1.} In $x$~years a man will be $40$~years old; what is his +present age? + +\Item{2.} How old will a man be in $y$~years, if his present age +is $a$~years? + +\Item{3.} What is the value of~$x$ if $7x$~equals~$28$? + +\Item{4.} If it takes $3$~men $4$~days to reap a field, how many +days will it take one man to reap it? +\Ans{$3 × 4$.} + +\Item{5.} If it takes $a$~men $b$~days to reap a field, how many +days will it take one man to reap it? + +\Item{6.} What is the excess of~$5x$ over~$3x$? + +\Item{7.} By how much does $20 - 3$ exceed $(10 + 1)$? +\Ans{$20 - 3 - (10 + 1)$.} + +\Item{8.} By how much does $2x - 3$ exceed $(x + 1)$? + +\Item{9.} If $x$~stands for~$10$, find the value of~$4(3x - 20)$. + +\Item{10.} If $a$~stands for~$10$, and $b$~for~$2$, find the value of +$2(a - 2b)$. + +\Item{11.} How many cents in $a$~dollars, $b$~quarters, and $c$~dimes? + +\Item{12.} A book-shelf contains French, Latin, and Greek +books. There are $100$~books in all, and there are $x$~Latin +and $y$~Greek books. How many French books are there? + +\Item{13.} A regiment of men is drawn up in $10$~ranks of $80$~men +each, and there are $15$~men over. How many men +are there in the regiment? +\Ans{$10 × 80 + 15$.} + +\Item{14.} A regiment of men is drawn up in $x$~ranks of $y$~men +each, and there are $c$~men over. How many men are there +in the regiment? +%% -----File: 023.png---Folio 17------- + +\Exercise{8.} + +\Item{1.} A room is $10$~yards long and $8$~yards wide. In the +middle there is a carpet $6$~yards square. How many +square yards of oilcloth will be required to cover the rest +of the floor? +\Ans{$10 × 8 - 6^{2}$.} + +\Item{2.} A room is $x$~yards long and $y$~yards wide. In the +middle there is a carpet $a$~yards square. How many +square yards of oilcloth will be required to cover the rest +of the floor? + +\Item{3.} How many rolls of paper $g$~feet long and $k$~feet +wide will be required to paper a room, the perimeter of +which, after proper allowance is made for doors and windows, +is $p$~feet and the height $h$~feet? + +\Item{4.} Write six times the square of~$m$, plus five~$c$ times +the expression $d$~plus $b$~minus~$a$. + +\Item{5.} Write five times the expression two~$n$ plus one, +diminished by six times the expression $c$~minus $a$~plus~$b$. + +\Item{6.} A lady bought a dress for $a$~dollars, a cloak for $b$~dollars, +two pairs of gloves for $c$~dollars a pair. She gave +a hundred-dollar bill in payment. How much money +should be returned to her? + +\Item{7.} If a man can perform a piece of work in $4$~days, how +much of it can he do in one day? +\Ans{$\frac{1}{4}$.} + +\Item{8.} If a man can perform a piece of work in $x$~days, +how much of it can he do in one day? + +\Item{9.} If A~can do a piece of work in $x$~days, B~in $y$~days, +C~in $z$~days, how much of it can they all do in one day, +working together? + +\Item{10.} Write an expression for the sum, and also for the +product, of three consecutive numbers of which the least is~$n$. +%% -----File: 024.png---Folio 18------- + +\Item{11.} The product of two factors is~$36$; if one of the +factors is~$x$, what is the other factor? + +\Item{12.} If $d$~is the divisor and $q$~the quotient, what is the +dividend? + +\Item{13.} If $d$~is the divisor, $q$~the quotient, and $r$~the remainder, +what is the dividend? + +\Item{14.} If $x$~oranges can be bought for $50$~cents, how many +oranges can be bought for $100$~cents? + +\Item{15.} What is the price in cents of $x$~apples, if they are +ten cents a dozen? + +\Item{16.} If $b$~oranges cost $6$~cents, what will $a$~oranges cost? + +\Item{17.} How many miles between two places, if a train +travelling $m$~miles an hour requires $4$~hours to make the +journey? + +\Item{18.} If a man was $x$~years old $10$~years ago, how many +years old will he be $7$~years hence? + +\Item{19.} If a man was $x$~years old $y$~years ago, how many +years old will he be $c$~years hence? + +\Item{20.} If a floor is $3x$~yards long and $12$~yards wide, how +many square yards does the floor contain? + +\Item{21.} How many hours will it take to walk $c$~miles, at +the rate of one mile in $15$~minutes? + +\Item{22.} Write three consecutive numbers of which $x$~is the +middle number. + +\Item{23.} If an odd number is represented by~$2n + 1$, what +will represent the next odd number? +%% -----File: 025.png---Folio 19------- + + +\Chapter{II.}{Simple Equations.} + +\Paragraph{43. Equations.} An equation is a statement in symbols +that two expressions stand for the same number. + +\begin{Remark} +Thus, the equation $3x + 2 = 8$ states that $3x + 2$ and~$8$ stand for +the same number. +\end{Remark} + +\Paragraph{44.} That part of the equation which precedes the sign +of equality is called the \Defn{first member}, or \Defn{left side}, and that +which follows the sign of equality is called the \Defn{second member}, +or \Defn{right side}. + +\Paragraph{45.} The statement of equality between two algebraic +expressions, if true for all values of the letters involved, is +called an \Defn{identical equation}; but if true only for certain +particular values of the letters involved, it is called an +\Defn{equation of condition}. + +\begin{Remark} +Thus, $a + b = b + a$, which is true for \emph{all values} of $a$~and~$b$, is an +\emph{identical equation}, and $3x + 2 = 8$, which is true only when $x$~stands +for~$2$, is an \emph{equation of condition}\Add{.} +\end{Remark} + +For brevity, an identical equation is called an \Defn{identity}, +and an equation of condition is called simply an \Defn{equation}. + +\Paragraph{46.} We often employ an equation to discover an \emph{unknown +number} from its relation to known numbers. We usually +represent the unknown number by one of the \emph{last} letters +of the alphabet, as $x$,~$y$,~$z$; and by way of distinction, we +use the \emph{first} letters, $a$,~$b$, $c$,~etc., to represent numbers that +%% -----File: 026.png---Folio 20------- +are supposed to be known, though not expressed in the +number-symbols of Arithmetic. + +\begin{Remark} +Thus, in the equation $ax + b = c$, $x$~is supposed to represent an +unknown number, and $a$,~$b$, and~$c$ are supposed to represent known +numbers. +\end{Remark} + +\Paragraph{47. Simple Equations.} An equation which contains the +first power of~$x$, the symbol for the unknown number, and +no higher power, is called a \Defn{simple equation}, or an \Defn{equation +of the first degree}. + +\begin{Remark} +Thus, $ax + b = c$ is a simple equation, or an equation of the first +degree \emph{in~$x$}. +\end{Remark} + +\Paragraph{48. Solution of an Equation.} To solve an equation is to +find the unknown number; that is, the number which, when +substituted for its symbol in the given equation, renders the +equation an identity. This number is said to \emph{satisfy} the +equation, and is called the \Defn{root} of the equation. + +\Paragraph{49. Axioms.} In solving an equation, we make use of the +following axioms: + +\Ax{1.} If equal numbers be added to equal numbers, +the sums will be equal. + +\Ax{2.} If equal numbers be subtracted from equal numbers, +the remainders will be equal. + +\Ax{3.} If equal numbers be multiplied by equal numbers, +the products will be equal. + +\Ax{4.} If equal numbers be divided by equal numbers, +the quotients will be equal. + +\begin{Theorem} +If, therefore, the two sides of an equation be increased by, +diminished by, multiplied by, or divided by equal numbers, +the results will be equal. +\end{Theorem} + +%[** TN: Next paragraph set in normal-size type in the original] +\begin{Remark} +Thus, if $8x = 24$, then $8x + 4 = 24 + 4$, $8x - 4 = 24 - 4$, +$4 × 8x = 4 × 24$, and $8x ÷ 4 = 24 ÷ 4$. +\end{Remark} +%% -----File: 027.png---Folio 21------- + +\Paragraph{50. Transposition of Terms.} It becomes necessary in solving +an equation to bring all the terms that contain the +symbol for the unknown number to one side of the equation, +and all the other terms to the other side. This is +called \Defn{transposing the terms}. We will illustrate by examples: + +\Item{1.} Find the number for which $x$~stands when +\[ +14x - 11 = 5x + 70. +\] + +The first object to be attained is to get all the terms +which contain~$x$ on the left side of the equation, and all the +other terms on the right side. This can be done by first +subtracting~$5x$ from both sides (Ax.~2), which gives +\[ +9x - 11 = 70, +\] +and then adding~$11$ to these equals (Ax.~1), which gives +\begin{DPalign*} +9x + 11 - 11 &= 70 + 11. \\ +\lintertext{\indent Combine,} +9x &= 81. \\ +\lintertext{\indent Divide by~$9$,} +x &= 9. +\end{DPalign*} + +\Item{2.} Find the number for which $x$~stands when $x + b = a$. +\begin{DPalign*}[m] +\lintertext{\indent The equation is} +x + b &= a. \\ +\lintertext{\indent Subtract~$b$ from each side,} +x + b - b &= a - b. +\rintertext{(Ax.~2)} +\end{DPalign*} + +Since $+b$~and~$-b$ in the left side cancel each other +(§~30), we have $x = a - b$. + +\Item{3.} Find the number for which $x$~stands when $x - b = a$. +\begin{DPalign*} +\lintertext{\indent The equation is} +x - b &= a. \\ +\lintertext{Add $+b$ to each side,} +x + b - b &= a + b. +\rintertext{(Ax.~1)} +\end{DPalign*} + +Since $+b$~and~$-b$ in the left side cancel each other +(§~30), we have $x = a + b$. +%% -----File: 028.png---Folio 22------- + +\Paragraph{51.} The effect of the operation in the preceding equations, +when Axioms (1)~and~(2) are used, is to take a term +from one side and put it on the other side with its sign +changed. We can proceed in a like manner in any other +case. Hence the general rule: + +\Paragraph{52.} \begin{Theorem}[nopar] Any term may be transposed from one side of an +equation to the other, provided its sign is changed. +\end{Theorem} + +\Paragraph{53.} Any term, therefore, which occurs on both sides +with \emph{the same sign} may be removed from both without +affecting the equality; and the sign of every term of an +equation may be changed without affecting the equality. + +\Paragraph{54. Verification.} When the root is substituted for its +symbol in the given equation, and the equation reduces to +an \emph{identity}, the root is said to be \Defn{verified}. We will illustrate +by examples: + +\Item{1.} What number added to twice itself gives~$24$? + +Let $x$~stand for the number; \\ +then $2x$~will stand for twice the number, \\ +and the number added to twice itself will be $x + 2x$. + +But the number added to twice itself is~$24$. +\begin{DPalign*} +\therefore x + 2x &= 24. \\ +\lintertext{\indent Combine $x$~and~$2x$,} +3x &= 24. \\ +\intertext{\indent Divide by~$3$, the coefficient of~$x$,} +x &= 8. +\rintertext{(Ax.~4)} +\end{DPalign*} + +Therefore the required number is~$8$. + +\begin{DPalign*} +\lintertext{\indent\textsc{Verification.}} +x + 2x &= 24, \\ +8 + 2 × 8 &= 24, \\ +8 + 16 &= 24, \\ +24 &= 24. +\end{DPalign*} +%% -----File: 029.png---Folio 23------- + +\ScreenBreak +\Item{2.} If $4x - 5$ stands for~$19$, for what number does $x$~stand? + +We have the equation +\begin{DPalign*} +4x - 5 &= 19. \\ +\lintertext{\indent Transpose $-5$,} +4x &= 19 + 5. \\ +\lintertext{\indent Combine,} +4x &= 24. \\ +\lintertext{\indent Divide by~$4$,} +x &= 6. +\rintertext{(Ax.~4)} \displaybreak[1] \\ +\lintertext{\indent \textsc{Verification.}} +4x - 5 &= 19, \\ +4 × 6 - 5 &= 19, \\ +24 - 5 &= 19, \\ +19 &= 19. +\end{DPalign*} + +\Item{3.} If $3x - 7$ stands for the same number as $14 - 4x$, +what number does $x$~stand for? + +We have the equation +\[ +3x - 7 = 14 - 4x. +\] + +Transpose $-4x$ to the left side, and $-7$ to the right side, +\begin{DPalign*} +3x + 4x &= 14 + 7. \\ +\lintertext{\indent Combine,} +7x &= 21. \\ +\lintertext{\indent Divide by~$7$,} +x &= 3. \displaybreak[1] \\ +\lintertext{\indent \textsc{Verification.}} +3x - 7 &= 14 - 4x, \\ +3 × 3 - 7 &= 14 - 4 × 3, \\ +2 &= 2. +\end{DPalign*} + +\Item{4.} Solve the equation +\[ +7(x - 1) - 30 = 4(x - 4). +\] +We have the equation +\[ +7(x - 1) - 30 = 4(x - 4). +\] +%% -----File: 030.png---Folio 24------- +%[** TN: Equation repeated at page break in the original] +% 7(x - 1) - 30 = 4(x - 4). + +Remove the parentheses, +\begin{DPalign*} +7x - 7 - 30 &= 4x - 16. \\ +\lintertext{\indent Then} +7x - 4x &= 7 + 30 - 16. \\ +\lintertext{\indent Combine,} +3x &= 21. \\ +\lintertext{\indent Divide by~$3$,} +x &= 7. \displaybreak[1] \\ +\lintertext{\indent \textsc{Verification.}} +7(7 - 1) - 30 &= 4(7 - 4), \\ +7 × 6 - 30 &= 4 x 3, \\ +42 - 30 &= 12, \\ +12 &= 12. +\end{DPalign*} + +\Exercise{9.} + +Find the number that $x$~stands for, if: +\begin{multicols}{2} +\Item{1.} $3x = x + 8$. + +\Item{2.} $3x = 2x + 5$. + +\Item{3.} $3x + 4 = x + 10$. + +\Item{4.} $4x + 6 = x + 9$. + +\Item{5.} $7x - 19 = 5x + 7$. + +\Item{6.} $3(x - 2) = 2(x - 3)$. + +\Item{7.} $8x + 7 = 4x + 27$. + +\Item{8.} $3x + 10 = x + 20$. + +\Item{9.} $5(x - 2) = 3x + 4$. + +\Item{10.} $3(x - 2) = 2(x - 1)$. + +\Item{11.} $2x + 3 = 16 - (2x - 3)$. + +\Item{12.} $19x - 3 = 2(7 + x)$. + +\Item{13.} $7x - 70 = 5x - 20$. + +\Item{14.} $2x - 22 = 108 - 2x$. + +\Item{15.} $2(x + 5) + 5(x - 4) = 32$. + +\Item{16.} $2(3x - 25) = 10$. + +\Item{17.} $33x - 70 = 3x + 20$. + +\Item{18.} $4(1 + x) + 3(2 + x) = 17$. + +\Item{19.} $8x - (x + 2) = 47$. + +\Item{20.} $3(x - 2) = 50 - (2x - 9)$. +\end{multicols} +%% -----File: 031.png---Folio 25------- + +\Item{21.} $2x - (3 + 4x - 3x + 5) = 4$. + +\Item{22.} $5(2 - x) + 7x - 21 = x + 3$. + +\Item{23.} $3(x - 2) + 2(x - 3) + (x - 4) = 3x + 5$. + +\Item{24.} $x + 1 + x + 2 + x + 4 = 2x + 12$. + +\Item{25.} $(2x - 5) - (x - 4) + (x - 3) = x - 4$. + +\Item{26.} $4 - 5x - (1 - 8x) = 63 - x$. + +\Item{27.} $3x - (x + 10) - (x - 3) = 14 - x$. + +\Item{28.} $x^{2} - 2x - 3 = x^{2} - 3x + 1$. + +\Item{29.} $(x^{2} - 9) - (x^{2} - 16) + x = 10$. + +\Item{30.} $x^{2} + 8x - (x^{2} - x - 2) = 5(x + 3) + 3$. + +\Item{31.} $x^{2} + x - 2 + x^{2} + 2x - 3 = 2x^{2} - 7x - 1$. + +\Item{32.} $10x - (x - 5) = 2x + 47$. + +\Item{33.} $7x - 5 - (6 - 8x) + 2 = 3x - 7 + 106$. + +\Item{34.} $6x + 3 - (3x + 2) = (2x - 1) + 9$. + +\Item{35.} $3(x + 10) + 4(x + 20) + 5x - 170 = 15 - 3x$. + +\Item{36.} $20 - x + 4(x - 1) - (x - 2) = 30$. + +\Item{37.} $5x + 3 - (2x - 2) + (1 - x) = 6(9 - x)$. + + +\Paragraph{55. Statement and Solution of Problems.} The difficulties +which the beginner usually meets in stating problems will +be quickly overcome if he will observe the following directions: + +Study the problem until you clearly understand its meaning +and just what is required to be found. + +Remember that $x$~must not be put for money, length, +time, weight,~etc., but for the \textbf{required number} of \emph{specified +units} of money, length, time, weight,~etc. + +Express each statement carefully in algebraic language, +and write out in full just what each expression stands for. +%% -----File: 032.png---Folio 26------- + +Do not attempt to form the equation until all the statements +are made in symbols. + +We will illustrate by examples: + +\Item{1.} John has three times as many oranges as James, and +they together have~$32$. How many has each? +\begin{Soln} +\begin{DPgather*} +\lintertext{\indent Let} +\text{$x$~stand for the \emph{number} of oranges James has;} \\ +\lintertext{then} +\text{$3x$~is the number of oranges John has;} \\ +\lintertext{and} +\text{$x + 3x$ is the number of oranges they together have.} +\end{DPgather*} + +But $32$~is the number of oranges they together have. +\begin{DPalign*} +\therefore x + 3x &= 32; \\ +\lintertext{or,} +4x &= 32, \\ +\lintertext{and} +x &= 8. \\ +\lintertext{\indent Since $x = 8$,} +3x &= 24. +\end{DPalign*} +\end{Soln} + +Therefore James has $8$~oranges, and John has $24$~oranges. + +\begin{Remark}[Note.] Beginners in stating the preceding problem generally write: +\[ +\text{Let $x = {}$\emph{what} James had.} +\] + +Now, we know \emph{what} James had. He had oranges, and we are to +discover simply the \emph{number} of oranges he had. +\end{Remark} + +\Item{2.} James and John together have~\$$24$, and James has +\$$8$~more than John. How many dollars has each? +\begin{Soln} +\begin{DPgather*} +\lintertext{\indent Let} +\text{$x$~stand for the number of dollars John has;} \\ +\lintertext{then} +\text{$x + 8$ is the number of dollars James has;} \\ +\lintertext{and} +\text{$x + (x + 8)$ is the number of dollars they both have.} +\end{DPgather*} + +But $24$~is the number of dollars they both have. +\[ +\therefore x + (x + 8) = 24. +\] + +Removing the parenthesis, +\[ +x + x + 8 = 24. +\] +\begin{DPalign*} +\therefore 2x &= 16. \\ +\lintertext{\indent Dividing by~$2$,} +x &= 8. \\ +\lintertext{\indent Since $x = 8$,} +x + 8 &= 16. +\end{DPalign*} +\end{Soln} + +Therefore John has~\$$8$, and James has~\$$16$. +%% -----File: 033.png---Folio 27------- + +\begin{Remark}[Note\Add{.}] The beginner must avoid the mistake of writing +\[ +\text{Let $x = {}$John's money\Add{.}} +\] + +We are required to find the \emph{number} of dollars John has, and therefore +$x$~must represent this required number. +\end{Remark} + +\Item{3.} The sum of two numbers is~$18$, and three times the +greater number exceeds four times the less by~$5$. Find the +numbers. + +\begin{Soln} +Let $x = {}$the greater number. + +Then, since $18$~is the sum and $x$~is one of the numbers, the other +number must be the sum minus~$x$. Hence +\[ +18 - x = \text{the smaller number}\Add{.} +\] + +Now, three times the greater number is~$3x$, and four times the less +number is~$4(18 - x)$\Add{.} +\begin{DPalign*} +\lintertext{\indent Hence,} +&3x - 4(18 - x) = \text{the excess}\Add{.} \\ +\lintertext{\indent But} +&5 = \text{the excess}\Add{,} \\ +\therefore\ &3x - 4(18 - x) = 5 \\ +\therefore\ &3x - (72 - 4x) = 5, \\ +\lintertext{or} +&3x - 72 + 4x = 5. \\ +\therefore\ &7x = 77, \\ +\lintertext{and} +&\Z x = 11\Add{.} +\end{DPalign*} +\end{Soln} + +Therefore the numbers are $11$~and~$7$. + +\Exercise{10.} + +\Item{1.} If a number is multiplied by~$9$, the product is~$270$. +Find the number. + +\Item{2.} If the sum of the ages of a father and son is $60$~years, +and the father is $5$~times as old as the son, what is the +age of each? + +\Item{3.} The sum of two numbers is~$91$, and the greater is $6$~times +the less. Find the numbers. +%% -----File: 034.png---Folio 28------- + +\Item{4.} A tree $90$~feet high was broken so that the part +broken off was $8$~times the length of the part left standing. +Find the length of each part. + +\Item{5.} The difference of two numbers is~$7$, and their sum is~$53$. +Find the numbers. + +\Item{6.} The difference of two numbers is~$12$, and their sum is~$84$. +Find the numbers. + +\Item{7.} Divide $35$ into two parts so that one part shall be +greater by~$5$ than the other part. + +\Item{8.} Three times a given number is equal to the number +increased by~$40$. Find the number. + +\Item{9.} Three times a given number diminished by~$24$ is +equal to the given number. Find the number. + +\Item{10.} One number is $4$~times another, and their difference +is~$30$. Find the numbers. + +\Item{11.} The sum of two numbers is~$36$, and one of them +exceeds twice the other by~$6$. Find the numbers. + +\begin{Remark}[Hint.] Let $x$~equal the greater number: then $36 - x$ will equal the +smaller. +\end{Remark} + +\Item{12.} The sum of two numbers is~$40$, and $5$~times the +smaller exceeds $2$~times the greater by~$25$. Find the +numbers. + +\Item{13.} The number $30$ is divided into two parts such that +$4$~times the greater part exceeds $5$~times the smaller part +by~$30$. Find the parts. + +\Item{14.} The sum of two numbers is~$27$, and twice the greater +number increased by $3$~times the less is~$61$. Find the +numbers. + +\Item{15.} The sum of two numbers is~$32$, and five times the +smaller is $3$~times the greater number. Find the numbers. +%% -----File: 035.png---Folio 29------- + +\Exercise{11.} + +\Item{1.} A farmer sold a horse and a cow for~\$$210$. He +sold the horse for four times as much as the cow. How +much did he get for each? + +\Item{2.} Three times the excess of a certain number over~$6$ +is equal to the number plus~$144$. Find the number. + +\Item{3.} Thirty-one times a certain number is as much above~$40$ +as nine times the number is below~$40$. Find the number. + +\Item{4.} Two numbers differ by~$10$, and their sum is equal to +seven times their difference. Find the numbers. + +\Item{5.} Find three consecutive numbers, $x$, $x + 1$, and~$x + 2$, +whose sum is~$78$. + +\Item{6.} Find five consecutive numbers whose sum is~$35$. + +\Item{7.} The sum of the ages of A~and~B is $40$~years, and $10$~years +hence A~will be twice as old as~B\@. Find their +present ages. + +\Item{8.} A father is four times as old as his son, and in $5$~years +he will be only three times as old. Find their present ages. + +\Item{9.} One man is $60$~years old, and another man is $50$~years. +How many years ago was the first man twice as +old as the second? + +\Item{10.} A man $50$~years old has a son $10$~years old. In +how many years will the father be three times as old as +the son? + +\Item{11.} A~has~\$$100$, and B~has~\$$20$. How much must A +give B in order that they may each have the same sum? + +\Item{12.} A banker paid \$$63$ in $5$-dollar bills and $2$-dollar +bills. He paid just as many $5$-dollar bills as $2$-dollar bills. +How many bills of each kind did he pay? +%% -----File: 036.png---Folio 30------- + +\Exercise{12.} + +\Item{1.} In a company of $90$~persons, composed of men, +women, and children, there are three times as many children +as men, and twice as many women as men. How many +are there of each? + +\Item{2.} Find the number whose double exceeds~$70$ by as +much as the number itself is less than~$80$. + +\Item{3.} A farmer employed two men to build $112$~rods of +wall. One of them built on the average $4$~rods a day, and +the other $3$~rods a day. How many days did they work? + +\Item{4.} Two men travel in \emph{opposite} directions, one $30$~miles +a day, and the other $20$~miles a day. In how many days +will they be $350$~miles apart? + +\Item{5.} Two men travel in the same direction, one $30$~miles +a day, and the other $20$~miles a day. In how many days +will they be $350$~miles apart? + +\Item{6.} A man bought $3$~equal lots of hay for~\$$408$. For +the first lot he gave \$$17$~a ton, for the second~\$$16$, for the +third~\$$18$. How many tons did he buy in all? + +\Item{7.} A farmer sold a quantity of wood for~\$$84$, one half +of it at \$$3$~a cord, and the other half at \$$4$~a cord. How +many cords did he sell? + +\begin{Remark}[Hint.] +Let $2x$~equal the number of cords. +\end{Remark} + +\Item{8.} If $2x - 3$ stands for~$29$, for what number will +$4 + x$ stand? + +\Item{9.} At an election two opposing candidates received +together $2044$~votes, and one received $104$~more votes than +the other. How many votes did each candidate receive? +%% -----File: 037.png---Folio 31------- + +\Exercise{13.} + +\Item{1.} A man walks $4$~miles an hour for $x$~hours, and +another man walks $3$~miles an hour for $x + 2$~hours. If +they each walk the same distance, how many miles does +each walk? + +\Item{2.} A has twice as much money as~B; but if A gives~B +\$$30$, it will take twice as much as A has left to equal~B's. +How much money has each? + +\Item{3.} I have \$$12.75$ in two-dollar bills and twenty-five +cent pieces, and I have twice as many bills as twenty-five +cent pieces. How many have I of each? + +\Item{4.} I have in mind a certain number. If this number +is diminished by~$8$ and the remainder multiplied by~$8$, the +result is the same as if the number was diminished by~$6$ +and the remainder multiplied by~$6$. What is the number? + +\Item{5.} I have five times as many half-dollars as quarters, and +the half-dollars and quarters amount to~\$$11$. How many +of each have I? + +\Item{6.} A man pays a debt of~\$$91$ with ten-dollar bills and +one-dollar bills, paying three times as many one-dollar bills +as ten-dollar bills. How many bills of each kind does he +pay? + +\Item{7.} A father is four times as old as his son, but $4$~years +hence he will be only three times as old as his son. How +old is each? + +\Item{8.} A workman was employed for $24$~days. For every +day he worked he was to receive~\$$1.50$, and for every day +he was idle he was to pay \$$0.50$ for his board. At the end +of the time he received~\$$28$. How many days did he +work? +%% -----File: 038.png---Folio 32------- + +\Exercise{14.} + +\Item{1.} A boy has $4$~hours at his disposal. How far can he +ride into the country at the rate of $9$~miles an hour and +walk back at the rate of $3$~miles an hour, if he returns just +on time? + +\begin{Remark}[Hint.] Let $x = {}$the number of hours he rides. + +Then $4 - x = {}$the number he walks. +\end{Remark} + +\Item{2.} A has~\$$180$, and B has~\$$80$. How much must A +give B in order that six times B's money shall be equal to +$7$~times~A's? + +\Item{3.} A grocer has two kinds of tea, one kind worth $45$~cents +a pound, and the other worth $65$~cents a pound. +How many pounds of each kind must he take to make $80$~pounds, +worth $50$~cents a pound? + +\Item{4.} A tank holding $1200$~gallons has three pipes. The +first lets in $8$~gallons a minute, the second $10$~gallons, and +the third $12$~gallons a minute. In how many minutes will +the tank be filled? + +\Item{5.} The fore and hind wheels of a carriage are $10$~feet +and $12$~feet respectively in circumference. How many +feet will the carriage have passed over when the fore wheel +has made $250$~revolutions more than the hind wheel? + +\Item{6.} Divide a yard of tape into two parts so that one part +shall be $6$~inches longer than the other part. + +\Item{7.} A boy bought $7$~dozen oranges for~\$$1.50$. For a part +he paid $20$~cents a dozen; and for the remainder, $25$~cents +a dozen. How many dozen of each kind did he buy? + +\Item{8.} How can a bill of~\$$3.30$ be paid in quarters and ten-cent +pieces so as to pay three times as many ten-cent +pieces as quarters? +%% -----File: 039.png---Folio 33------- + + +\Chapter{III.}{Positive and Negative Numbers.} + +\Paragraph{56. Quantities Opposite in Kind.} If a person is engaged +in trade, his capital will be \emph{increased} by his \emph{gains}, and +\emph{diminished} by his \emph{losses}. + +\emph{Increase} in temperature is measured by the number of +degrees the mercury rises in a thermometer, and \emph{decrease} +in temperature by the number of degrees the mercury \emph{falls}. + +In considering any quantity whatever, a quantity that +\emph{increases} the quantity considered is called a \emph{positive quantity}; +and a quantity that \emph{decreases} the quantity considered +is called a \emph{negative quantity}. + +\Paragraph{57. Positive and Negative Numbers.} If from a given +point, marked~$0$, we draw a straight line to the right, and +beginning from the \emph{zero} point lay off units of length on this +line, the successive repetitions of the unit will be expressed +by the \emph{natural series of numbers}, $1$,~$2$, $3$, $4$,~etc. Thus: +\Graphic{1} + +If we wish to \emph{add} $2$~to~$5$, we begin at~$5$, count $2$~units +\emph{forwards}, and arrive at~$7$, the sum required. If we wish +to \emph{subtract} $2$~from~$5$, we begin at~$5$, count $2$~units \emph{backwards}, +and arrive at~$3$, the difference required. If we wish +to subtract $5$~from~$5$, we count $5$~units backwards, and arrive +at~$0$. If we wish to subtract $5$~from~$2$, we cannot do it, +because when we have counted backwards from~$2$ as far as~$0$, +\emph{the natural series of numbers comes to an end}. +%% -----File: 040.png---Folio 34------- + +In order to subtract a greater number from a smaller, it +is necessary to \emph{assume} a new series of numbers, beginning +at zero and extending to the left of zero. The series to the +left of zero must proceed from zero by \emph{the repetitions of the +unit}, precisely like the natural series to the right of zero; +and the \emph{opposition} between the right-hand series and the +left-hand series must be clearly marked. This opposition +is indicated by calling every number in the right-hand +series a \Defn{positive number}, and prefixing to it, when written, +the sign~$+$; and by calling every number in the left-hand +series a \Defn{negative number}, and prefixing to it the sign~$-$. +The two series of numbers may be called the \Defn{algebraic series +of numbers}, and written thus: +\Graphic{2} + +If, now, we wish to subtract $7$~from~$4$, we begin at~$4$ in +the positive series, count $7$~units in the \emph{negative direction} +(to the left), and arrive at~$-3$ in the negative series; that +is, $4 - 7 = -3$. + +The result obtained by subtracting a greater number from +a less, when both are positive, is \emph{always a negative number}. + +In general, if $a$~and~$b$ represent any two numbers of the +positive series, the expression $a - b$ will be a positive number +when $a$~is greater than~$b$; will be zero when $a$~is equal +to~$b$; will be a negative number when $a$~is less than~$b$. + +In counting from left to right in the algebraic series, numbers +\emph{increase} in magnitude; in counting from right to left, +numbers \emph{decrease} in magnitude. Thus $-3$,~$-1$, $0$, $+2$,~$+4$, +are arranged in \emph{ascending} order of magnitude. + +\Paragraph{58.} Every algebraic number, as $+4$~or~$-4$, consists of a +\emph{sign} $+$~or~$-$ and the \emph{absolute value} of the number. The +sign shows whether the number belongs to the positive or +%% -----File: 041.png---Folio 35------- +negative series of numbers; the absolute value shows the +place the number has in the positive or negative series. + +When no sign stands before a number, the sign~$+$ is +always understood. Thus $4$~means the same as~$+4$, $a$~means +the same as~$+a$. But \emph{the sign~$-$ is never omitted}. + +\Paragraph{59.} Two algebraic numbers which have, one the sign~$+$, +and the other the sign~$-$, are said to have \emph{unlike signs}. + +Two algebraic numbers which have the same absolute +values, but unlike signs, always cancel each other when +combined. Thus $+4 - 4 = 0$; $+a - a = 0$. + +\Paragraph{60. Double Meanings of the Signs $+$~and~$-$.} The use of +the signs $+$~and~$-$ to indicate addition and subtraction +must be carefully distinguished from the use of the signs $+$~and~$-$ +to indicate in which series, the positive or the negative, +a given number belongs. In the first sense they are +signs of \emph{operations}, and are common to Arithmetic and +Algebra; in the second sense they are signs of \emph{opposition}, +and are employed in Algebra alone. + +\begin{Remark}[Note.] In Arithmetic, if the things counted are \emph{whole units}, the +numbers which count them are called \Defn{whole numbers}, \Defn{integral numbers}, +or \Defn{integers}, where the adjective is transferred from the things counted +to the numbers which count them. But if the things counted are +only \emph{parts of units}, the numbers which count them are called \Defn{fractional +numbers}, or simply \Defn{fractions}, where again the adjective is transferred +from the things counted to the numbers which count them. + +Likewise in Algebra, if the units counted are \emph{negative}, the numbers +which count them are called \Defn{negative numbers}, where the adjective +which defines the nature of the units counted is transferred to the +numbers that count them. + +A whole number means a number of whole units, a fractional number +means a number of parts of units, and a negative number means +a number of negative units. +\end{Remark} +%% -----File: 042.png---Folio 36------- + +\Paragraph{61. Addition and Subtraction of Algebraic Numbers.} An +algebraic number which is to be added or subtracted is +often \DPtypo{inclosed}{enclosed} in a parenthesis, in order that the signs $+$~and~$-$, +which are used to distinguish positive and negative +numbers, may not be confounded with the $+$~and~$-$ signs +that denote the operations of addition and subtraction. +Thus $+4 + (-3)$ expresses the sum, and $+4 - (-3)$ expresses +the difference, of the numbers $+4$~and~$-3$. + +\Paragraph{62. Addition.} In order to add two algebraic numbers, we +begin at the place in the series which the first number occupies, +and count, \emph{in the direction indicated by the sign of the +second number}, as many units as there are in the absolute +value of the second number. +\Graphic{3} + +Thus the sum of $+4 + (+3)$ is found by counting from +$+4$ three units in \emph{the positive direction}; that is, to the +right, and is, therefore,~$+7$. + +The sum of $+4 + (-3)$ is found by counting from $+4$ +three units in \emph{the negative direction}; that is, to the left, and +is, therefore,~$+1$. + +The sum of $-4 + (+3)$ is found by counting from $-4$ +three units in the positive direction, and is, therefore,~$-1$. + +The sum of $-4 + (-3)$ is found by counting from $-4$ +three units in the negative direction, and is, therefore,~$-7$. + +\Paragraph{63. Subtraction.} In order to subtract one algebraic number +from another, we begin at the place in the series which +the minuend occupies, and count, \emph{in the direction opposite to +that indicated by the sign of the subtrahend}, as many units +as there are in the absolute value of the subtrahend. + +Thus the result of subtracting $+3$~from~$+4$ is found by +%% -----File: 043.png---Folio 37------- +counting from $+4$ three units in the \emph{negative direction}; +that is, in the direction \emph{opposite to that indicated by the sign~$+$ +before~$3$}, and is, therefore,~$+1$. + +The result of subtracting $-3$~from~$+4$ is found by counting +from $+4$ three units in the \emph{positive direction}, and is, +therefore,~$+7$. + +The result of subtracting $+3$~from~$-4$ is found by counting +from $-4$ three units in the \emph{negative direction}, and is, +therefore,~$-7$. + +The result of subtracting $-3$~from~$-4$ is found by counting +from $-4$ three units in the \emph{positive direction}, and is, +therefore,~$-1$. + +\Paragraph{64.} Collecting the results obtained in addition and subtraction, +we have: +\[ +\begin{array}{c>{\quad}c} +\textsc{Addition.} & \textsc{Subtraction.} \\ + +4 + (-3) = +4 - 3 = +1. & +4 - (+3) = +4 - 3 = +1. \\ + +4 + (+3) = +4 + 3 = +7. & +4 - (-3) = +4 + 3 = +7. \\ +- 4 + (-3) = -4 - 3 = -7. & -4 - (+3) = -4 - 3 = -7. \\ +- 4 + (+3) = -4 + 3 = -1. & -4 - (-3) = -4 + 3 = -1. \\ +\end{array} +\] + +\Paragraph{65.} From these four cases of addition, therefore, + +% [**** TN: Book uses commas elsewhere] +\Dictum{To Add Algebraic Numbers}\DPtypo{:}{,} +\begin{Theorem}[I.] If the numbers have like signs, find the sum of their +absolute values, and prefix the common sign to the result. +\end{Theorem} + +\begin{Theorem}[II.] If the numbers have unlike signs, find the difference +of their absolute values, and prefix the sign of the greater +number to the result. +\end{Theorem} + +\begin{Theorem}[III.] If there are more than two numbers, find the sum +of the positive numbers and the sum of the negative numbers, +%% -----File: 044.png---Folio 38------- +take the difference between the absolute values of these two +sums, and prefix the sign of the greater sum to the result. +\end{Theorem} + +\begin{Remark}[Note.] Since the order in which numbers are added is immaterial, +we may add any two of the numbers, and then this sum to +any third number, and so on. +\end{Remark} + +\Paragraph{66.} The result is generally called the \Defn{algebraic sum}, in +distinction from the arithmetical sum; that is, the sum of +the absolute values of the numbers. + +\Paragraph{67.} From the four cases of subtraction in §~64, we see +that \textit{subtracting a positive number is equivalent to adding +an equal negative number, and subtracting a negative number +is equivalent to adding an equal positive number}. + + +\Dictum{To Subtract One Algebraic Number from Another}, +\begin{Theorem} +Change the sign of the subtrahend, and add the subtrahend +to the minuend. +\end{Theorem} + +\Paragraph{68. Examples.} + +\Item{1.} Find the sum of $3a$, $2a$, $a$, $5a$, $7a$. + +The sum of the coefficients is $3 + 2 + 1 + 5 + 7 = 18$. + +Hence the sum of the numbers is~$18a$. + +\Item{2.} Find the sum of $-5c$, $-c$, $-3c$, $-4c$, $-2c$. + +The sum of the coefficients is $-5 - 1 - 3 - 4 - 2 = -15$. + +Hence the sum of the numbers is~$-15c$. + +\Item{3.} Find the sum of $8x$, $-9x$, $-x$, $3x$, $4x$, $-12x$, $x$. + +The sum of the positive coefficients is $8 + 3 + 4 + 1 = 16$. + +The sum of the negative coefficients is $-9 - 1 - 12 = -22$. + +The difference between $16$~and~$22$ is~$6$, and the sign of +the greater is negative. + +Hence the required sum is~$-6x$. +%% -----File: 045.png---Folio 39------- + +\PrintBreak +\Exercise{15.} + +Find the sum of: +\begin{multicols}{2} +\Item{1.} $5c$, $23c$, $c$, $11c$. + +\Item{2.} $4a$, $3a$, $7a$, $10a$. + +\Item{3.} $7x$, $12x$, $11x$, $9x$. + +\Item{4.} $6y$, $8y$, $2y$, $35y$. + +\Item{5.} $-3a$, $-5a$, $-18a$. + +\Item{6.} $-5x$, $-6x$, $-18x$, $-11x$. + +\Item{7.} $-3b$, $-b$, $-9b$, $-4b$. + +\Item{8.} $-z$, $-2z$, $-10z$, $-53z$. + +\Item{9.} $-11m$, $-3m$, $-5m$, $-m$. + +\Item{10.} $5d$, $-d$, $-4d$, $2d$. +\end{multicols} + +\Item{11.} $13n$, $13n$, $-11n$, $-6n$, $-9n$, $n$, $2n$, $-3n$. + +\Item{12.} $5g$, $-3g$, $-6g$, $-4g$, $20g$, $-5g$, $-11g$, $-14g$. + +\Item{13.} $-9a^{2}$, $5a^{2}$, $6a^{2}$, $a^{2}$, $2a^{2}$, $-a^{2}$, $-3a^{2}$. + +\Item{14.} $3x^{3}$, $-2x^{3}$, $-5x^{3}$, $-7x^{3}$, $-x^{3}$, $2x^{3}$, $-10x^{3}$, $-x^{3}$. + +\Item{15.} $4a^{2}b^{2}$, $-a^{2}b^{2}$, $-6a^{2}b^{2}$, $4a^{2}b^{2}$, $-2a^{2}b^{2}$, $a^{2}b^{2}$. + +\Item{16.} $6mn$, $-5mn$, $mn$, $-3mn$, $4mn$. + +\Item{17.} $3xyz$, $-2xyz$, $5xyz$, $-7xyz$, $xyz$. + +\Item{18.} $5a^{3}b^{3}c^{3}$, $-7a^{3}b^{3}c^{3}$, $-3a^{3}b^{3}c^{3}$, $2a^{3}b^{3}c^{3}$. + +\Item{19.} $11abcd$, $-10abcd$, $-9abcd$, $-abcd$. + +\Item{20.} Subtract $-a$ from $-b$, and find the value of the +result if $a = -4$, $b = -5$. + +When $a = 4$, $b = -2$, $c = -3$, find the difference in +the values of: + +\Item{21.} $a - b + c$ and $-a + b + c$. + +\Item{22.} $a + (-b) + c$ and $a - (-b) + c$. + +\Item{23.} $-a - (-b) + c$ and $-(-a) + (-b) - c$. + +\Item{24.} $a - b + (-c)$ and $a - (-b) - (-c)$. +%% -----File: 046.png---Folio 40------- + + +\Section{MULTIPLICATION AND DIVISION OF ALGEBRAIC +NUMBERS} + +\Paragraph{69. Multiplication.} Multiplication is generally defined +in Arithmetic as the process of finding the result when one +number (the multiplicand) is taken as many times as there +are units in another number (the multiplier). This definition +fails when the \emph{multiplier is a fraction}. In multiplying +by a fraction, we divide the multiplicand into as many +equal parts as there are units in the denominator, and take +as many of these parts as there are units in the numerator. + +If, for example, we multiply $6$~by~$\frac{2}{3}$, we divide $6$ into +\emph{three} equal parts and take \emph{two} of these parts, obtaining $4$ +for the product. The multiplier,~$\frac{2}{3}$, is~$\frac{2}{3}$ of~$1$, and the +product,~$4$, is~$\frac{2}{3}$ of~$6$, in other words, \emph{the product is obtained +from the multiplicand precisely as the multiplier is obtained +from~$1$}. + +This statement is also true when the multiplier is a whole +number. Thus in $5 × 7 = 35$, the multiplier,~$5$, is equal to +\[ +1 + 1 + 1 + 1 + 1, +\] +and the product,~$35$, is equal to +\[ +7 + 7 + 7 + 7 + 7. +\] + +\Paragraph{70.} \Dictum{Multiplication may be defined}, therefore, + +As the operation of finding from two given numbers, +called \emph{multiplicand} and \emph{multiplier}, a third number called +\emph{product}, which is \emph{formed from the multiplicand as the multiplier +is formed from unity}. + +\Paragraph{71.} According to this definition of multiplication, +\begin{DPalign*}[m] +\lintertext{since} ++3 &= + 1 + 1 + 1, \\ +3 × (+8) &= +8 + 8 + 8 +\Tag{(1)} \\ +&= +24, \displaybreak[1] \\ +%% -----File: 047.png---Folio 41------- +\lintertext{and} +3 × (-8) &= -8 - 8 - 8 +\Tag{(2)} \\ +&= -24. \displaybreak[1] \\ +\lintertext{\indent Again, since} +-3 &= -1 - 1 - 1; \\ +(-3) × 8 &= -8 - 8 - 8 +\Tag{(3)} \\ +&= -24, \displaybreak[1] \\ +\lintertext{and} +(-3) × (-8) &= -(-8) - (-8) - (-8) +\Tag{(4)} \\ +&= +8 + 8 + 8 \\ +&= +24. +\end{DPalign*} + +\Paragraph{72.} From these four cases it follows that in finding +the product of two algebraic numbers, if the two numbers +have \emph{like} signs, the product will have the \emph{plus} sign, and if +\emph{unlike} signs, the product will have the \emph{minus} sign. + +Hence the \Defn{Law of Signs in Multiplication} is: +\begin{Theorem} +Like signs give~$+$, and unlike signs give~$-$. +\end{Theorem} + +If $a$~and~$b$ stand for any two numbers, we have +\begin{align*} +(+a) × (+b) &= +ab, \\ +(+a) × (-b) &= -ab, \\ +(-a) × (+b) &= -ab, \\ +(-a) × (-b) &= +ab. +\end{align*} + +\Paragraph{73. The Index Law in Multiplication.} +\begin{DPalign*} +\lintertext{\indent Since} +a^{2} &= aa, \quad\text{and}\quad a^{3} = aaa, \\ +a^{2} × a^{3} &= aa × aaa = aaaaa = a^{5} = a^{2 + 3}; \\ +a^{4} × a &= aaaa × a = aaaaa = a^{5} = a^{4 + 1}. +\end{DPalign*} + +If $a$~stands for any number, and $m$~and~$n$ for any integers, +\[ +a^{m} × a^{n} = a^{m + n}. \EqText{Hence,} +\] +\begin{Theorem} +The index of the product of two powers of the same number +is equal to the sum of the indices of the factors. +\end{Theorem} +%% -----File: 048.png---Folio 42------- + +\Paragraph{74. Examples.} + +\Item{1.} Find the product of $6a^{2}b^{2}$ and $7ab^{2}c^{3}$. + +Since the order of the factors is immaterial, +\begin{align*} +6a^{2}b^{3} × 7ab^{2}c^{3} + &= 6 × 7 × a^{2} × a × b^{3} × b^{2} × c^{3} \\ + &= 42a^{3}b^{5}c^{3}. +\end{align*} + +\Item{2.} Find the product of $-3ab$ and $7ab^{3}$. +\begin{align*} +-3ab × 7ab^{3} + &= -3 × 7 × a × a × b × b^{3} \\ + &= -21a^{2}b^{4}. +\end{align*} + +\Paragraph{75. To Find the Product of Simple Expressions}, therefore, +\begin{Theorem} +Take the product of the coefficients and the sum of the +indices of the like letters. +\end{Theorem} + +\Exercise{16.} + +Find the product of: +\begin{multicols}{2} +\Item{1.} $5a^{2}$ and $6a^{3}$. + +\Item{2.} $8ab$ and $5a^{3}b^{2}$. + +\Item{3.} $9xy$ and $7xy$. + +\Item{4.} $2a^{2}b$ and $a^{3}b^{4}c^{2}$. + +\Item{5.} $3a^{3}b^{7}c^{8}$ and $3a^{4}b^{2}c$. + +\Item{6.} $2a$ and $-5a$. + +\Item{7.} $-3a$ and $-4b$. + +\Item{8.} $-ab$ and $a^{3}b^{2}$. + +\Item{9.} $-2ab^{4}$ and $-5a^{4}bc$. + +\Item{10.} $-2x^{6}y^{3}z$ and $-6xy^{2}z$. +\end{multicols} + +\Item{11.} $3a^{2}b$, $-5ab^{2}$, and $-7a^{4}b^{2}$. + +\Item{12.} $2a^{2}bc^{3}$, $-3a^{3}b^{2}c$, and $-ab^{2}c^{3}$. + +\Item{13.} $2b^{2}c^{2}x^{2}$, $2a^{2}b^{2}c^{3}$, and $-3a^{3}bx^{3}$. + +\Item{14.} $2a^{3}b^{2}c$, $-3a^{2}b^{3}c$, and $-4a^{2}bc^{3}$. + +\Item{15.} $7am^{2}x^{3}$, $3a^{4}m^{2}x^{3}$, and $-2amx$. + +\Item{16.} $-3x^{2}y^{2}z^{2}$, $2x^{2}yz^{3}$, and $-5x^{4}yz$. +%% -----File: 049.png---Folio 43------- + +If $a = -2$, $b = 3$, and $c = -1$, find the value of: +\begin{multicols}{2} +\Item{17.} $2ab^{2} - 3bc^{2} + c$. + +\Item{18.} $4a^{2} - 2b^{2} - c^{2}$. + +\Item{19.} $5a + 2b - 4c^{4}$. + +\Item{20.} $2a^{3} - 3b + 8c^{2}$. + +\Item{21.} $-a + 3b - 2c^{2}$. + +\Item{22.} $-a^{3} - 2b - 10c$. + +\Item{23.} $3a^{3} - 3b^{3} - 3c^{3}$. + +\Item{24.} $2ab^{2} - 3bc^{2} + 2ac$. + +\Item{25.} $3abc + 5a^{2}b^{2} - 2a^{2}b$. + +\Item{26.} $ab^{2}c^{2} + 2abc^{2} + a^{2}b^{2}c^{2}$. + +\Item{27.} $2a^{2}bc + 3abc + a^{2}b^{2}c^{2}$. + +\Item{28.} $6a^{2} + 8a^{2}b^{2} - 5a^{2}bc$. +\end{multicols} + +\Paragraph{76. Division.} To divide $48$~by~$8$ is to find the number +of times it is necessary to take $8$ to make~$48$. Here the +\emph{product} and \emph{one factor} are given, and \emph{the other factor} is +required. We may, therefore, take for the definition of +division\Add{:} + +The operation by which when \emph{the product} and \emph{one factor} +are given, \emph{the other factor is found}. + +With reference to this operation the product is called +the \Defn{dividend}, the given factor the \Defn{divisor}, and the required +factor the \Defn{quotient}. + +\Paragraph{77. Law of Signs in Division.} +\begin{alignat*}{2} +&\text{Since } (+a) × (+b) = +ab,\quad && \therefore +ab ÷ (+a) = +b. \\ +&\text{Since } (+a) × (-b) = -ab, && \therefore -ab ÷ (+a) = -b. \\ +&\text{Since } (-a) × (+b) = -ab, && \therefore -ab ÷ (-a) = +b. \\ +&\text{Since } (-a) × (-b) = +ab, && \therefore +ab ÷ (-a) = -b. +\end{alignat*} + +That is, if the dividend and divisor have like signs, the +quotient has the plus sign; and if they have unlike signs, +the quotient has the minus sign. Hence, in division, +\begin{Theorem} +Like signs give~$+$, and unlike signs give~$-$. +\end{Theorem} +%% -----File: 050.png---Folio 44------- + +\Paragraph{78. Index Law in Division.} + +The dividend contains all the factors of the divisor and +of the quotient, and therefore the quotient contains the +factors of the dividend that are not found in the divisor. + +Thus, $\dfrac{abc}{bc} = a$, $\dfrac{aabx}{ab} = ax$, $\dfrac{124abc}{-4ab} = -31c$. + +Divide $a^{5}$~by~$a^{2}$, $a^{6}$~by~$a^{4}$, $a^{4}$~by~$a$\DPtypo{,}{.} +\begin{alignat*}{4} +\frac{a^{5}}{a^{2}} + &= \frac{aaaaa}{aa} &&= aaa &&= a^{3} &&= a^{5-2}; \\ +\frac{a^{6}}{a^{4}} + &= \frac{aaaaaa}{aaaa} &&= aa &&= a^{2} &&= a^{6-4}\DPtypo{,}{;} \\ +\frac{a^{4}}{a} + &= \frac{aaaa}{a} &&= aaa &&= a^{3} &&= a^{4-1}\DPtypo{,}{.} +\end{alignat*} + +If $m$~and~$n$ stand for any integers, and $m$~is greater than~$n$\Add{,} +\[ +a^{m} - a^{n} = a^{m - n}\Add{.} +\] + +\begin{Theorem} +The index of the quotient of two powers of the same letter +is equal to the index of the letter in the dividend diminished +by the index of the letter in the divisor. +\end{Theorem} + +\Paragraph{79. Examples.} + +\Item{1.} Divide $15xy$~by~$5x$\Add{.} +\[ +\frac{15xy}{5x} = \frac{3 × 5xy}{5x} = 3y. +\] + +Here we cancel the factors $5$~and~$x$, which are common +to the dividend and divisor\Add{.} + +\Item{2.} Divide $-21a^{2}b^{3}$~by~$3ab^{2}$\Add{.} +\[ +\frac{-21a^{2}b^{3}}{3ab^{2}} = -7ab. +\] +%% -----File: 051.png---Folio 45------- + +\Item{3.} Divide $54a^{5}b^{3}c$~by~$-6ab^{2}c$. +\[ +\frac{54a^{5}b^{3}c}{-6ab^{2}c} = -9a^{4}b. +\] + +\Item{4.} Divide $-45x^{4}y^{5}z^{7}$~by~$-15x^{4}y^{5}z^{5}$. +\[ +\frac{-45x^{4}y^{5}z^{7}}{-15x^{4}y^{5}z^{5}} = 3z^{2}. +\] + +\Item{5.} Divide $-15a^{3}b^{2}c^{3}$~by~$-60a^{2}bc^{3}$. +\[ +\frac{-15a^{3}b^{2}c^{3}}{-60a^{2}bc^{3}} = \frac{ab}{4}. +\] + +\Exercise{17.} + +Divide: +\begin{multicols}{2} +\Item{1.} $x^{3}$~by~$x$. + +\Item{2.} $21x^{5}$~by~$7x^{3}$. + +\Item{3.} $35x^{2}$~by~$-5x^{2}$. + +\Item{4.} $-42x^{2}$~by~$6x^{2}$. + +\Item{5.} $-63x^{5}$~by~$-9x$. + +\Item{6.} $-72x^{3}$~by~$-8x^{2}$. + +\Item{7.} $-32a^{2}b^{2}$~by~$8ab^{2}$. + +\Item{8.} $-16x^{3}y^{3}$~by~$-4xy$. + +\Item{9.} $18x^{2}y$~by~$-2xy$. + +\Item{10.} $-25x^{4}y^{2}$~by~$-5x^{3}y^{2}$. + +\Item{11.} $-51x^{2}y^{3}$~by~$-17x^{2}y$. + +\Item{12.} $-28a^{4}b^{3}$~by~$7a^{3}b$. + +\Item{13.} $-36x^{2}y^{6}$~by~$-3xy^{2}$. + +\Item{14.} $-3x^{4}y^{6}$~by~$-5xy^{3}$. + +\Item{15.} $-12a^{2}b^{3}$~by~$8ab^{3}$. + +\Item{16.} $-abcd$~by~$ac$. + +\Item{17.} $-a^{2}b^{3}c^{4}d^{5}$~by~$-ab^{3}c^{3}d^{3}$. + +\Item{18.} $2x^{2}y^{2}z^{3}$~by~$-3xyz^{3}$. + +\Item{19.} $-5a^{5}b^{3}c^{7}$~by~$-a^{4}b^{2}c^{7}$. + +\Item{20.} $52a^{2}m^{3}n^{4}$~by~$13a^{2}m^{2}n^{3}$. + +\Item{21.} $13xy^{2}z^{4}$~by~$39xyz$. + +\Item{22.} $68xc^{2}d^{3}$~by~$-4xcd^{2}$. + +\Item{23.} $-8m^{5}n^{3}p^{2}$~by~$-4m^{5}np$. + +\Item{24.} $-6pqr^{3}$~by~$-2p^{2}qr$. + +\Item{25.} $26a^{2}g^{2}t^{5}$~by~$-2agt^{4}$. + +\Item{26.} $-a^{4}b^{2}c^{3}$~by~$-a^{5}b^{3}c^{4}$. + +\Item{27.} $-3x^{2}y^{2}z^{2}$~by~$-2x^{3}y^{4}z^{5}$. + +\Item{28.} $-6mnp$~by~$-3m^{2}n^{2}p^{2}$. + +\Item{29.} $-17a^{2}b^{3}c^{4}$~by~$51ab^{5}c^{4}$. + +\Item{30.} $-19mg^{2}t^{3}$~by~$57m2^gt^{4}$. +\end{multicols} +%% -----File: 052.png---Folio 46------- + + +\Chapter{IV.}{Addition and Subtraction.} + +\Section{Integral Compound Expressions.} + +\Paragraph{80.} If an algebraic expression contains only \emph{integral +forms}, that is, contains \emph{no letter in the denominator of +any of its terms}, it is called an \Defn{integral expression}. + +Thus, $x^{3} + 7cx^{2} - c^{3} - 5c^{2}x$, is an integral expression. + +Integral and fractional expressions are so named on +account of the \emph{form of the expressions}, and with no reference +whatever to the \emph{numerical value} of the expressions +when definite numbers are put in place of the letters. + +\Paragraph{81. Addition of Integral Compound Expressions.} The addition +of two algebraic expressions can be represented by +connecting the second expression with the first by the sign~$+$. +If there are no like terms in the two expressions, the +operation is \emph{algebraically complete} when the two expressions +are thus connected. + +If, for example, it is required to add $m + n - p$ to +$a + b + c$, the result will be $a + b + c + (m + n - p)$; or, +removing the parenthesis (§~37), $a + b + c + m + n - p$. + +\Paragraph{82.} If there are like terms in the expressions, the like +terms can be \emph{collected}; that is, every set of like terms can +be replaced by a single term with a coefficient equal to +the algebraic sum of the coefficients of the like terms. +%% -----File: 053.png---Folio 47------- + +\Item{1.} Add $6x^{2} + 5x + 4$ to $x^{2} - 4x - 5$. +\begin{DPalign*} +\lintertext{\indent The sum} +&= x^{2} - 4x - 5 + (6x^{2} + 5x + 4) \\ +&= x^{2} - 4x - 5 + 6x^{2} + 5x + 4 +\rintertext{§~37} \\ +&= x^{2} + 6x^{2} - 4x + 5x - 5 + 4 \\ +&= 7x^{2} + x - 1. +\end{DPalign*} + +This process is more conveniently represented by arranging +the terms in columns, so that like terms shall stand in +the same column, as follows: +\[ +\begin{array}{r*{2}{cr}} + x^{2} &-& 4x &-& 5 \\ +6x^{2} &+& 5x &+& 4 \\ +\hline +7x^{2} &+& x &-& 1 \\ +\end{array} +\] + +The coefficient of~$x^{2}$ in the result will be $6 + 1$, or~$7$; the +coefficient of~$x$ will be $-4 + 5$, or~$1$; and the last term is +$-5 + 4$, or~$-1$. + +\begin{Remark}[Note.] When the coefficient of a term is~$1$, it is not written, but +understood. +\end{Remark} + +\Item{2.} Add $2c^{3} - 5c^{2}d + 6cd^{2} + d^{3}$; $c^{3} + 6c^{2}d - 5cd^{2} - 2d^{3}$; +and $3c^{3} - c^{2}d - 7cd^{2} - 3d^{3}$. +\[ +\begin{array}{r*{3}{cr}} +2c^{3} &-& 5c^{2}d &+& 6cd^{2} &+& d^{3} \\ + c^{3} &+& 6c^{2}d &-& 5cd^{2} &-& 2d^{3} \\ +3c^{3} &-& c^{2}d &-& 7cd^{2} &-& 3d^{3} \\ +\hline +6c^{3} & & &-& 6cd^{2} &-& 4d^{3} \\ +\end{array} +\] + +The coefficient of~$c^{3}$ in the result will be $2 + 1 + 3$, or~$6$; +the coefficient of~$c^{2}d$ will be $-5 + 6 - 1$, or~$0$; therefore +$c^{2}d$ will not appear in the result; the coefficient of~$cd^{2}$ will +be $6 - 5 - 7$, or~$-6$; and the coefficient of~$d^{3}$ will be +$1 - 2 - 3$, or~$-4$. +%% -----File: 054.png---Folio 48------- + +\Exercise{18.} + +Find the sum of: + +\Item{1.} $a^{2} - ab + b^{2}$; $a^{2} + ab + b^{2}$. + +\Item{2.} $3a^{2} + 5a-7$; $6a^{2} - 7a + 13$. + +\Item{3.} $x + 2y - 3z$; $-3x + y + 2z$; $2x - 3y + z$. + +\Item{4.} $3x + 2y - z$; $-x + 3y + 2z$; $2x - y + 3z$. + +\Item{5.} $-3a + 2b + c$; $a - 3b + 2c$; $2a + 3b - c$. + +\Item{6.} $-a + 3b + 4c$; $3a - b + 2c$; $2a + 2b - 2c$. + +\Item{7.} $4a^{2} + 3a + 5$; $-2a^{2} + 3a - 8$; $a^{2} - a + 1$. + +\Item{8.} $5ab + 6bc - 7ac$; $3ab - 9bc + 4ac$; $3bc + 6ac$. + +\Item{9.} $x^{3} + x^{2} + x$; $2x^{3} + 3x^{2} - 2x$; $3x^{3} - 4x^{2} + x$. + +\Item{10.} $3y^{2} - x^{2} - 3xy$; $5x^{2} + 6xy - 7y^{2}$; $x^{2} + 2y^{2}$. + +\Item{11.} $2a^{2} - 2ab + 3b^{2}$; $4b^{2} + 5ab - 2a^{2}$; $a^{2} - 3ab - 9b^{2}$. + +\Item{12.} $a^{3} - a^{2} + a - 1$; $a^{2} - 2a + 2$; $3a^{3} + 7a + 1$. + +\Item{13.} $2m^{3} - m^{2} - m$; $4m^{3} + 8m^{2} - 7$; $-3m^{3} + m + 9$. + +\Item{14.} $x^{3} - 3x + 6y$; $x^{2} + 2x - 5y$; $x^{3} - 3x^{2} + 5x$. + +\Item{15.} $6x^{3} - 5x + 1$; $x^{3} + 3x + 4$; $7x^{2} + 2x - 3$. + +\Item{16.} $a^{3} + 3a^{2}b - 3ab^{2}$; $-3a^{2}b - 6ab^{2} - b^{3}$; $3a^{2}b + 4ab^{2}$. + +\Item{17.} $a^{3} - 2a^{2}b - 2ab^{2}$; $a^{2}b - 3ab^{2} - b^{3}$; $3ab^{2} - 2a^{3} - b^{3}$. + +\Item{18.} $7x^{3} - 2x^{2}y + 9xy^{2} + 13y^{3}$; $5x^{2}y - 4xy^{2} - 2x^{3} - 3y^{3}$; +$y^{3} - x^{3} - 3x^{2}y - 5xy^{2}$; $2x^{2}y - 5y^{3} - 2x^{3} - xy^{2}$. + +\Item{19.} Show that $x + y + z = 0$, if $x = a - b - c$, +$y = 2b + 2c - 3a$, and $z = 2a - b - c$. + +\Item{20.} Show that $x + y = 3z$, if $x = 3a^{2} - 6a + 12$, +$y = 9a^{2} + 12a - 21$, and $z = 4a^{2} + 2a - 3$. +%% -----File: 055.png---Folio 49------- + +\Paragraph{83. Subtraction of Integral Compound Expressions.} The +subtraction of one expression from another, if none of the +terms are alike, can be represented only by connecting the +subtrahend with the minuend by means of the sign~$-$. + +If, for example, it is required to subtract $a + b + c$ from +$m + n - p$, the result will be represented by +\[ +m + n - p - (a + b + c); +\] +or, removing the parenthesis (§~38), +\[ +m + n - p - a - b - c. +\] + +If, however, some of the terms in the two expressions are +alike, we can replace two like terms by a single term. + +Thus, suppose it is required to subtract $a^{3} + 2a^{2} + 3a - 5$ +from $2a^{3} - 3a^{2} + 2a - 1$; the result may be expressed as +follows: +\[ +2a^{3} - 3a^{2} + 2a - 1 - (a^{3} + 2a^{2} + 3a - 5); +\] +or, removing the parenthesis (§~38), +\begin{align*} +&2a^{3} - 3a^{2} + 2a - 1 - a^{3} - 2a^{2} - 3a + 5 \\ +&\quad= 2a^{3} - a^{3} - 3a^{2} - 2a^{2} + 2a - 3a - 1 + 5 \\ +&\quad= a^{3} - 5a^{2} - a + 4. +\end{align*} + +This process is more easily performed by writing the subtrahend +below the minuend, \emph{mentally} changing the sign of +each term in the subtrahend, and adding. +\[ +\begin{array}{r*{3}{cr}} +2a^{3} &-& 3a^{2} &+& 2a &-& 1 \\ + a^{3} &+& 2a^{2} &+& 3a &-& 5 \\ +\hline + a^{3} &-& 5a^{2} &-& a &+& 4 \\ +\end{array} +\] + +By changing the sign of each term in the subtrahend, +the coefficient of~$a^{3}$ will be $2 - 1$, or~$1$; the coefficient of~$a^{2}$ +will be $-3 - 2$, or~$-5$; the coefficient of~$a$ will be~$2 - 3$, +or~$-1$; the last term will be $-1 + 5$, or~$4$. +%% -----File: 056.png---Folio 50------- + +Again, suppose it is required to subtract $x^{5} - 2ax^{4} - +3a^{2} x^{3} + 4a^{3} x^{2}$ from $4a^{3} x^{2} - 2a^{2} x^{3} - 5ax^{4}$. Here terms +which are alike can be written in columns, as before: +\[ +\begin{array}{r*{3}{cr}} + &-& 5ax^{4} &-& 2a^{2} x^{3} &+& 4a^{3} x^{2} \\ + x^{5} &-& 2ax^{4} &-& 3a^{2} x^{3} &+& 4a^{3} x^{2} \\ +\hline +-x^{5} &-& 3ax^{4} &+& a^{2} x^{3} & & +\end{array} +\] + +There is no term of~$x^{5}$ in the minuend, hence the coefficient +of~$x^{5}$ in the result will be~$0 - 1$, or~$-1$; the coefficient of~$ax^{4}$ +will be $-5 + 2$, or~$-3$; the coefficient\DPtypo{,}{} of~$a^{2}x^{3}$ will be +$-2 + 3$, or~$+1$; the coefficient of~$a^{3}x^{2}$ will be $-4 + 4$, or~$0$, +and therefore the term~$a^{3}x^{2}$ will not appear in the result. + +\Exercise{19.} + +Subtract: + +\Item{1.} $a - 2b + 3c$ from $2a - 3b + 4c$. + +\Item{2.} $a - 3b - 5c$ from $3a - 5b + c$. + +\Item{3.} $2x - 4y + 6z$ from $4x - y - 2z$. + +\Item{4.} $5x - 11y - 3z$ from $6x - 7y + 2z$. + +\Item{5.} $ab - ac - bc + bd$ from $ab + ac + bc + bd$. + +\Item{6.} $3ab + 2ac - 3bc + bd$ from $5ab - ac + bc + bd$. + +\Item{7.} $2x^{3} - x^{2} - 5x + 3$ from $3x^{3} + 2x^{2} - 3x - 5$. + +\Item{8.} $7x^{2} - 5x + 1 - a$ from $x^{3} - x + 1 - a$. + +\Item{9.} $7b^{3} + 8c^{3} - 15abc$ from $9b^{3} + 3abc - 7c^{3}$. + +\Item{10.} $x^{4} + x - 5x^{3} + 5$ from $7 - 2x^{2} - 3x^{3} + x^{4}$. + +\Item{11.} $a^{3} + b^{3} + c^{3} - 3abc$ from $3abc + a^{3} - 2b^{3} - 3c^{3}$. + +\Item{12.} $2x^{4} - 5x^{2} + 7x - 3$ from $x^{4} + 2 - 2x^{3} - x^{2}$. + +\Item{13.} $1 - x^{5} - x + x^{4} - x^{3}$ from $x^{4} + 1 + x + x^{2}$. + +\Item{14.} $a^{3} - b^{3} + 3a^{2}b - 3ab^{2}$ from $a^{3} + b^{3} - a^{2}b - ab^{2}$. + +\Item{15.} $a^{2} b - ab^{2} - 3a^{3} b^{3} - b^{4}$ from $b^{4} - 5a^{3} b^{3} - 2ab^{2} + a^{2} b$. + +\Item{16.} $-x^{3} + 7x^{2} y - 2y^{3} + 3xy^{2}$ from $3x^{3} + 5y^{3} - xy^{2} + 4x^{2}y$. +%% -----File: 057.png---Folio 51------- + +\Paragraph{84. Parentheses or Brackets.} We have for positive numbers +(§§~37,~38): +\begin{alignat*}{2} +a + (b + c) &= a + b + c,\qquad & \therefore a + b + c &= a + (b + c); \\ +a + (b - c) &= a + b - c, & \therefore a + b - c &= a + (b - c); \\ +a - (b + c) &= a - b - c, & \therefore a - b - c &= a - (b + c); \\ +a - (b - c) &= a - b + c, & \therefore a - b + c &= a - (b - c). +\end{alignat*} + +That is, a parenthesis preceded by~$+$ may be removed +\emph{without changing the sign of any term within the parenthesis}; +and any number of terms may be enclosed within a parenthesis +preceded by the sign~$+$, \emph{without changing the sign +of any term}. + +A parenthesis preceded by the sign~$-$ may be removed, +\emph{provided the sign of every term within the parenthesis is +changed}, namely, $+$~to~$-$, and $-$~to~$+$; and any number +of terms may be enclosed within a parenthesis preceded +by the sign~$-$, \emph{provided the sign of every term enclosed is +changed}. + +The same laws hold for \emph{negative numbers}. + +\Paragraph{85.} Expressions may occur having a parenthesis within +a parenthesis. In such cases parentheses of different shapes +are used, and the beginner when he meets with a branch +of a parenthesis~$($, or bracket~$[$, or brace~$\{$, must look carefully +for the other part, whatever may intervene; and all +that is included between the two parts of each parenthesis +must be treated as the sign before it directs, without regard +to other parentheses. It is best to remove each parenthesis +in succession, \emph{beginning with the innermost}. +\begin{align*} +a - &\bigl\{b - [c - (d - e) + f]\bigr\} \\ + &= a - \bigl\{b - [c - d + e + f]\bigr\} \\ + &= a - \bigl\{b - c + d - e - f\bigr\} \\ + &= a - b + c - d + e + f. +\end{align*} +%% -----File: 058.png---Folio 52------- + +\Exercise{20.} + +Remove the brackets and collect the like terms: + +\Item{1.} $a - b - (b - c) - a + 2b$. + +\Item{2.} $x - [x - (a - b) + a - y]$. + +\Item{3.} $3x - \bigl\{2y - [-7c - 2x] + y\bigr\}$. + +\Item{4.} $5a - [7 - (2b + 5) - 2a]$. + +\Item{5.} $x - [2x + (3a - 2x) - 5a]$. + +\Item{6.} $x - [15y - (13z + 12x)]$. + +\Item{7.} $2a - b + [4c - (b + 2c)]$. + +\Item{8.} $5a - \bigl\{b + [3c - (2b - c)]\bigr\}$. + +\Item{9.} $7x - \bigl\{5y - [3z - (3x + z)]\bigr\}$. + +\Item{10.} $(a - b + c) - (b - a - c) + (a + b - 2c)$. + +\Item{11.} $3x - [-2y - (2y - 3x) + z] + [x - (y - 2z - x)]$. + +\Item{12.} $x - [2x + (x - 2y) + 2y] - 3x - \bigl\{4x - [(x + 2y) - y]\bigr\}$. + +\Item{13.} $x - [y + z - x - (x + y) - z] + (3 x - \Vinc{2y + z})$. + +\begin{Remark}[Note.] +The expression $-\Vinc{2y + z}$ is equivalent to~$-(2y + z)$. +\end{Remark} + +Consider \emph{all the factors} that precede $x$,~$y$, and~$z$, respectively, +as the \emph{coefficients} of these letters, and collect in +brackets the coefficients of each of these letters: + +\Item{14.} $ax + by + cz - ay + az - bx += (a - b)x - (a - b)y + (a + c)z$. + +\Item{15.} $ax + az + by - cz - ay + cx$. + +\Item{16.} $2ax - 3ay - 4by + 5cx - 6bz - 7cz$. + +\Item{17.} $az - bmy + 3 cz - anx - cny + acx$. + +\Item{18.} $mnx - x - mny - y + mnz + z$. +%% -----File: 059.png---Folio 53------- + + +\Chapter{V.}{Multiplication and Division.} + +\Section{Compound Integral Expressions.} + +\Paragraph{86. Multiplication. Polynomials by Monomials.} + +We have for positive numbers (§~39), +\begin{align*} +a(b + c) &= ab + ac, \\ +a(b - c) &= ab - ac. +\end{align*} + +The same law holds for negative numbers. + +\Dictum{To multiply a polynomial by a monomial}, therefore, +\begin{Theorem} +Multiply each term of the polynomial by the monomial, +and add the partial products. +\end{Theorem} + +\Item{1.} Find the product of $ab + ac - bc$ and~$abc$. +\[ +\begin{array}{rcr} +ab + ac - bc && \\ + abc && \\ +\hline +a^{2}b^{2}c + a^{2}bc^{2} &-& ab^{2}c^{2} +\end{array} +\] + +\begin{Remark}[Note.] +We multiply~$ab$, the first term of the multiplicand, by~$abc$, +and work to the right. +\end{Remark} + +\Exercise{21.} + +Find the product of: +\begin{multicols}{2} +\Item{1.} $x + 7$ and $x$. + +\Item{2.} $2x - 3y$ and $4x$. + +\Item{3.} $2x - 3y$ and $7y$. + +\Item{4.} $x - 2a$ and $2a$. + +\Item{5.} $-x + 3b$ and $-b$. + +\Item{6.} $2a^{2} - 3ab$ and $-3a$. + +\Item{7.} $2x^{2} + 3xz$ and $5z$. + +\Item{8.} $a^{2} - 5ab$ and $5ab$. +%% -----File: 060.png---Folio 54------- + +\Item{9.} $x^{2} - 3 xy$ and $-y^{2}$. + +\Item{10.} $2 x^{3} - 3x^{2}$ and $2x^{2}$. + +\Item{11.} $x^{2} - 3y^{2}$ and $4y$. + +\Item{12.} $x^{2} - 3 y^{2}$ and $-x^{2}$. + +\Item{13.} $b^{3} - a^{2}b^{2}$ and $-a^{3}$. + +\Item{14.} $-a^{2}b^{2} - a^{3}$ and $-a^{2}$. + +\Item{15.} $2x^{3} - 3x^{2} + x$ and $2x^{2}$. + +\Item{16.} $a^{2} - 5ab - b^{2}$ and $5ab$. +\end{multicols} + +\Item{17.} $a^{3} + 2a^{2}b + 2ab^{2}$ and $a^{2}$. + +\Item{18.} $a^{3} + 2a^{2}b + 2ab^{2}$ and $b^{3}$. + +\Item{19.} $4x^{2} - 6xy - 9y^{2}$ and $2x$. + +\Item{20.} $-x^{2} - 2xy + y^{2}$ and $-y$. + +\Item{21.} $-a^{3} - a^{2}b^{2} - b^{3}$ and $-a^{2}$. + +\Item{22.} $-x^{2} + 2xy - y^{2}$ and $-y^{2}$. + +\Item{23.} $3 a^{2}b^{2} - 4 ab^{3} + a^{3}b$ and $5 a^{2}b^{2}$. + +\Item{24.} $-ax^{2} + 3axy^{2} - ay^{4}$ and $-3ay^{2}$. + +\Item{25.} $x^{12} - x^{10}y^{3} - x^{3}y^{10}$ and $x^{3}y^{2}$. + +\Item{26.} $-2x^{3} + 3x^{2}y^{2} - 2xy^{3}$ and $-2x^{2}y^{3}$. + +\Item{27.} $a^{3}x^{2}y^{5} - a^{2}xy^{4} - ay^{3}$ and $a^{7}x^{3}y^{5}$. + +\Item{28.} $3a^{2}b^{2} - 2ab^{3} + 5a^{3}b$ and $5a^{2}b^{3}$. + +\Paragraph{87. Multiplication. Polynomials by Polynomials.} + +If we have $m + n +p$ to be multiplied by $a + b + c$, we +may substitute~$M$ for the multiplier $a + b + c$. Then +\[ +M(m + n + p) = Mm + Mn + Mp. +\] + +If now we substitute $a + b + c$ for~$M$, we shall have +\begin{align*} +&(a + b + c) m + (a + b + c) n + (a + b + c) p \\ +&= am + bm + cm + an + bn + cn + ap + bp + cp\DPtypo{.}{} \\ +&= am + an + ap + bm + bn + bp + cm + cn + cp. +\end{align*} + +\Dictum{To find the product of two polynomials}, therefore, +\begin{Theorem} +Multiply every term of the multiplicand by each term of +the multiplier, and add the partial products. +\end{Theorem} +%% -----File: 061.png---Folio 55------- + +\Paragraph{88.} In multiplying polynomials, it is a convenient +arrangement to write the multiplier under the multiplicand, +and place like terms of the partial products in +columns. + +\Item{1.} Multiply $2x - 3y$ by $5x - 4y$. +\[ +\begin{array}{ccrrcr} +2x &-& 3&y && \\ +5x &-& 4&y && \\ +\cline{1-4} +10x^{2} &-& 15&xy & & \\ + &-& 8&xy &+& 12y^{2} \\ +\hline +10x^{2} &-& 23&xy &+& 12y^{2} \\ +\end{array} +\] + +We multiply~$2x$, the first term of the multiplicand, by~$5x$, +the first term of the multiplier, and obtain~$10x^{2}$, then~$-3y$, +the second term of the multiplicand, by~$5x$, and obtain~$-15xy$. +The first line of partial products is $10x^{2} - 15xy$. +In multiplying by~$-4y$, we obtain for a second line of partial +products $-8xy + 12y^{2}$, which is put one place to the +right, so that the like terms $-15xy$~and~$-8xy$ may stand +in the same column. We then add the coefficients of the +like terms, and obtain the complete product in its simplest +form. + +\Item{2.} Multiply $2a + 3 - 4a^{2}$ by $3 - 2a^{2} - 3a$. + +Arrange both multiplicand and multiplier according to +the \emph{ascending} powers of~$a$. +\[ +\begin{array}{r*{4}{cr}} +3 &+& 2a &-& 4a^{2} & & && \\ +3 &-& 3a &-& 2a^{2} & & && \\ +\cline{1-5} +9 &+& 6a &-&12a^{2} & & && \\ + &-& 9a &-& 6a^{2} &+&12a^{3} & & \\ + & & &-& 6a^{2} &-& 4a^{3} &+& 8a^{4} \\ +\hline +9 &-& 3a &-&24a^{2} &+& 8a^{3} &+& 8a^{4} \\ +\end{array} +\] +%% -----File: 062.png---Folio 56------- + +\Item{3.} Multiply $3x + x^{4} - 2x^{2}$ by $x^{3} - 2 - x$. + +Arrange according to the \emph{descending} powers of~$x$. +\[ +\begin{array}{r*{5}{cr}} +x^{4} &-& 2x^{2} &+& \PadTo[l]{3x^4}{3x} && && && \\ +x^{3} &-& \PadTo[c]{2x^2}{x} &-& \PadTo[l]{3x^4}{2}&& && && \\ +\cline{1-5} +x^{7} &-& 2x^{5} &+& 3x^{4} && && && \\ + &-& x^{5} & & &+& 2x^{3} &-& 3x^{2} && \\ + & & &-& 2x^{4} & & &+& 4x^{2} &-& 6x \\ +\hline +x^{7} &-& 3x^{5} &+& x^{4} &+& 2x^{3} &+& x^{2} &-& 6x \\ +\end{array} +\] + +\Item{4.} Multiply $a^{2} + b^{2} + c^{2} - ab - bc - ac$ by $a + b + c$. + +Arrange according to descending powers of~$a$. +\[ +\begin{array}{l*{8}{cr}} +a^{2} &-& ab &-& ac &+& b^{2} &-& bc &+& c^{2} \\ +a &+& b &+& c \\ +\cline{1-11} +a^{3} &-& a^{2}b &-& a^{2}c &+& ab^{2} &-& abc &+& ac^{2} \\ + &+& a^{2}b & & &-& ab^{2} &-& abc & & &+& b^{3} &-& b^{2}c + bc^{2} \\ + & & &+& a^{2}c & & &-& abc &-& ac^{2} & & &+& b^{2}c - bc^{2} &+& c^{3} \\ +\hline +a^{3} & & & & & & &-&3abc & & &+& b^{3} & & &+& c^{3} \\ +\end{array} +\] + +\begin{Remark}[Note.] +The pupil should observe that, with a view to bringing +like terms of the partial products in columns, the terms of the multiplicand +and multiplier are arranged in the \emph{same order}. +\end{Remark} + +\ScreenBreak +\Exercise{22.} + +Find the product of: +\begin{multicols}{2} +\Item{1.} $x + 7$ and $x + 6$. + +\Item{2.} $x - 7$ and $x + 6$. + +\Item{3.} $x + 7$ and $x - 6$. + +\Item{4.} $x - 7$ and $x - 6$. + +\Item{5.} $x + 8$ and $x - 5$. + +\Item{6.} $2x + 3$ and $2x + 3$. + +\Item{7.} $2x - 3$ and $2x - 3$. + +\Item{8.} $2x + 3$ and $2x - 3$. + +\Item{9.} $3x - 2$ and $2 - 3x$. + +\Item{10.} $5x - 3$ and $4x - 7$. + +\Item{11.} $a - 2b$ and $a + 3b$. + +\Item{12.} $a - 7b$ and $a - 5b$. +%% -----File: 063.png---Folio 57------- + +\Item{13.} $5x - 3y$ and $5x - 3y$. + +\Item{14.} $x - b$ and $x - c$. + +\Item{15.} $2m - p$ and $4m - 3p$. + +\Item{16.} $a + b + c$ and $a - c$. + +\Item{17.} $a^{2} - ab + b^{2}$ and $a^{2} + b^{2}$. + +\Item{18.} $x^{3} - 3x^{2} + 7$ and $x^{2} - 3$. + +\Item{19.} $a^{2} + ab + b^{2}$ and $a - b$. + +\Item{20.} $a^{2} - ab + b^{2}$ and $a + b$. +\end{multicols} + +\Item{21.} $x^{2} + 5x - 10$ and $2x^{2} + 3x - 4$. + +\Item{22.} $3x^{3} - 2x^{2} + x$ and $3x^{2} + 2x - 2$. + +\Item{23.} $x^{3} + 2x^{2}y + 3xy^{2}$ and $x^{2} - 2xy + y^{2}$. + +\Item{24.} $a^{2} - 3ab - b^{2}$ and $-a^{2} + ab + 2b^{2}$. + +\Item{25.} $3a^{2}b^{2} + 2 ab^{3} - 5a^{3}b$ and $5a^{2}b^{2} - ab^{3} - b^{4}$. + +\Item{26.} $a^{2} - 2ab + b^{2}$ and $a^{2} + 2ab + b^{2}$. + +\Item{27.} $ab + ac + cd$ and $ab - ac + cd$. + +\Item{28.} $3x^{2}y^{2} + xy^{3} - 2x^{3}y$ and $x^{2}y^{2} + xy^{3} - 3y^{4}$. + +\Item{29.} $x^{2} + 2xy - y^{2}$ and $x^{2} - 2xy + y^{2}$. + +\Item{30.} $3x^{2} + xy - y^{2}$ and $x^{2} - 2xy - 3y^{2}$. + +\Item{31.} $a^{2} - 2ab - b^{2}$ and $b^{2} - 2ab - a^{2}$. + +\Item{32.} $a^{2} + b^{2} + c^{2} - ac$ and $a^{2} - b^{2} - c^{2}$. + +\Item{33.} $a^{2} + 4abx - 4a^{2}b^{2}x^{2}$ and $a^{2} - 4abx + 4a^{2}b^{2}x^{2}$. + +\Item{34.} $3 a^{2} - 2abx + b^{2}x^{2}$ and $2a^{2} + 3abx - 2b^{2}x^{2}$. + +\Item{35.} $2x^{3}y + 4x^{2}y^{2} - 8xy^{3}$ and $2xy^{3} - 3x^{2}y^{2} + 5x^{3}y$. + +\Paragraph{89. Division. Polynomials by Monomials.} +\begin{DPalign*} +\lintertext{\indent Since} +a(b + c - d) &= ab + ac - ad, \\ +\therefore \frac{ab + ac - ad}{a} + &= \frac{ab}{a} + \frac{ac}{a} - \frac{ad}{a} \\ + &= b + c - d. +\end{DPalign*} +%% -----File: 064.png---Folio 58------- + +\Dictum{To divide a polynomial by a monomial}, therefore, +\begin{Theorem} +Divide each term of the dividend by the divisor, and add +the partial quotients. +\end{Theorem} + +Divide $3a^{4}b^{2}c - 9a^{3}bc^{2} - 6a^{2}c^{3}$ by $3a^{2}c$. +\begin{align*} +\frac{3a^{4}b^{2}c - 9a^{3}bc^{2} - 6a^{2}c^{3}}{3a^{2}c} + &= \frac{3a^{4}b^{2}c}{3a^{2}c} + - \frac{9a^{3}bc^{2}}{3a^{2}c} + - \frac{6a^{2}c^{3}}{3a^{2}c} \\ + &= a^{2}b^{2} - 3abc - 2c^{2}. +\end{align*} + +\Exercise{23.} + +Divide: +\begin{multicols}{2} +\Item{1.} $2a^{3} - a^{2}$ by $a$. + +\Item{2.} $42a^{5} - 6a^{2}$ by $6a$. + +\Item{3.} $21x^{4} + 3x^{2}$ by $3x^{2}$. + +\Item{4.} $35m^{4} - 7p^{2}$ by $7$. + +\Item{5.} $27x^{5} - 45x^{4}$ by $9x^{2}$. + +\Item{6.} $24x^{6} - 8x^{3}$ by $-8x^{3}$. + +\Item{7.} $34x^{3} - 51x^{2}$ by $17x$. + +\Item{8.} $5x^{5} - 10x^{3}$ by $-5x^{3}$. + +\Item{9.} $-3a^{2} - 6ac$ by $-3a$. + +\Item{10.} $-5x^{3} + x^{2}y$ by $-x^{2}$. + +\Item{11.} $2a^{5}x^{3} - 2a^{4}x^{2}$ by $2a^{4}x^{2}$. + +\Item{12.} $-x^{2}y - x^{2}y^{2}$ by $-xy$. + +\Item{13.} $9a - 12b + 6c$ by $-3$. + +\Item{14.} $a^{3}b^{2} - a^{2}b^{5} - a^{4}b^{2}$ by $a^{2}b$. + +\Item{15.} $3x^{3} - 6x^{2}y - 9xy^{2}$ by $3x$. + +\Item{16.} $x^{2}y^{2} - x^{3}y - xy^{3}$ by $xy$. + +\Item{17.} $a^{3} - a^{2}b - ab^{2}$ by $-a$. + +\Item{18.} $a^{2}b - ab + ab^{2}$ by $-ab$. + +\Item{19.} $xy - x^{2}y^{2} + x^{3}y^{3}$ by $-xy$. + +\Item{20.} $-x^{6} - 2x^{5} - x^{4}$ by $-x^{4}$. +\end{multicols} + +\Item{21.} $a^{2}x - abx - acx$ by $ax$. + +\Item{22.} $3x^{5}y^{2} - 3x^{4}y^{3} - 3x^{2}y^{4}$ by $3x^{2}y^{2}$. + +\Item{23.} $a^{2}b^{2} - 2ab - 3ab^{3}$ by $ab$. + +\Item{24.} $3a^{3}c^{3} + 3a^{2}c - 3ac^{2}$ by $3ac$. +%% -----File: 065.png---Folio 59------- + +\Paragraph{90. Division. Polynomials by Polynomials.} +\[ +\begin{array}{l*{4}{cr}} +\text{If the divisor (one factor)} + &=& & & a &+& b &+& c, \\ +\text{and the quotient (other factor)} + &=& & & n &+& p &+& q, \\ +\cline{5-9} + & & & &an &+& bn&+& cn \\ +\text{then the dividend (product)} + &=& \smash{\left\{\threelines\right.}\kern-4pt + &+&ap &+& bp&+& cp \\ + & & &+&aq &+& bq&+& cq\rlap{.} \\ +\end{array} +\] + +The first term of the dividend is~$an$; that is, the product +of~$a$, the first term of the divisor, by~$n$, the first term of the +quotient. The first term~$n$ of the quotient is therefore +found by dividing~$an$, the first term of the dividend, by~$a$, +the first term of the divisor. + +If the partial product formed by multiplying the entire +divisor by~$n$ be subtracted from the dividend, the first term +of the remainder~$ap$ is the product of~$a$, the first term of +the divisor, by~$p$, the second term of the quotient; that is, +the second term of the quotient is obtained by dividing the +first term of the remainder by the first term of the divisor. +In like manner, the third term of the quotient is obtained +by dividing the first term of the new remainder by the first +term of the divisor; and so on. + +\Dictum{To divide one polynomial by another}, therefore, +\begin{Theorem} +Arrange both the dividend and divisor in ascending or +descending powers of some common letter. + +Divide the first term of the dividend by the first term of +the divisor. + +Write the result as the first term of the quotient. + +Multiply all the terms of the divisor by the first term of +the quotient. + +Subtract the product from the dividend. + +If there is a remainder, consider it as a new dividend, +and proceed as before. +\end{Theorem} +%% -----File: 066.png---Folio 60------- + +\Paragraph{91.} It is of fundamental importance to arrange the dividend +and divisor \emph{in the same order} with respect to a common +letter, and \emph{to keep this order throughout the operation}. + +The beginner should study carefully the processes in the +following examples: + +\Item{1.} Divide $x^{2} + 18x + 77$ by $x + 7$. +\[ +\begin{array}{r*{2}{cr}|l} +x^{2} &+& 18x &+& 77 & x + 7 \\ +\cline{6-6} +x^{2} &+& 7x & & & x + 11 \\ +\cline{1-5} + & & 11x &+& \NoBar{77} \\ + & & 11x &+& \NoBar{77} \\ +\cline{3-5} +\end{array} +\] + +\begin{Remark}[Note.] +The pupil will notice that by this process we have in +effect separated the dividend into two parts, $x^{2} + 7x$ and $11x + 77$, +and divided each part by $x + 7$, and that the complete quotient is the +sum of the partial quotients $x$~and~$11$. Thus, +\begin{align*} +x^{2} + 18x + 77 + &= x^{2} + 7x + 11x + 77 = (x^{2} + 7x) + (11x + 77). \\ +\therefore \frac{x^{2} + 18x + 77}{x + 7} + &= \frac{x^{2} + 7x}{x + 7} + \frac{11x + 77}{x + 7} = x + 11. +\end{align*} +\end{Remark} + +\Item{2.} Divide $a^{2} - 2ab + b^{2}$ by $a - b$. +\[ +\begin{array}{r*{2}{cr}|l} +a^{2} &-& 2ab &+& b^{2} & a - b \\ +\cline{6-6} +a^{2} &-& ab & & & a - b \\ +\cline{1-5} + &-& ab &+& \NoBar{b^{2}} \\ + &-& ab &+& \NoBar{b^{2}} \\ +\cline{2-5} +\end{array} +\] + +\Item{3.} Divide $a^{4} - ab^{3} + b^{4} + 2a^{2}b^{2} - a^{3}b$ by $a^{2} + b^{2}$. + +Arrange according to the descending powers of~$a$. +\[ +\begin{array}{r*{4}{cr}|l} +a^{4} &-& a^{3}b &+& 2a^{2}b^{2} &-& ab^{3} &+& b^{4} & a^{2} + b^{2} \\ +\cline{10-10} +a^{4} & & &+& a^{2}b^{2} & & & & & a^{2} - ab + b^{2} \\ +\cline{1-9} + &-& a^{3}b &+& a^{2}b^{2} &-& ab^{3} &+& \NoBar{b^{4}} \\ + &-& a^{3}b & & &-& ab^{3} \\ +\cline{2-9} + & & &+& a^{2}b^{2} & & &+& \NoBar{b^{4}} \\ + & & &+& a^{2}b^{2} & & &+& \NoBar{b^{4}} \\ +\cline{4-9} +\end{array} +\] +%% -----File: 067.png---Folio 61------- + +\Item{4.} Divide $10a^{2}b^{2} - 20b^{4} - 17a^{3}b + 6a^{4} + ab^{3}$ +by $2a^{2} - 4b^{2} - 3ab$. + +Arrange according to descending powers of~$a$. +\[ +\begin{array}{r*{4}{cr}|l} +6a^{4} &-&17a^{3}b &+& 10a^{2}b^{2} &+& ab^{3} &-& 20b^{4} & 2a^{2} - 3ab - 4b^{2} \\ +\cline{10-10} +6a^{4} &-& 9a^{3}b &-& 12a^{2}b^{2} & & & & & 3a^{2} - 4ab + 5b^{2} \\ +\cline{1-9} + &-& 8a^{3}b &+& 22a^{2}b^{2} &+& ab^{3} &-& \NoBar{20b^{4}} \\ + &-& 8a^{3}b &+& 12a^{2}b^{2} &+&16ab^{3} \\ +\cline{2-9} + & & & & 10a^{2}b^{2} &-&15ab^{3} &-& \NoBar{20b^{4}} \\ + & & & & 10a^{2}b^{2} &-&15ab^{3} &-& \NoBar{20b^{4}} \\ +\cline{5-9} +\end{array} +\] + +\Item{5.} Divide $5x^{3} - 3x^{4} - 4x^{2} + 1 + x$ by $1 + 2x - 3x^{2}$. + +Arrange according to ascending powers of~$x$. +\[ +\begin{array}{r*{4}{cr}|l*{2}{cr}} +1 &+& x &-& 4x^{2} &+& 5x^{3} &-& 3x^{4} & 1 &+& 2x &-& 3x^{2} \\ +\cline{10-14} +1 &+& 2x &-& 3x^{2} & & & & & 1 &-& x &+& x^{2} \\ +\cline{1-9} + &-& x &-& x^{2} &+& 5x^{3} &-& \NoBar{3x^{4}} \\ + &-& x &-& 2x^{2} &+& 3x^{3} \\ +\cline{2-9} + & & & & x^{2} &+& 2x^{3} &-& \NoBar{3x^{4}} \\ + & & & & x^{2} &+& 2x^{3} &-& \NoBar{3x^{4}} \\ +\end{array} +\] + +\Item{6.} Divide $a^{3} + b^{3} + c^{3} - 3abc$ by $a + b + c$. + +Arrange according to descending powers of~$a$. +\[ +%[** TN: Re-formatted slightly from the original] +\begin{array}{r*{4}{cr}clcr|l} +a^{3} & & & & &-& 3abc & & &+& b^{3} &+& c^{3} & a + b + c \\ +\cline{14-14} +a^{3} &+& a^{2}b &+& a^{2}c & & & & & & & & & +\smash[b]{\begin{aligned}[t] + a^{2} &- ab - ac \\ + &+b^{2} - bc + c^{2} +\end{aligned}} \\ +\cline{1-13} + &-& a^{2}b &-& a^{2}c &-& 3abc & & &+& b^{3} &+& \NoBar{c^{3}} \\ + &-& a^{2}b &-& ab^{2} &-& abc \\ +\cline{2-13} + &-& a^{2}c &+& ab^{2} &-& 2abc & & &+& b^{3} &+& \NoBar{c^{3}} \\ + &-& a^{2}c & & &-& abc &-& ac^{2} \\ +\cline{2-13} + & & & & ab^{2} &-& abc &+& ac^{2} &+& b^{3} &+& \NoBar{c^{3}} \\ + & & & & ab^{2} & & & & &+& b^{3} &+& \NoBar{b^{2}c} \\ +\cline{4-13} + & & & & &-& abc &+& ac^{2} &-& b^{2}c&+& \NoBar{c^{3}} \\ + & & & & &-& abc & & &-& b^{2}c&-& \NoBar{bc^{2}} \\ +\cline{6-13} + & & & & & & & & ac^{2} &+& bc^{2}&+& \NoBar{c^{3}} \\ + & & & & & & & & ac^{2} &+& bc^{2}&+& \NoBar{c^{3}} \\ +\cline{9-13} +\end{array} +\] +%% -----File: 068.png---Folio 62------- + +\Exercise{24.} + +Divide: +\begin{multicols}{2} +\Item{1.} $x^{2} + 15x + 56$ by $x + 7$. + +\Item{2.} $x^{2} - 15x + 56$ by $x - 7$. + +\Item{3.} $x^{2} + x-56$ by $x - 7$. + +\Item{4.} $x^{2} - x-56$ by $x + 7$. + +\Item{5.} $2a^{2} + 11a + 5$ by $2a + 1$. + +\Item{6.} $6a^{2} - 7a-3$ by $2a - 3$. + +\Item{7.} $4a^{2} + 23a + 15$ by $4a + 3$. + +\Item{8.} $3a^{2} - 4a-4$ by $2 - a$. + +\Item{9.} $x^{4} + x^{2} + 1$ by $x^{2} + x + 1$. + +\Item{10.} $x^{8} + x^{4} + 1$ by $x^{4} - x^{2} + 1$. + +\Item{11.} $1 - a^{3}b^{3}$ by $1 - ab$. + +\Item{12.} $x^{3} - 8x-3$ by $x - 3$. +\end{multicols} + +\Item{13.} $a^{2} - 2ab + b^{2} - c^{2}$ by $a - b - c$. + +\Item{14.} $a^{2} + 2ab + b^{2} - c^{2}$ by $a + b + c$. + +\Item{15.} $x^{2} - y^{2} + 2yz - z^{2}$ by $x - y + z$. + +\Item{16.} $c^{4} + 2c^{2} - c + 2$ by $c^{2} - c + 1$. + +\Item{17.} $x^{2} - 4y^{2} - 4yz - z^{2}$ by $x + 2y + z$. + +Arrange and divide: + +\Item{18.} $x^{3} - 6a^{3} + 11a^{2}x - 6ax^{2}$ by $x^{2} + 6a^{2} - 5ax$. + +\Item{19.} $a^{2} - 4b^{2} - 9c^{2} + 12bc$ by $a - 3c + 2b$. + +\Item{20.} $2a^{3} - 8a + a^{4} + 12-7a^{2}$ by $2 + a^{2} - 3a$. + +\Item{21.} $q^{4} + 6q^{3} + 4 + 12q + 13q^{2}$ by $3q + 2 + q^{2}$. + +\Item{22.} $27a^{3} - 8b^{3}$ by $3a - 2b$. + +Find the remainder when: + +\Item{23.} $a^{4} + 9a^{2} + 15-11a - 7a^{3}$ is divided by $a - 5$. + +\Item{24.} $7 - 8c^{2} + 5c^{3} + 8c$ is divided by $5c - 3$. + +\Item{25.} $3 + 11a^{3} + 30a^{4} - 82a^{2} - 5a$ is divided by $3a^{2} - 4 + 2a$. + +\Item{26.} $2x^{3} - 16x + 10-39x^{2} + 17x^{4}$ is divided by $2 - 5x^{2} - 4x$. +%% -----File: 069.png---Folio 63------- + +\Exercise[Miscellaneous Examples.]{25.} + +\Item{1.} Add $2a^{2} - 3ac - 3ab$; $2b^{2} + 3ac + a^{2}$; $-a^{2} - 2b^{2} + 3ab$. + +\Item{2.} Subtract $3a^{4} - 2 a^{3}b + 4 a^{2}b^{2}$ from $4b^{4} - 2 ab^{3} + 4 a^{2}b^{2}$. + +\Item{3.} Simplify $x - y - \{z - x - (y - x + z)\}$. + +\Item{4.} Multiply $a^{2} + b^{2} + c^{2} - d^{2}$ by $a^{2} + b^{2} - c^{2} + d^{2}$. + +\Item{5.} Divide $10y^{6} + 2 - 12y^{5}$ by $1 + y^{2} - 2y$. + +\Item{6.} If $a = 1$, $b = 2$, and $c = -3$, find the value of + $a^{2} + b^{2} + c^{2} + 2ab + 2ac + 2bc$. + +\Item{7.} Simplify $x - (y - z) - \bigl\{4y + [2y - (z - x)]\bigr\}$. + +\Item{8.} Multiply $a^{2} + b^{2} + c^{2} - ab - ac - bc$ by $a + b + c$. + +\Item{9.} Divide $16y^{4} - 21x^{2}y^{2} + 21x^{3}y - 10x^{4}$ by $4y^{2} - 5x^{2} + 3xy$. + +\Item{10.} Add $-2a^{4} + 3a^{3}b - 4a^{2}b^{2}$; $2a^{3}b - 3a^{2}b^{2}$; $7a^{2}b^{2} + 2a^{4} - b^{4}$. + +\Item{11.} From $3x^{3} + 5x - 1$ take the sum of $x - 5 + 5x^{2}$ and +$3 + 4x -3x^{2}$. + +\Item{12.} The minuend is $9c^{2} + 11c - 5$, and the remainder is +$6c^{2} - 13c + 7$. What is the subtrahend? + +\Item{13.} Find the remainder when $a^{4} + 6b^{4}$ is divided by +$a^{2} + 2ab + 2b^{2}$. + +\Item{14.} Multiply $2 - 5x^{2} - 4x$ by $5 + 2x - 3x^{2}$. + +\Item{15.} Divide $a^{6} + a^{5}x + a^{4}x^{2} - a^{3}x^{3} + x^{6}$ by $a^{2} + ax + x^{2}$. + +Bracket the coefficients of the different powers of~$x$: + +\Item{16.} $ax^{3} - cx + bx^{2} - bx^{3} + cx^{2} - x$. + +\Item{17.} $ax^{4} - 2x + bx^{4} - cx - ax^{3} + bx^{3}$. + +\Item{18.} $x^{3} - bx^{2} - cx + bx - cx^{2} + ax^{3}$. +%% -----File: 070.png---Folio 64------- + + +\Chapter[Special Rules in Multiplication and Division.] +{VI.}{Multiplication and Division.} + +\Section{Special Rules.} + +\Paragraph{92. Special Rules of Multiplication.} Some results of multiplication +are of so great utility in shortening algebraic +work that they should be carefully noticed and remembered. +The following are important: + +\Paragraph{93. Square of the Sum of Two Numbers.} +\begin{align*} +(a + b)^{2} + &= (a + b)(a + b) \\ + &= a(a + b) + b(a + b) \\ + &= a^{2} + ab + ab + b^{2} \\ + &= a^{2} + 2ab + b^{2}. +\end{align*} + +Since $a$~and~$b$ stand for any two numbers, we have +\begin{Theorem}[\textsc{Rule 1.}] The square of the sum of two numbers is the +sum of their squares plus twice their product. +\end{Theorem} + +\Paragraph{94. Square of the Difference of Two Numbers.} +\begin{align*} +(a - b)^{2} + &= (a - b) (a - b) \\ + &= a(a - b) - b(a - b) \\ + &= a^{2} - ab - ab + b^{2} \\ + &= a^{2} - 2ab + b^{2}. +\end{align*} + +Hence we have +\begin{Theorem}[\textsc{Rule 2.}] The square of the difference of two numbers is +the sum of their squares minus twice their product. +\end{Theorem} +%% -----File: 071.png---Folio 65------- + +\Paragraph{95. Product of the Sum and Difference of Two Numbers.} +\begin{align*} +(a + b)(a - b) + &= a(a - b) + b(a - b) \\ + &= a^{2} - ab + ab - b^{2} \\ + &= a^{2} - b^{2}. +\end{align*} + +Hence, we have +\begin{Theorem}[\textsc{Rule 3.}] The product of the sum and difference of two +numbers is the difference of their squares. +\end{Theorem} + +If we put $2x$~for~$a$, and $3$~for~$b$, we have +\begin{DPalign*} +\lintertext{\indent Rule 1,} &(2x + 3)^{2} = 4x^{2} + 12x + 9. \\ +\lintertext{\indent Rule 2,} &(2x - 3)^{2} = 4x^{2} - 12x + 9. \\ +\lintertext{\indent Rule 3,} &(2x + 3)(2x - 3) = 4x^{2} - 9. +\end{DPalign*} + +\Exercise{26.} + +Write by inspection the value of: +\begin{multicols}{2} +\Item{1.} $(m + n)^{2}$. + +\Item{2.} $(c - a)^{2}$. + +\Item{3.} $(a + 2c)^{2}$. + +\Item{4.} $(3a - 2b)^{2}$. + +\Item{5.} $(2a + 3b)^{2}$. + +\Item{6.} $(a - 3b)^{2}$. + +\Item{7.} $(2x - y)^{2}$. + +\Item{8.} $(y - 2x)^{2}$. + +\Item{9.} $(a + 5b)^{2}$. + +\Item{10.} $(2a - 5c)^{2}$. + +\Item{11.} $(x + y)(x - y)$. + +\Item{12.} $(4a - b)(4a + b)$. + +\Item{13.} $(2b - 3c)(2b + 3c)$. + +\Item{14.} $(x + 5b)(x + 5b)$. + +\Item{15.} $(y - 2z)(y - 2z)$. + +\Item{16.} $(y + 3z)(y - 3z)$. + +\Item{17.} $(2a - 3b)(2a + 3b)$. + +\Item{18.} $(2a - 3b)(2a - 3b)$. + +\Item{19.} $(2a + 3b)(2a + 3b)$. + +\Item{20.} $(5x + 3a)(5x - 3a)$. +\end{multicols} +%% -----File: 072.png---Folio 66------- + +\Paragraph{96. Product of Two Binomials of the Form $x + a$, $x + b$.} +The product of two binomials which have the form $x + a$, +$x + b$, should be carefully noticed and remembered. +\begin{DPalign*} +\lintertext{\Item{1.}} +(x + 5)(x + 3) + &= x(x + 3) + 5(x + 3) \\ + &= x^{2} + 3x + 5x + 15 \\ + &= x^{2} + 8x + 15. \displaybreak[1] \\ +% +\lintertext{\Item{2.}} +(x - 5)(x - 3) + &= x(x - 3) - 5(x - 3) \\ + &= x^{2} - 3x - 5x + 15 \\ + &= x^{2} - 8x + 15. \displaybreak[1] \\ +% +\lintertext{\Item{3.}} +(x + 5)(x - 3) + &= x(x - 3) + 5(x - 3) \\ + &= x^{2} - 3x + 5x - 15 \\ + &= x^{2} + 2x - 15. \displaybreak[1] \\ +% +\lintertext{\Item{4.}} +(x - 5)(x + 3) + &= x(x + 3) - 5(x + 3) \\ + &= x^{2} + 3x - 5x - 15 \\ + &= x^{2} - 2x - 15. +\end{DPalign*} + +\Item{1.} Each of these results has three terms. + +\Item{2.} The first term of each result is the product of the first +terms of the binomials. + +\Item{3.} The last term of each result is the product of the +second terms of the binomials. + +\Item{4.} The middle term of each result has for a coefficient +the \emph{algebraic sum} of the second terms of the binomials. + +\Paragraph{97.} The intermediate step given above may be omitted, +and the products written at once by \emph{inspection}. Thus, + +\Item{1.} Multiply $x + 8$ by $x + 7$. +\begin{align*} +&8 + 7 = 15,\quad 8 × 7 = 56. \\ +\therefore\ &(x + 8)(x + 7) = x^{2} + 15x + 56. +\end{align*} +%% -----File: 073.png---Folio 67------- + +\Item{2.} Multiply $x - 8$ by $x - 7$. +\begin{align*} +&(-8) + (-7) = -15,\quad (-8)(-7) = +56. \\ +\therefore\ &(x - 8)(x - 7) = x^{2} - 15x + 56. +\end{align*} + +\Item{3.} Multiply $x - 7y$ by $x + 6y$. +\begin{align*} +&-7 + 6 = -1,\quad (-7) × 6y = -42y^{2}. \\ +\therefore\ &(x - 7y)(x + 6y) = x^{2} - xy - 42y^{2}. +\end{align*} + +\Item{4.} Multiply $x + 6y$ by $x - 5y$. +\begin{align*} +&+6 - 5 = 1,\quad 6y × (-5y) = -30y^{2}. \\ +\DPtypo{}{\therefore}\ &(x + 6y)(x - 5y) = x^{2} + xy - 30y^{2}. +\end{align*} + +\Exercise{27.} + +Write by inspection the product of: +\begin{multicols}{2} +\Item{1.} $(x + 7)(x + 4)$. + +\Item{2.} $(x - 3)(x + 7)$. + +\Item{3.} $(x - 2)(x - 4)$. + +\Item{4.} $(x - 6)(x - 10)$. + +\Item{5.} $(x + 7)(x - 4)$. + +\Item{6.} $(x + a)(x - 2a)$. + +\Item{7.} $(x + 3a)(x - a)$. + +\Item{8.} $(a + 3c)(a + 3c)$. + +\Item{9.} $(a + 2x)(a - 4x)$. + +\Item{10.} $(a - 3b)(a - 4b)$. + +\Item{11.} $(a^{2} - c)(a^{2} + 2c)$. + +\Item{12.} $(x - 17)(x - 3)$. + +\Item{13.} $(x + 6y)(x - 5y)$. + +\Item{14.} $(3 + 2x)(3 - x)$. + +\Item{15.} $(5 + 2x)(1 - 2x)$. + +\Item{16.} $(a - 2b)(a + 3b)$. + +\Item{17.} $(a^{2}b^{2} - x^{2})(a^{2}b^{2} - 5x^{2})$. + +\Item{18.} $(a^{3}b - ab^{3})(a^{3}b + 5ab^{3})$. + +\Item{19.} $(x^{2}y - xy^{2})(x^{2}y - 3xy^{2})$. + +\Item{20.} $(x^{2}y + xy^{2})(x^{2}y + xy^{2})$. + +\Item{21.} $(x + a)(x + b)$. + +\Item{22.} $(x + a)(x - b)$. + +\Item{23.} $(x - a)(x + b)$. + +\Item{24.} $(x - a)(x - b)$. + +\Item{25.} $(x + 2a)(x + 2b)$. + +\Item{26.} $(x - 2a)(x + 2b)$. + +\Item{27.} $(x + 2a)(x - 2b)$. + +\Item{28.} $(x - 2a)(x - 2b)$. + +\Item{29.} $(x - a)(x + 3a)$. + +\Item{30.} $(x - 2a)(x + 3a)$. +\end{multicols} +%% -----File: 074.png---Folio 68------- + +\Paragraph{98. Special Rules of Division.} Some results in division +are so important in abridging algebraic work that they +should be carefully noticed and remembered. + +\Paragraph{99. Difference of Two Squares.} + +Since $(a + b)(a - b) = a^{2} - b^{2}$, +\[ +\therefore +\frac{a^{2} - b^{2}}{a + b} = a - b;\quad\text{and}\quad +\frac{a^{2} - b^{2}}{a - b} = a + b. \EqText{Hence\Add{,}} +\] +\begin{Theorem}[\textsc{Rule 1.}] The difference of the squares of two numbers is +divisible by the sum, and by the difference, of the numbers. +\end{Theorem} + +\ScreenBreak +\Exercise{28.} + +Write by inspection the quotient of: +\begin{multicols}{3} +\Item{1.} $\dfrac{x^{2} - 4}{x - 2}$. + +\Item{2.} $\dfrac{x^{2} - 4}{x + 2}$. + +\Item{3.} $\dfrac{a^{2} - 9}{a - 3}$. + +\Item{4.} $\dfrac{a^{2} - 9}{a + 3}$. + +\Item{5.} $\dfrac{c^{2} - 25}{c - 5}$. + +\Item{6.} $\dfrac{c^{2} - 25}{c + 5}$. + +\Item{7.} $\dfrac{49x^{2} - y^{2}}{7x - y}$. + +\Item{8.} $\dfrac{49x^{2} - y^{2}}{7x + y}$. + +\Item{9.} $\dfrac{9b^{2} - 1}{3b - 1}$. + +\Item{10.} $\dfrac{9b^{2} - 1}{3b + 1}$. + +\Item{11.} $\dfrac{16x^{4} - 25a^{2}}{4x^{2} - 5a}$. + +\Item{12.} $\dfrac{16x^{4} - 25a^{2}}{4x^{2} + 5a}$. +\end{multicols} + +\begin{multicols}{2} +\Item{13.} $\dfrac{9x^{2} - 25y^{2}}{3x - 5y}$. + +\Item{14.} $\dfrac{a^{2}-(b - c)^{2}}{a-(b - c)}$. + +\Item{15.} $\dfrac{a^{2}-(b - c)^{2}}{a + (b - c)}$. + +\Item{16.} $\dfrac{a^{2}-(2b - c)^{2}}{a-(2b - c)}$. + +\Item{17.} $\dfrac{(5a - 7b)^{2} - 1}{(5a - 7b) - 1}$. + +\Item{18.} $\dfrac{(5a - 7b)^{2} - 1}{(5a - 7b) + 1}$. + +\Item{19.} $\dfrac{z^{2}-(x - y)^{2}}{z-(x - y)}$. + +\Item{20.} $\dfrac{z^{2}-(x - y)^{2}}{z + (x - y)}$. +%% -----File: 075.png---Folio 69------- + +\Item{21.} $\dfrac{a^{2}-(2b - c)^{2}}{a + (2b - c)}$. + +\Item{22.} $\dfrac{(x + 3y)^{2} - z^{2}}{(x + 3y) - z}$. + +\Item{23.} $\dfrac{(x + 3y)^{2} - z^{2}}{x + 3y + z}$. + +\Item{24.} $\dfrac{(a + 2b)^{2} - 4c^{2}}{(a + 2b) - 2c}$. + +\Item{25.} $\dfrac{(a + 2b)^{2} - 4c^{2}}{(a + 2b) + 2c}$. + +\Item{26.} $\dfrac{1 - (3x - 2y)^{2}}{1 + (3x - 2y)}$. +\end{multicols} + +\Paragraph{100. Difference of Two Cubes.} By performing the division +we have +\[ +\frac{a^{3} - b^{3}}{a - b} = a^{2} + ab + b^{2}. \EqText{Hence,} +\] +\begin{Theorem}[\textsc{Rule 2.}] The difference of the cubes of two numbers is +divisible by the difference of the numbers, and the quotient +is the sum of the squares of the numbers plus their product. +\end{Theorem} + +\Exercise{29.} + +Write by inspection the quotient of: +\begin{multicols}{3} +\Item{1.} $\dfrac{1 - x^{3}}{1 - x}$. + +\Item{2.} $\dfrac{1 - 8a^{3}}{1 - 2a}$. + +\Item{3.} $\dfrac{1 - 27c^{3}}{1 - 2c}$. + +\Item{4.} $\dfrac{8a^{3} - b^{3}}{2a - b}$. + +\Item{5.} $\dfrac{64b^{3} - 27c^{3}}{4b - 3c}$. + +\Item{6.} $\dfrac{27x^{3} - 8y^{3}}{3x - 2y}$. + +\Item{7.} $\dfrac{x^{3}y^{3} - z^{3}}{xy - z}$. + +\Item{8.} $\dfrac{a^{3}b^{3} - 8}{ab - 2}$. + +\Item{9.} $\dfrac{125a^{3} - b^{3}}{5a - b}$. + +\Item{10.} $\dfrac{a^{3} - 8b^{3}}{a - 2b}$. + +\Item{11.} $\dfrac{a^{3} - 64}{a - 4}$. + +\Item{12.} $\dfrac{a^{9} - 27}{a^{3} - 3}$. + +\Item{13.} $\dfrac{a^{12} - x^{6}y^{6}}{a^{4} - x^{2}y^{2}}$. + +\Item{14.} $\dfrac{x^{15} - a^{9}b^{9}}{x^{5} - a^{3}b^{3}}$. + +\Item{15.} $\dfrac{27x^{3}y^{3} - z^{12}}{3xy - z^{4}}$. + +\Item{16.} $\dfrac{x^{3}y^{3}z^{3} - 1}{xyz - 1}$. + +\Item{17.} $\dfrac{8a^{3}b^{3}c^{3} - 27}{2abc - 3}$. + +\Item{18.} $\dfrac{1 - 64x^{3}y^{3}z^{3}}{1 - 4xyz}$. +\end{multicols} +%% -----File: 076.png---Folio 70------- + +\Paragraph{101. Sum of Two Cubes.} By performing the division, +we find that +\[ +\frac{a^{3} + b^{3}}{a + b} = a^{2} - ab + b^{2}. \EqText{Hence,} +\] +\begin{Theorem}[\textsc{Rule 3.}] The sum of the cubes of two numbers is divisible +by the sum of the numbers, and the quotient is the sum +of the squares of the numbers minus their product. +\end{Theorem} + +\Exercise{30.} + +Write by inspection the quotient of: +\begin{multicols}{3} +\Item{1.} $\dfrac{1 + x^{3}}{1 + x}$. + +\Item{2.} $\dfrac{1 + 8a^{3}}{1 + 2a}$. + +\Item{3.} $\dfrac{1 + 27c^{3}}{1 + 3c}$. + +\Item{4.} $\dfrac{8a^{3} + b^{3}}{2a + b}$. + +\Item{5.} $\dfrac{64b^{3} + 27c^{3}}{4b + 3c}$. + +\Item{6.} $\dfrac{27x^{3} + 8y^{3}}{3x + 2y}$. + +\Item{7.} $\dfrac{8x^{3} + 125y^{3}}{2x + 5y}$. + +\Item{8.} $\dfrac{x^{3}y^{3} + z^{3}}{xy + z}$. + +\Item{9.} $\dfrac{a^{3}b^{3} + 8}{ab + 2}$. + +\Item{10.} $\dfrac{125a^{3} + b^{3}}{5a + b}$. + +\Item{11.} $\dfrac{a^{3} + 8b^{3}}{a + 2b}$. + +\Item{12.} $\dfrac{a^{6} + 64}{a^{2} + 4}$. + +\Item{13.} $\dfrac{a^{9} + 27}{a^{3} + 3}$. + +\Item{14.} $\dfrac{8a^{6} + b^{3}}{2a^{2} + b}$. + +\Item{15.} $\dfrac{a^{12} + x^{6}y^{6}}{a^{4} + x^{2}y^{2}}$. + +\Item{16.} $\dfrac{x^{15} + a^{9}b^{9}}{x^{5} + a^{3}b^{3}}$. + +\Item{17.} $\dfrac{27x^{3}y^{3} + z^{12}}{3xy + z^{4}}$. + +\Item{18.} $\dfrac{x^{3}y^{3}z^{3} + 1}{xyz + 1}$. + +\Item{19.} $\dfrac{8a^{3}b^{3}c^{3} + 27}{2abc + 3}$. + +\Item{20.} $\dfrac{1 + 64x^{3}y^{3}z^{3}}{1 + 4xyz}$. + +\Item{21.} $\dfrac{1 + 27a^{6}b^{3}c^{3}}{1 + 3a^{2}bc}$. +\end{multicols} + +Find by division the quotient of: +\begin{multicols}{3} +\Item{22.} $\dfrac{x^{4} - y^{4}}{x - y}$. + +\Item{23.} $\dfrac{x^{4} - y^{4}}{x + y}$. + +\Item{24.} $\dfrac{x^{5} - y^{5}}{x - y}$. + +\Item{25.} $\dfrac{x^{5} + y^{5}}{x + y}$. + +\Item{26.} $\dfrac{x^{6} - y^{6}}{x - y}$. + +\Item{27.} $\dfrac{x^{6} - y^{6}}{x + y}$. +\end{multicols} +%% -----File: 077.png---Folio 71------- + + +\Chapter{VII.}{Factors.} + +\Paragraph{102. Rational Expressions.} An expression is \emph{rational} when +none of its terms contain square or other roots. + +\Paragraph{103. Factors of Rational and Integral Expressions.} By factors +of a given integral number in arithmetic we mean +integral numbers that will divide the given number without +remainder. Likewise by factors of a rational and integral +expression in algebra we mean rational and integral +expressions that will divide the given expression without +remainder. + +\Paragraph{104. Factors of Monomials.} The factors of a monomial +may be found by inspection. Thus, the factors of~$21a^{2}b$ +are $3$, $7$, $a$, $a$, and~$b$. + +\Paragraph{105. Factors of Polynomials.} The form of a polynomial +that can be resolved into factors often suggests the process +of finding the factors. + + +\Section{Case I.} + +\Paragraph{106. When all the terms have a common factor.} + +\Item{1.} Resolve into factors $3a^{2} - 6ab$. + +Since $3a$~is seen to be a factor of each term, we have +\begin{align*} +\frac{3a^{2} - 6ab}{3a} &= \frac{3a^{2}}{3a} - \frac{6ab}{3a} = a - 2b. \\ +\therefore\ 3a^{2} - 6ab &= 3a(a - 2b). +\end{align*} + +Hence, the required factors are $3a$~and~$a - 2b$. +%% -----File: 078.png---Folio 72------- + +\Item{2.} Resolve into factors $4x^{3} + 12x^{2} - 8x$. + +Since $4x$ is seen to be a factor of each term, we have +\begin{align*} +\frac{4x^{3} + 12x^{2} - 8x}{4x} + &= \frac{4x^{3}}{4x} + \frac{12x^{2}}{4x} - \frac{8x}{4x} \\ + &= x^{2} + 3x - 2. \\ +\therefore\ +4x^{3} + 12x^{2} - 8x &= 4x(x^{2} + 3x - 2). +\end{align*} + +Hence the required factors are $4x$~and~$x^{2} + 3x - 2$. + +\Exercise{31.} + +Resolve into two factors: +\begin{multicols}{2} +\Item{1.} $2x^{2} - 4x$. + +\Item{2.} $3a^{3} - 6a$. + +\Item{3.} $5a^{2}b^{2} - 10a^{3}b^{3}$. + +\Item{4.} $3x^{2}y + 4xy^{2}$. + +\Item{5.} $8a^{3}b^{2} + 4a^{2}b^{3}$. + +\Item{6.} $3a^{4} - 12a^{2} - 6a^{3}$. + +\Item{7.} $4x^{2} - 8x^{4} - 12x^{5}$. + +\Item{8.} $5 - 10x^{2}y^{2} + 15x^{2}y$. + +\Item{9.} $7a^{2} + 14a - 21a^{3}$. + +\Item{10.} $3x^{3}y^{3} - 6x^{4}y^{4} - 9x^{2}y^{2}$. +\end{multicols} + +\Section{Case II.} + +\Paragraph{107. When the terms can be grouped so as to show a common +factor in each group.} + +\Item{1.} Resolve into factors $ac + ad + bc + bd$. +\begin{align*} +ac + ad + bc + bd + &= (ac + ad) + (bc + bd) + \Tag{(1)} \\ + &= a(c + d) + b(c + d) + \Tag{(2)} \\ + &= (a + b)(c + d). + \Tag{(3)} +\end{align*} + +\begin{Remark}[Note.] The first two terms of $ac + ad + bc + bd$ are seen to +have the common factor~$a$, and the last two terms, the common factor~$b$. +Hence we bracket the first two terms and also the last two +terms. Then we take out the factor~$a$ from $(ac + ad)$ and $b$~from +$(bc + bd)$, and get equation~(2). Since one factor is seen in~(2) to be +$c + d$, dividing by $c + d$, we obtain the other factor, $a + b$. +\end{Remark} +%% -----File: 079.png---Folio 73------- + +\Item{2.} Find the factors of $ac + ad - bc - bd$. +\begin{align*} +ac + ad - bc - bd + &= (ac + ad) - (bc + bd) \\ + &= a(c + d) - b(c + d) \\ + &= (a - b)(c + d). +\end{align*} + +\begin{Remark}[Note.] Here the last two terms, $-bc - bd$, being put within a +parenthesis preceded by the sign~$-$, have their signs changed to~$+$. +\end{Remark} + +\Item{3.} Resolve into factors $2x^{3} - 3x^{2} - 4x + 6$. +\begin{align*} +2x^{3} - 3x^{2} - 4x + 6 + &= (2x^{3} - 3x^{2}) - (4x - 6) \\ + &= x^{2}(2x - 3) - 2(2x - 3) \\ + &= (x^{2} - 2)(2x - 3). +\end{align*} + +\Item{4.} Resolve into factors $x^{3} + x^{2} - ax - a$. +\begin{align*} +x^{3} + x^{2} - ax - a + &= (x^{3} + x^{2}) - (ax + a) \\ + &= x^{2}(x + 1) - a(x + 1) \\ + &= (x^{2} - a)(x + 1). +\end{align*} + +\Item{5.} Resolve into factors $x^{3} + 3ax^{2} + x + 3a$. +\begin{align*} +x^{3} + 3ax^{2} + x + 3a + &= (x^{3} + 3ax^{2}) + (x + 3a) \\ + &= x^{2}(x + 3a) + 1(x + 3a) \\ + &= (x^{2} + 1)(x + 3a). +\end{align*} + +\Exercise{32.} + +Resolve into factors: +\begin{multicols}{2} +\Item{1.} $x^{3} + x^{2} + x + 1$. + +\Item{2.} $x^{3} - x^{2} + x - 1$. + +\Item{3.} $x^{2} + xy + xz + yz$. + +\Item{4.} $ax - bx - ay + by$. + +\Item{5.} $a^{2} - ac + ab - bc$. + +\Item{6.} $x^{2} - bx + 3x - 3b$. + +\Item{7.} $2x^{3} - x^{2} + 4x - 2$. + +\Item{8.} $a^{2} - 3a - ab + 3b$. + +\Item{9.} $6a^{2} + 2ab - 3ac - bc$. + +\Item{10.} $abxy + cxy + abc + c^{2}$. + +\Item{11.} $ax - ay - bx + cy - cx + by$. + +\Item{12.} $(a - b)^{2} - 2c(a - b)$. +\end{multicols} +%% -----File: 080.png---Folio 74------- + + +\Section{Case III.} + +\Paragraph{108. When a binomial is the difference of two squares.} + +\Item{1.} Resolve into factors $x^{2} - y^{2}$. +\begin{DPgather*} +\lintertext{\indent Since,} +(x + y)(x - y) = x^{2} - y^{2}, +\end{DPgather*} +the factors of $x^{2} - y^{2}$ are $x + y$ and~$x - y$. + +\Dictum{To find the factors of a binomial when it is the difference of +two squares}, therefore, +\begin{Theorem} +Take the square root of the first term and the square root +of the second term. + +The sum of these roots will form the first factor; + +The difference of these roots will form the second factor. +\end{Theorem} + +\Paragraph{109.} The \Defn{square root} of a \emph{monomial} is one of the \textbf{two equal +factors} of the monomial. + +Thus $9x^{8}y^{2} = 3x^{4}y × 3x^{4}y$; and $3x^{4}y$ is the square root +of~$9x^{8}y^{2}$. + +The rule for extracting the square root of a \emph{monomial, +when a perfect square}, is as follows: +\begin{Theorem} +Extract the square root of the coefficient, and divide the +index of each letter by~$2$. +\end{Theorem} + +\Exercise{33.} + +Resolve into factors: +\begin{multicols}{3} +\Item{1.} $4 - x^{2}$. + +\Item{2.} $9 - x^{2}$. + +\Item{3.} $9a^{2} - x^{2}$. + +\Item{4.} $25 - x^{2}$. + +\Item{5.} $25x^{2} - a^{2}$. + +\Item{6.} $16a^{4} - 121$. + +\Item{7.} $121a^{4} - 16$. + +\Item{8.} $4a^{2}b^{2} - c^{2}d^{2}$. + +\Item{9.} $1 - x^{2}y^{2}$. + +\Item{10.} $81x^{2}y^{2} - 1$. + +\Item{11.} $49a^{2}b^{2} - 4$. + +\Item{12.} $25a^{4}b^{4} - 9$. +\end{multicols} +%% -----File: 081.png---Folio 75------- + +\begin{multicols}{2} +\Item{13.} $9a^{8}b^{6} - 16x^{10}$. + +\Item{14.} $144x^{2}y^{2} - 1$. + +\Item{15.} $100x^{6}y^{2}z^{4} - 1$. + +\Item{16.} $1 - 121a^{4}b^{8}c^{12}$. + +\Item{17.} $25a^{2} - 64x^{6}y^{6}$. + +\Item{18.} $16x^{16}-25y^{18}$. +\end{multicols} + +Find, by resolving into factors, the value of: +\begin{multicols}{2} +\Item{19.} $(375)^{2} - (225)^{2}$. + +\Item{20.} $(579)^{2} - (559)^{2}$. + +\Item{21.} $(873)^{2} - (173)^{2}$. + +\Item{22.} $(101)^{2} - (99)^{2}$. + +\Item{23.} $(7244)^{2} - (7242)^{2}$. + +\Item{24.} $(3781)^{2} - (219)^{2}$. +\end{multicols} + +\Paragraph{110.} If the squares are compound expressions, the same +method may be employed. + +\Item{1.} Resolve into factors $(x + 3y)^{2} - 16a^{2}$. +\begin{Soln} +The square root of the first term is~$x + 3y$. + +The square root of the second term is~$4a$. + +The sum of these roots is~$x + 3y - 4a$. + +The difference of these roots is $x + 3y - 4a$. + +Therefore $(x + 3y)^{2} - 16a^{2} = (x + 3y + 4a)(x + 3y - 4a)$. +\end{Soln} + +\Item{2.} Resolve into factors $a^{2} - (3b - 5c)^{2}$. +\begin{Soln} +The square roots of the terms are $a$~and~$(3b - 5c)$. + +The sum of these roots is $a + (3b - 5c)$, or $a + 3b - 5c$. + +The difference of these roots is $a - (3b - 5c)$, or $a - 3b + 5c$. + +Therefore $a^{2} - (3b - 5c)^{2} = (a + 3b - 5c)(a - 3b + 5c)$. +\end{Soln} + +\Exercise{34.} + +\DPtypo{}{Resolve into factors:} +\begin{multicols}{2} +\Item{1.} $(x + y)^{2} - z^{2}$. + +\Item{2.} $(x - y)^{2} - z^{2}$. + +\Item{3.} $z^{2} - (x + y)^{2}$. + +\Item{4.} $z^{2} - (x - y)^{2}$. + +\Item{5.} $(x + y)^{2} - 4z^{2}$. + +\Item{6.} $4z^{2} - (x - y)^{2}$. + +\Item{7.} $(a + 2b)^{2} - c^{2}$. + +\Item{8.} $(a - 2b)^{2} - c^{2}$. + +\Item{9.} $c^{2} - (a - 2b)^{2}$. + +\Item{10.} $(2a + 5c)^{2} - 1$. +%% -----File: 082.png---Folio 76------- + +\Item{11.} $1 - (2a - 5c)^{2}$. + +\Item{12.} $(a + 3b)^{2} - 16c^{2}$. + +\Item{13.} $(a - 5b)^{2} - 9c^{2}$. + +\Item{14.} $16c^{2} - (a - 5b)^{2}$. + +\Item{15.} $4a^{2} - (x + y)^{2}$. + +\Item{16.} $b^{2} - (a - 2x)^{2}$. + +\Item{17.} $4z^{2} - (x + 3y)^{2}$. + +\Item{18.} $9 - (3a - 7b)^{2}$. + +\Item{19.} $16a^{2} - (2b + 5c)^{2}$. + +\Item{20.} $25c^{2} - (3a - 2x)^{2}$. + +\Item{21.} $9a^{2} - (3b - 5c)^{2}$. + +\Item{22.} $16y^{2} - (a - 3c)^{2}$. + +\Item{23.} $49m^{2} - (p + 2q)^{2}$. + +\Item{24.} $36n^{2} - (d - 2c)^{2}$. + +\Item{25.} $(x + y)^{2} - (a + b)^{2}$. + +\Item{26.} $(x - y)^{2} - (a - b)^{2}$. + +\Item{27.} $(2x + 3)^{2} - (2a + b)^{2}$. + +\Item{28.} $(b - c)^{2} - (a - 2x)^{2}$. + +\Item{29.} $(3x - y)^{2} - (2a - b)^{2}$. + +\Item{30.} $(x - 3y)^{2} - (a + 2b)^{2}$. + +\Item{31.} $(x + 2y)^{2} - (a + 3b)^{2}$. + +\Item{32.} $(x + y)^{2} - (a - z)^{2}$. +\end{multicols} + + +\Section{Case IV.} + +\Paragraph{111. When a binomial is the difference of two cubes.} +\begin{DPgather*} +\lintertext{\indent Since} +\frac{a^{3} - b^{3}}{a - b} = a^{2} + ab + b^{2}, +\end{DPgather*} +the factors of $a^{3} - b^{3}$ are $a - b$ and $a^{2} + ab + b^{2}$. + +In like manner we can resolve into factors any expression +which can be written as the difference of two cubes. + +\Paragraph{112.} The rule for extracting the cube root of a \emph{monomial, +when the monomial is a perfect cube}, is, +\begin{Theorem} +Extract the cube root of the coefficient, and divide the index +of each letter by~$3$. +\end{Theorem} + +\ScreenBreak +\Item{1.} Resolve into factors $8a^{3} - 27b^{6}$. + +Since $8a^{3} = (2a)^{3}$, and $27b^{6} = (3b^{2})^{3}$, we can write +$8a^{3} - 27b^{6}$ as $(2a)^{3} - (3b^{2})^{3}$. +%% -----File: 083.png---Folio 77------- +\begin{DPgather*} +\lintertext{\indent Since} +a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2}), +\end{DPgather*} +we have, by putting $2a$ for~$a$ and $3b^{2}$ for~$b$, +\begin{align*} +(2a)^{3} - (3b^{2})^{3} + &= (2a - 3b^{2})[(2a)^{2} + 2a × 3b^{2} + (3b^{2})^{2}] \\ + &= (2a - 3b^{2})(4a^{2} + 6ab^{2} + 9b^{4}). +\end{align*} + +\Item{2.} Resolve into factors $64x^{3} - 1$. +\begin{align*} +64x^{3} - 1 + &= (4x)^{3} - 1 \\ + &= (4x - 1)[(4x)^{2} + 4x +1] \\ + &= (4x - 1)(16x^{2} + 4x + 1). +\end{align*} + +\Dictum{To find the factors of a binomial when it is the difference of +two cubes}, therefore, +\begin{Theorem} +Take the difference of the cube roots of the terms for one +factor, and the sum of the squares of the cube roots of the +terms plus their product for the other factor. +\end{Theorem} + +\ScreenBreak +\Exercise{35.} + +Resolve into factors: +\begin{multicols}{3} +\Item{1.} $8x^{3} - y^{3}$. + +\Item{2.} $x^{3} - 1$. + +\Item{3.} $x^{3}y^{3} - z^{3}$. + +\Item{4.} $x^{3} - 64$. + +\Item{5.} $125a^{3} - b^{3}$. + +\Item{6.} $a^{3} - 343$. + +\Item{7.} $a^{3}b^{3} - 27c^{3}$. + +\Item{8.} $x^{3}y^{3}z^{3} - 8$. + +\Item{9.} $8a^{3}b^{3} - 27y^{6}$. + +\Item{10.} $64x^{3} - y^{9}$. + +\Item{11.} $27a^{3} - 64c^{6}$. + +\Item{12.} $x^{3}y^{3} - 216z^{3}$. + +\Item{13.} $64x^{3} - 729y^{3}$. + +\Item{14.} $27a^{3} - 512c^{3}$. + +\Item{15.} $8x^{6} - 125y^{3}$. + +\Item{16.} $64x^{12} - 27y^{15}$. + +\Item{17.} $216 - 8a^{3}$. + +\Item{18.} $343 - 27y^{3}$. +\end{multicols} + + +\Section{Case V.} + +\Paragraph{113. When a binomial is the sum of two cubes.} +\begin{DPgather*} +\lintertext{\indent Since} +\frac{a^{3} + b^{3}}{a + b} = a^{2} - ab + b^{2}, +\end{DPgather*} +the factors of $a^{3} + b^{3}$ are $a + b$ and $a^{2} - ab + b^{2}$. +%% -----File: 084.png---Folio 78------- + +In like manner we can resolve into factors any expression +which can be written as the sum of two cubes. + +\Item{1.} Resolve into factors $8x^{3} + 27y^{3}$. + +Since by §~112, $8x^{3} = (2x)^{3}$ and $27y^{3} = (3y)^{3}$, we can +write $8x^{3} + 27y^{3}$ as $(2x)^{3} + (3y)^{3}$. + +\begin{DPgather*} +\lintertext{\indent Since} +a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2}), +\end{DPgather*} +we have, by putting $2x$ for~$a$, and $3y$ for~$b$, +\begin{align*} +(2x)^{3} + (3y)^{3} + &= (2x + 3y)[(2x)^{2} - 2x × 3y + (3y)^{2}] \\ + &= (2x + 3y)(4x^{2} - 6xy + 9y^{2}). +\end{align*} + +\Item{2.} Resolve into factors $125a^{3} + 64x^{6}$\Add{.} +\begin{gather*} +125a^{3} = (5a)^{3},\quad 64x^{6} = (4x^{2})^{3}; \\ +\begin{aligned} +\therefore 125a^{3} + 64x^{6} + &= (5a + 4x^{2})[(5a)^{2} - 5a × 4x^{2} + (4x^{2})^{2}] \\ + &= (5a + 4x^{2})(25a^{2} - 20ax^{2} + 16x^{4}) +\end{aligned} +\end{gather*} + +\Dictum{To find the factors of a binomial when it is the sum of two +cubes}, therefore, +\begin{Theorem} +Take the sum of the cube roots of the terms for one factor, +and the sum of the squares of the cube roots of the terms +minus their product for the other factor. +\end{Theorem} + +\PrintBreak +\Exercise{36.} + +Resolve into factors: +\begin{multicols}{3} +\Item{1.} $x^{3} + 1$. + +\Item{2.} $8x^{3} + y^{3}$. + +\Item{3.} $x^{3} + 125$. + +\Item{4.} $64a^{3} + 27$. + +\Item{5.} $x^{3}y^{3} + z^{3}$. + +\Item{6.} $a^{3} + 64$. + +\Item{7.} $8a^{6} + b^{3}$. + +\Item{8.} $x^{3} + 343$. + +\Item{9.} $8 + x^{3}y^{3}z^{3}$. + +\Item{10.} $y^{9} + 64x^{3}$. + +\Item{11.} $a^{3}b^{3} + 27x^{3}$. + +\Item{12.} $8y^{3}z^{3} + x^{6}$. + +\Item{13.} $y^{9} + 64x^{6}$. + +\Item{14.} $64a^{12} + x^{15}$. + +\Item{15.} $27x^{15} + 8a^{6}$. + +\Item{16.} $27x^{9} + 512$. + +\Item{17.} $343 + 64x^{3}$. + +\Item{18.} $125 + 27y^{3}$. +\end{multicols} +%% -----File: 085.png---Folio 79------- + + +\Section{Case VI.} + +\Paragraph{114. When a trinomial is a perfect square.} +\begin{DPgather*} +\lintertext{\indent Since} +(x + y)^{2} = x^{2} + 2xy + y^{2}, +\end{DPgather*} +the factors of $x^{2} + 2xy + y^{2}$ are $x + y$ and $x + y$. + +\begin{DPgather*} +\lintertext{\indent Since} +(x - y)^{2} = x^{2} - 2xy + y^{2}, +\end{DPgather*} +the factors of $x^{2} - 2xy + y^{2}$ are $x - y$ and $x - y$. + +Therefore, a trinomial is a perfect square, if its first and +last terms are perfect squares and positive, and its middle +term is twice the product of their square roots. + +\Dictum{To find the factors of a trinomial when it is a perfect square}, +therefore, +\begin{Theorem} +Extract the square roots of the first and last terms, and +connect these square roots by the sign of the middle term. +\end{Theorem} + +Thus, if we wish to find the square root of +\[ +16a^{2} - 24ab + 9b^{2}, +\] +we take the square roots of $16a^{2}$ and $9b^{2}$, which are $4a$ +and~$3b$, respectively, and connect these square roots by +the minus sign, the sign of the middle term. The square +root is therefore +\[ +4a - 3b. +\] + +Again, if we wish to find the square root of +\[ +25x^{2} + 40xy + 16y^{2}, +\] +we take the square roots of $25x^{2}$ and $16y^{2}$ and connect these +roots by the plus sign, the sign of the middle term. The +square root is therefore +\[ +5x + 4y. +\] +%% -----File: 086.png---Folio 80------- + +\Exercise{37.} + +Resolve into factors: +\begin{multicols}{2} +\Item{1.} $4x^{2} + 4xy + y^{2}$. + +\Item{2.} $x^{2} + 6xy + 9y^{2}$. + +\Item{3.} $x^{2} + 16x + 64$. + +\Item{4.} $x^{2} + 10ax + 25a^{2}$. + +\Item{5.} $a^{2} - 16a + 64$. + +\Item{6.} $a^{2} - 10ab + 25b^{2}$. + +\Item{7.} $c^{2} - 6cd + 9d^{2}$. + +\Item{8.} $4x^{2} - 4x + 1$. + +\Item{9.} $4a^{2} - 12ab + 9b^{2}$. + +\Item{10.} $9a^{2} - 24ab + 16b^{2}$. + +\Item{11.} $x^{2} + 8xy + 16y^{2}$. + +\Item{12.} $x^{2} - 8xy + 16y^{2}$. + +\Item{13.} $4x^{2} - 20xy + 25y^{2}$. + +\Item{14.} $1 + 20a + 100a^{2}$. + +\Item{15.} $49a^{2} - 28a + 4$. + +\Item{16.} $36a^{2} + 60ab + 25b^{2}$. + +\Item{17.} $81x^{2} - 36bx + 4b^{2}$. + +\Item{18.} $m^{2}n^{2} + 14mnx^{2} + 49x^{2}$. +\end{multicols} + + +\PrintBreak +\Section{Case VII.} + +\Paragraph{115. When a trinomial has the form $x^{2} + ax + b$.} + +Where $a$ is the \emph{algebraic sum} of two numbers, and is +either positive or negative; and $b$~is the \emph{product} of these +two numbers, and is either positive or negative. +\begin{DPgather*} +\lintertext{\indent Since} +(x + 5)(x + 3) = x^{2} + 8x + 15, +\end{DPgather*} +the factors of $x^{2} + 8x + 15$ are $x + 5$ and $x + 3$. +\begin{DPgather*} +\lintertext{\indent Since} +(x + 5)(x - 3) = x^{2} + 2x- 15, +\end{DPgather*} +the factors of $x^{2} + 2x - 15$ are $(x + 5)$ and $(x - 3)$. + +Hence, if a trinomial of the form $x^{2} + ax + b$ is such an +expression that it can be resolved into two binomial factors, +it is obvious that the first term of each factor will be~$x$, +and that the second terms of the factors will be two +numbers whose product is~$b$, the last term of the trinomial, +and whose algebraic sum is~$a$, the coefficient of~$x$ in the +middle term of the trinomial. +%% -----File: 087.png---Folio 81------- + +\Item{1.} Resolve into factors $x^{2} + 11x + 30$. +\begin{Soln} +We are required to find two numbers whose product is~$30$ and +whose sum is~$11$. + +Two numbers whose product is $30$ are $1$~and~$30$, $2$~and~$15$, $3$~and~$10$, +$5$~and~$6$, and the sum of the last two numbers is~$11$. Hence, +\[ +x^{2} + 11x + 30 = (x + 5)(x + 6). +\] +%[** TN: Solutions sometimes printed in normal-size type; using smaller type] +\end{Soln} + +\Item{2.} Resolve into factors $x^{2} - 7x + 12$. +\begin{Soln} +We are required to find two numbers whose product is~$12$ and +whose algebraic sum is~$-7$. + +Since the product is~$+12$, the two numbers are \emph{both positive} or \emph{both +negative}, and since their sum is~$-7$, they must both be negative. + +Two negative numbers whose product is~$12$ are $-12$~and~$-1$, $-6$ +and~$-2$, $-4$~and~$-3$, and the sum of the last two numbers is~$-7$. +Hence, +\[ +x^{2} - 7x + 12 = (x - 4)(x - 3). +\] +\end{Soln} + +\Item{3.} Resolve into factors $x^{2} + 2x - 24$. +\begin{Soln} +We are required to find two numbers whose product is~$-24$ and +whose algebraic sum is~$2$. + +Since the product is~$-24$, one of the numbers is positive and the +other negative, and since their sum is~$+2$, the larger number is +positive. + +Two numbers whose product is~$-24$, and the larger number positive, +are $24$~and~$-1$, $12$~and~$-2$, $8$~and~$-3$, $6$~and~$-4$, and the sum +of the last two numbers is~$+2$. Hence, +\[ +x^{2} + 2x - 24 = (x + 6)(x - 4). +\] +\end{Soln} + +\Item{4.} Resolve into factors $x^{2} - 3x - 18$. +\begin{Soln} +Since the product is~$-18$, one of the numbers is positive and the +other negative, and since their sum is~$-3$, the larger number is +negative. + +Two numbers whose product is~$-18$, and the larger number negative, +are $-18$~and~$1$, $-9$~and~$2$, $-6$~and~$3$, and the sum of the last +two numbers is~$-3$. Hence, +\[ +x^{2} - 3x - 18 = (x - 6)(x + 3). +\] +\end{Soln} +%% -----File: 088.png---Folio 82------- + +\Exercise{38.} + +Resolve into factors: +\begin{multicols}{2} +\Item{1.} $a^{2} + 5a + 6$. + +\Item{2.} $a^{2} - 5a + 6$. + +\Item{3.} $a^{2} + 6a + 5$. + +\Item{4.} $a^{2} - 6a + 5$. + +\Item{5.} $a^{2} + 4a - 5$. + +\Item{6.} $a^{2} - 4a - 5$. + +\Item{7.} $c^{2} - 9c + 18$. + +\Item{8.} $c^{2} + 9c + 18$. + +\Item{9.} $c^{2} + 3c - 18$. + +\Item{10.} $c^{2} - 3c - 18$. + +\Item{11.} $x^{2} + 9x + 14$. + +\Item{12.} $x^{2} - 9x + 14$. + +\Item{13.} $x^{2} - 5x - 14$. + +\Item{14.} $x^{2} - 9x + 20$. + +\Item{15.} $x^{2} - x - 20$. + +\Item{16.} $x^{2} + x - 20$. + +\Item{17.} $x^{2} - 10x + 21$. + +\Item{18.} $x^{2} - 4x - 21$. + +\Item{19.} $x^{2} + 4x - 21$. + +\Item{20.} $x^{2} - 15x + 56$. + +\Item{21.} $x^{2} - x - 56$. + +\Item{22.} $x^{2} - 10x + 9$. + +\Item{23.} $x^{2} + 13x + 30$. + +\Item{24.} $x^{2} + 7x - 30$. + +\Item{25.} $x^{2} - 7x - 30$. + +\Item{26.} $a^{2} + ab - 6b^{2}$. + +\Item{27.} $a^{2} - ab - 6b^{2}$. + +\Item{28.} $a^{2} + 3ab - 4b^{2}$. + +\Item{29.} $a^{2} - 3ab - 4b^{2}$. + +\Item{30.} $a^{2}x^{2} - 2ax - 63$. + +\Item{31.} $a^{2} + 2ax - 63x^{2}$. + +\Item{32.} $a^{2} - 9ab + 20b^{2}$. + +\Item{33.} $x^{2}y^{2} - 19xyz + 48z^{2}$. + +\Item{34.} $a^{2}b^{2} + 15abc + 44c^{2}$. + +\Item{35.} $x^{2} - 13xy + 36y^{2}$. + +\Item{36.} $x^{2} + 19xy + 84y^{2}$. + +\Item{37.} $a^{2}x^{2} - 23axy + 102y^{2}$. + +\Item{38.} $x^{4} - 9x^{2}y^{2} + 20y^{4}$. + +\Item{39.} $a^{4}x^{4} - 24a^{2}x^{2}y^{2} + 143y^{4}$. + +\Item{40.} $a^{6}b^{6} - 23a^{3}b^{3}c^{2} + 132c^{4}$. + +\Item{41.} $a^{2} - 20abc - 96b^{2}c^{2}$. + +\Item{42.} $a^{2} - 4abc - 96b^{2}c^{2}$. + +\Item{43.} $a^{2} - 10abc - 96b^{2}c^{2}$. + +\Item{44.} $a^{2} + 29abc - 96b^{2}c^{2}$. + +\Item{45.} $a^{2} - 46abc - 96b^{2}c^{2}$. + +\Item{46.} $a^{2} + 49abc + 48b^{2}c^{2}$. + +\Item{47.} $x^{2} - 18xyz - 243y^{2}z^{2}$. + +\Item{48.} $x^{2}y^{2} - xyz - 182z^{2}$. +\end{multicols} +%% -----File: 089.png---Folio 83------- + +\Exercise{39.} + +\Section{Examples For Review.} + +Resolve into factors: +\begin{multicols}{2} +\Item{1.} $a^{3} - 7a$. + +\Item{2.} $3a^{2}b^{2} - 2a^{3}b + 3ab^{3}$. + +\Item{3.} $(a - b)^{2} + (a - b)$. + +\Item{4.} $(a + b)^{2} - 1$. + +\Item{5.} $a^{3} + 8b^{3}$. + +\Item{6.} $(x^{2} - 4y^{2}) + (x - 2y)$. + +\Item{7.} $(a^{3} - b^{3}) + (a - b)$. + +\Item{8.} $a^{2} - 6ab + 9b^{2}$. + +\Item{9.} $x^{2} - x -2$. + +\Item{10.} $x^{2} - 2x - 3$. + +\Item{11.} $x^{2} + 4x - 21$. + +\Item{12.} $a^{2} - 11a - 26$. + +\Item{13.} $ax^{2} + bx^{2} + 3a + 3b$. + +\Item{14.} $x^{2} - 3x- xy + 3y$. + +\Item{15.} $x^{2} - 7x + 12$. + +\Item{16.} $a^{2} + 5ab + 6b^{2}$. + +\Item{17.} $x^{4} + 10x^{2} + 25$. + +\Item{18.} $x^{2} - 18x + 81$. + +\Item{19.} $x^{2} - 21x + 110$. + +\Item{20.} $x^{2} + 19x + 88$. + +\Item{21.} $x^{2} - 19x + 88$. + +\Item{22.} $x^{3} - x^{2} + x - 1$. + +\Item{23.} $9x^{4} - x^{2}$. + +\Item{24.} $1 - (a - b)^{2}$. + +\Item{25.} $(a^{3} + b^{3}) + (a + b)$. + +\Item{26.} $m^{2}x - n^{2}x + m^{2}y - n^{2}y$. + +\Item{27.} $(x - y)^{2} - z^{2}$. + +\Item{28.} $z^{2} - (x - y)^{2}$. + +\Item{29.} $4a^{4} - (3a - 1)^{2}$. + +\Item{30.} $8x^{3} - y^{3}$. + +\Item{31.} $x^{3} - 3x^{2}y$. + +\Item{32.} $x^{3} - 27y^{3}$. + +\Item{33.} $x^{2} + 3x - 40$. + +\Item{34.} $x^{2} + 3xy - 10y^{2}$. + +\Item{35.} $1 - 16x^{2}$. + +\Item{36.} $a^{6} - 9a^{2}b^{4}$. + +\Item{37.} $x^{3} + 3x^{2}y + 2xy^{2}$. + +\Item{38.} $x^{4} + 4x^{3}y + 3x^{2}y^{2}$. + +\Item{39.} $x^{2} - 4xy^{2} + 4y^{4}$. + +\Item{40.} $16x^{4} + 8x^{2} + 1$. + +\Item{41.} $9a^{4} - 4a^{2}c^{2}$. + +\Item{42.} $a^{3}b - a^{2}b^{2} - 2ab^{3}$. + +\Item{43.} $x^{4} - x^{3} + 8x - 8$. + +\Item{44.} $a^{4} - a^{3}x + ay^{3} - xy^{3}$. +\end{multicols} +%% -----File: 090.png---Folio 84------- + + +\Chapter{VIII.}{Common Factors and Multiples.} + +\Paragraph{116. Common Factors.} A \Defn{common factor} of two or more +\emph{integral numbers} is an integral number which divides each +of them without a remainder. + +\Paragraph{117.} A \Defn{common factor} of two or more integral and rational +\emph{expressions} is an integral and rational expression which +divides each of them without a remainder. + +\begin{Remark} +Thus $5a$~is a common factor of $20a$~and~$25a$, $3x^{2}y^{2}$ is a common +factor of~$12x^{2}y^{2}$ and~$15x^{3}y^{3}$. +\end{Remark} + +\Paragraph{118.} Two \emph{numbers} are said to be \Defn{prime} to each other +when they have no common factor except~$1$. + +\Paragraph{119.} Two \emph{expressions} are said to be \Defn{prime} to each other +when they have no common factor except~$1$. + +\Paragraph{120.} The \Defn{highest common factor} of two or more integral +\emph{numbers} is the greatest number that will divide each of +them without a remainder. + +\Paragraph{121.} The \Defn{highest common factor} of two or more integral +and rational \emph{expressions} is an integral and rational expression +of highest degree that will divide each of them without +a remainder. + +\begin{Remark} +Thus $3a^{2}$ is the highest common factor of $3a^{2}$, $6a^{3}$, and~$12a^{4}$, +$5x^{2}y^{2}$ is the highest common factor of $10x^{3}y^{2}$ and~$15x^{2}y^{2}$. +\end{Remark} + +For brevity, we use \HCF\ for ``highest common factor.'' +%% -----File: 091.png---Folio 85------- + +\Paragraph{122. To Find the Highest Common Factor of Two or More +Algebraic Expressions.} + +\Item{1.} Find the \HCF\ of $42a^{3}b^{2}$ and~$30a^{2}b^{4}$. +\begin{alignat*}{2} +&42a^{3}b^{2} &&= 2 × 3 × 7 × aaa × bb; \\ +&30a^{2}b^{4} &&= 2 × 3 × 5 × aa × bbbb. \\ +\therefore\ &\text{the \HCF} &&= 2 × 3 × aa × bb, +\quad\text{or}\quad 6a^{2}b^{2}. +\end{alignat*} + +\Item{2.} Find the \HCF\ of $x^{2} - 9y^{2}$ and $x^{2} + 6xy + 9y^{2}$. +\begin{alignat*}{2} +&x^{2} - 9y^{2} &&= (x + 3y)(x - 3y); \\ +&x^{2} + 6xy + 9y^{2} &&= (x + 3y)(x + 3y). \\ +\therefore\ &\text{the \HCF} &&= (x + 3y). +\end{alignat*} + +\Item{3.} Find the \HCF\ of $4x^{2} - 4x - 80$, $2x^{2} - 18x + 40$. +\begin{alignat*}{3} +&4x^{2} -& 4x &- 80 &&= 4(x^{2} - x-20) \\ +&&& &&= 4(x - 5)(x + 4); \\ +&2x^{2} -& 18x &+ 40 &&= 2(x^{2} - 9x + 20) \\ +&&& &&= 2(x - 5)(x - 4). \\ +\therefore\ &\rlap{\text{the \HCF}} &&&&= 2(x - 5). +\end{alignat*} + +\Dictum{To find the \HCF\ of two or more expressions}, therefore, +\begin{Theorem} +Resolve each expression into its simplest factors. + +Find the product of all the common factors, taking each +factor the least number of times it occurs in any of the given +expressions. +\end{Theorem} + +\begin{Remark}[Note.] The \emph{highest common factor} in Algebra corresponds to the +\emph{greatest common measure}, or \emph{greatest common divisor} in Arithmetic. +We cannot apply the terms \emph{greatest} and \emph{least} to an algebraic expression +in which particular values have not been given to the letters +contained in the expression. Thus $a$~is \emph{greater} than~$a^{2}$, if $a$~stands +for~$\frac{1}{4}$. +\end{Remark} +%% -----File: 092.png---Folio 86------- + +\Exercise{40.} + +Find the \HCF\ of: +\begin{multicols}{2} +\Item{1.} $330$ and $546$. + +\Item{2.} $20x^{3}$ and $15x^{4}$. + +\Item{3.} $42ax^{2}$ and $60a^{2}x$. + +\Item{4.} $35a^{2}b^{2}$ and $49ab^{3}$. + +\Item{5.} $28x^{4}$ and $63y^{4}$. + +\Item{6.} $54a^{2}b^{2}$ and $56a^{3}b^{3}$. + +\Item{7.} $x^{3} + 3x^{2}y$ and $x^{3} + 27y^{3}$. + +\Item{8.} $x^{2} + 3x$ and $x^{2} - 9$. + +\Item{9.} $2ax^{3} + x^{3}$ and $8a^{3} + 1$. + +\Item{10.} $(x + y)^{2}$ and $x^{2} - y^{2}$. + +\Item{11.} $a^{3} + a^{2}x$ and $a^{2} - x^{2}$. + +\Item{12.} $a^{2} - 4b^{2}$ and $a^{2} + 2ab$. +\end{multicols} + +\Item{13.} $x^{2} - 1$ and $x^{2} + 2x - 3$. + +\Item{14.} $x^{2} + 5x + 6$ and $x^{2} + 4x + 3$. + +\Item{15.} $x^{2} - 9x + 18$ and $x^{2} - 10x + 24$. + +\Item{16.} $x^{3} + 1$ and $x^{2} - x + 1$. + +\Item{17.} $x^{2} - 3x + 2$ and $x^{2} - 4x + 3$. + +\Item{18.} $x^{2} - 3xy + 2y^{2}$ and $x^{2} - 2xy + y^{2}$. + +\Item{19.} $x^{2} - 4x - 5$ and $x^{2} - 25$. + +\Item{20.} $(a - b)^{2} - c^{2}$ and $ab - b^{2} - bc$. + +\Item{21.} $x^{2} + xy -2y^{2}$ and $x^{2} + 5xy + 6y^{2}$. + +\Item{22.} $x^{2} + 7xy + 12y^{2}$ and $x^{2} + 3xy - 4y^{2}$. + +\Item{23.} $x^{3} - 8y^{3}$ and $x^{2} + 2xy + 4y^{2}$. + +\Item{24.} $x^{3} - 2x^{2} - x + 2$ and $x^{2} - 4x + 4$. + +\Item{25.} $1 - 5a + 6a^{2}$ and $1 - 7a + 12a^{2}$. + +\Item{26.} $x^{2} - 8xy + 7y^{2}$ and $x^{2} - 3xy - 28y^{2}$. + +\Item{27.} $8a^{3} + b^{3}$ and $4a^{2} + 4ab + b^{2}$. + +\Item{28.} $x^{2} - (y - z)^{2}$ and $(x + y)^{2} - z^{2}$. +%% -----File: 093.png---Folio 87------- + +\Paragraph{123. Common Multiples.} A \Defn{common multiple} of two or +more integral \emph{numbers} is a number which is exactly divisible +by each, of the numbers. + +A \Defn{common multiple} of two or more \emph{expressions} is an expression +which is exactly divisible by each of the expressions. + +\Paragraph{124.} The \Defn{lowest common multiple} of two or more \emph{numbers} +is the least number that is exactly divisible by each of the +given numbers. + +The \Defn{lowest common multiple} of two or more \emph{expressions} is +the expression of lowest degree that is exactly divisible by +each of the given expressions. + +We use \LCM\ for ``lowest common multiple.'' + +\Paragraph{To find the lowest common multiple of two or more algebraic +expressions.} + +\Item{1.} Find the \LCM\ of $42a^{3}b^{2}$, $30a^{2}b^{4}$, and~$66ab^{3}$. +\begin{align*} +42a^{3}b^{2} &= 2 × 3 × 7 × a^{3} × b^{2}; \\ +30a^{2}b^{4} &= 2 × 3 × 5 × a^{2} × b^{4}; \\ +66ab^{3} &= 2 × 3 × 11 × a × b^{3}. +\end{align*} + +\begin{Soln} +The \LCM\ must evidently contain each factor the greatest number +of times that it occurs in any expression. +\begin{align*} +\therefore\ \text{\LCM} + &= 2 × 3 × 7 × 5 × 11a^{3} × b^{4}, \\ + &= 2310a^{3}b^{4}. +\end{align*} +\end{Soln} + +\Item{2.} Find the \LCM\ of $4x^{2} - 4x - 80$ and $2x^{2} - 18x + 40$. +\begin{Soln} +\begin{alignat*}{3} +&4x^{2} -& 4x &- 80 &&= 4(x^{2} - x - 20) = 4(x - 5)(x + 4); \\ +&2x^{2} -& 18x &+ 40 &&= 2(x^{2} - 9x + 20) = 2(x - 5)(x - 4). \\ +\therefore\ &\rlap{\text{\LCM}}&&& &= 4(x - 5)(x + 4)(x - 4). +\end{alignat*} +\end{Soln} + +\Dictum{To find the \LCM\ of two or more expressions}, therefore, +%% -----File: 094.png---Folio 88------- +\begin{Theorem} +Resolve each expression into its simplest factors. + +Find the product of all the different factors, taking each +factor the greatest number of times it occurs in any of the +given expressions. +\end{Theorem} + +\Exercise{41.} + +Find the \LCM\ of: +\begin{multicols}{2} +\Item{1.} $9xy^{3}$ and $6x^{2}y$. + +\Item{2.} $3abc^{2}$ and $2a^{2}bc^{3}$. + +\Item{3.} $4a^{3}b$ and $10ab^{3}$. + +\Item{4.} $6a^{3}b^{3}$ and $15a^{2}b^{4}$. + +\Item{5.} $21xy^{3}$ and $27x^{3}y^{5}$. + +\Item{6.} $xy^{3}z^{2}$ and $x^{2}y^{2}z^{3}$. + +\Item{7.} $a^{2}$ and $a^{2} + a$. + +\Item{8.} $x^{2}$ and $x^{3} - 3x^{2}$. + +\Item{9.} $x^{2} - 1$ and $x^{2} + x$. + +\Item{10.} $x^{2} - 1$ and $x^{2} - x$. + +\Item{11.} $x^{2} + xy$ and $xy + y^{2}$. + +\Item{12.} $x^{2} + 2x$ and $(x + 2)^{2}$. +\end{multicols} + +\Item{13.} $a^{2} + 4a + 4$ and $a^{2} + 5a + 6$. + +\Item{14.} $c^{2} + c - 20$ and $c^{2} - c - 30$. + +\Item{15.} $b^{2} + b - 42$ and $b^{2} - 11b + 30$. + +\Item{16.} $y^{2} - 10y + 24$ and $y^{2} + y -20$. + +\Item{17.} $z^{2} + 2z - 35$ and $z^{2} - 11z + 30$. + +\Item{18.} $x^{2} - 64; x^{3} - 64$; and $x + 8$. + +\Item{19.} $a^{2} - b^{2}; (a + b)^{2}$; and $(a - b)^{2}$. + +\Item{20.} $4ab(a + b)^{2}$ and $2a^{2}(a^{2} - b^{2})$. + +\Item{21.} $y^{2} + 7y + 12; y^{2} + 6y + 8$; and $y^{2} + 5y +6$. + +\Item{22.} $x^{2} - 1; x^{3} + x^{2} + x + 1$; and $x^{3} - x^{2} + x - 1$. + +\Item{23.} $1 - x^{2}; 1 - x^{3}$; and $1 + x$. + +\Item{24.} $x^{2} + 2xy + y^{2}; x^{2} - y^{2}$; and $x^{2} - 2xy + y^{2}$. + +\Item{25.} $x^{3} - 27; x^{2} + 2x - 15; x^{2} + 5x$. + +\Item{26.} $(a + b)^{2} - c^{2}; (a + b + c)^{2}$; and $a + b - c$. + +\Item{27.} $x^{2} - (a + b)x + ab$ and $x^{2} - (a + c)x + ac$. + +\Item{28.} $(a + b)^{2} - c^{2}$ and $a^{2} + ab + ac$. +%% -----File: 095.png---Folio 89------- + + +\Chapter{IX.}{Fractions.} + +\Paragraph{125.} An \Defn{algebraic fraction} is the indicated quotient of +two expressions, written in the form~$dfrac{a}{b}$. + +The dividend~$a$ is called the \Defn{numerator}, and the divisor~$b$ +is called the \Defn{denominator}; and the numerator and denominator +are called the \Defn{terms} of the fraction. + +\Paragraph{126.} The introduction of the same factor into the dividend +and divisor does not alter the value of the quotient, +and the rejection of the same factor from the dividend and +divisor does not alter the value of the quotient. +\begin{DPgather*} +\lintertext{\indent Thus} +\frac{12}{4} = 3;\quad +\frac{2 × 12}{2 × 4} = 3; \frac{12÷2}{4÷2} = 3. \EqText{Hence,} +\end{DPgather*} +\begin{Theorem} +The value of a fraction is not altered if the numerator and +denominator are both multiplied, or both divided, by the +same factor. +\end{Theorem} + + +\Section{Reduction of Fractions.} + +\Paragraph{127.} To reduce a fraction is to change its \emph{form} without +altering its \emph{value}. + + +\Section{Case I.} + +\Paragraph{128. To Reduce a Fraction to its Lowest Terms.} + +A fraction is in its \emph{lowest terms} when the numerator and +denominator have no common factor. We have, therefore, +the following rule: +%% -----File: 096.png---Folio 90------- +\begin{Theorem} +Resolve the numerator and denominator into their prime +factors, and cancel all the common factors. +\end{Theorem} + +Reduce the following fractions to their lowest terms: + +\Item{1.} $\dfrac{38a^{2}b^{3}c^{4}}{57a^{3}bc^{2}} = \dfrac{2 × 19a^{2}b^{3}c^{4}}{3 × 19a^{3}bc^{2}} = \dfrac{2b^{2}c^{2}}{3a}$. + +\Item{2.} $\dfrac{a^{3} - x^{3}}{a^{2} - x^{2}} = \dfrac{(a - x)(a^{2} + ax + x^{2})}{(a - x)(a + x)} = \dfrac{a^{2} + ax + x^{2}}{a + x}$. + +\Item{3.} $\dfrac{a^{2} + 7a + 10}{a^{2} + 5a + 6} = \dfrac{(a + 5)(a + 2)}{(a + 3)(a + 2)} = \dfrac{a + 5}{a + 3}$. + +\Exercise{42.} + +Reduce to lowest terms: +\begin{multicols}{3} +\Item{1.} $\dfrac{2a}{6ab}$. + +\Item{2.} $\dfrac{12m^{2}n}{15mn^{2}}$. + +\Item{3.} $\dfrac{21m^{2}p^{2}}{28mp^{4}}$. + +\Item{4.} $\dfrac{3x^{3}y^{2}z}{6xy^{3}z^{2}}$. + +\Item{5.} $\dfrac{5a^{3}b^{3}c^{3}}{15c^{5}}$. + +\Item{6.} $\dfrac{34x^{3}y^{4}z^{5}}{51x^{2}y^{3}z^{5}}$. + +\Item{7.} $\dfrac{46m^{2}np^{3}}{69mnp^{4}}$. + +\Item{8.} $\dfrac{39a^{2}b^{3}c^{4}}{52a^{5}bc^{3}}$. + +\Item{9.} $\dfrac{58xy^{4}z^{6}}{87xy^{2}z^{2}}$. +\end{multicols} + +\begin{multicols}{2} +\Item{10.} $\dfrac{abx - bx^{2}}{acx - cx^{2}}$. + +\Item{11.} $\dfrac{4a^{2} - 9b^{2}}{4a^{2} + 6ab}$. + +\Item{12.} $\dfrac{3a^{2} + 6a}{a^{2} + 4a + 4}$. + +\Item{13.} $\dfrac{x^{2} + 5x}{x^{2} + 4x-5}$. + +\Item{14.} $\dfrac{xy - 3y^{2}}{x^{3} - 27y^{3}}$. + +\Item{15.} $\dfrac{x^{2} + 5x + 4}{x^{2} - x-20}$. + +\Item{16.} $\dfrac{x^{2} + 2x + 1}{x^{2} - x-2}$. + +\Item{17.} $\dfrac{(a + b)^{2} - c^{2}}{a^{2} + ab-ac}$. + +\Item{18.} $\dfrac{x^{2} + 9x + 20}{x^{2} + 7x + 12}$. + +\Item{19.} $\dfrac{x^{2} - 14x - 15}{x^{2} - 12x - 45}$. +\end{multicols} +%% -----File: 097.png---Folio 91------- + + +\PrintBreak +\Section{Case II.} + +\Paragraph{129. To Reduce a Fraction to an Integral or Mixed Expression.} + +\Item{1.} Reduce $\dfrac{x^{3} - 1}{x - 1}$ to an integral or mixed expression. + +By division, $\dfrac{x^{3} - 1}{x - 1} = x^{2} + x + 1$. + +\Item{2.} Reduce $\dfrac{x^{3} - 1}{x + 1}$ to an integral or mixed expression. + +By division, $\dfrac{x^{3} - 1}{x + 1} = x^{2} - x + 1 - \dfrac{2}{x + 1}$. + +\ScreenBreak +\Dictum{To reduce a fraction to an integral or mixed expression}, +therefore, +\begin{Theorem} +Divide the numerator by the denominator. +\end{Theorem} + +\begin{Remark}[Note.] If there is a remainder, this remainder must be written +as the numerator of a fraction of which the divisor is the denominator, +and this fraction with its proper sign must be annexed to the integral +part of the quotient. +\end{Remark} + +\Exercise{43.} + +Reduce to integral or mixed expressions: +\begin{multicols}{2} +\Item{1.} $\dfrac{a^{2} - b^{2} + 2}{a - b}$. + +\Item{2.} $\dfrac{a^{2} - b^{2} - 2}{a + b}$. + +\Item{3.} $\dfrac{a^{3} - 2a^{2} + 2a + 1}{a^{2} - a - 1}$. + +\Item{4.} $\dfrac{2x^{2} - 2x + 1}{x + 1}$. + +\Item{5.} $\dfrac{8x^{3}}{2x + 1}$. + +\Item{6.} $\dfrac{5x^{3} + 9x^{2} + 3}{x^{2} + x - 1}$. + +\Item{7.} $\dfrac{a^{3} + a^{2} + 7a - 2}{a^{2} + a + 2}$. + +\Item{8.} $\dfrac{y^{4} + y^{2}x^{2} + x^{4}}{y^{2} + yx + x^{2}}$. + +\Item{9.} $\dfrac{x^{4} - 3x^{3} + x - 1}{x^{2} + x + 1}$. + +\Item{10.} $\dfrac{x^{5} - x^{4} + 1}{x^{2} - x - 1}$. +\end{multicols} +%% -----File: 098.png---Folio 92------- + + +\ScreenBreak +\Section{Case III.} + +\Paragraph{130. To Reduce a Mixed Expression to a Fraction.} + +The process is precisely the same as in Arithmetic. Hence, +\begin{Theorem} +Multiply the integral expression by the denominator, to +the product add the numerator, and under the result write +the denominator. +\end{Theorem} + +Reduce to a fraction $a - b - \dfrac{a^{2} - ab - b^{2}}{a + b}$. +\begin{align*} +a - b - \frac{a^{2} - ab - b^{2}}{a + b} + &= \frac{(a - b)(a + b) - (a^{2} - ab - b^{2})}{a + b} \\ + &= \frac{a^{2} - b^{2} - a^{2} + ab + b^{2}}{a + b} \\ + &= \frac{ab}{a + b}. +\end{align*} + +\begin{Remark}[Note.] The dividing line between the terms of a fraction has the +force of a vinculum affecting the numerator. If, therefore, a \emph{minus +sign} precedes the dividing line, as in the preceding Example, and +this line is removed, the numerator of the given fraction must be +enclosed in a parenthesis preceded by the minus sign, or the sign of +every term of the numerator must be changed. +\end{Remark} + +\ScreenBreak +\Exercise{44.} + +Reduce to a fraction: +\begin{multicols}{2} +\Item{1.} $x - y + \dfrac{2xy}{x - y}$. + +\Item{2.} $x + y - \dfrac{2xy}{x + y}$. + +\Item{3.} $1 - \dfrac{x - y}{x + y}$. + +\Item{4.} $a - x - \dfrac{a^{2} + x^{2}}{a - x}$. + +\Item{5.} $x + 2 - \dfrac{x^{2} - 4}{x - 3}$. + +\Item{6.} $\dfrac{x - 3}{x - 2} - 2x + 1$. + +\Item{7.} $\dfrac{x + 3}{x + 2} + x^{2} - x - 1$. + +\Item{8.} $2a - 1 + \dfrac{3 - 4a}{a - 3}$. + +\Item{9.} $1 - 2a^{2} - \dfrac{a^{2} - a + 2}{a - 1}$. + +\Item{10.} $a^{2} + 2a - 5 - \dfrac{2a - 1}{3a^{2} + 1}$. +\end{multicols} +%% -----File: 099.png---Folio 93------- + + +\Section{Case IV.} + +\Paragraph{131. To Reduce Fractions to their Lowest Common Denominator.} + +The process is the same as in Arithmetic. Hence: +\begin{Theorem} +Find the lowest common multiple of the denominators; +this will be the required denominator. Divide this denominator +by the denominator of each fraction. + +Multiply the first numerator by the first quotient, the second +numerator by the second quotient, and so on. + +The products will be the respective numerators of the +equivalent fractions. +\end{Theorem} + +\begin{Remark}[Note.] Every fraction should be in its lowest terms before the +common denominator is found. +\end{Remark} + +\Item{1.} Reduce $\dfrac{3x}{4a^{2}}$, $\dfrac{2y}{3a}$, and $\dfrac{5}{6a^{3}}$ to equivalent fractions +having the lowest common denominator. +\begin{Soln} +The \LCM\ of $4a^{2}$, $3a$, and $6a^{3} = 12a^{3}$. + +The respective quotients are $3a$, $4a^{2}$, and~$2$. + +The products are $9ax$, $8a^{2}y$, and~$10$. + +Hence, the required fractions are +\[ +\frac{9ax}{12a^{3}},\quad +\frac{8a^{2}y}{12a^{3}}, \quad\text{and}\quad +\frac{10}{12a^{3}}. +\] +\end{Soln} + +\Item{2.} Express $\dfrac{1}{x^{2} + 5x + 6}$ and $\dfrac{1}{x^{2} + 4x + 3}$ with lowest +common denominator. +\begin{Soln} +The factors of the denominators are $x + 3$, $x + 2$; and $x + 3$, $x + 1$. + +Hence the lowest common denominator (\LCD) is $(x + 3)(x + 2)(x + 1)$, +and the required numerators are $x + 1$ and $x + 2$. Hence the required +fractions are +\[ +%[** TN: Equations not displayed in the original] +\frac{x + 1}{(x + 3)(x + 2)(x + 1)} \quad\text{and}\quad +\frac{x + 2}{(x + 3)(x + 2)(x + 1)}. +\] +\end{Soln} +%% -----File: 100.png---Folio 94------- + +\ScreenBreak +\Exercise{45.} + +Express with lowest common denominator: + +\begin{multicols}{2} +\Item{1.} $\dfrac{x}{x - a}$, $\dfrac{x^{2}}{x^{2} - a^{2}}$. + +\Item{2.} $\dfrac{a}{a + b}$, $\dfrac{a^{2}}{a^{2} - b^{2}}$. + +\Item{3.} $\dfrac{1}{1 + 2a}$, $\dfrac{1}{1 - 4a^{2}}$. + +\Item{4.} $\dfrac{9}{16 - x^{2}}$, $\dfrac{4 - x}{4 + x}$. + +\Item{5.} $\dfrac{a^{2}}{27 - a^{3}}$, $\dfrac{a}{3 - a}$. + +\Item{6.} $\dfrac{1}{x^{2} - 5x + 6}$, $\dfrac{1}{x^{2} - x-6}$. +\end{multicols} + + +\Section{Addition and Subtraction of Fractions.} + +\Paragraph{132.} The algebraic sum of two or more fractions which +have the same denominator, is a fraction whose numerator +is the algebraic sum of the numerators of the given fractions, +and whose denominator is the common denominator +of the given fractions. Hence, + +\Dictum{To add fractions}, +\begin{Theorem} +Reduce the fractions to equivalent fractions having the +same denominator; and write the algebraic sum of the +numerators of these fractions over the common denominator. +\end{Theorem} + +\Paragraph{133. When the denominators are simple expressions.} + +\Item{1.} Simplify $\dfrac{3a - 4b}{4} - \dfrac{2a - b + c}{3} + \dfrac{a - 4c}{12}$. +\begin{Soln} +The $\text{\LCD} = 12$. + +The multipliers, that is, the quotients obtained by dividing $12$ by +$4$, $3$, and $12$, are $3$,~$4$, and~$1$. + +Hence the sum of the fractions equals +\begin{align*} +&\frac{9a - 12b}{12} - \frac{8a - 4b + 4c}{12} + \frac{a - 4c}{12} \\ +&\quad= \frac{9a - 12b - 8a + 4b - 4c + a - 4c}{12} \\ +&\quad= \frac{2a - 8b-8c}{12} = \frac{a - 4b - 4c}{6}. +\end{align*} +\end{Soln} +%% -----File: 101.png---Folio 95------- + +The preceding work may be arranged as follows: +\begin{Soln} +The $\text{\LCD} = 12$. + +The multipliers are $3$,~$4$, and~$1$, respectively. +\begin{gather*} +\begin{array}{l*{5}{cr}cl} +3(3a &-& 4b & & ) &=& 9a &-&12b & & &=& \text{1st numerator.} \\ +\llap{$-$} +4(2a &-& b &+& c) &=&-8a &+& 4b &-& 4c &=& \text{2d numerator.} \\ +1( a &-& & &4c) &=& a & & &-& 4c &=& \text{3d numerator.} \\ +\cline{7-11} + & & & & & & 2a &-& 8b &-& 8c \\ + & & & & &\rlap{or}& 2(a &-& 4b &-& 4c) &=& \text{the sum of the numerators.} +\end{array} \\ +\therefore\ \text{sum of fractions} = \frac{2(a - 4b - 4c)}{12} = \frac{a - 4b - 4c}{6}. +\end{gather*} +\end{Soln} + +\Exercise{46.} + +Find the sum of: + +\Item{1.} $\dfrac{x + 1}{2} + \dfrac{x - 3}{5} + \dfrac{x + 5}{10}$. + +\Item{2.} $\dfrac{2x - 1}{3} + \dfrac{x + 5}{4} + \dfrac{x - 4}{6}$. + +\Item{3.} $\dfrac{7x - 1}{6} - \dfrac{3x - 2}{7} + \dfrac{x - 5}{3}$. + +\Item{4.} $\dfrac{3x - 2}{9} - \dfrac{x - 2}{6} + \dfrac{5x + 3}{4}$. + +\Item{5.} $\dfrac{x - 1}{6} - \dfrac{x - 3}{3} + \dfrac{x - 5}{2}$. + +\Item{6.} $\dfrac{x - 2y}{2x} + \dfrac{x + 5y}{4x} - \dfrac{x + 7y}{8x}$. + +\Item{7.} $\dfrac{5x - 11}{3} - \dfrac{2x - 1}{10} - \dfrac{11x - 5}{15}$. + +\Item{8.} $\dfrac{x - 3}{3x} - \dfrac{x^{2} - 6x}{5x^{2}} - \dfrac{7x^{2} - x^{3}}{15x^{3}}$. + +\Item{9.} $\dfrac{ac - b^{2}}{ac} - \dfrac{ab - c^{2}}{ab} + \dfrac{a^{2} - bc}{bc}$. +%% -----File: 102.png---Folio 96------- + +\PrintBreak +\Paragraph{134. When the denominators have compound expressions, +arranged in the same order.} + +\Item{1.} Simplify $\dfrac{a + b}{a - b} - \dfrac{a - b}{a + b} - \dfrac{4ab}{a^{2} - b^{2}}$. +\begin{Soln} +The \LCD\ is $(a - b)(a + b)$. + +The multipliers are $a + b$, $a - b$, and~$1$, respectively. +\begin{gather*} +\begin{array}{l*{3}{cr}cl} + (a + b)(a + b) &=& a^{2} &+& 2ab &+& b^{2} &=& \text{1st numerator.} \\ +\llap{$-$} + (a - b)(a - b) &=&-a^{2} &+& 2ab &-& b^{2} &=& \text{2d numerator.} \\ +\llap{$-$} + 1(4ab) &=& &-& 4ab & & &=& \text{3d numerator.} \\ +\cline{3-7} + & & \multicolumn{5}{c}{0} &=& \text{sum of numerators.} +\end{array} \\ +\therefore\ \text{Sum of fractions${} = 0$.} +\end{gather*} +\end{Soln} + +\Exercise{47.} + +Find the sum of: +\begin{multicols}{2} +\Item{1.} $\dfrac{1}{x + 3} + \dfrac{1}{x - 2}$. + +\Item{2.} $\dfrac{1}{x + 1} + \dfrac{1}{x - 1}$. + +\Item{3.} $\dfrac{4}{x - 8} - \dfrac{1}{x + 2}$. + +\Item{4.} $\dfrac{a + x}{a - x} - \dfrac{a - x}{a + x}$. + +\Item{5.} $\dfrac{x}{x - a} - \dfrac{x^{2}}{x^{2} - a^{2}}$. + +\Item{6.} $\dfrac{4a^{2} + b^{2}}{4a^{2} - b^{2}} - \dfrac{2a + b}{2a - b}$. + +\Item{7.} $\dfrac{7}{9 - a^{2}} - \dfrac{1}{3 + a} - \dfrac{1}{3 - a}$. + +\Item{8.} $\dfrac{1}{a - b} - \dfrac{1}{a + b} - \dfrac{b}{a^{2} - b^{2}}$. +\end{multicols} +%[** TN: Adjusted layout] + +\Item{9.} $\dfrac{2}{x - 2} - \dfrac{2}{x + 2} + \dfrac{5x}{x^{2} - 4}$. + +\Item{10.} $\dfrac{3 - x}{1 - 3x} - \dfrac{3 + x}{1 + 3x} - \dfrac{15x - 1}{1 - 9x^{2}}$. + +\begin{multicols}{2} +\Item{11.} $\dfrac{1}{a} - \dfrac{1}{a + 3} + \dfrac{3}{a + 1}$. + +\Item{12.} $\dfrac{x}{x - 1} - \dfrac{1} - \dfrac{1}{x + 1}$. +\end{multicols} + +\Item{13.} $\dfrac{x + 1}{x + 2} + \dfrac{x - 2}{x - 3} + \dfrac{2x + 7}{x^{2} - x - 6}$. + +\Item{14.} $\dfrac{1}{x(x - 1)} - \dfrac{2}{x^{2} - 1} + \dfrac{1}{x(x + 1)}$. +%% -----File: 103.png---Folio 97------- + +\Exercise{48.} + +Find the sum of: + +\Item{1.} $\dfrac{1}{2x + 1} + \dfrac{1}{2x - 1} - \dfrac{4x}{4x^{2} - 1}$. + +\Item{2.} $\dfrac{a^{2} + b^{2}}{a^{2} - b^{2}} + \dfrac{a}{a + b} - \dfrac{b}{a - b}$. + +\Item{3.} $\dfrac{3a}{1 - a^{2}} + \dfrac{2}{1 - a} - \dfrac{2}{1 + a}$. + +\Item{4.} $\dfrac{1}{2x + 5y} - \dfrac{3x}{4x^{2} - 25y^{2}} + \dfrac{1}{2x + 5y}$. + +\Item{5.} $\dfrac{1}{x + 4y} - \dfrac{8y}{x^{2} - 16y^{2}} + \dfrac{1}{x - 4y}$. + +\Item{6.} $\dfrac{3}{2x - 3} - \dfrac{2}{2x + 3} - \dfrac{3}{4x^{2} - 9}$. + + +\Section{Multiplication and Division of Fractions.} + +\Paragraph{135.} Find the product of $\dfrac{a}{b} × \dfrac{c}{d}$. + +Let $\dfrac{a}{b} = x$, and $\dfrac{c}{d} = y$. + +Then $a = bx$, and $c = dy$. + +The product of these two equations is +\begin{DPalign*} +ac &= bdxy. \\ +\lintertext{\indent Divide by~$bd$,} +\frac{ac}{bd} &= xy. \displaybreak[1] \\ +\lintertext{\indent But} +\frac{a}{b} × \frac{c}{d} = xy. \displaybreak[1] \\ +\lintertext{\indent Therefore} +\frac{a}{b} × \frac{c}{d} = \frac{ac}{bd}. +\end{DPalign*} +%% -----File: 104.png---Folio 98------- + +\Dictum{To find the product of two fractions}, therefore, +\begin{Theorem} +Find the product of the numerators for the required +numerator, and the product of the denominators for the +required denominator. +\end{Theorem} + +In like manner, +\[ +\frac{a}{b} × \frac{c}{d} × \frac{e}{f} = \frac{ac}{bd} × \frac{e}{f} = \frac{ace}{bdf}. +\] + +\Paragraph{136. Reciprocals.} If the product of two numbers is equal +to~$1$, each of the numbers is called the \Defn{reciprocal} of the +other. + +The reciprocal of $\dfrac{a}{b}$ is $\dfrac{b}{a}$, for $\dfrac{b}{a} × \dfrac{a}{b} = \dfrac{ba}{ab} = 1$. + +The reciprocal of a fraction, therefore, is the fraction +inverted. + +%[** TN: Next line broken in the original] +Since $\dfrac{a}{b} ÷ \dfrac{a}{b} = 1$, and $\dfrac{b}{a} × \dfrac{a}{b} = 1$, +it follows that +\begin{Theorem} +To divide by a fraction is the same as to multiply by its +reciprocal. +\end{Theorem} + +\Paragraph{137.} \Dictum{To Divide by a Fraction}, therefore, +\begin{Theorem} +Invert the divisor and multiply. +\end{Theorem} + +\begin{Remark}[Note.] Every mixed expression should first be reduced to a fraction, +and every integral expression should be written as a fraction +having $1$~for the denominator. Both terms of each fraction should +be expressed in their prime factors, and if a factor is common to a +numerator and denominator, it should be cancelled, as the cancelling +of a common factor \emph{before} the multiplication is evidently equivalent +to cancelling it \emph{after} the multiplication. +\end{Remark} +%% -----File: 105.png---Folio 99------- + +\Item{1.} Find the product of $\dfrac{3a^{2}b}{2x^{2}y} × \dfrac{6xy^{2}}{7ab} × \dfrac{7abc}{9a^{2}by^{2}}$. +\[ +\frac{3a^{2}b}{2x^{2}y} × \frac{6xy^{2}}{7ab} × \frac{7abc}{9a^{2}by^{2}} + = \frac{3 × 6 × 7a^{3}b^{2}cxy^{2}}{2 × 7 × 9a^{3}b^{2}x^{2}y^{3}} + = \frac{c}{xy}. +\] + +\Item{2.} Find the product of $\dfrac{ab - b^{2}}{a + b} × \dfrac{ab + b^{2}}{a^{2} - b^{2}}$. +\[ +\frac{ab - b^{2}}{a + b} × \frac{ab + b^{2}}{a^{2} - b^{2}} + = \frac{b(a - b)}{(a + b)} × \frac{b(a + b)}{(a - b)(a + b)} + = \frac{b^{2}}{a + b}. +\] + +\Item{3.} Find quotient of $\dfrac{ab}{(a - b)^{2}} ÷ \dfrac{ac}{a^{2} - b^{2}}$. +\begin{align*} +\frac{ab}{(a - b)^{2}} ÷ \frac{ac}{a^{2} - b^{2}} + &= \frac{ab}{(a - b)(a - b)} × \frac{(a - b)(a + b)}{ac} \\ + &= \frac{b(a + b)}{c(a - b)}. +\end{align*} + +\Item{4.} Find the result of $\dfrac{1}{x} × \dfrac{x^{2} - 1}{x^{2} - 4x - 5} ÷ \dfrac{x^{2} + 2x - 3}{x^{2} - 25}$. +\begin{align*} +&\frac{1}{x} × \frac{x^{2} - 1}{x^{2} - 4x - 5} ÷ \frac{x^{2} + 2x - 3}{x^{2} - 25} \\ +&\qquad= \frac{1}{x} × \frac{x^{2} - 1}{x^{2} - 4x - 5} × \frac{x^{2} - 25}{x^{2} + 2x - 3} \\ +&\qquad= \frac{1}{x} × \frac{(x - 1)(x + 1)}{(x - 5)(x + 1)} × \frac{(x - 5)(x + 5)}{(x + 3)(x - 1)} \\ +&\qquad= \frac{x + 5}{x(x + 3)}. +\end{align*} +%% -----File: 106.png---Folio 100------- + +\Exercise{49.} + +Express in the simplest form: +\begin{multicols}{2} +\Item{1.} $\dfrac{15a^{2}}{7b^{2}} × \dfrac{28ab}{9a^{3}c}$. + +\Item{2.} $\dfrac{3x^{2}y^{2}z^{3}}{4a^{2}b^{2}c^{2}} × \dfrac{8a^{3}b^{2}c^{2}}{9x^{2}yz^{3}}$. + +\Item{3.} $\dfrac{5m^{2}n^{2}p^{4}}{3x^{2}yz^{3}} × \dfrac{21xyz^{2}}{20m^{2}n^{2}p^{2}}$. + +\Item{4.} $\dfrac{16a^{4}b^{2}c^{3}}{21m^{2}x^{3}y^{4}} × \dfrac{3m^{3}x^{3}y^{4}}{8a^{2}b^{2}c^{2}}$. + +\Item{5.} $\dfrac{2a}{bc} × \dfrac{3b}{ac} × \dfrac{5c}{ab}$. + +\Item{6.} $\dfrac{2a^{3}}{3bc} × \dfrac{3b^{3}}{5ac} × \dfrac{5c^{3}}{2ab}$. + +\Item{7.} $\dfrac{5abc^{3}}{3x^{2}} ÷ \dfrac{10ac^{3}}{6bx^{2}}$. + +\Item{8.} $\dfrac{x^{2} - a^{2}}{x^{2} - 4a^{2}} × \dfrac{x + 2a}{x - a}$. + +\Item{9.} $\dfrac{x^{2}y^{2} + 3xy}{4c^{2} - 1} × \dfrac{2c + 1}{xy + 3}$. + +\Item{10.} $\dfrac{a^{2} - 100}{a^{2} - 9} × \dfrac{a - 3}{a - 10}$. + +\Item{11.} $\dfrac{9x^{2} - 4y^{2}}{x^{2} - 4} × \dfrac{x + 2}{3x - 2y}$. + +\Item{12.} $\dfrac{25a^{2} - b^{2}}{16a^{2} - 9b^{2}} ÷ \dfrac{5a - b}{4a - 3b}$. +\end{multicols} +%[** TN: Moved end of two-column layout up two questions] + +\Item{13.} $\dfrac{x^{2} - 49}{(a + b)^{2} - c^{2}} ÷ \dfrac{x + 7}{(a + b) - c}$. + +\Item{14.} $\dfrac{x^{2} + 2x + 1}{x^{2} - 25} ÷ \dfrac{x + 1}{x^{2} + 5x}$. + +\Item{15.} $\dfrac{a^{2} + 3a + 2}{a^{2} + 5a + 6} × \dfrac{a^{2} + 7a + 12}{a^{2} + 9a + 20}$. + +\Item{16.} $\dfrac{y^{2} - y-30}{y^{2} - 36} × \dfrac{y^{2} - y-2}{y^{2} + 3y-10} × \dfrac{y^{2} + 6y}{y^{2} + y}$. + +\Item{17.} $\dfrac{x^{2} - 2x + 1}{x^{2} - y^{2}} × \dfrac{x^{2} + 2xy + y^{2}}{x - 1} ÷ \dfrac{x^{2} - 1}{x^{2} - xy}$. + +\Item{18.} $\dfrac{a^{2} - b^{2}}{a^{2} - 3ab + 2b^{2}} × \dfrac{ab - 2b^{2}}{a^{2} + ab} ÷ \dfrac{(a - b)^{2}}{a(a - b)}$. + +\Item{19.} $\dfrac{(a + b)^{2} - c^{2}}{a^{2} + ab-ac} × \dfrac{a^{2}b^{2}c^{2}}{a^{2} + ab + ac} ÷ \dfrac{b^{2}c^{2}}{abc}$. + +\Item{20.} $\dfrac{x^{2} + 7xy + 10y^{2}}{x^{2} + 6xy + 5y^{2}} × \dfrac{x + 1}{x^{2} + 4x + 4} ÷ \dfrac{1}{x + 2}$. +%% -----File: 107.png---Folio 101------- + +\Paragraph{138. Complex Fractions.} A complex fraction is one that +has a fraction in the numerator, or in the denominator, or +in both. + +The shortest way to reduce to its simplest form a complex +fraction is to multiply both terms of the fraction by +the \LCD\ of the fractions contained in the numerator and +denominator. + +\Item{1.} Simplify $\dfrac{3x}{x - \frac{1}{4}}$. + +Multiply both terms by~$4$, and we have +\[ +\frac{12x}{4x - 1}. +\] + +\Item{2.} Simplify $\dfrac{\dfrac{a + x}{a - x} - \dfrac{a - x}{a + x}}{\dfrac{a + x}{a - x} + \dfrac{a - x}{a + x}}$. + +The \LCD\ of the fractions in the numerator and denominator +is +\[ +(a - x)(a + x). +\] + +Multiply by $(a - x)(a + x)$, and the result is +\begin{align*} +&\frac{(a + x)^{2} - (a - x)^{2}}{(a + x)^{2} + (a - x)^{2}} \\ +&\qquad= \frac{(a^{2} + 2ax + x^{2}) - (a^{2} - 2ax + x^{2})} + {(a^{2} + 2ax + x^{2}) + (a^{2} - 2ax + x^{2})} \\ +&\qquad= \frac{a^{2} + 2ax + x^{2} - a^{2} + 2ax - x^{2}} + {a^{2} + 2ax + x^{2} + a^{2} - 2ax + x^{2}} \\ +&\qquad= \frac{4ax}{2a^{2} + 2x^{2}} \\ +&\qquad= \frac{2ax}{a^{2} + x^{2}}. +\end{align*} +%% -----File: 108.png---Folio 102------- + +\Exercise{50.} + +Reduce to the simplest form: +\begin{multicols}{2} +\Item{1.} $\dfrac{\dfrac{x}{b} + \dfrac{y}{b}}{\dfrac{z}{b}}$. + +\Item{2.} $\dfrac{x + \dfrac{y}{4}}{x - \dfrac{y}{3}}$. + +\Item{3.} $\dfrac{\dfrac{ab}{7} - 3d}{3c - \dfrac{ab}{d}}$. + +\Item{4.} $\dfrac{1 + \dfrac{1}{x + 1}}{1 - \dfrac{1}{x - 1}}$. + +\Item{5.} $\dfrac{\dfrac{2m + x}{m + x} - 1}{1 - \dfrac{x}{m + x}}$. + +\Item{6.} $\dfrac{\dfrac{x + y}{x^{2} - y^{2}}}{\dfrac{x - y}{x + y}}$. + +\Item{7.} $\dfrac{a + \dfrac{ab}{a - b}}{a - \dfrac{ab}{a + b}}$. + +\Item{8.} $\dfrac{9a^{2} - 64}{a - 1 - \dfrac{a + 4}{4}}$. + +\Item{9.} $\dfrac{\dfrac{1}{x} + \dfrac{1}{y}}{\dfrac{1}{x} - \dfrac{1}{y}}$. + +\Item{10.} $\dfrac{x + 3 + \dfrac{2}{x}}{1 + \dfrac{3}{x} + \dfrac{2}{x^{2}}}$. + +\Item{11.} $\dfrac{\dfrac{1}{x} - \dfrac{2}{x^{2}} + \dfrac{1}{x^{3}}}{\dfrac{(1 - x)^{2}}{x^{2}}}$. + +\Item{12.} $\dfrac{x^{2} - x-6}{1 - \dfrac{4}{x^{2}}}$. + +\Item{13.} $\dfrac{a^{2} - a + \dfrac{a - 1}{a + 1}}{a + \dfrac{1}{a + 1}}$. + +\Item{14.} $\dfrac{\dfrac{4a(a - x)}{a^{2} - x^{2}}}{\dfrac{a - x}{a + x}}$. +\end{multicols} +%% -----File: 109.png---Folio 103------- + + +\Chapter{X.}{Fractional Equations.} + +\Paragraph{139. To Reduce Equations containing Fractions.} + +\Item{1.} Solve $\dfrac{x}{3} - \dfrac{x - 1}{11} = x - 9$. +\begin{Soln} +Multiply by~$33$, the \LCM\ of the denominators. +\begin{DPalign*} +\lintertext{\indent Then,} +11x - 3x + 3 &= 33x - 297. \\ +\lintertext{\indent Transpose,} +11x - 3x - 33x &= - 297 - 3. \\ +\lintertext{\indent Combine,} +- 25x &= - 300. \\ +\lintertext{\indent Divide by~$-25$,} +x &= 12. +\end{DPalign*} +\end{Soln} + +\begin{Remark}[Note.] Since the minus sign precedes the second fraction, in removing +the denominator, the~$+$ (understood) before~$x$, the first term +of the numerator, is changed to~$-$, and the~$-$ before~$1$, the second +term of the numerator, is changed to~$+$. +\end{Remark} + +\Dictum{To clear an equation of fractions}, therefore, +\begin{Theorem} +Multiply each term by the \LCM\ of the denominators. +\end{Theorem} + +If a fraction is preceded by a \textbf{minus sign}, \emph{the sign of +every term of the numerator must be changed when the +denominator is removed}. + +\Item{2.} Solve $\dfrac{x + 1}{4} - \frac{1}{5}(x - 1) = 1$. +\begin{Soln} +Multiply by~$20$, the \LCD +\begin{DPalign*} +5x + 5 - 4(x - 1) &= 20. \\ +5x + 5 - 4x + 4 &= 20. \\ +\lintertext{\indent Transpose,} +5x - 4x &= 20 - 5 - 4. \\ +\lintertext{\indent Combine,} +x &= 11. +\end{DPalign*} +\end{Soln} +%% -----File: 110.png---Folio 104------- + +\Item{3.} Solve +\[ +7x - \frac{(2x - 3)(3x - 5)}{5} = \frac{153}{10} - \frac{(4x - 5)(3x - 1)}{10}. +\] +\begin{Soln} +Multiply by~$10$, the \LCD, and we have +\[ +70x - 2(2x - 3)(3x - 5) = 153 - (4x - 5) (3x - 1). +\] + +Find the products of $(2x - 3)(3x - 5)$ and $(4x - 5)(3x - 1)$. +\[ +70x - 2(6x^{2} - 19x + 15) = 153 - (12x^{2} - 19x + 5). +\] + +Remove the parentheses, +\[ +70x - 12x^{2} + 38x - 30 = 153 - 12x^{2} + 19x - 5. +\] + +Cancel the~$-12x^{2}$ on each side and transpose, +\begin{DPalign*} +70x + 38x - 19x &= 153 + 30 - 5. \\ +\lintertext{\indent Combine,} +89x &= 178. \\ +\lintertext{\indent Divide by~$89$,} +x &= 2. +\end{DPalign*} +\end{Soln} + +\Item{4.} Solve $\dfrac{2x + 1}{2x - 1} - \dfrac{2x - 1}{2x + 1} = \dfrac{8}{4x^{2} - 1}$. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Since} +4x^{2} - 1 &= (2x + 1)(2x - 1), \\ +\lintertext{the \LCD} + &= (2x + 1)(2x - 1). +\end{DPalign*} +Multiply by the \LCD, and we have, +\begin{DPalign*} +4x^{2} + 4x + 1-(4x^{2} - 4x + 1) &= 8. \\ +\therefore 4x^{2} + 4x + 1 - 4x^{2} + 4x - 1 &= 8. \\ +\therefore 8x &= 8. \\ +\therefore x &= 1. +\end{DPalign*} +\end{Soln} + +\Item{5.} Solve $\dfrac{4}{x + 1} - \dfrac{x + 1}{x - 1} + \dfrac{x^{2} - 3}{x^{2} - 1} = 0$. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Since} +x^{2} - 1 &= (x + 1)(x - 1) \\ +\lintertext{the \LCD} + &= (x + 1)(x - 1). +\end{DPalign*} + +Multiply by the \LCD, $x^{2} - 1$, and we have, +\begin{DPalign*} +4(x - 1) - (x + 1)(x + 1) + x^{2} - 3 &= 0. \\ +\therefore 4x - 4 - x^{2} - 2x - 1 + x^{2} - 3 &= 0. \\ +\therefore 2x &= 8. \\ +\therefore x &= 4. +\end{DPalign*} +\end{Soln} +%% -----File: 111.png---Folio 105------- + +%[** TN: Force page break in both print and screen layout] +\newpage +\Exercise{51.} + +Solve: +\begin{multicols}{2} +\Item{1.} $\dfrac{x - 1}{2} = \dfrac{x + 1}{3}$. + +\Item{2.} $\dfrac{3x - 1}{4} = \dfrac{2x + 1}{3}$. + +\Item{3.} $\dfrac{6x - 19}{2} = \dfrac{2x - 11}{3}$. + +\Item{4.} $\dfrac{7x - 40}{8} = \dfrac{9x - 80}{10}$. +\end{multicols} +%[** TN: Move end of two-column layout up six questions] + +\Item{5.} $\dfrac{3x - 116}{4} + \dfrac{180 - 5x}{6} = 0$. + +\Item{6.} $\dfrac{3x - 4}{2} - \dfrac{3x - 1}{16} = \dfrac{6x - 5}{8}$. + +\Item{7.} $\dfrac{x - 1}{8} - \dfrac{x + 1}{18} = 1$. + +\Item{8.} $\dfrac{60 - x}{14} - \dfrac{3x - 5}{7} = \dfrac{3x}{4}$. + +\Item{9.} $\dfrac{3x - 1}{11} - \dfrac{2 - x}{10} = \dfrac{6}{5}$. + +\Item{10.} $\dfrac{4x}{x + 1} - \dfrac{x}{x - 2} = 3$. + +\Item{11.} $\dfrac{2x + 1}{4} - \dfrac{4x - 1}{10} + 1 - \frac{1}{4} = 0$. + +\Item{12.} $\dfrac{x - 1}{5} - \dfrac{43 - 5x}{6} - \dfrac{3x - 1}{8} = 0$. + +\Item{13.} $\dfrac{1}{x + 7} = \dfrac{2}{x + 1} - \dfrac{1}{x + 3}$. + +\Item{14.} $\dfrac{1}{x + 4} + \dfrac{2}{x + 6} - \dfrac{3}{x + 5} = 0$. + +\Item{15.} $\dfrac{4}{x^{2} - 1} + \dfrac{1}{x - 1} + \dfrac{1}{x + 1} = 0$. + +\Item{16.} $\dfrac{3x + 1}{4} - \dfrac{5x - 4}{7} = 12 - 2x - \dfrac{x - 2}{3}$. + +\Item{17.} $\frac{1}{8}(5x + 3) - \frac{1}{3}(3 - 4x) + \frac{1}{6}(9 - 5x) = \frac{1}{2}(31 - x)$. + +\Item{18.} $\frac{1}{15} (34x - 56) - \frac{1}{5}(7x - 3) - \frac{1}{3}(7x - 5) = 0$. +%% -----File: 112.png---Folio 106------- + +\Exercise{52.} + +Solve: + +\Item{1.} $\frac{2}{3}(x + 1) - \frac{1}{7}(x + 5) = 1$. + +\Item{2.} $\frac{6}{7}(x - 9) - \frac{1}{3}(5 - x) + 3x + 1 = 0$. + +\Item{3.} $\frac{1}{3}(5x - 24) + \frac{1}{7}(x - 2) - 2(x - 1) = 0$. + +\Item{4.} $\dfrac{x + 3}{4} + \dfrac{7x - 2}{5} = \dfrac{5x - 1}{4} + \dfrac{5x + 4}{9}$. + +\Item{5.} $\dfrac{x + 1}{3} - \dfrac{x - 1}{4} = \dfrac{x - 2}{5} - \dfrac{x - 3}{6} + \dfrac{31}{60}$. + +\Item{6.} $\dfrac{(2x - 1)(2 - x)}{2} + x^{2} - \dfrac{1 + 3x}{2} = 0$. + +\Item{7.} $\dfrac{6x - 11}{4} - \dfrac{3 - 4x}{6} = \dfrac{4}{3} - \dfrac{x}{8}$. + +\Item{8.} $\dfrac{x + 6}{4} - \dfrac{16 - 3x}{12} = 4\frac{1}{6}$. + +\Item{9.} $x - \dfrac{x - 2}{3} = \dfrac{x + 23}{4} - \dfrac{10 + x}{5}$. + +\Item{10.} $\dfrac{5x + 3}{x - 1} + \dfrac{2x - 3}{2x - 1} = 6$. + +\Item{11.} $\dfrac{3x}{4x + 1} + 1 = 2 - \dfrac{x}{2(2x - 1)}$. + +\Item{12.} $\dfrac{8x + 7}{5x + 4} - 1 = 1 - \dfrac{2x}{5x + 1}$. + +\Item{13.} $\dfrac{x + 1}{2(x - 1)} - \dfrac{x - 1}{x + 1} = \dfrac{17 - x^{2}}{2(x^{2} - 1)}$. +%% -----File: 113.png---Folio 107------- + +\Paragraph{140.} If the denominators contain both simple and compound +expressions, it is generally best to remove the simple +expressions first, and then the compound expressions. After +each multiplication the result should be reduced to the +simplest form. + +\Item{1.} Solve $\dfrac{4x + 3}{10} - \dfrac{2x + 3}{5x - 1} = \dfrac{2x - 1}{5}$. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Multiply by~$10$,} +4x + 3 - \frac{10(2x + 3)}{5x - 1} &= 4x - 2. \\ +\lintertext{\indent Transpose,} +4x + 3 - 4x + 2 &= \frac{10(2x + 3)}{5x - 1}. \\ +\lintertext{\indent Combine,} +5 &= \frac{10(2x + 3)}{5x - 1}. \displaybreak[1] \\ +\lintertext{\indent Divide by~$5$,} +1 &= \frac{2(2x + 3)}{5x - 1}. \\ +\lintertext{\indent Multiply by $5x - 1$,} +5x - 1 &= 4x + 6. \\ +\lintertext{\indent Transpose and combine,} +x &= 7. +\end{DPalign*} +\end{Soln} + +\Exercise{53.} + +Solve: + +\Item{1.} $\dfrac{10x + 13}{18} - \dfrac{x + 2}{x - 3} = \dfrac{5x - 4}{9}$. + +\Item{2.} $\dfrac{6x + 7}{10} - \dfrac{3x + 1}{5} = \dfrac{x - 1}{3x - 4}$. + +\Item{3.} $\dfrac{11x - 12}{14} - \dfrac{11x - 7}{19x + 7} = \dfrac{22x - 36}{28}$. + +\Item{4.} $\dfrac{2x - 1}{5} + \dfrac{2x - 3}{17x - 12} = \dfrac{4x - 3}{10}$. + +\Item{5.} $\dfrac{11x - 13}{7} - \dfrac{13x + 7}{3x + 7} = \dfrac{22x - 75}{14}$. + +\Item{6.} $\dfrac{6x - 13}{2x + 3} + \dfrac{6x + 7}{9} - \dfrac{2x + 4}{3} = 0$. +%% -----File: 114.png---Folio 108------- + +\Paragraph{141. Literal Equations.} Literal equations are equations +in which some or all of the known numbers are represented +by letters; the numbers regarded as known numbers are +usually represented by the \emph{first} letters of the alphabet. + +\Item{1.} Solve $\dfrac{x + a}{x - b} + \dfrac{x + b}{x - a} = 2$. +\begin{Soln} +Multiply by $(x - a)(x - b)$, +\begin{DPalign*} +\llap{$(x + a)(x - a) + (x + b)(x - b)$} &= 2(x - a)(x - b), \\ +\lintertext{or} +x^{2} - a^{2} + x^{2} - b^{2} &= 2x^{2} - 2ax - 2bx + 2ab. \displaybreak[1] \\ +\lintertext{\indent Transpose,} +x^{2} + x^{2} - 2x^{2} + 2ax + 2bx &= a^{2} + 2ab + b^{2}\Add{.} \displaybreak[1] \\ +\lintertext{\indent Combine,} +2ax + 2bx &= a^{2} + 2ab + b^{2}, \\ +\lintertext{or} +2(a + b)x &= a^{2} + 2ab + b^{2}. \displaybreak[1] \\ +\lintertext{\indent \rlap{Divide by~$a + b$,}} +2x &= a + b\Add{,} \\ +\therefore x &= \frac{a + b}{2}. +\end{DPalign*} +\end{Soln} + +\Exercise{54.} + +Solve: + +\Item{1.} $a(x - a) = b(x - b)$. + +\Item{2.} $(a + b)x + (a - b)x = a^{2}$. + +\Item{3.} $(a + b)x - (a - b)x = b^{2}$. + +\Item{4.} $(2x - a) + (x - 2a) = 3a$. + +\Item{5.} $(x + a + b) + (x + a - b) = 2b$. + +\Item{6.} $(x - a)(x - b) = x(x + c)$. + +\Item{7.} $x^{2} + b^{2} = (a - x)(a - x)$. + +\Item{8.} $(a + b)(2 - x) = (a - b)(2 + x)$. + +\Item{9.} $(x - a)(2x - a) = 2(x - b)^{2}$. + +\Item{10.} $(a + bx)(c + d) = (a + b)(c + dx)$. + +\Item{11.} $\dfrac{x}{a - b} - \dfrac{3a}{a + b} = \dfrac{bx}{a^{2} - b^{2}}$. +%% -----File: 115.png---Folio 109------- + +\PrintBreak +\Paragraph{142. Problems involving Fractional Equations.} + +\Exercise{55.} + +Ex. The sum of the third and fifth parts of a certain +number exceeds two times the difference of the fourth and +sixth parts by~$22$. Find the number. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{the number\Add{.}} \\ +\lintertext{\indent Then} +\frac{x}{3} + \frac{x}{5} &= \text{the sum of its third and fifth parts,} \\ +\frac{x}{4} - \frac{x}{6} &= \text{the difference of its fourth and sixth parts,} \\ +2\left(\frac{x}{4} - \frac{x}{6}\right) + &= \text{$2$~times the difference of} \\ + &\qquad\text{its fourth and sixth parts,} +\end{DPalign*} +\begin{DPalign*} +\frac{x}{3} + \frac{x}{5} - 2\left(\frac{x}{4} - \frac{x}{6}\right) + &= \text{the given excess.} \\ +\lintertext{\indent But} +22 &= \text{the given excess.} \\ +\therefore \frac{x}{3} + \frac{x}{5} - 2\left(\frac{x}{4} - \frac{x}{6}\right) + &= 22. +\end{DPalign*} + +Multiply by~$60$ the \LCD\ of the fractions. +\begin{DPalign*} +20x + 12x - 30x + 20x &= 60 × 22. \\ +\lintertext{\indent Combining,} +22x &= 60 × 22\Add{,} \\ +\therefore x &= 60. +\end{DPalign*} + +The required number, therefore, is~$60$. +\end{Soln} + +\Item{1.} The difference between the fifth and seventh parts of +a certain number is~$2$. Find the number. + +\Item{2.} One-half of a certain number exceeds the sum of its +fifth and seventh parts by~$11$. Find the number. + +\Item{3.} The sum of the third and sixth parts of a certain +number exceeds the difference of its sixth and ninth parts +by~$16$. Find the number. + +\Item{4.} There are two consecutive numbers, $x$~and~$x + 1$, such +that one-half the larger exceeds one-third the smaller +number by~$10$. Find the numbers. +%% -----File: 116.png---Folio 110------- + +\Exercise{56.} + +Ex. The sum of two numbers is~$63$, and if the greater +is divided by the smaller number, the quotient is~$2$ and the +remainder~$3$. Find the numbers. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{the greater number.} \\ +\lintertext{\indent Then} +63 - x &= \text{the smaller number.} \\ +\text{Since the quotient} + &= \frac{\text{Dividend} - \text{Remainder}}{\text{Divisor}}, +\end{DPalign*} +and since, in this problem, the dividend is~$x$, the remainder is~$3$, +and the divisor is~$63 - x$, we have +\begin{DPalign*} +\frac{x - 3}{63 - x} = 2. \\ +\lintertext{\indent Solving,} +x &= 43. +\end{DPalign*} + +The two numbers, therefore, are $43$~and~$20$. +\end{Soln} + +\Item{1.} The sum of two numbers is~$100$, and if the greater is +divided by the smaller number, the quotient is~$4$ and the +remainder~$5$. Find the numbers. + +\Item{2.} The sum of two numbers is~$124$, and if the greater is +divided by the smaller number, the quotient is~$4$ and the +remainder~$4$. Find the numbers. + +\Item{3.} The difference of two numbers is~$49$, and if the greater +is divided by the smaller, the quotient is~$4$ and the remainder~$4$. +Find the numbers. + +\Item{4.} The difference of two numbers is~$91$, and if the +greater is divided by the smaller, the quotient is~$8$ and the +remainder~$7$. Find the numbers. + +\Item{5.} Divide $320$ into two parts such that the smaller part +is contained in the larger part $11$~times, with a remainder +of~$20$. +%% -----File: 117.png---Folio 111------- + +\Exercise{57.} + +Ex. Eight years ago a boy was one-fourth as old as he +will be one year hence. How old is he now? +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{the number of years old he is now.} \\ +\lintertext{\indent Then} +x - 8 &= \text{the number of years old he was eight years ago,} \\ +\lintertext{and} +x + 1 &= \text{the number of years old he will be one year hence.} +\end{DPalign*} +\begin{DPalign*} +\therefore x - 8 &= \tfrac{1}{4}(x + 1). \\ +\lintertext{\indent Solving,} +x &= 11. +\end{DPalign*} + +Therefore the boy is $11$~years old. +\end{Soln} + +\Item{1.} A son is one-fourth as old as his father. In $24$~years +he will be one-half as old. Find the age of the son. + +\Item{2.} B's~age is one-sixth of A's~age. In $15$~years B's~age +will be one-third of A's~age. Find their ages. + +\Item{3.} The sum of the ages of A~and~B is $30$~years, and $5$~years +hence B's~age will be one-third of~A's. Find their +ages. + +\Item{4.} A father is $35$~years old, and his son is one-fourth of +that age. In how many years will the son be half as old +as his father? + +\Item{5.} A is $60$~years old, and B's~age is two-thirds of~A's. +How many years ago was B's~age one-fifth of~A's? + +\Item{6.} A son is one-third as old as his father. Four years +ago he was only one-fourth as old as his father. What is +the age of each? + +\Item{7.} A is $50$~years old, and B~is half as old as~A\@. In how +many years will B be two-thirds as old as~A? + +\Item{8.} B~is one-half as old as~A\@. Ten years ago he was +one-fourth as old as~A\@. What are their present ages? + +\Item{9.} The sum of the ages of a father and his son is $80$~years. +The son's age increased by $5$~years is one-fourth of +the father's age. Find their ages. +%% -----File: 118.png---Folio 112------- + +\Exercise{58.} + +Ex. A~can do a piece of work in $2$~days, and B~can do +it in $3$~days. How long will it take both together to do +the work? +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{the number of days it will take both together.} \\ +\lintertext{\indent Then} +\frac{1}{x} &= \text{\emph{the part both together can do in one day},} \\ +\tfrac{1}{2} &= \text{the part A can do in one day,} \\ +\tfrac{1}{3} &= \text{the part B can do in one day,} \\ +\lintertext{and} +\tfrac{1}{2} + \tfrac{1}{3} + &= \text{\emph{the part both together can do in one day}.} +\end{DPalign*} +\begin{DPalign*} +\therefore \frac{1}{2} + \frac{1}{3} &= \frac{1}{x}. \\ +\lintertext{\indent Solving,} +x &= 1\tfrac{1}{5}. +\end{DPalign*} + +Therefore they together can do the work in $1\frac{1}{5}$~days. +\end{Soln} + +\Item{1.} A~can do a piece of work in $3$~days, B~in $5$~days, and +C~in $6$~days. How long will it take them to do it working +together? + +\Item{2.} A~can do a piece of work in $5$~days, B~in $4$~days, and +C~in $3$~days. How long will it take them together to do +the work? + +\Item{3.} A~can do a piece of work in $2\frac{1}{2}$~days, B~in $3\frac{1}{2}$~days, +and C~in $3\frac{3}{4}$~days. How long will it take them together to +do the work? + +\Item{4.} A~can do a piece of work in $10$~days, B~in $12$~days; +A~and~B together, with the help of~C, can do the work in +$4$~days. How long will it take C~alone to do the work? + +\Item{5.} A~and~B together can mow a field in $10$~hours, A~and~C +in $12$~hours, and A~alone in $20$~hours. In what time +can B~and~C together mow the field? + +\Item{6.} A~and~B together can build a wall in $12$~days, A~and~C +in $15$~days, B~and~C in $20$~days. In what time can they +build the wall if they all work together? + +\begin{Remark}[Hint.] By working \emph{$2$~days each} they build $\frac{1}{12} + \frac{1}{15} + \frac{1}{20}$ of it. +\end{Remark} +%% -----File: 119.png---Folio 113------- + +\Exercise{59.} + +Ex. A cistern can be filled by three pipes in $15$, $20$, and +$30$~hours, respectively. In what time will it be filled by +all the pipes together? +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{the number of hours it will take all together.} \\ +\lintertext{\indent Then} +\frac{1}{x} &= \text{the part all together can fill in one hour\Add{.}} \\ +\llap{$\tfrac{1}{15} + \tfrac{1}{20}$} + \tfrac{1}{30} + &= \text{the part all together can fill in one hour\Add{.}} +\end{DPalign*} +\begin{DPalign*} +\frac{1}{15} + \frac{1}{20} + \frac{1}{30} &= \frac{1}{x}\Add{.} \\ +\lintertext{\indent Solving,} +x &= 6 \tfrac{2}{3}\Add{.} +\end{DPalign*} + +Therefore the pipes together can fill it in $6 \frac{2}{3}$~hours. +\end{Soln} + +\Item{1.} A cistern can be filled by three pipes in $16$, $24$, and +$32$~hours, respectively. In what time will it be filled by +all the pipes together? + +\Item{2.} A tank can be filled by two pipes in $3$~hours and $4$~hours, +respectively, and can be emptied by a third pipe in +$6$~hours. In what time will the cistern be filled if the pipes +are all running together? + +\Item{3.} A tank can be filled by three pipes in $1$~hour and $40$ +minutes, $3$~hours and $20$ minutes, and $5$~hours, respectively. +In what time will the tank be filled if all three pipes are +running together? + +\Item{4.} A cistern can be filled by three pipes in $2\frac{1}{3}$~hours, $3\frac{1}{2}$~hours, +and $4\frac{2}{3}$~hours, respectively. In what time will the +cistern be filled if all the pipes are running together? + +\Item{5.} A cistern has three pipes. The first pipe will fill the +cistern in $12$~hours, the second in $20$~hours, and all three +pipes together will fill it in $6$~hours. How long will it take +the third pipe alone to fill it? +%% -----File: 120.png---Folio 114------- + +\Exercise{60.} + +Ex. A courier who travels $6$~miles an hour is followed, +after $2$~hours, by a second courier who travels $7\frac{1}{2}$~miles an +hour. In how many hours will the second courier overtake +the first? +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{the number of hours the first travels.} \\ +\lintertext{\indent Then} +x - 2 &= \text{the number of hours the second travels,} \\ +6x &= \text{the number of miles the first travels,} \\ +\lintertext{and} +(x - 2) 7\tfrac{1}{2} &= \text{the number of miles the second travels.} +\displaybreak[1] \\ +\intertext{\indent They both travel the same distance.} +\therefore 6x &= (x - 2) 7\tfrac{1}{2}, \\ +\lintertext{or} +12x &= 15x - 30. \\ +\therefore x &= 10. +\end{DPalign*} + +Therefore the second courier will overtake the first in $10 - 2$, or +$8$~hours. +\end{Soln} + +\Item{1.} A sets out from Boston and walks towards Portland +at the rate of $3$~miles an hour. Three hours afterward B +sets out from the same place and walks in the same direction +at the rate of $4$~miles an hour. How far from Boston +will B~overtake~A? + +\Item{2.} A courier who goes at the rate of $6\frac{1}{2}$~miles an hour is +followed, after $4$~hours, by another who goes at the rate of +$7\frac{1}{2}$~miles an hour. In how many hours will the second +overtake the first? + +\Item{3.} A person walks to the top of a mountain at the rate +of two miles an hour, and down the same way at the rate +of $4$~miles an hour. If he is out $6$~hours, how far is it to +the top of the mountain? + +\Item{4.} In going a certain distance, a train travelling at the +rate of $40$~miles an hour takes $2$~hours less than a train +travelling $30$~miles an hour. Find the distance. +%% -----File: 121.png---Folio 115------- + +\Exercise{61.} + +Ex. A hare takes $4$~leaps to a greyhound's~$3$; but $2$~of +the greyhound's leaps are equivalent to $3$~of the hare's. +The hare has a start of $50$~leaps. How many leaps must +the greyhound take to catch the hare? +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +3x &= \text{the number of leaps taken by the greyhound.} \\ +\lintertext{\indent Then} +4x &= \text{the number of leaps of the hare in the same time.} \\ +%[** TN: Hack to coax spacing] +\lintertext{\indent Also, l\rlap{et}} +a &= \text{the number of feet in one leap of the hare.} \displaybreak[1] \\ +\lintertext{\indent Then} +\frac{3a}{2} &= \text{the number of feet in one leap of the hound.} +\displaybreak[1] \\ +\lintertext{\indent \rlap{Therefore,}} +&\quad 3x x \frac{3a}{2} \quad\text{or}\quad +\frac{9ax}{2} = \text{the whole distance}. +\end{DPalign*} + +As the hare has a start of $50$~leaps, and takes $4x$~leaps more before +she is caught, and as each leap is $a$~feet, +\begin{DPalign*} +(50 + 4x)a &= \text{the whole distance.} \\ +\therefore \frac{9ax}{2} &= (50 + 4x)a. \\ +\lintertext{\indent Multiply by~$2$,} +9ax &= (100 + 8x)a, \\ +\lintertext{\indent Divide by~$a$,} +9x &= 100 + 8x, \\ +x &= 100, \\ +\therefore 3x &= 300. +\end{DPalign*} + +Therefore the greyhound must take $300$~leaps. +\end{Soln} + +\Item{1.} A hound makes $3$~leaps while a rabbit makes~$5$; but +$1$~of the hound's leaps is equivalent to $2$~of the rabbit's. +The rabbit has a start of $120$~leaps. How many leaps +will the rabbit take before she is caught? + +\Item{2.} A rabbit takes $6$~leaps to a dog's~$5$, and $7$~of the dog's +leaps are equivalent to $9$~of the rabbit's. The rabbit has +a start of~$60$ of her own leaps. How many leaps must the +dog take to catch the rabbit? + +\Item{3.} A dog makes $4$~leaps while a rabbit makes~$5$; but $3$~of +the dog's leaps are equivalent to $4$~of the rabbit's. The +rabbit has a start of $90$~of the \emph{dog's leaps}. How many +leaps will each take before the rabbit is caught? +%% -----File: 122.png---Folio 116------- + +\Exercise{62.} + +Ex. Find the time between $2$ and $3$~o'clock when the +hands of a clock are together. +\begin{Soln} +At 2 o'clock the hour-hand is 10 minute-spaces ahead of the +minute-hand. +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{\emph{the number of spaces the minute-hand moves over}.} \\ +\lintertext{\indent The\rlap{n}} +x - 10 &= \text{the number of spaces the hour-hand moves over.} +\displaybreak[1] \\ +\intertext{\indent Now, as the minute-hand moves $12$~times as fast as the hour-hand,} +\llap{$12(x - 10)$} &= \text{\emph{the number of spaces the minute-hand moves over}.} +\end{DPalign*} +\begin{DPalign*} +\therefore 12(x - 10) &= x, \\ +\lintertext{and} +11x &= 120. \\ +\therefore x &= 10\tfrac{10}{11}. +\end{DPalign*} + +Therefore the time is $10\frac{10}{11}$~minutes past $2$~o'clock. +\end{Soln} + +\Item{1.} Find the time between $5$ and $6$~o'clock when the +hands of a clock are together. + +\Item{2.} Find the time between $2$ and $3$~o'clock when the +hands of a clock are at right angles to each other. + +\begin{Remark}[Hint.] In this case the minute-hand is $15$~minutes ahead of the +hour-hand. +\end{Remark} + +\Item{3.} Find the time between $2$ and $3$~o'clock when the +hands of a clock point in opposite directions. + +\begin{Remark}[Hint.] In this case the minute-hand is $30$~minutes ahead of the +hour-hand. +\end{Remark} + +\Item{4.} Find the time between $1$ and $2$~o'clock when the +hands of a clock are at right angles to each other. + +\Item{5.} Find the time between $1$ and $2$~o'clock when the +hands of a clock point in opposite directions. + +\Item{6.} At what time between $7$ and $8$~o'clock are the hands +of a watch together? +%% -----File: 123.png---Folio 117------- + +\Exercise{63.} + +Ex. A rectangle has its length $6$~feet more and its width +$5$~feet less than the side of its equivalent square. Find the +dimensions of the rectangle. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{the number of feet in a side of the square.} \\ +\lintertext{\indent Then} +x + 6 &= \text{the number of feet in the length of the rectangle,} \\ +\lintertext{and} +x - 5 &= \text{the number of feet in the width of the rectangle.} +\intertext{Since the area of a rectangle is equal to the product of the number +of units of length in the length and width of the rectangle,} +\llap{$(x + 6)(x - 5)$} &= \text{the area of the rectangle in square feet,} \\ +\lintertext{and} +x × x &= \text{the area of the square in square feet.} +\end{DPalign*} + +But these areas are equal. +\begin{DPalign*} +\therefore (x + 6)(x - 5) &= x^{2}. \\ +\lintertext{\indent Solving,} +x &= 30. +\end{DPalign*} + +Therefore the dimensions of the rectangle are $36$~feet and $25$~feet. +\end{Soln} + +\Item{1.} A rectangle has its length and breadth respectively $7$~feet +longer and $6$~feet shorter than the side of the equivalent +square. Find its area. + +\Item{2.} The length of a floor exceeds the breadth by $5$~feet. +If each dimension were $1$~foot more, the area of the floor +would be $42$~sq.~ft.\ more. Find its dimensions. + +\Item{3.} A rectangle whose length is $6$~feet more than its +breadth, would have its area $35$~sq.~ft.\ more, if each dimension +were $1$~foot more. Find its dimensions. + +\Item{4.} The length of a rectangle exceeds its width by $3$~feet. +If the length is increased by $3$~feet and the width diminished +by $2$~feet, the area will not be altered. Find its +dimensions. + +\Item{5.} The length of a floor exceeds its width by $10$~feet +If each dimension were $2$~feet more, the area would be $144$~sq.~ft.\ +more. Find its dimensions. +%% -----File: 124.png---Folio 118------- + +\Paragraph{143. Formulas and Rules.} When the \emph{given} numbers of a +problem are represented by letters, the result obtained from +solving the problem is a general expression which includes +all problems of that class. Such an expression is called a +\Defn{formula}, and the translation of this formula into words is +called a \Defn{rule}. + +\Paragraph{144.} We will illustrate by examples. + +\Item{1.} The sum of two numbers is~$s$, and their difference~$d$. +Find the numbers. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{the smaller number;} \\ +\lintertext{then} +x + d &= \text{the larger number.} \displaybreak[1] \\ +\lintertext{\indent Hence} +x + x + d &= s, \\ +\lintertext{or} +2x &= s - d. \\ +\therefore x &= \frac{s - d}{2}, \displaybreak[1] \\ +\lintertext{and} +x + d &= \frac{s - d}{2} + d = \frac{s - d + 2d}{2}, \\ + &= \frac{s + d}{2}. +\end{DPalign*} + +Therefore the numbers are $\dfrac{s + d}{2}$ and $\dfrac{s - d}{2}$. +\end{Soln} + +As these formulas hold true whatever numbers $s$~and~$d$ +stand for, we have the general rule for finding two numbers +when their sum and difference are given: +\begin{Theorem} +Add the difference to the sum and take half the result for +the greater number. + +Subtract the difference from the sum and take half the +result for the smaller number. +\end{Theorem} + +\Item{2.} If A~can do a piece of work in $a$~days, and B~can +do the same work in $b$~days, in how many days can both +together do it? +%% -----File: 125.png---Folio 119------- +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{the required number of days\Add{.}} \\ +\lintertext{\indent Then,} +\frac{1}{x} &= \text{\emph{the part both together can do in one day}\Add{.}} +\displaybreak[1] \\ +\lintertext{\indent Now} +\frac{1}{a} &= \text{the part A can do in one day,} \\ +\lintertext{and} +\frac{1}{b} &= \text{the part B can do in one day\Add{,}} \\ +\lintertext{therefore} +\frac{1}{a} + \frac{1}{b} + &= \text{\emph{the part both together can do in one day}} +\end{DPalign*} +\begin{DPalign*} +\frac{1}{a} + \frac{1}{b} &= \frac{1}{x}\Add{.} \\ +\lintertext{\indent Whence} +x &= \frac{ab}{a + b}\Add{.} +\end{DPalign*} +\end{Soln} + +The translation of this formula gives the following rule +for finding the time required by two agents together to +produce a given result when the time required by each +agent separately is known. +\begin{Theorem} +Divide the product of the numbers which express the units +of time required by each to do the work by the sum of these +numbers, the quotient is the time required by both together. +\end{Theorem} + +\Paragraph{145. Interest Formulas.} The elements involved in computation +of interest are the \emph{principal}, \emph{rate}, \emph{time}, \emph{interest}, +and \emph{amount}. +\begin{DPalign*} +\lintertext{\indent Let} +p &= \text{the principal,} \\ +r &= \text{the interest of \$$1$ for $1$~year, at the given rate,} \\ +t &= \text{the time expressed in years,} \\ +i &= \text{the interest for the given time and rate,} \\ +a &= \text{the amount (sum of principal and interest).} +\end{DPalign*} + +\Paragraph{146. Given the Principal, Rate, and Time. Find the Interest.} + +Since $r$ is the interest of~\$$1$ for $1$~year, $pr$~is the interest +of~\$$p$ for $1$~year, and $prt$~is the interest of~\$$p$ for $t$~years +\begin{DPgather*} +i = prt. +\rintertext{(Formula 1.)} +\end{DPgather*} + +\begin{Theorem}[\textsc{Rule.}] Find the product of the principal, rate, and time\Add{.} +\end{Theorem} +%% -----File: 126.png---Folio 120------- + +\Paragraph{147. Given the Interest, Rate, and Time. Find the Principal.} +\begin{DPalign*} +\lintertext{\indent By formula 1,} +prt &= i. \\ +\lintertext{\indent Divide by~$rt$,} +p &= \frac{i}{rt}. +\rintertext{(Formula 2.)} +\end{DPalign*} + +\begin{Theorem}[\textsc{Rule.}] Divide the interest by the product of the rate and +time. +\end{Theorem} + +\Paragraph{148. Given the Amount, Rate, and Time. Find the Principal.} +\begin{DPalign*} +\lintertext{\indent From formula 1,} +p + prt &= a, \\ +\lintertext{or} +p(1 + rt) &= a. \\ +\lintertext{\indent Divide by $1 + rt$,} +p &= \frac{a}{1 + rt}. +\rintertext{(Formula 3.)} +\end{DPalign*} + +\begin{Theorem}[\textsc{Rule.}] Divide the amount by $1$~plus the product of the +rate and time. +\end{Theorem} + +\Paragraph{149. Given the Amount, Principal, and Rate. Find the Time.} +\begin{DPalign*} +\lintertext{\indent From formula 1,} +p + prt &= a. \\ +\lintertext{\indent Transpose~$p$,} +prt &= a - p. \\ +\lintertext{\indent Divide by~$pr$,} +t &= \frac{a - p}{pr}. +\rintertext{(Formula 4.)} +\end{DPalign*} + +\begin{Theorem}[\textsc{Rule.}] Subtract the principal from the amount, and +divide the result by the product of the principal and rate. +\end{Theorem} + +\Paragraph{150. Given the Amount, Principal, and Time. Find the Rate.} +\begin{DPalign*} +\lintertext{\indent From formula 1,} +p + prt &= a. \\ +\lintertext{\indent Transpose~$p$,} +prt &= a - p. \\ +\lintertext{\indent Divide by~$pt$,} +r &= \frac{a - p}{pt}. +\rintertext{(Formula 5.)} +\end{DPalign*} + +\begin{Theorem}[\textsc{Rule.}] Subtract the principal from the amount, and +divide the result by the product of the principal and time. +\end{Theorem} +%% -----File: 127.png---Folio 121------- + +\Exercise{64.} + +Solve the following examples by the preceding formulas: + +\Item{1.} The sum of two angles is $120°\, 30'\, 30''$ and their difference +$59°\, 30'\, 30''$. Find the angles. + +\Item{2.} Find the interest of \$$1000$ for $3$~years and $4$~months +at~$4$\%. + +\Item{3.} Find the principal that will amount to \$$2280$ in $3$~years +and $6$~months at~$4$\%. + +\Item{4.} Find the principal that will produce \$$280$ interest in +$2$~years and $4$~months at~$3$\%. + +\Item{5.} Find the principal that will produce \$$270$ interest +in $1$~year and $6$~months at~$6$\%. + +\Item{6.} Find the principal that will amount to \$$590$ in $4$~years +at~$4\frac{1}{2}$\%. + +\Item{7.} Find the rate if the amount of \$$250$ for $4$~years +is~\$$300$. + +\Item{8.} Find the rate if \$$1000$ amounts to \$$2000$ in $16$~years +and $8$~months. + +\Item{9.} Find the time required for the interest on \$$400$ to +be \$$54$ at~$4\frac{1}{2}$\%. + +\Item{10.} Find the time required for \$$160$ to amount to \$$250$ +at~$6$\%. + +\Item{11.} How much money must be invested at~$5$\% to yield +an annual income of~\$$1250$? + +\Item{12.} Find the principal that will produce \$$100$ a month +if invested at $6$\%~per annum. + +\Item{13.} Find the rate if the interest on \$$1000$ for $8$~months +is~\$$40$. + +\Item{14.} Find the time for a sum of money on interest at~$5$\% +to double itself. +%% -----File: 128.png---Folio 122------- + + +\Chapter{XI.}{Simultaneous Equations of the First +Degree.} + +\Paragraph{151.} If we have two unknown numbers and but one relation +between them, we can find an unlimited number of +pairs of values for which the given relation will hold true. +Thus, if $x$~and~$y$ are unknown, and we have given only the +one relation $x + y = 10$, we can \emph{assume} any value for~$x$, +and then from the relation $x + y = 10$ find the corresponding +value of~$y$. For from $x + y = 10$ we find $y = 10 - x$. +If $x$~stands for~$1$, $y$~stands for~$9$; if $x$~stands for~$2$, $y$~stands +for~$8$; if $x$~stands for~$-2$, $y$~stands for~$12$; and so on without +end. + +\Paragraph{152.} We may, however, have two equations that express +\emph{different} relations between the two unknown numbers. +Such equations are called \Defn{independent equations}. Thus, +$x + y = 10$ and $x - y = 2$ are independent equations, for +they evidently express \emph{different} relations between $x$~and~$y$. + +\Paragraph{153.} Independent equations involving the \emph{same} unknown +numbers are called \Defn{simultaneous equations}. + +If we have two unknown numbers, and two independent +equations involving them, there is but \emph{one} pair of values +which will hold true for both equations. Thus, if besides +the relation $x + y = 10$, we have also the relation $x - y = 2$, +the only pair of values for which both equations will hold +true is the pair $x = 6$, $y = 4$. + +Observe that in this problem $x$~stands for the same number +in \emph{both} equations; so also does~$y$. +%% -----File: 129.png---Folio 123------- + +\Paragraph{154.} Simultaneous equations are solved by combining +the equations so as to obtain a single equation with one +unknown number. + +This process is called \Defn{Elimination}. + +\Paragraph{155. Elimination by Addition or Subtraction.} + +%[** TN: Omitted large brace for grouping, horizontal bar indicating summation] +\Item{1.} Solve: +\begin{alignat*}{2} +5x &- 3y &&= 20 +\Tag{(1)} \\ +2x &+ 5y &&= 39 +\Tag{(2)} +\end{alignat*} +\begin{Soln} +Multiply (1) by~$5$, and (2) by~$3$, +\begin{DPalign*} +25x - 15y &= 100 +\Tag{(3)} \\ + 6x + 15y &= 117 +\Tag{(4)} \\ +\lintertext{\indent Add (3) and (4),} +31x \PadTo{{} + 15y}{} &= 217 \\ +\therefore x &= 7. +\intertext{\indent Substitute the value of~$x$ in~(2),} +14 + 5y &= 39. \\ +5y &= 25. \\ +\therefore y &= 5. +\end{DPalign*} + +In this solution $y$~is eliminated by \emph{addition}. +\end{Soln} + +\Item{2.} Solve: +\begin{alignat*}{2} +6x &+ 35y &&= 177 +\Tag{(1)} \\ +8x &- 21y &&= \Z33 +\Tag{(2)} +\end{alignat*} +\begin{Soln} +Multiply (1) by~$4$, and (2) by~$3$, +\begin{DPalign*} +24x + 140y &= 708 +\Tag{(3)} \\ +24x - \Z63y &= \Z99 +\Tag{(4)} \\ +\lintertext{\indent Subtract,} +203y &= 609 \\ +\therefore y &= 3. \\ +\intertext{\indent Substitute the value of~$y$ in (2),} +8x - 63 &= 33. \\ +8x &= 96. \\ +\therefore x &= 12. +\end{DPalign*} + +In this solution $x$~is eliminated by \emph{subtraction}. +\end{Soln} +%% -----File: 130.png---Folio 124------- + +\Paragraph{156.} \Dictum{To eliminate by addition or subtraction}, therefore, +\begin{Theorem} +Multiply the equations by such numbers as will make the +coefficients of one of the unknown numbers equal in the +resulting equations. + +Add the resulting equations if these equal coefficients have +unlike signs; subtract one from the other if these equal coefficients +have like signs. +\end{Theorem} + +\begin{Remark}[Note.] It is generally best to select the letter to be eliminated +which requires the smallest multipliers to make its coefficients equal; +and the smallest multiplier for each equation is found by dividing +the \LCM\ of the coefficients of this letter by the given coefficient +in that equation. Thus, in example~2, the \LCM\ of $6$~and~$8$ (the +coefficients of~$x$) is~$24$, and hence the smallest multipliers of the two +equations are $4$~and~$3$, respectively. +\end{Remark} + +Sometimes the solution is simplified by first adding the +given equations, or by subtracting one from the other. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Ex.} +x + 49y &= 51 +\Tag{(1)} \\ +49x + \PadTo{49y}{y} &= 99 +\Tag{(2)} \displaybreak[1] \\ +\lintertext{\indent Add (1) and (2),} +50x + 50y &= 150 +\Tag{(3)} \displaybreak[1] \\ +\lintertext{\indent Divide (3) by~$50$,} +x + y &= 3. +\Tag{(4)} \displaybreak[1] \\ +\lintertext{\indent \rlap{Subtract (4) from (1),}} +48y &= 48. \\ +\therefore y &= 1. \displaybreak[1] \\ +\lintertext{\indent \rlap{Subtract (4) from (2),}} +48x &= 96. \\ +\therefore x &= 2. +\end{DPalign*} +\end{Soln} + +\Exercise{65.} + +Solve by addition or subtraction: +\begin{multicols}{2} +\Item{1.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + 5x &+ 4y &&= 14 \\ +17x &- 3y &&= 31 +\end{alignedat}\right\}$} + +\Item{2.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + 3x &- 2y &&= \Z5 \\ + 2x &+ 5y &&= 16 +\end{alignedat}\right\}$} + +\Item{3.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + 2x &- 3y &&= \Z7 \\ + 5x &+ 2y &&= 27 +\end{alignedat}\right\}$} + +\Item{4.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + 7x &+ 6y &&= 20 \\ + 2x &+ 5y &&= \Z9 +\end{alignedat}\right\}$} + +\Item{5.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + x &+ 5y &&= 11 \\ + 3x &+ 2y &&= \Z7 +\end{alignedat}\right\}$} + +\Item{6.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + 3x &- 5y &&= 13 \\ + 4x &- 7y &&= 17 +\end{alignedat}\right\}$} +%% -----File: 131.png---Folio 125------- + +\Item{7.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + 8x &- \Z y &&= \Z3 \\ + 7x &+ 2y &&= 63 +\end{alignedat}\right\}$} + +\Item{8.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + 5x &- 4y &&= \Z7 \\ + 7x &+ 3y &&= 70 +\end{alignedat}\right\}$} + +\Item{9.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + x &+ 21y &&= \Z2 \\ + 2x &+ 27y &&= 19 +\end{alignedat}\right\}$} + +\Item{10.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + 6x &- 13y &&= -1 \\ + 5x &- 12y &&= -2 +\end{alignedat}\right\}$} + +\Item{11.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + 7x &+ \Z y &&= 265 \\ + 3x &- 5y &&= \Z\Z5 +\end{alignedat}\right\}$} + +\Item{12.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + 2x &+ 3y &&= \Z7 \\ + 8x &- 5y &&= 11 +\end{alignedat}\right\}$} + +\Item{13.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + 5x &+ 7y &&= 19 \\ + 7x &+ 4y &&= 15 +\end{alignedat}\right\}$} + +\Item{14.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} +11x &- 12y &&= \Z9 \\ + 4x &+ \Z5y &&= 22 +\end{alignedat}\right\}$} + +\Item{15.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + x &+ 8y &&= 17 \\ + 7x &- 3y &&= \Z1 +\end{alignedat}\right\}$} + +\Item{16.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + 4x &+ 3y &&= 25 \\ + 5x &- 4y &&= \Z8 +\end{alignedat}\right\}$} +\end{multicols} + +Clear of fractions and solve: +\begin{multicols}{2} +\Item{17.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + \frac{2x}{3} &- \frac{5y}{4} &&= 3 \\ + \frac{7x}{4} &- \frac{5y}{3} &&= \frac{43}{3} +\end{alignedat}\right\}$} + +\Item{18.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + \frac{7x}{6} &+ \frac{6y}{7} &&= 32 \\ + \frac{5x}{4} &- \frac{2y}{3} &&= 1 +\end{alignedat}\right\}$} + +\Item{19.} \raisebox{-0.5\baselineskip}{$\left.\begin{aligned} + \frac{x + y}{4} - \frac{7x - 5y}{11} &= 3 \\ + \frac{x}{5} - \frac{2y}{7} + 1 &= 0 +\end{aligned}\right\}$} + +\Item{20.} \raisebox{-0.5\baselineskip}{$\left.\begin{aligned} + \frac{ 6x + 7y}{2} &= 22 \\ + \frac{55y - 2x}{5} &= 20 +\end{aligned}\right\}$} +\end{multicols} + +\Item{21.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + \frac{x + y}{2} &- \frac{x - y}{3} &&= \Z8 \\ + \frac{x + y}{3} &+ \frac{x - y}{4} &&= 11 +\end{alignedat}\right\}$} + +\Item{22.} \raisebox{-0.5\baselineskip}{$\left.\begin{aligned} + \frac{8x - 5y}{7} + \frac{11y - 4x}{5} &= 4 \\ + \frac{17x - 13y}{5} + \frac{2x}{3} &= 7 +\end{aligned}\right\}$} +%% -----File: 132.png---Folio 126------- + +\Item{23.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + \frac{ 5x - 3y}{3} &+ \frac{7x - 5y}{11} &&= 4 \\ + \frac{15y - 3x}{7} &+ \frac{7y - 3x}{5} &&= 4 + \end{alignedat}\right\}$} + +\Item{24.} \raisebox{-0.5\baselineskip}{$\left.\begin{aligned} + \frac{2x - 3}{4} - \frac{y - 8}{5} &= \frac{y + 3}{4} \\ + \frac{x - 7}{3} + \frac{4y + 1}{11} &= 3 + \end{aligned}\right\}$} + +\Item{25.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + \frac{x - 2y}{6} &- \frac{x + 3y}{4} &&= \frac{3}{2} \\ + \frac{2x - y}{6} &- \frac{3x + y}{4} &&= \frac{5y}{4} + \end{alignedat}\right\}$} + +\Item{26.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + \frac{x}{a + b} &+ \frac{y}{a - b} &&= \frac{1}{a - b} \\ + \frac{x}{a + b} &- \frac{y}{a - b} &&= \frac{1}{a + b} + \end{alignedat}\right\}$} + +\begin{Remark}[Note.] To find $x$ in problem~26, add the equations; to find~$y$, +subtract one from the other. Do not clear of fractions. +\end{Remark} + +\Paragraph{157. Problems involving Two Unknown Numbers.} + +Ex. If A gives B \$$10$, B~will have three times as much +money as~A\@. If B gives A \$$10$, A~will have twice as +much money as~B\@. How much has each? +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{the number of dollars A has,} \\ +\lintertext{and} +y &= \text{the number of dollars B has.} \displaybreak[1] \\ +\intertext{\indent Then, after A gives B \$$10,$} +x - 10 &= \text{the number of dollars A has,} \\ +y + 10 &= \text{the number of dollars B has.} \displaybreak[1] \\ +\intertext{\indent Since B's money is now $3$~times A's, we have,} +y + 10 &= 3(x - 10). +\Tag{(1)} \displaybreak[1] \\ +%% -----File: 133.png---Folio 127------- +\intertext{\indent If B gives A \$$10$,} +x + 10 &= \text{the number of dollars A has,} \\ +y - 10 &= \text{the number of dollars B has.} \displaybreak[1] \\ +\intertext{\indent Since A's money is now $2$~times B's, we have} +x + 10 &= 2(y - 10). +\Tag{(2)} +\end{DPalign*} + +From the solution of equations (1) and (2), $x = 22$, and $y = 26$. + +Therefore A has \$$22$, and B has~\$$26$. +\end{Soln} + +\Exercise{66.} + +\Item{1.} If A gives B \$$200$, A~will then have half as much +money as~B; but if B gives A \$$200$, B~will have one-third +as much as A\@. How much has each? + +\Item{2.} Half the sum of two numbers is~$20$, and three times +their difference is~$18$. Find the numbers. + +\Item{3.} The sum of two numbers is~$36$, and their difference +is equal to one-eighth of the smaller number increased by~$2$. +Find the numbers. + +\Item{4.} If $4$~yards of velvet and $3$~yards of silk are sold for~\$$33$, +and $5$~yards of velvet and $6$~yards of silk for~\$$48$, +what is the price per yard of the velvet and of the silk? + +\Item{5.} If $7$~bushels of wheat and $10$~of rye are sold for~\$$15$, +and $4$~bushels of wheat and $5$~of rye are sold for~\$$8$, what +is the price per bushel of the wheat and of the rye? + +\Item{6.} If $12$~pounds of tea and $4$~pounds of coffee cost~\$$7$, +and $4$~pounds of tea and $12$~pounds of coffee cost~\$$5$, what is +the price per pound of tea and of coffee? + +\Item{7.} Six horses and $7$~cows can be bought for~\$$1000$, +and $11$~horses and $13$~cows for~\$$1844$. Find the value of +a horse and of a cow. +%% -----File: 134.png---Folio 128------- + +\Exercise{67.} + +Ex. A certain fraction becomes equal to~$\frac{1}{2}$ if $2$~is added +to its numerator, and equal to~$\frac{1}{3}$ if $3$~is added to its denominator. +Find the fraction. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +\frac{x}{y} &= \text{the required fraction.} \\ +\lintertext{\indent Then} +\frac{x + 2}{y} &= \frac{1}{2}, \\ +\lintertext{and} +\frac{x}{y + 3} &= \frac{1}{3}. +\end{DPalign*} + +The solution of these equations gives $7$ for~$x$, and $18$ for~$y$. + +Therefore the required fraction is~$\frac{7}{18}$. +\end{Soln} + +\Item{1.} If the numerator of a certain fraction is increased +by~$2$ and its denominator diminished by~$2$, its value will +be~$1$. If the numerator is increased by the denominator +and the denominator is diminished by~$5$, its value will be~$5$. +Find the fraction. + +\Item{2.} If $1$~is added to the denominator of a fraction, its +value will be~$\frac{1}{2}$. If $2$~is added to its numerator, its value +will be~$\frac{3}{5}$. Find the fraction. + +\Item{3.} If $1$~is added to the numerator of a fraction, its value +will be~$\frac{1}{5}$. If $1$~is added to its denominator, its value will +be~$\frac{1}{7}$. Find the fraction. + +\Item{4.} If the numerator of a fraction is doubled and its +denominator diminished by~$1$, its value will be~$\frac{1}{2}$. If its +denominator is doubled and its numerator increased by~$1$, +its value will be~$\frac{1}{7}$. Find the fraction. + +\Item{5.} In a certain proper fraction the difference between +the numerator and the denominator is~$15$. If the numerator +is multiplied by~$4$ and the denominator increased by~$6$, +its value will be~$1$. Find the fraction. +%% -----File: 135.png---Folio 129------- + +\Exercise{68.} + +The expression $64$ means $60 + 4$, that is, $10$~\emph{times} $6 + 4$, +and has for its \emph{digits} $6$~and~$4$. If the digits were unknown +and represented by $x$~and~$y$, the number would be represented +by~$10x + y$. + +Ex. The sum of the two digits of a number is~$10$, and if +$18$~is added to the number, the digits will be reversed. +Find the number. +\begin{Soln} +\begin{DPalign*}[m] +\lintertext{\indent Let} +x &= \text{the tens' digit,} \\ +\lintertext{and} +y &= \text{the units' digit.} \\ +\lintertext{\indent Then} +10x + y &= \text{the number.} \displaybreak[1] \\ +\lintertext{\indent Hence} +x + y &= 10, +\Tag{(1)} \\ +\lintertext{and} +10x + y + 18 &= 10y + x. +\Tag{(2)} \displaybreak[1] \\ +\lintertext{\indent From (2),} +9x - 9y &= -18, \\ +\lintertext{or} +x - y &= -2. +\Tag{(3)} \displaybreak[1] \\ +\lintertext{\indent \rlap{Add (1) and (3),}} +2x &= 8, \\ +\lintertext{and therefore} +x &= 4. \\ +\lintertext{\indent \rlap{Subtract (3) from (1),}} +2y &= 12, \\ +\lintertext{and therefore} +y &= 6. +\end{DPalign*} + +Therefore the number is~$46$. +\end{Soln} + +\Item{1.} The sum of the two digits of a number is~$9$, and if $9$ +is added to the number, the digits will be reversed. Find +the number. + +\Item{2.} A certain number of two digits is equal to eight times +the sum of its digits, and if $45$~is subtracted from the +number, the digits will be reversed. Find the number. + +\Item{3.} The sum of a certain number of two digits and the +number formed by reversing the digits is~$132$, and the +difference of these numbers is~$18$. Find the numbers. + +\Item{4.} The sum of the two digits of a number is~$9$, and if +the number is divided by the sum of its digits, the quotient +is~$6$. Find the number. +%% -----File: 136.png---Folio 130------- + +\Exercise{69.} + +Ex. A sum of money, at simple interest, amounted to +\$$2480$ in $4$~years, and to \$$2600$ in $5$~years. Find the sum +and the rate of interest. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{the number of dollars in the principal,} \\ +\lintertext{and} +y &= \text{the rate of interest.} +\end{DPalign*} + +The interest for one year is $\dfrac{y}{100}$ of the principal; that is, $\dfrac{xy}{100}$. +For $4$~years the interest is~$\dfrac{4xy}{100}$, and for $5$~years~$\dfrac{5xy}{100}$. The amount +is principal $+$~interest, +\begin{DPalign*} +\lintertext{or} +x + \frac{4xy}{100} &= 2480. \\ +x + \frac{5xy}{100} &= 2600. \\ +\lintertext{\indent Hence} +100x + 4xy &= 248,000. +\Tag{(1)} \\ +100x + 5xy &= 260,000. +\Tag{(2)} \displaybreak[1] \\ +\intertext{\indent Divide (1) by~$4$ and (2) by~$5$, and we have} +25x + xy &= 62,000 \\ +20x + xy &= 52,000 \displaybreak[1] \\ +\lintertext{\indent Subtract,} +5x \PadTo{{}+ xy}{} &= 10,000. \displaybreak[1] \\ +\lintertext{\indent Therefore} +\PadTo{5x}{x} \PadTo{{}+ xy}{} &= \PadTo{10,000}{2000}. +\end{DPalign*} + +Substitute the value of~$x$ in~(1), $y = 6$. + +Therefore the sum is \$$2000$, and the rate~$6$%. +\end{Soln} + +\Item{1.} A sum of money, at simple interest, amounted in $5$~years +to~\$$3000$, and in $6$~years to~\$$3100$. Find the sum +and the rate of interest. + +\Item{2.} A sum of money, at simple interest, amounted in $10$~months +to~\$$1680$, and in $18$~months to~\$$1744$. Find the +sum and the rate of interest. + +\Item{3.} A man has \$$10,000$ invested, a part at~$4$\%, and the +remainder at~$5$\%. The annual income from his $4$\%~investment +is \$$40$~more than from his $5$\%~investment. Find the +sum invested at~$4$\% and at~$5$\%. +%% -----File: 137.png---Folio 131------- + +\Exercise{70.} + +\Section{Miscellaneous Examples.} + +\Item{1.} Half the sum of two numbers is~$20$; and $5$~times +their difference is~$20$. Find the numbers. + +\Item{2.} A certain number when divided by a second number +gives $7$ for a quotient and $4$ for a remainder. If three +times the first number is divided by twice the second +number, the quotient is~$11$ and the remainder~$4$. Find +the numbers. + +\Item{3.} A fraction becomes $\frac{4}{5}$ in value by the addition of~$2$ to +its numerator and $3$~to its denominator. If $2$~is subtracted +from its numerator and $1$~from its denominator, the value +of the fraction is~$\frac{3}{4}$. Find the fraction. + +\Item{4.} A farmer sold $50$~bushels of wheat and $30$~of barley +for $74$~dollars; and at the same prices he sold $30$~bushels +of wheat and $50$~bushels of barley for $70$~dollars. What +was the price of the wheat and of the barley per bushel? + +\Item{5.} If A gave \$$10$ to~B, he would then have three times +as much money as~B; but if B gave \$$5$ to~A, A~would +have four times as much as~B\@. How much has each? + +\Item{6.} A and B have together \$$100$. If A~were to spend +one-half of his money, and B~one-third of his, they would +then have only \$$55$ between them. How much money +has each? + +\Item{7.} A fruit-dealer sold $6$~lemons and $3$~oranges for $21$~cents, +and $3$~lemons and $8$~oranges for $30$~cents. What +was the price of each? + +\Item{8.} If A gives me $10$~apples, he will have just twice as +many as~B\@. If he gives the $10$~apples to~$B$ instead of to +me, A~and~B will each have the same number. How +many apples has each? +%% -----File: 138.png---Folio 132------- + + +\Chapter{XII.}{Quadratic Equations.} + +\Paragraph{158.} An equation which contains the \emph{square} of the +unknown number, but no higher power, is called a \Defn{quadratic +equation}. + +\Paragraph{159.} A quadratic equation which involves but one unknown +number as~$x$, can contain only: + +1. Terms involving the square of~$x$. + +2. Terms involving the first power of~$x$. + +3. Terms which do not involve~$x$. + +Collecting similar terms, every quadratic equation can be +made to assume the form +\[ +ax^{2} + bx + c = 0, +\] +where $a$,~$b$, and~$c$ are known numbers, and $x$~the unknown +number. + +If $a$,~$b$,~$c$ are numbers expressed by figures, the equation +is a \Defn{numerical quadratic}. If $a$,~$b$,~$c$ are numbers represented +wholly or in part by letters, the equation is a \Defn{literal quadratic}. + +\Paragraph{160.} In the equation $ax^{2} + bx + c = 0$, $a$,~$b$, and~$c$ are +called the \Defn{coefficients} of the equation. The third term,~$c$, is +called the \Defn{constant term}. + +If the first power of~$x$ is wanting, the equation is a \Defn{pure +quadratic}; in this case $b = 0$. + +If the first power of~$x$ is present, the equation is an +\Defn{affected} or \Defn{complete quadratic}. +%% -----File: 139.png---Folio 133------- + + +\Section{Pure Quadratic Equations.} + +\Paragraph{161. Examples.} + +\Item{1.} Solve the equation $5x^{2} - 48 = 2x^{2}$. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent We have} +5x^{2} - 48 &= 2x^{2}. \\ +\lintertext{\indent Collect the terms,} +3x^{2} &= 48. \displaybreak[1] \\ +\lintertext{\indent Divide by~$3$,} +x^{2} &= 16. \\ +\lintertext{\indent Extract the square root,} +x &= ±4. +\end{DPalign*} + +The sign~$±$ before the~$4$, read \emph{plus or minus}, shows that the root +is either $+$~or~$-$. For $(+4) × (+4) = 16$, and $(-4) × (-4) = 16$ + +The square root of any number is positive or negative. Hitherto +we have given only the positive value. In this chapter we shall +give both values. This sign~$\surd$, called the \Defn{radical sign}, is used to +indicate that a root is to be extracted. Thus $\sqrt{4}$~means the square +root of~$4$ is required. $\sqrt[3]{4}$~means the third root of~$4$ is required; the +small figure placed over the radical sign is called the \Defn{index} of the +root, and shows the root required. +\end{Soln} + +\Item{2.} Solve the equation $3x^{2} - 15 = 0$. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent We have} +3x^{2} &= 15, \\ +\lintertext{or} +x^{2} &= 5. \\ +\lintertext{\indent Extract the square root,} +x &= ±\sqrt{5}. +\end{DPalign*} + +The roots cannot be found exactly, since the square root of~$5$ cannot +be found exactly; it can, however, be determined approximately +to any required degree of accuracy; for example, the roots lie between +$2.23606$ and $2.23607$; and between $-2.23606$ and~$-2.23607$. +\end{Soln} + +\Item{3.} Solve the equation $3x^{2} + 15 = 0$. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent We have} +3x^{2} &= -15, \\ +\lintertext{or} +x^{2} &= -5. \\ +\lintertext{\indent Extract the square root,} +x &= ±\sqrt{-5}. +\end{DPalign*} + +There is no square root of a negative number, since the square of +any number, positive or negative, is positive; $(-5) × (-5) = +25$. + +The square root of~$-5$ differs from the square root of~$+5$ in that +the latter can be found as accurately as we please, while the former +cannot be found at all. +\end{Soln} +%% -----File: 140.png---Folio 134------- + +\Paragraph{162.} A root which can be found exactly is called an \Defn{exact} +or \Defn{rational} root. Such roots are either whole numbers or +fractions. + +A root which is indicated but can be found only approximately +is called a \Defn{surd}. Such roots involve the roots of +imperfect powers. + +Rational and surd roots are together called \Defn{real} roots. + +A root which is indicated but cannot be found, either +exactly or approximately, is called an \Defn{imaginary} root. Such +roots involve the even roots of negative numbers. + +\Exercise{71.} + +Solve: +\begin{multicols}{2} +\Item{1.} $5x^{2} - 2 = 3x^{2} + 6$. + +\Item{2.} $3x^{2} + 1 = 2x^{2} + 10$. + +\Item{3.} $4x^{2} - 50 = x^{2} + 25$. + +\Item{4.} $(x - 6)(x + 6) = 28$. +\end{multicols} + +\Item{5.} $(x - 5)(x + 5) = 24$. + +\Item{6.} $3(x^{2} - 11) + 2(x^{2} - 5) = 82$. + +\Item{7.} $11(x^{2} + 5) + 6(3 - x^{2}) = 198$. + +\Item{8.} $5x^{2} + 3 - 2(17 - x^{2}) = 32$. + +\Item{9.} $4(x + 1) - 4(x - 1) = x^{2} - 1$. + +\Item{10.} $86 - 52x = 2(8 - x)(2 - 3x)$. + +\Item{11.} Find two numbers that are to each other as $3$~to~$4$, +and the difference of whose squares is~$112$. + +\begin{Remark}[Hint.] Let $3x$~stand for the smaller and $4x$~for the larger number. +\end{Remark} + +\Item{12.} A boy bought a number of oranges for $36$~cents. +The price of an orange was to the number bought as $1$~to~$4$. +How many oranges did he buy, and how many cents +did each orange cost? + +\Item{13.} A certain street contains $144$~square rods, and the +length is $16$~times the width. Find the width. +%% -----File: 141.png---Folio 135------- + +\Item{14.} Find the number of rods in the length, and in the +width of a rectangular field containing $3\frac{3}{5}$~acres, if the +length is $4$~times the width. + + +\Section{Affected Quadratic Equations.} + +\Paragraph{163.} Since +\[ +(x + b)^{2} = x^{2} + 2bx + b^{2}, \quad\text{and}\quad +(x - b)^{2} = x^{2} - 2bx + b^{2}, +\] +it is evident that the expression $x^{2} + 2bx$ or $x^{2} - 2bx$ lacks +only the \emph{third term},~$b^{2}$, of being a perfect square. + +This third term is the square of half the coefficient of~$x$. + +Every affected quadratic may be made to assume the +form $x^{2} + 2bx = c$ or $x^{2} - 2bx = c$, by dividing the equation +through by the coefficient of~$x^{2}$. + +To \textbf{solve} such an equation: + +The first step is to add to both members \emph{the square of +half the coefficient of~$x$}. This is called \emph{completing the square}. + +The second step is to \emph{extract the square root} of each member +of the resulting equation. + +The third step is to \emph{reduce} the two resulting simple +equations. + +\Item{1.} Solve the equation $x^{2} - 8x = 20$. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent We have} +x^{2} - 8x &= 20. \\ +\lintertext{\indent Complete the square,} +x^{2} - 8x + 16 &= 36. \\ +\lintertext{\indent Extract the square root,} +x - 4 &= ±6. \displaybreak[1] \\ +\lintertext{\indent Reduce, using the upper \rlap{sign,}} +x &= 4 + 6 = 10, \\ +\lintertext{or using the lower sign,} +x &= 4 - 6 = -2. +\end{DPalign*} + +The roots are $10$ and~$-2$. + +Verify by putting these numbers for~$x$ in the given equation. +\[ +\begin{array}{rcl<{\qquad}|>{\qquad}rcl} + x &=& 10, & x &=& -2, \\ +10^{2} - 8(10) &=& 20, & (-2)^{2} - 8(-2) &=& 20, \\ + 100 - 80 &=& 20. & 4 + 16 &=& 20. \\ +\end{array} +\] +\end{Soln} +%% -----File: 142.png---Folio 136------- + +\Item{2.} Solve the equation $\dfrac{x + 1}{x - 1} = \dfrac{4x - 3}{x + 9}$. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Free from fractions,} +(x + 1)(x + 9) &= (x - 1)(4x - 3). \\ +\lintertext{\indent Therefore,} +-3x^{2} + 17x &= -6. +\end{DPalign*} + +Since the square root of a negative number cannot be taken, the +coefficient of~$x^{2}$ must be changed to~$+$. +\begin{DPalign*} +\lintertext{\indent Divide by~$-3$,} +x^{2} - \tfrac{17}{3}x &= 2. \displaybreak[1] \\ +\intertext{\indent +Half the coefficient of~$x$ is $\frac{1}{2}$~of $-\frac{17}{3} = -\frac{17}{6}$, and the square of~$-\frac{17}{6}$ +is~$\frac{289}{36}$. Add the square of~$-\frac{17}{6}$ to both sides, and we have} +x^{2} - \frac{17x}{3} + \left(\frac{17}{6}\right)^{2} &= 2 + \frac{289}{36}. \displaybreak[1] \\ +\lintertext{\indent Now} +2 + \frac{289}{36} = \frac{72}{36} + \frac{289}{36} &= \frac{361}{36}, \\ +\lintertext{therefore,} +x^{2} - \tfrac{17}{3}x + \left(\frac{17}{6}\right)^{2} &= \frac{361}{36}. \displaybreak[1] \\ +\lintertext{\indent \rlap{Extract the root,}} +x - \frac{17}{6} &= ±\frac{19}{6}. \displaybreak[1] \\ +\lintertext{\indent Reduce,} +x - \frac{17}{6} &= ±\frac{19}{6}. \\ +\therefore x &= \frac{17}{6} + \frac{19}{6} = \frac{36}{6} = 6, \\ +\lintertext{or} +x &= \frac{17}{6} - \frac{19}{6} = -\frac{2}{6} = -\frac{1}{3}. +\end{DPalign*} + +The roots are $6$~and~$-\dfrac{1}{3}$. + +Verify by putting these numbers for~$x$ in the original equation. +\[ +\begin{array}[t]{rcl<{\qquad}|} +x &=& 6. \\ +\dfrac{6 + 1}{6 - 1} &=& \dfrac{24 - 3}{6 + 9}. \\ +\dfrac{7}{5} &=& \dfrac{21}{15} \\ +\dfrac{7}{5} &=& \dfrac{7}{5} \\ +\end{array} +\begin{array}[t]{>{\qquad}rcl} +x &=& -\dfrac{1}{3} \\ +\dfrac{-\dfrac{1}{3} + 1}{-\dfrac{1}{3} - 1} + &=& \dfrac{-\dfrac{4}{3} - 3}{-\dfrac{1}{3} + 9}. \\ +-\dfrac{2}{4} &=& -\dfrac{13}{26}. +\end{array} +\] +\end{Soln} +%% -----File: 143.png---Folio 137------- + +\PrintBreak +\Exercise{72.} + +Solve: +\begin{multicols}{2} +\Item{1.} $x^{2} - 12x + 27 = 0$. + +\Item{2.} $x^{2} - 6x + 8 = 0$. + +\Item{3.} $x^{2} - 4 = 4x - 7$. + +\Item{4.} $5x^{2} - 4x-1 = 0$. + +\Item{5.} $4x - 3 = 2x - x^{2}$. + +\Item{6.} $9x^{2} - 24x + 16 = 0$. + +\Item{7.} $6x^{2} - 5x-1 = 0$. + +\Item{8.} $4x + 3 = x^{2} + 2x$. + +\Item{9.} $16x^{2} - 16x + 3 = 0$. + +\Item{10.} $3x^{2} - 10x + 3 = 0$. + +\Item{11.} $x^{2} - 14x - 51 = 0$. + +\Item{12.} $34x - x^{2} - 225 = 0$. + +\Item{13.} $x^{2} + x - 20 = 0$. + +\Item{14.} $x^{2} - x - 12 = 0$. + +\Item{15.} $2x^{2} - 12x = - 10$. + +\Item{16.} $3x^{2} + 12x - 36 = 0$. + +\Item{17.} $(2x - 1)^{2} + 9 = 6(2x - 1)$. + +\Item{18.} $6(9x^{2} - x) = 55(x^{2} - 1)$. + +\Item{19.} $32 - 3x^{2} - 10x = 0$. + +\Item{20.} $9x^{2} - 6x - 143 = 0$. + +\Item{21.} $\dfrac{x}{x - 1} - \dfrac{x - 1}{x} = \dfrac{3}{2}$. + +\Item{22.} $\dfrac{1}{x - 2} + \dfrac{2}{x + 2} = \dfrac{5}{6}$. + +\Item{23.} $\dfrac{5x + 7}{x - 1} = 3x + 11$. + +\Item{24.} $\dfrac{7}{x + 4} - \dfrac{1}{4 - x} = \dfrac{2}{3}$. + +\Item{25.} $\dfrac{2}{x + 3} + \dfrac{x + 3}{2} = \dfrac{10}{3}$. + +\Item{26.} $\dfrac{2x}{x + 2} + \dfrac{x + 2}{2x} = 2$. + +\Item{27.} $\dfrac{3(x - 1)}{x + 1} - \dfrac{2(x + 1)}{x - 1} = 5$. + +\Item{28.} $\dfrac{2x + 5}{2x - 5} = \dfrac{7x - 5}{2x}$. + +\Item{29.} $\dfrac{3x - 1}{4x + 7} = \dfrac{x + 1}{x + 7}$. + +\Item{30.} $\dfrac{2x - 1}{x + 3} = \dfrac{x + 3}{2x - 1}$. + +\Item{31.} $\dfrac{x + 4}{x - 4} - \dfrac{x + 2}{x - 3} = 1$. + +\Item{32.} $\dfrac{4}{x - 1} - \dfrac{5}{x + 2} = \dfrac{1}{2}$. + +\Item{33.} $\dfrac{2}{x - 1} = \dfrac{3}{x - 2} + \dfrac{2}{x - 4}$. + +\Item{34.} $\dfrac{5}{x - 2} - \dfrac{3}{x - 1} = \dfrac{1}{2}$. + +\Item{35.} $\dfrac{x}{7 - x} + \dfrac{7 - x}{x} = \dfrac{29}{10}$. + +\Item{36.} $\dfrac{2x - 1}{x - 1} + \dfrac{1}{6} = \dfrac{2x - 3}{x - 2}$. +\end{multicols} +%% -----File: 144.png---Folio 138------- + +\ScreenBreak +\Paragraph{164. Problems involving Quadratics.} Problems which involve +quadratic equations apparently have two solutions, +since a quadratic equation has two roots. + +When both roots of the quadratic equation are positive +integers, they will, in general, both be admissible solutions. +Fractional and negative roots will in some problems give +admissible solutions; in other problems they will not give +admissible solutions. + +The reason that every root of the equation will not +always satisfy the conditions of the problem is that the +problem may have certain restrictions, expressed or implied, +that cannot be expressed in the equation. + +No difficulty will be found in selecting the result which +belongs to the particular problem we are solving. Sometimes, +by a change in the statement of the problem, we +may form a new problem which corresponds to the result +that was inapplicable to the original problem. + +Here as in simple equations x stands for an unknown +\emph{number}. + +\Item{1.} The sum of the squares of two consecutive numbers is~$41$. +Find the numbers. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{one number,} \\ +\lintertext{and} +x + 1 &= \text{the other.} \\ +\lintertext{\indent Then} +x^{2} + (x + 1)^{2} &= \text{the sum of the squares;} \\ +\lintertext{but} +41 &= \text{the sum of the squares.} +\end{DPalign*} +\begin{align*} +\therefore x^{2} + (x + 1)^{2} &= 41. \\ +x^{2} + x^{2} + 2x + 1 &= 41. \\ +2x^{2} + 2x &= 40. \\ +x^{2} + x &= 20. +\end{align*} + +The solution of this equation gives $x = 4$, or~$-5$. + +The positive root~$4$ gives for the numbers $4$~and~$5$. +\end{Soln} + +The negative root~$-5$ is inapplicable to the problem, as +\emph{consecutive numbers} are understood to be integers which +follow each other in the common scale: $1$,~$2$, $3$, $4\dots$. +%% -----File: 145.png---Folio 139------- + +\Item{2.} In a certain nest seven times the number of birds in +the nest is equal to twice the square of the number increased +by~$3$. Find the number. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{the number of birds\Add{.}} \\ +\lintertext{\indent Then} +7x &= \text{$7$~times the number} \\ +\lintertext{and} +2x^{2} + 3 &= \text{twice the square of the number plus~$3$.} +\intertext{\indent As these two expressions are equal we have} +2x^{2} + 3 &= 7x\Add{.} +\end{DPalign*} + +The solution of this \emph{equation} gives $x = 3$ or $x = \frac{1}{2}$. + +The value $\frac{1}{2}$ is not applicable to the \emph{problem} for the number of +birds must be a whole number. +\end{Soln} + +\Item{3.} A cistern has two pipes. By one of them it can be +filled $6$~hours sooner than by the other, and by both +together in $4$~hours. Find the time it will take each pipe +alone to fill it. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{the number of hours it takes the smaller pipe.} \\ +\lintertext{\indent Then} +x - 6 &= \text{the number of hours it takes the larger pipe.} +\end{DPalign*} +\begin{DPalign*} +\lintertext{\indent Therefore} +\frac{1}{x} + \frac{1}{x - 6} &= \text{the part both can fill in one hour.} \\ +\lintertext{\indent But} +\frac{1}{4} &= \text{the part both can fill in one hour.} \\ +\frac{1}{x} + \frac{1}{x - 6} &= \frac{1}{4} +\end{DPalign*} + +The solution of this \emph{equation} gives $x = 12$ or $x = 2$. + +The value~$2$ is not applicable to the \emph{problem}. + +Therefore it takes one pipe $12$~hr.\ and the other $6$~hr. +\end{Soln} + +\Item{4.} A rug is $1$~yard longer than it is broad. The number +of sq.~yds.\ in the rug is~$12$. Find its length and breadth. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{the number of yards in the breadth\Add{.}} \\ +\lintertext{\indent Then} +x + 1 &= \text{the number of yards in the length} \\ +\lintertext{and} +x(x + 1) &= \text{the number of sq yds in the rug\Add{.}} \\ +\lintertext{\indent Hence} +x(x + 1) &= 12\Add{.} +\end{DPalign*} + +The solution of this equation gives $x = 3$ or $x = -4$. + +The dimensions are therefore $3$~yards and $4$~yards. +\end{Soln} +%% -----File: 146.png---Folio 140------- + +\Exercise{73.} + +\Item{1.} Find two numbers whose sum is~$11$, and whose +product is~$30$. + +\Item{2.} Find two numbers whose difference is~$10$, and the +sum of whose squares is~$250$. + +\Item{3.} A man is five times as old as his son, and the square +of the son's age diminished by the father's age is~$24$. Find +their ages. + +\Item{4.} A number increased by its square is equal to nine +times the next higher number. Find the number. + +\Item{5.} The square of the sum of any two consecutive numbers +lacks~$1$ of being twice the sum of the squares of the +numbers. Show that this statement is true. + +\Item{6.} The length of a rectangular court exceeds its breadth +by $2$~rods. If the length and breadth were each increased +by $3$~rods, the area of the court would be $80$~square rods. +Find the dimensions of the court. + +\Item{7.} The area of a certain square will be doubled, if its +dimensions are increased by $6$~feet and $4$~feet respectively. +Find its dimensions. + +\Item{8.} The perimeter of a rectangular floor is $76$~feet and +the area of the floor is $360$~square feet. Find the dimensions +of the floor. + +\Item{9.} The length of a rectangular court exceeds its breadth +by $2$~rods, and its area is $120$~square rods. Find the +dimensions of the court. + +\Item{10.} The combined ages of a father and son amount to +$64$~years. Twice the father's age exceeds the square of the +son's age by $8$~years. Find their respective ages. +%% -----File: 147.png---Folio 141------- + +\Exercise{74.} + +Ex. A boat sails $30$~miles at a uniform rate. If the +rate had been $1$~mile an hour more, the time of the sailing +would have been $1$~hour less. Find the rate of the sailing. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{the rate in miles per hour.} \\ +\lintertext{\indent Then} +\frac{30}{x} &= \text{the number of hours.} +\end{DPalign*} + +On the other supposition the rate would have been $x + 1$~miles +an hour and the time~$\dfrac{30}{x + 1}$. +\begin{DPalign*} +\lintertext{\indent Hence} +\frac{30}{x} - \frac{30}{x + 1} &= \text{the difference in hours for the sailing.} \\ +\lintertext{\indent But} +1 &= \text{the difference in hours for the sailing.} \\ +\frac{30}{x} - \frac{30}{x + 1} &= 1\Add{.} +\end{DPalign*} + +The solution of this equation gives $x = 5$, or $x = -6$. + +Therefore, the rate of the sailing is $5$~miles an hour. +\end{Soln} + +\Item{1.} A boat sails $30$~miles at a uniform rate. If the rate +had been $1$~mile an hour less, the time of the sailing would +have been $1$~hour more. Find the rate of the sailing. + +\Item{2.} A laborer built $35$~rods of stone wall. If he had +built $2$~rods less each day, it would have taken him $2$~days +longer. How many rods did he build a day on the +average? + +\Item{3.} A man bought flour for~\$$30$. Had he bought $1$~barrel +more for the same sum, the flour would have cost +him \$$1$~less per barrel. How many barrels did he buy? + +\Item{4.} A man bought some knives for~\$$6$. Had he bought +$2$~less for the same money, he would have paid $25$~cents +more for each knife. How many knives did he buy? + +\Item{5.} What number exceeds its square root by~$30$? + +\begin{Remark}[Hint.] Let $x^{2}$ denote the number. +\end{Remark} +%% -----File: 148.png---Folio 142------- + + +\Chapter{XIII.}{Arithmetical Progression.} + +\Paragraph{165.} A series of numbers is said to form an \Defn{Arithmetical +Progression} if the difference between any term and the preceding +term is the same throughout the series. + +\begin{Remark} +Thus $a$, $b$, $c$, $d$, etc., are in arithmetical progression if $b - a$, $c - b$, +$d - c$,~etc., are all equal. +\end{Remark} + +\Paragraph{166.} This difference is called the \Defn{common difference} of the +progression, and is represented by~$d$. If $d$~is positive, the +progression is an \emph{increasing} series; if $d$~is negative, the progression +is a \emph{decreasing} series. + +What is the common difference in each of the following +series? +\[ +\begin{array}{r*{4}{>{\quad}r}} + 1, & \Neg4, & 7, & 10, & \dots \\ + 5, & 7, & 9, & 11, & \dots \\ +10, & 9, & 8, & 7, & \dots \\ + 7, & 3, & -1, & -5, & \dots \\ +\end{array} +\] + +\Paragraph{167.} If the first term of an arithmetical progression is +represented by~$a$ and the common difference by~$d$, then +\begin{alignat*}{2} +&\text{the \emph{second} term will be } && a + d, \\ +&\text{the \emph{third} term will be } && a + 2d, \\ +&\text{the \emph{fourth} term will be } && a + 3d, +\end{alignat*} +and so on, the coefficient of~$d$ in each term being always +less by~$1$ than the \emph{number of the term}. + +Hence the $n$th~term will be $a + (n - 1)d$. + +If we represent the $n$th~term by~$l$, we have +\[ +l = a + (n - 1)d. +\Tag{Formula (1)} +\] +%% -----File: 149.png---Folio 143------- + +\Paragraph{168.} We can, therefore, find any term of an arithmetical +progression if the first term and common difference are +given, or if any \emph{two} terms are given. + +\Item{1.} Find the 10th~term of an arithmetical progression if +the 1st~term is~$3$ and the common difference is~$4$. +\begin{Soln} +By formula~(1), the 10th~term is~$3 + (10 - 1)4$, or~$39$. +\end{Soln} + +\Item{2.} If the 8th~term of an arithmetical progression is~$25$, +and the 23d~term~$70$, find the series. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent By formula (1),} +&\text{the 23d term is $a + 22d$,} \\ +\lintertext{and} +&\text{the 8th term is $a + 7d$.} +\end{DPalign*} +\begin{DPalign*} +\lintertext{\indent Therefore,} +a + 22d &= 70 \\ +\lintertext{and} +a + \Z7d &= 25\Add{.} \displaybreak[1] \\ +\lintertext{\indent Subtract,} +15d &= 45 \\ +\lintertext{and} +d &= 3, \\ +\lintertext{whence} +a &= 4. +\end{DPalign*} + +The series is therefore $4$, $7$, $10$, $13$,~etc. +\end{Soln} + +\Paragraph{169. Arithmetical Mean.} If three numbers are in arithmetical +progression, the middle number is called the arithmetical +mean of the other two numbers. + +If $a$, $A$, $b$ are in arithmetical progression, $A$~is the arithmetical +mean of $a$~and~$b$. Hence, by the definition of an +arithmetical series, +\begin{DPalign*} +A - a &= b - A, \\ +\lintertext{whence} +A &= \frac{a + b}{2}. +\rintertext{Formula (2)} +\end{DPalign*} +\begin{Theorem} +Hence, the arithmetical mean of any two numbers is found +by taking half their sum. +\end{Theorem} + +\Paragraph{170.} Sometimes it is required to insert several arithmetical +means between two numbers. +%% -----File: 150.png---Folio 144------- + +If $m = {}$the number of means, and $n = {}$the whole number +of terms, then $m + 2 = n$. If $m + 2$ is substituted for~$n$ +in formula~(1), +\begin{DPalign*} +l &= a + (n - 1)d, \\ +\lintertext{the result is} +l &= a + (m + 1)d. \displaybreak[1] \\ +\lintertext{\indent By transposing~$a$,} +l - a &= (m + 1) d. \\ +\therefore \frac{l - a}{m + 1} &= d. +\Tag{\llap{Formula (3)}} +\end{DPalign*} +\begin{Remark} +Thus, if it be required to insert six means between $3$~and~$17$, +the value of~$d$ is found to be $\dfrac{17 - 3}{6 + 1} = 2$; and the series will be $3$,~$5$, +$7$, $9$, $11$, $13$, $15$,~$17$. +\end{Remark} + +\Exercise{75.} + +\Item{1.} Find the 25th~term in the series $3$, $6$, $9$,~$\dots$. + +\Item{2.} Find the 13th~term in the series $50$, $49$, $48$,~$\dots$. + +\Item{3.} Find the 15th~term in the series $\frac{1}{7}$, $\frac{3}{7}$, $\frac{5}{7}$,~$\dots$. + +\Item{4.} Find the 19th~term in the series $\frac{1}{4}$, $-\frac{1}{4}$, $-\frac{3}{4}$,~$\dots$. + +\Item{5.} Find the 10th~term in an arithmetical progression +whose 1st~term is~$5$ and 3d~term~$9$. + +\Item{6.} Find the 11th~term in an arithmetical progression +whose 1st~term is~$10$ and whose 6th~term is~$5$. + +\Item{7.} If the 3d~term of an arithmetical progression is~$20$ +and the 13th~term is~$100$, what is the 20th~term? + +\Item{8.} Which term of the series $5$, $7$, $9$, $11$,~$\dots$, is~$43$? + +\Item{9.} Which term of the series $\frac{4}{3}$, $\frac{3}{2}$, $\frac{5}{3}$,~$\dots$, is~$18$? + +\Item{10.} What is the arithmetical mean of $20$~and~$32$? + +\Item{11.} What is the arithmetical mean of $a + b$ and $a - b$? + +\Item{12.} Insert $8$ arithmetical means between $20$~and~$29$. +%% -----File: 151.png---Folio 145------- + +\Paragraph{171. To Find the Sum of Any Number of Terms of an Arithmetical +Series.} + +If $l$ denote the last term, $a$~the first term, $n$~the number +of terms, $d$~the common difference, and $s$~the sum of the +terms, it is evident that the series beginning with the first +term will be $a$, $a + d$, $a + 2d$,~etc., and beginning with the +last term will be $l$, $l - d$, $l - 2d$,~etc. Therefore, +\begin{gather*} +\begin{array}{r*{12}{c}} +s &=& a &+& (a + d) &+& (a + 2d) &+& \Add{\dots} &+& (l - d) &+& l, \rlap{\quad\text{or}} \\ +s &=& l &+& (l - d) &+& (l - 2d) &+& \Add{\dots} &+& (a + d) &+& a \\ +\hline +2s &=& (a + l) &+& (a + l) &+& (a + l) &+& \Add{\dots} &+& (a + l) &+& (a + l) \\ +2s &=& \multicolumn{11}{l}{\text{$(a + l)$ taken as many times as there are \emph{terms},}} +\end{array} \displaybreak[1] \\ +\begin{aligned}[b] +2s &= n(a + l), \\ +\text{and } s &= \frac{n}{2}(a + l)\Add{.} +\end{aligned} +\Tag{Formula (4)} +\end{gather*} + +Putting for $l$~its value $a + (n - 1)d$, in formula~(4), we +have +\begin{align*} +s &= \frac{n}{2}\bigl\{a + a + (n - 1)d\bigr\} \\ + &= \frac{n}{2}\bigl\{2a + (n - 1)d\bigr\} +\Tag{Formula (5)} +\end{align*} + +\Item{1.} Find the sum of the first $16$~terms of the series $5$, $7$, +$9$,~$11$\DPtypo{,}{.} +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Here} +a &= 5,\quad d = 2,\quad n = 16\Add{.} \\ +\intertext{\indent Putting these values in formula~(5) we have} +s &= \tfrac{16}{2}(10 + 15 × 2) \\ + &= 320 +\end{DPalign*} +\end{Soln} + +\Item{2.} Show that the sum of any number of odd numbers, +beginning with~$1$, is a square number. +\begin{Soln} +The series of odd numbers is $1$, $3$, $5$, $7$,~$\dots$. +%% -----File: 152.png---Folio 146------- +\begin{DPalign*} +\lintertext{\indent Here} +a &= 1 \quad\text{and}\quad d = 2\Add{.} +\intertext{\indent Putting these values in formula (5) we have} +s &= \frac{n}{2} \bigl\{2 + (n - 1)2\bigr\} \\ + &= \frac{n}{2} × 2n \\ + &= n^{2}\Add{.} +\end{DPalign*} + +Therefore the sum of the first $5$~odd numbers is~$5^{2}$ or~$25$\Add{,} of the first +$8$ odd numbers is~$8^{2}$ or~$64$, and so on. +\end{Soln} + +\Item{3.} The sum of $20$~terms of an arithmetical progression +is~$420$, and the first term is~$2$. Find the common difference. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Here} +s &= 420,\quad n = 20, \quad\text{and}\quad a = 2\Add{.} \\ +\intertext{\indent Putting these values in formula~(5), we have} +420 &= \tfrac{20}{2}(4 + 19d) \\ + &= 40 + 190d \\ +190d &= 380 \\ +\Add{\therefore} d &= 2\Add{.} +\end{DPalign*} + +Therefore the common difference is~$2$. +\end{Soln} + +\Exercise{76.} + +\Item{1.} Find the sum of $3$, $5$, $7$, $\dots$, to $20$~terms. + +\Item{2.} Find the sum of $14$, $14\frac{1}{2}$, $15$, $\dots$, to $12$~terms. + +\Item{3.} Find the sum of $\frac{7}{6}$, $1$, $\frac{5}{6}$, $\dots$, to $10$~terms. + +\Item{4.} Find the sum of $-7$, $-5$, $-3$, $\dots$, to $16$~terms. + +\Item{5.} Find the sum of $12$, $9$, $6$, $\dots$, to $21$~terms. + +\Item{6.} Find the sum of $-10\frac{1}{2}$, $-9$, $-7\frac{1}{2}$, $\dots$, to $25$~terms. + +\Item{7.} The sum of three numbers in arithmetical progression +is~$9$, and the sum of their squares is~$35$. Find the numbers. + +\begin{Remark}[Hint.] Let $x - y$, $x$, $x + y$, stand for the numbers. +\end{Remark} +%% -----File: 153.png---Folio 147------- + +\Item{8.} A common clock strikes the hours from $1$ to~$12$. +How many times does it strike every $24$~hours? + +\Item{9.} The Greenwich clock strikes the hours from $1$ to~$24$. +How many times does it strike in $24$~hours? + +\Item{10.} In a potato race each man picked up $50$~potatoes +placed in line a yard apart, and the first potato one yard +from the basket, picking up one potato at a time and bringing +it to the basket. How many yards did each man run, +the start being made from the basket? + +\Item{11.} A heavy body falling from a height falls $16.1$~feet +the first second, and in each succeeding second $32.2$~feet +more than in the second next preceding. How far will a +body fall in $19$~seconds? + +\Item{12.} A stone dropped from a bridge reached the water in +just $3$~seconds. Find the height of the bridge. (See Ex.~11.) + +\Item{13.} The arithmetical mean between two numbers is~$13$, +and the mean between the double of the first and the triple +of the second is~$33\frac{1}{2}$. Find the numbers. + +\Item{14.} Find three numbers of an arithmetical series whose +sum shall be~$27$, and the sum of the first and second shall +be~$frac{4}{5}$ of the sum of the second and third. + +\Item{15.} A travels uniformly $20$~miles a day; B~travels $8$~miles +the first day, $12$~the second, and so on, in arithmetical +progression. If they start Monday morning from the same +place and travel in the same direction, how far apart will +they be Saturday night? + +\Item{16.} The sum of three terms of an arithmetical progression +is~$36$, and the square of the mean exceeds the product of +the other two terms by~$49$. Find the numbers. +%% -----File: 154.png---Folio 148------- + + +\Chapter{XIV.}{Geometrical Progression.} + +\Paragraph{172.} A series of numbers is said to be in \Defn{Geometrical Progression} +when the quotient of any term divided by the +preceding term is the same throughout the series. + +\begin{Remark} +Thus $a$, $b$, $c$, $d$, etc., are in geometrical progression if $\dfrac{b}{a} = \dfrac{c}{b} = \dfrac{d}{c}$,~etc. +\end{Remark} + +\Paragraph{173.} This quotient is called the \Defn{common ratio}, and is represented +by~$r$. + +State the common ratio of the following series: +\[ +\begin{array}{r*{4}{>{\quad}r}} +1, & 3, & \Neg9, &27, & \dots \\ +2, & 4, & 8, &16, & \dots \\ +16,& 8, & 4, &2, & \dots \\ +\frac{2}{3}, & 1, & \frac{3}{2}, & \frac{9}{4}, & \dots \\ +4, & -2, & 1, & -\frac{1}{2}, & \dots \\ +\end{array} +\] + +\Paragraph{174.} If the first term of a geometrical progression is represented +by~$a$, and the common ratio by~$r$, then +\begin{align*} +&\text{the \emph{second} term will be~$ar$,} \\ +&\text{the \emph{third} term will be~$ar^{2}$,} \\ +&\text{the \emph{fourth} term will be~$ar^{3}$,} +\end{align*} +and so on, the index of~$r$ being always less by~$1$ than the +\emph{number of the term in the series}. + +Hence the $n$th~term will be~$ar^{n - 1}$. +%% -----File: 155.png---Folio 149------- + +If we denote the $n$th~term by~$l$, we have +\[ +l = ar^{n - 1}. +\Tag{Formula (1)} +\] + +\Paragraph{175.} If the first term and common ratio are given, or if +any \emph{two terms} are given, we can find the series. + +\Item{1.} Find the 5th~term of a geometrical progression if the +first is~$3$ and the common ratio~$2$. +\begin{Soln} +In formula~(1), put $5$~for~$n$, $3$~for~$a$, and $2$~for~$r$. +\begin{DPalign*} +\lintertext{\indent Then} +l &= 3 × 2^{4} = 48. +\end{DPalign*} + +Therefore the 5th~term is~$48$. +\end{Soln} + +\Item{2.} Find the geometrical series if the 5th~term is~$48$ and +the 7th~term is~$192$. +\begin{Soln} +The 5th and 7th~terms are $ar^{4}$~and~$ar^{6}$, respectively. +\begin{DPalign*} +\lintertext{\indent Whence} +ar^{4} &= 48, +\Tag{(1)} \\ +\lintertext{and} +ar^{6} &= 192. +\Tag{(2)} \\ +\lintertext{\indent Divide (2) by~(1),} +r^{2} &= 4. \\ +\therefore r &= ±2. \\ +\lintertext{\indent From~(1),} +a = \tfrac{48}{16} &= 3. +\end{DPalign*} + +Therefore the series is $3$, $±6$, $12$, $±24$,~$\dots$. +\end{Soln} + +\Paragraph{176. Geometrical Mean.} If three numbers are in geometrical +progression, the middle number is called the \emph{geometrical +mean} of the other two numbers. Hence, if +$a$,~$G$,~$b$ are in geometrical progression, $G$~is the geometrical +mean of $a$~and~$b$. + +By the definition of a geometrical progression, +\begin{DPalign*} +\frac{G}{a} &= \frac{b}{G}. \\ +\therefore G^{2} &= ab, \\ +\lintertext{and} +G &= ± \sqrt{ab}. +\Tag{Formula (2)} +\end{DPalign*} +\begin{Theorem}[Hence], the geometrical mean of any two numbers is the +square root of their product. +\end{Theorem} +%% -----File: 156.png---Folio 150------- + +\PrintBreak +\Paragraph{177. To Find the Sum of Any Number of Terms of a Geometrical +Progression.} + +If $l$~denote the last term, $a$~the first term, $n$~the number +of terms, $r$~the common ratio, and $s$~the sum of the $n$~terms, +then +\begin{DPalign*} +s &= a + ar + ar^{2} + ar^{3} + \dots+ ar^{n - 1}. \\ +\lintertext{\indent Multiply by~$r$,} +rs &= ar + ar^{2} + ar^{3} + \dots + ar^{n - 1} + ar^{n}. +\end{DPalign*} + +Therefore, by subtracting the first equation from the +second, +\begin{DPalign*} +rs - s &= ar^{n} - a, \\ +\lintertext{or} +(r - 1)s &= a(r^{n} - 1). \\ +\therefore s &= \frac{a(r^{n} - 1)}{r - 1}. +\Tag{Formula (3)} +\end{DPalign*} + +\Paragraph{178.} When $r$~is $< 1$, this formula will be more convenient +if written +\[ +s = \frac{a(1 - r^{n})}{1 - r}. +\] + +\Item{1.} Find the sum of $8$~terms of the series +\[ +1,\quad 2,\quad 4,\quad \dots\Add{.} +\] +\begin{DPalign*} +\lintertext{\indent Here} +a &= 1,\quad r = 2,\quad n = 8. \\ +\lintertext{\indent From formula~(3),} +s &= 1(2^{8} - 1) = 255. +\end{DPalign*} + +\Item{2.} Find the sum of $6$~terms of the series +\[ +2,\quad 3,\quad \tfrac{9}{2},\quad \dots\Add{.} +\] +\begin{DPalign*} +\lintertext{\indent Here} +a &= 2,\quad r = \tfrac{3}{2},\quad n = 6. \\ +\lintertext{\indent From formula~(3),} +s &= \frac{2\bigl\{(\frac{3}{2})^{6} - 1\bigr\}}{\frac{3}{2} - 1} \\ + &= \frac{2\bigl\{\frac{729}{64} - 1\bigr\}}{\frac{1}{2}} \\ + &= \frac{4\{729 - 64\}}{64} \\ + &= 41\tfrac{9}{16}. +\end{DPalign*} +%% -----File: 157.png---Folio 151------- + +\Exercise{77.} + +\Item{1.} Find the 5th~term of $3$, $9$, $27$\Add{,}~$\dots$. + +\Item{2.} Find the 7th~term of $3$, $6$, $12$\Add{,}~$\dots$. + +\Item{3.} Find the 8th~term of $6$, $3$, $\frac{3}{2}$\Add{,}~$\dots$. + +\Item{4.} Find the 9th~term of $1$, $-2$, $4$\Add{,}~$\dots$. + +\Item{5.} Find the geometrical mean between $2$~and~$8$. + +\Item{6.} Find the common ratio if the 1st~and 3d~terms are +$2$~and~$32$. + +Find the sum of the series: + +\Item{7.} $3$, $9$, $27$, $\dots$ to $6$~terms. + +\Item{8.} $3$, $6$, $12$, $\dots$ to $8$~terms. + +\Item{9.} $6$, $3$, $\frac{3}{2}$, $\dots$ to $7$~terms. + +\Item{10.} $8$, $4$, $2$, $\dots$ to $8$~terms. + +\Item{11.} $64$, $32$, $16$, $\dots$ to $9$~terms. + +\Item{12.} $64$, $-32$, $16$, $\dots$ to $5$~terms. + +\Item{13.} $\frac{1}{2}$, $\frac{1}{3}$, $\frac{2}{9}$, $\dots$ to $4$~terms. + +\Item{14.} If a blacksmith uses seven nails in putting a shoe on +a horse's foot, and receives $1$~cent for the first nail, $2$~cents +for the second nail, and so on, what does he receive for +putting on the shoe? + +\Item{15.} If a boy receives $2$~cents for his first day's work, +$4$~cents for his second day, $8$~cents for the third day, and +so on for $12$~days, what will his wages amount to? + +\Item{16.} If the population of a city is $10,000$, and increases +$10$\%~a year for four years, what will be its population at +the end of the four years? (Here $l = ar^{4}$.) +%% -----File: 158.png---Folio 152------- + + +\Chapter{XV.}{Square and Cube Roots.} + +\Section{Square Roots of Compound Expressions.} + +\Paragraph{179.} Since the square of $a + b$ is $a^{2} + 2ab + b^{2}$, the square +root of $a^{2} + 2ab + b^{2}$ is $a + b$. + +It is required to find a method of extracting the root +$a + b$ when $a^{2} + 2ab + b^{2}$ is given. +\begin{Soln} +Ex. The first term,~$a$, of the root is obviously the square root of +the first term,~$a^{2}$, in the expression. +\[ +\begin{array}{r*{3}{cr}l} + & & a^{2} &+& 2ab &+& \TbBar{b^{2}} & a + b \\ +\cline{8-8} + & & a^{2} \\ +\cline{3-3} +2a &+& \TbBar{b} & & 2ab &+& b^{2} \\ + & & \TbBar{ } & & 2ab &+& b^{2} \\ +\cline{4-7} +\end{array} +\] + +If the $a^{2}$ is subtracted from the given +expression, the remainder is $2ab + b^{2}$\Add{.} +Therefore the second term,~$b$, of the root +is obtained when the first term of this +remainder is divided by~$2a$; that is, by +\emph{double the part of the root already found}. Also, since +\[ +2ab + b^{2} = (2a + b)b, +\] +the divisor is \emph{completed by adding to the trial-divisor the new term of +the root}\Add{.} +\end{Soln} + +\ScreenBreak +Ex. Find the square root of $25x^{2} - 20x^{3}y + 4x^{4}y^{2}$. +\begin{Soln} +\[ +\begin{array}{rcccrll} +& && & 25x^{2} & \TbBar{-20 x^{3}y + 4x^{4}y^{2}} & 5x - 2x^{2}y \\ +\cline{7-7} +\text{Here $a^{2}$} + &=& (5x)^{2} &=& 25x^{2} \\ +\cline{5-5} +2a + b &=& 10x &-& \TbBar{2x^{2}y} & -20x^{3}y + 4x^{4}y^{2} \\ + & & & & \TbBar{ } & -20x^{3}y + 4x^{4}y^{2} \\ +\cline{6-6} +\end{array} +\] + +The expression is \emph{arranged} according to the ascending powers of~$x$\Add{.} + +The square root of the first term is~$5x$, hence $5x$~is the first term +of the root. $(5x)^{2}$ or $25x^{2}$ is subtracted, and the remainder is +\[ +-20x^{3}y + 4x^{4}y^{2}. +\] + +The second term of the root, $-2x^{2}y$, is obtained by dividing +$-20x^{3}y$ by~$10x$, the double of~$5x$, and this new term of the root is +also annexed to the divisor,~$10x$, to complete the divisor. +\end{Soln} +%% -----File: 159.png---Folio 153------- + +\Paragraph{180.} The same method will apply to longer expressions, +if care be taken to obtain the \emph{trial-divisor} at each stage of +the process, \emph{by doubling the part of the root already found}, +and to obtain the \emph{complete divisor by annexing the new term +of the root to the trial-divisor}. + +\ScreenBreak +Ex. Find the square root of +\[ +1 + 10x^{2} + 25x^{4} + 16x^{6} - 24x^{5} - 20x^{3} - 4x. +\] +\begin{Soln} +\[ +%[** TN: Spacing hack] +\qquad\makebox[0pt][c]{$\begin{array}{r*{2}{cr}lll} +16x^{6} &-& 24x^{5} &+& 25x^{4} &-20x^{3} + 10x^{2} &\TbBar{- 4x + 1} & 4x^{3} - 3x^{2} + 2x - 1 \\ +\cline{8-8} +16x^{6} \\ +\cline{1-1} +\TbBar{\llap{$8x^{3} - 3x^{2}$}} + &-& 24x^{5} &+& 25x^{4} \\ +\TbBar{}&-& 24x^{5} &+& 9x^{4} \\ +\cline{2-5} +\multicolumn{4}{r|}{8x^{3} - 6x^{2} + 2x} & + 16x^{4} &-20x^{3} + 10x^{2} \\ + & & & \TbBar{ }& 16x^{4} &-12x^{3} + \Z4x^{2} \\ +\cline{5-6} +\multicolumn{5}{r|}{8x^{3} - 6x^{2} + 4x - 1} + &-\Z8x^{3} + \Z6x^{2} & -4x + 1 \\ + & & & & \TbBar{}&-\Z8x^{3} + \Z6x^{2} & -4x + 1 \\ +\cline{6-7} +\end{array}$} +\] + +The expression is arranged according to the descending powers of~$x$. + +It will be noticed that each successive trial-divisor may be obtained +by taking the preceding complete divisor with its \emph{last term doubled}. +\end{Soln} + +\Exercise{78.} + +Find the square root of: +\Item{1.} $a^{2} + 2ab + 2ac + b^{2} + 2bc + c^{2}$. + +\Item{2.} $x^{4} + 2x^{3} + 3x^{2} + 2x + 1$. + +\Item{3.} $x^{4} - 4x^{3}y + 6x^{2}y^{2} - 4xy^{3} + y^{4}$. + +\Item{4.} $4a^{4} - 12a^{3}b + 29a^{2}b^{2} - 30ab^{3} + 25b^{4}$. + +\Item{5.} $16x^{6} + 24x^{5}y + 9x^{4}y^{2} - 16x^{3}y^{3} - 12x^{2}y^{4} + 4y^{6}$. + +\Item{6.} $4x^{6} - 4x^{4}y^{2} + 12x^{3}y^{3} + x^{2}y^{4} - 6xy^{5} + 9y^{6}$. + +\Paragraph{181. Arithmetical Square Roots.} In the general method +of extracting the square root of a number expressed by +figures, the first step is to mark off the figures into \emph{groups}. +%% -----File: 160.png---Folio 154------- + +Since $1 = 1^{2}$, $100 = 10^{2}$, $10,000 = 100^{2}$, and so on, it is +evident that the square root of a number between $1$~and +$100$ lies between $1$~and~$10$; of a number between $100$ and +$10,000$ lies between $10$~and~$100$. In other words, the +square root of a number expressed by \emph{one} or \emph{two} figures is +a number of \emph{one} figure, of a number expressed by \emph{three} or +\emph{four} figures is a number of \emph{two} figures, and so on. + +If, therefore, an integral square number is divided into +groups of two figures each, from the right to the left, the +number of figures in the root will be equal to the number +of groups of figures. The last group to the left may have +only one figure. + +Ex. Find the square root of~$3249$. +%[** TN: Tabulated calculation inset in the original] +\begin{Soln} +\[ +\begin{array}{rc@{}ll} + &3&249&(57 \\ + &2&5 \\ +\cline{2-3} +107\,\rlap{)}&&749 \\ + &&749 \\ +\cline{3-3} +\end{array} +\] + +In this case, $a$~in the typical form $a^{2} + 2ab + b^{2}$ +represents $5$~tens, that is,~$50$, and $b$~represents~$7$\Add{.} The +$25$ subtracted is really~$2500$, that is,~$a^{2}$, and the complete +divisor $2a + b$ is $2 × 50 + 7 = 107$. +\end{Soln} + +\Paragraph{182.} The same method will apply to numbers of more +than two groups of figures by considering $a$~in the typical +form to represent at each step \emph{the part of the root already +found}. + +It must be observed that \emph{$a$~represents so many tens with +respect to the next figure of the root}. + +Ex. Find the square root of~$94,249$. +\begin{Soln} +\[ +\begin{array}{r*{3}{c@{\,}}l} + &9& 42&49&(307 \\ + &9 \\ +\cline{2-4} +607\rlap{)} & & \multicolumn{2}{l}{4249} \\ + & & \multicolumn{2}{l}{4249} \\ +\cline{3-4} +\end{array} +\] +\end{Soln} + +\begin{Remark}[Note.] Since the first trial divisor,~$60$, is not contained in~$42$, we +put a zero in the root, and bring down the next group,~$49$. +\end{Remark} +%% -----File: 161.png---Folio 155------- + +\Paragraph{183.} If the square root of a number has decimal places, +the number itself will have \emph{twice} as many. Thus, if $0.21$ +is the square root of some number, this number will be +$(0.21)^{2} = 0.21 × 0.21 = 0.0441$, and if $0.111$ be the root, +the number will be $(0.111)^{2} = 0.111 x 0.111 = 0.012321$. + +Therefore, the number of \emph{decimal} places in every square +decimal will be \emph{even}, and the number of decimal places in +the root will be \emph{half} as many as in the given number itself. + +Hence, if a given number contain a decimal, we divide +it into groups of two figures each, by beginning at the +decimal point and marking toward the left for the integral +number, and toward the right for the decimal. We must +have the last group on the right of the decimal point contain +\emph{two} figures, annexing a cipher when necessary. + +Ex. Find the square roots of $41.2164$ and $965.9664$. +\begin{Soln} +\[ +\begin{array}{r@{}r@{}l@{\,}ll} + & 41&.21&64& (6.42 \\ + & 36 \\ +\cline{2-3} +124\rlap{)} & 5&21 \\ + & 4&96 \\ +\cline{3-4} +\multicolumn{2}{r}{1282\rlap{)}}&\multicolumn{2}{l}{2564} \\ + & &\multicolumn{2}{l}{2564} \\ +\cline{3-4} +\end{array}\qquad\qquad\qquad +\begin{array}{rl@{}l@{\,}ll} + 9&65&.96& 64&(31.08 \\ + 9 \\ +\cline{2-2} + 61\rlap{)}&65 \\ + &61 \\ +\cline{2-3} +6208\rlap{)}&49&664 \\ + &49&664 \\ +\cline{2-3} +\end{array} +\] +\end{Soln} + +\Paragraph{184.} If a number contain an \emph{odd} number of decimal +places, or if any number give a \emph{remainder} when as many +figures in the root have been obtained as the given number +has groups, then its exact square root cannot be found. We +may, however, approximate to its exact root as near as we +please by annexing ciphers and continuing the operation. + +The square root of a common fraction whose denominator +is not a perfect square can be found approximately by +reducing the fraction to a decimal and then extracting the +root; or by reducing the fraction to an equivalent fraction +whose denominator is a perfect square, and extracting the +square root of both terms of the fraction. +%% -----File: 162.png---Folio 156------- + +%[** Force page break in both print and screen layout] +\newpage +\Item{1.} Find the square roots of $3$ and $357.357$. +\begin{Soln} +\[ +\begin{array}{rcccl} + &3.&\multicolumn{3}{l}{(\rlap{$1.732\dots$}} \\ + &1 \\ +\cline{2-3} +27\rlap{)} + &2&00 \\ + &1&89 \\ +\cline{3-4} +\multicolumn{2}{r}{343\rlap{)}} + &11&00 \\ + &&10&29 \\ +\cline{4-5} +\multicolumn{3}{r}{3462\rlap{)}} + &71&00 \\ +&& &69&24 \\ +\cline{4-5} +\end{array}\qquad\qquad\qquad +\begin{array}{rcclll} + &3&57.&35&70&\rlap{$(18.903\dots$} \\ + &1 \\ +\cline{2-3} +28\rlap{)} + &2&57 \\ + &2&24 \\ +\cline{3-4} +\multicolumn{2}{r}{369\rlap{)}} + &33&35 \\ +& &33&21 \\ +\cline{4-6} +\multicolumn{3}{r}{37803\rlap{)}} + &14&70&00 \\ +&& &11&34&09 \\ +\cline{4-6} +\end{array} +\] +\end{Soln} + +\Item{2.} Find the square root of~$\frac{5}{8}$. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Since} +\frac{5}{8} &= 0.625, \\ +\lintertext{the square root of} +\frac{5}{8} &= \sqrt{0.625} \\ + &= 0.79057. \displaybreak[1] \\ +\lintertext{\indent Or,} +\frac{5}{8} &= \frac{10}{16}, \\ +\lintertext{and the square root of} +\frac{5}{8} &= \frac{\sqrt{10}}{\sqrt{16}} = \tfrac{1}{4}\sqrt{10} \\ + &= \tfrac{1}{4}(3.16227) \\ + &= 0.79057. +\end{DPalign*} +\end{Soln} + +\ScreenBreak +\Exercise{79.} + +Find the square root of: +\begin{multicols}{3} +\Item{1.} $324$. + +\Item{2.} $441$. + +\Item{3.} $529$. + +\Item{4.} $961$. + +\Item{5.} $10.24$. + +\Item{6.} $53.29$. + +\Item{7.} $53,824$. + +\Item{8.} $616,225$. + +\Item{9.} $1,500,625$. + +\Item{10.} $346,921$. + +\Item{11.} $31,371,201$. + +\Item{12.} $1,522,756$. +\end{multicols} + +\PrintBreak +Find to four decimal places the square root of: +\begin{multicols}{5} +\Item{13.} $2$. + +\Item{14.} $3$. + +\Item{15.} $5$. + +\Item{16.} $6$. + +\Item{17.} $0.5$. + +\Item{18.} $0.9$. + +\Item{19.} $\frac{2}{3}$. + +\Item{20.} $\frac{3}{4}$. + +\Item{21.} $\frac{4}{5}$. + +\Item{22.} $\frac{5}{8}$. +\end{multicols} +%% -----File: 163.png---Folio 157------- + + +\Section{Cube Roots of Compound Expressions.} + +\Paragraph{185.} Since the cube of $a + b$ is $a^{3} + 3a^{2}b + 3ab^{2} + b^{3}$, +the cube root of a$^{3} + 3a^{2}b + 3ab^{2} + b^{3}$ is $a + b$. + +It is required to devise a method for extracting the cube +root $a + b$ when $a^{3} + 3a^{2}b + 3ab^{2} + b^{3}$ is given. + +\Item{1.} Find the cube root of $a^{3} + 3a^{2}b + 3ab^{2} + b^{3}$. +\begin{Soln} +\[ +\begin{array}{rl<{\quad}rcll} + & & a^{3} &+& \TbBar{3a^{2}b + 3ab^{2} + b^{3}} & a + b \\ +\cline{6-6} +3a^{2} & & a^{3} \\ +\cline{3-3} + &+ 3ab + b^{2} &\TbBar{}& & 3a^{2}b + 3ab^{2} + b^{3} \\ +\cline{1-3} +3a^{2} &+ 3ab + b^{2} &\TbBar{}& & 3a^{2}b + 3ab^{2} + b^{3} \\ +\cline{4-5} +\end{array} +\] + +The first term $a$ of the root is obviously the cube root of the first +term~$a^{3}$ of the given expression. + +If $a^{3}$ be subtracted, the remainder is $3a^{2}b + 3ab^{2} + b^{3}$; therefore, +the second term~$b$ of the root is obtained by dividing the first term of +this remainder by \emph{three times the square of~$a$}. + +Also, since $3a^{2}b + 3ab^{2} + b^{3} = (3a^{2} + 3ab + b^{2})b$, the \emph{complete +divisor} is obtained by adding $3ab + b^{2}$ to the \emph{trial divisor~$3a^{2}$}. +\end{Soln} + +\Item{2.} Find the cube root of $8x^{3} + 36x^{2}y + 54xy^{2} + 27y^{3}$. +\begin{Soln} +\[ +\begin{array}{rlrlcll} + & & & 8x^{3} &+& \TbBar{36x^{2}y + 54xy^{2} + 27y^{3}} & 2x + 3y \\ +\cline{7-7} + & 12x^{2} & & 8x^{3} \\ +\cline{4-4} +(6x + 3y)3y =& & 18xy +& \TbBar{9y^{2}} & & 36x^{2}y + 54xy^{2} + 27y^{3} \\ +\cline{2-4} + & 12x^{2} &+ 18xy +& \TbBar{9y^{2}} & & 36x^{2}y + 54xy^{2} + 27y^{3} \\ +\cline{5-6} +\end{array} +\] + +The cube root of the first term is~$2x$, and this is therefore the first +term of the root. $8x^{3}$,~the cube of~$2x$, is subtracted. + +The second term of the root,~$3y$, is obtained by dividing $36x^{2}y$ by +$3(2x)^{2} = 12x^{2}$, which corresponds to $3a^{2}$ in the typical form, and the +divisor is completed by annexing to~$12x^{2}$ the expression +\[ +\bigl\{3(2x) + 3y\bigr\}3y = 18xy + 9y^{2}, +\] +which corresponds to $3ab + b^{2}$ in the typical form. +\end{Soln} +%% -----File: 164.png---Folio 158------- + +\Paragraph{186.} The same method may be applied to longer expressions +by considering~$a$ in the typical form $3a^{2} + 3ab + b^{2}$ +to represent at each stage of the process \emph{the part of the root +already found}. Thus, if the part of the root already found +is $x + y$, then $3a^{2}$~of the typical form will be represented +by $3(x + y)^{2}$; and if the third term of the root be~$+z$, \DPtypo{the}{then} +$3ab + b^{2}$ will be represented by $3(x + y)z + z^{2}$. So that +the complete divisor, $3a^{2} + 3ab + b^{2}$, will be represented +by $3(x + y)^{2} + 3(x + y)z + z^{2}$. + +Ex. Find the cube root of $x^{6} - 3x^{5} + 5x^{3} - 3x - 1$. +\begin{Soln} +%[** TN: Special spacing] +\ifthenelse{\not\boolean{ForPrinting}}{\footnotesize}{} +\[ +\ifthenelse{\not\boolean{ForPrinting}}{\quad}{} +\begin{array}{*{2}{r}@{}*{3}{r}*{4}{l}} + & & & \TbBar{}& x^{2}& \multicolumn{3}{@{}l}{{} - x - 1} \\ +\cline{5-6} + & & & & x^{6}& {} - 3x^{5} &\multicolumn{2}{@{}l}{\phantom{{}+ 3x^{4}} + 5x^{3} - 3x - 1} \\ + & 3x^{4}& & & x^{6} \\ +\cline{5-5} + (3x^{2} - x)(-x) =& &{} - 3x^{3}&+& \TbBar{x^{2}}&{} - 3x^{5}&\phantom{{}+ 3x^{4}} + 5x^{3} \\ +\cline{3-5} + & 3x^{4}&{} - 3x^{3}&+& \TbBar{x^{2}}&{} - 3x^{5}&{} + 3x^{4} - \Z x^{3} \\ +\cline{6-7} + & & & & &\TbBar{} &{} - 3x^{4} + 6x^{3} - 3x - 1 \\ + 3(x^{2} - x)^{2} =& 3x^{4}&{} - 6x^{3}&+& 3x^{2}&\TbBar{} \\ +\llap{$(3x^{2} - 3x - 1)$}(-1) =& & &-& 3x^{2}& \multicolumn{1}{l|}{{}+ 3x + 1} \\ +\cline{2-6} + & 3x^{4}&{} - 6x^{3}& & & \multicolumn{1}{l|}{{}+ 3x + 1}&{} - 3x^{4} + 6x^{3} - 3x - 1 \\ +\cline{7-7} +\end{array} +\] +\end{Soln} + +\begin{Remark}[Note.] The root is placed \emph{above} the given expression because +there is no room for it on the page at the right of the expression. + +The first term of the root,~$x^{2}$, is obtained by taking the cube root +of the first term of the given expression; and the first trial-divisor,~$3x^{4}$, +is obtained by taking three times the square of this term. + +The next term of the root is found by dividing~$-3x^{5}$, the first term +of the remainder after~$x^{6}$ is subtracted, by~$3x^{4}$, and the first complete +divisor, $3x^{4} - 3x^{3} + x^{2}$, is found by annexing to the trial divisor +$(3x^{2} - x)(-x)$, which expression corresponds to $(3a + b)b$ in the +typical form. + +\emph{The part of the root already found}~($a$) is now represented by $x^{2} - x$, +therefore $3a^{2}$~is represented by $3(x^{2} - x)^{2} = 3x^{4} - 6x^{3} + 3x^{2}$, the +second trial divisor\Add{,} and $(3a + b)b$ by $(3x^{2} - 3x - 1)(-1)$, since $b$~in +this case is found to be~$-1$, therefore, in the second complete +divisor, $3a^{2} + (3a + b)b$ is represented by +\[ +(3x^{4} - 6x^{3} + 3x^{2}) + (3x^{2} - 3x - 1)(-1) = 3x^{4} - 6x^{3} + 3x + 1. +\] +\end{Remark} +%% -----File: 165.png---Folio 159------- + +\Exercise{80.} + +Find the cube root of + +\Item{1.} $x^{3} + 3x^{2} y + 3xy^{2} + y^{3}$\Add{.} + +\Item{2.} $8x^{3} - 12x^{2} + 6x - 1$. + +\Item{3.} $8x^{3} - 36x^{2} y + 54xy^{2} - 27y^{3}$. + +\Item{4.} $64a^{3} - 144a^{2} x + 108ax^{2} - 27x^{3}$\Add{.} + +\Item{5.} $1 + 3x + 6x^{2} + 7x^{3} + 6x^{4} + 3x^{5} + x^{6}$. + +\Item{6.} $x^{6} - 3x^{5} + 6x^{4} - 7x^{3} + 6x^{2} - 3x + 1$. + +\ScreenBreak +\Paragraph{187. Arithmetical Cube Roots.} In extracting the cube root +of a number expressed by figures, the first step is to mark +it off into groups\Add{.} + +Since $1 = 1^{3}$, $1000 = 10^{3}$, $1,000,000 = 100^{3}$, and so on, it +follows that the cube root of any number between $1$~and +$1000$, that is, of any number which has \emph{one}, \emph{two}, or \emph{three} +figures, is a number of \emph{one} figure, and that the cube root +of any number between $1000$ and $1,000,000$, that is, of any +number which has \emph{four}, \emph{five}, or \emph{six} figures, is a number of +\emph{two} figures, and so on. + +If, therefore, an integral cube number be divided into +groups of three figures each, from right to left, the number +of figures in the root will be equal to the number of groups\Add{.} +The last group to the left may consist of one, two, or three +figures. + +\Paragraph{188.} If the cube root of a number have decimal places, +the number itself will have \emph{three times} as many. Thus, if +$0.11$ be the cube root of a number\Add{,} the number is $0.11 × 0.11 +x 0.11 = 0.001331$. Hence\Add{,} if a given number contain a +decimal, we divide the figures of the number into groups +of three figures each, by beginning at the decimal point +and marking toward the left for the integral number, and +%% -----File: 166.png---Folio 160------- +toward the right for the decimal. We must be careful to +have the last group on the right of the decimal point contain +\emph{three} figures, annexing ciphers when necessary. + +\ScreenBreak +Extract the cube root of~$42875$. +\begin{Soln} +\[ +\begin{array}{rrrlll} + & & & 42&875&(35 \\ + & & a^{3} =& 27 \\ +\cline{4-5} +3a^{2} =& 3 × 30^{2} =& \TbBar{2700} & 15&875 \\ + 3ab =& 3 × (30 × 5) =& \TbBar{450} \\ + b^{2} =& 5^{2} =& \TbBar{25} \\ +\cline{3-3} + & & \TbBar{3175} & 15&875 \\ +\cline{4-5} +\end{array} +\] + +Since $42875$ has two groups, the root will have two figures. + +The first group, $42$, contains the cube of the tens of the root. + +The greatest cube in~$42$ is~$27$, and the cube root of~$27$ is~$3$. Hence +$3$~is the tens' figure of the root. + +We subtract $27$ from~$42$, and bring down the next group,~$875$. +Since $a$~is $3$~tens or~$30$, $3a^{2} = 3 × 30^{2}$, or~$2700$. This trial-divisor is +contained $5$~times in~$15875$. The trial-divisor is completed by adding +$3ab + b^{2}$; that is, $450 + 25$, to the trial-divisor. +\end{Soln} + +\Paragraph{189.} The same method will apply to numbers of more +than two groups of figures, by considering in each case~$a$, +the part of the root already found, as so many tens with +respect to the next figure of the root. + +\PrintStretch{12pt} +Extract the cube root of~$57512456$. +\begin{Soln} +\[ +\begin{array}{*{3}{r}*{4}{l}} + & & & 57&512&456&(386 \\ + & & a^{3} =& 27 \\ +\cline{4-5} +3a^{2} =& 3 × 30^{2} =& \TbBar{2700} & 30&512 \\ + 3ab =& 3 × (30 × 8) =& \TbBar{720} \\ + b^{2} =& 8^{2} =& \TbBar{64} \\ +\cline{3-3} + & & \TbBar{3484} & 27&872 \\ +\cline{4-6} + & & \TbBar{} & \Z2&640&456 \\ +3a^{2} =& 3 × 380^{2} =& \TbBar{433200} \\ + 3ab =& 3 × (380 × 6) =& \TbBar{6840} \\ + b^{2} =& 6^{2} =& \TbBar{36} \\ +\cline{3-3} + & & \TbBar{440076} & \Z2&640&456 \\ +\cline{4-6} +\end{array} +\] +\end{Soln} +%% -----File: 167.png---Folio 161------- + +Extract the cube root of~$187.149248$. +\begin{Soln} +\[ +\begin{array}{*{3}{r}@{}*{4}{r}l} + & & & \multicolumn{2}{r}{187\rlap{.}}&149&248&(5.72 \\ + & & a^{3} =& \multicolumn{2}{r}{125} \\ +\cline{4-6} +3a^{2} =& 3 × 50^{2} =& \TbBar{7500}&& 62&149 \\ + 3ab =& 3 × (50 × 7) =& \TbBar{1050} \\ + b^{2} =& 7^{2} =& \TbBar{49} \\ +\cline{3-3} + & & \TbBar{8599}&& 60&193 \\ +\cline{4-7} + & & & \TbBar{}& 1&956&248 \\ +3a^{2} =& 3 × 570^{2} =& 9747&\TbBar{00} \\ + 3ab =& 3 × (570 × 2) =& 34&\TbBar{20} \\ + b^{2} =& 2^{2} =& & \TbBar{4} \\ +\cline{3-4} + & & 9781&\TbBar{24}& 1&956&248 \\ +\cline{5-7} +\end{array} +\] + +It will be seen from the groups of figures that the root will have +one integral and two decimal places. +\end{Soln} + +\Paragraph{190.} If the given number is not a perfect cube, ciphers +may be annexed, and a value of the root may be found as +near to the \emph{true} value as we please. + +Extract the cube root of~$1250.6894$. +\begin{Soln} +\[ +\begin{array}{rrrcclll} + & & & &1&\multicolumn{3}{l}{\rlap{250.689\,400\,(10.77}} \\ + & & a^{3} &=&1 \\ +\cline{4-6} +3a^{2} =& \PadTo{3 × (1070 × 7)}{3 × 10^{2}} =& \TbBar{\Z\Z300}& && 250 & \phantom{999} & \phantom{999} \\ +\end{array} +\] +Since $300$ is not contained in~$250$, the next figure of the root will +be~$0$. +\[ +\begin{array}{rrr@{}ccrll} +3a^{2} =& 3 × 100^{2} =& \TbBar{30000}& &\Z&250&689 \\ + 3ab =& 3 × (100 × 7) =& \TbBar{2100} \\ + b^{2} =& 7^{2} =& \TbBar{49} \\ +\cline{3-3} + & & \TbBar{32149}& && 225&043 \\ +\cline{4-8} + & & &\TbBar{}&& 25&646&400 \\ +3a^{2} =& 3 × 1070^{2} =& 34347&\TbBar{00} \\ + 3ab =& 3 × (1070 × 7) =& 224&\TbBar{70} \\ + b^{2} =& 7^{2} =& &\TbBar{49} \\ +\cline{3-4} + & & 34572&\TbBar{19}&& 24&200&533 \\ +\cline{5-8} + & & & && 1&445&867 \\ +\end{array} +\] +\end{Soln} +%% -----File: 168.png---Folio 162------- + +\Paragraph{191.} Notice that if $a$~denotes the first term, and $b$~the +second term of the root, the \emph{first complete divisor} is +\[ +3a^{2} + 3ab + b^{2}, +\] +and the \emph{second trial-divisor} is $3(a + b)^{2}$, that is, +\[ +3a^{2} + 6ab + 3b^{2}. +\] + +This expression may be obtained by adding to the preceding +complete divisor, $3a^{2} + 3ab + b^{2}$, \emph{its second term and +twice its third term}. Thus: +\[ +\begin{array}{rr} +3a^{2} + 3ab + & b^{2} \\ + 3ab + &2b^{2} \\ +\cline{1-2} +3a^{2} + 6ab + &3b^{2} \\ +\end{array} +\] + +This method of obtaining \emph{trial-divisors} is of great importance +for shortening numerical work, as may be seen in the +following example: + +Ex. Extract the cube root of~$5$ to five places of decimals. +\begin{Soln} +\[ +\begin{array}{rrr@{}r@{}rcrrl} + & & & &&5\rlap{.}&\multicolumn{3}{l}{000\,(1.70997} \\ + & && \multicolumn{3}{r}{a^{3} = 1} \\ +\cline{4-7} +3a^{2} =& 3 × 10^{2} =& 300&\TbBar{}&&4&000 \\ + 3ab =& 3(10 × 7) =& 210&\TbBar{} \\ + b^{2} =& 7^{2} =& 49&\TbBar{}\\ +\cline{3-3} + & & 559&\TbBar{\BB}&&3&913 \\ +\cline{5-9} + & & 259& & &\TbBar{}& 87& 000& 000 \\ +\cline{3-5} +3a^{2} =& 3 × 1700^{2} =& 867&00&00&\TbBar{} \\ + 3ab =& 3(1700 × 9) =& 4&59&00&\TbBar{} \\ + b^{2} =& 9^{2} =& & &81&\TbBar{} \\ +\cline{3-5} + & & 871&59&81&\TbBar{\BB}& 78& 443& 829 \\ +\cline{7-9} + & & 4&59&81&\TbBar{}& 8& 556& 1710 \\ +\cline{3-5} +3a^{2} =& 3 × 1709^{2} =& 876&20&43&\TbBar{}& 7& 885& 8387 \\ +\cline{7-9} + & & & & &\TbBar{}& & 670& 33230 \\ + & & & & &\TbBar{}& & 613& 34301 \\ +\cline{7-9} +\end{array} +\] +%% -----File: 169.png---Folio 163------- + +After the first two figures of the root are found, the next trial-divisor +is obtained by bringing down~$259$, the sum of the $210$ and $49$ +obtained in completing the preceding divisor, then \emph{adding the three +lines connected by the brace}, and annexing two ciphers to the result. + +This trial divisor is~$86,700$, and if we add $3ab + b^{2}$ to complete the +divisor, when $b = 1$, the complete divisor will be $86,700 + 511 = 87,211$, +and this is larger than the dividend~$87,000$. We therefore put $0$ for +the next figure of the root. We then bring down another group +and annex two more ciphers to the trial-divisor. + +The last two figures of the root are found by division. The rule +in such cases is, that two less than the number of figures already +obtained may be found without error by division, the divisor being +three times the square of the part of the root already found. +\end{Soln} + +\Paragraph{192.} The cube root of a common fraction whose denominator +is not a perfect cube can be found approximately by +reducing the fraction to a decimal, and then extracting +the root. + +\Exercise{81.} + +Find the cube root of: +\begin{multicols}{2} +\Item{1.} $46,656$. + +\Item{2.} $42,875$. + +\Item{3.} $91,125$. + +\Item{4.} $274,625$. + +\Item{5.} $110,592$. + +\Item{6.} $258,474,853$. + +\Item{7.} $109,215,352$. + +\Item{8.} $259,694,072$. + +\Item{9.} $127,263,527$. + +\Item{10.} $385,828,352$. + +\Item{11.} $1879.080904$. + +\Item{12.} $1838.265625$. +\end{multicols} + +Find to four decimal places the cube root of: +\begin{multicols}{4} +\Item{13.} $0.01$. + +\Item{14.} $0.05$. + +\Item{15.} $0.2$. + +\Item{16.} $4$. + +\Item{17.} $10$. + +\Item{18.} $87$. + +\Item{19.} $2.5$. + +\Item{20.} $2.05$. + +\Item{21.} $3.02$. + +\Item{22.} $\frac{2}{3}$. + +\Item{23.} $\frac{3}{4}$. + +\Item{24.} $\frac{9}{11}$. +\end{multicols} +%% -----File: 170.png---Folio 164------- +%[Blank Page] +%% -----File: 171.png---Folio 165------- + + +\Answers + +\AnsTo[7]{Exercise}{1.} % Page 10. + +\Item{1.} $14$. + +\Item{2.} $10$. + +\Item{3.} $13$. + +\Item{4.} $11$. + +\Item{5.} $13$. + +\Item{6.} $7$. + +\Item{7.} $9$. + +\Item{8.} $7$. + +\Item{9.} $6$. + +\Item{10.} $2$. + +\Item{11.} $3$. + +\Item{12.} $6$. + +\Item{13.} $2$. + +\Item{14.} $8$. + +\Item{15.} $4$. + +\Item{16.} $3$. + +\Item{17.} $1$. + +\Item{18.} $1$. + +\Item{19.} $3$. + +\Item{20.} $4$. + +\Item{21.} $10$. + + +\AnsTo[7]{Exercise}{2.} % Page 12. + +\Item{1.} $91$. + +\Item{2.} $21$. + +\Item{3.} $60$. + +\Item{4.} $24$. + +\Item{5.} $96$. + +\Item{6.} $16$. + +\Item{7.} $36$. + +\ResetCols[3] +\Item{8.} $4a + 4b$. + +\Item{9.} $4a - 4b$. + +\Item{10.} $2a^{2} + 2b^{2}$. + +\Item{11.} $2a^{2} - 2b^{2}$. + +\Item{12.} $3ab + 3c$. + +\Item{13.} $3ab - 3c$. + +\Item{14.} $3c - 3ab$. + +\Item{15.} $ab + ac$. + +\Item{16.} $ab - ac$. + +\Item{17.} $3ab + 3ac$. + +\Item{18.} $3ab - 3ac$. + +\Item{19.} $5ab^{2} + 5ac$. + +\Item{20.} $5ab^{2} - 5ac^{2}$. + +\Item{21.} $5a^{2}b^{2} - 5a^{2}c$. + + +\AnsTo[5]{Exercise}{3.} % Page 12. + +\Item{1.} $63$. + +\Item{2.} $280$. + +\Item{3.} $300$. + +\Item{4.} $98$. + +\Item{5.} $81$. + +\Item{6.} $1250$. + +\Item{7.} $105$. + +\Item{8.} $105$. + +\Item{9.} $315$. + +\Item{10.} $35$. + +\Item{11.} $105$. + +\Item{12.} $105$. + +\Item{13.} $0$. + +\Item{14.} $135$. + +\Item{15.} $120$. + +\Item{16.} $0$. + +\Item{17.} $1800$. + +\Item{18.} $540$. + +\Item{19.} $0$. + +\Item{20.} $270$. + +\Item{21.} $540$. + + +\AnsTo[6]{Exercise}{4.} % Page 13. + +\Item{1.} $21$. + +\Item{2.} $26$. + +\Item{3.} $72$. + +\Item{4.} $85$. + +\Item{5.} $30$. + +\Item{6.} $17$. + +\Item{7.} $8$. + +\Item{8.} $50$. + +\Item{9.} $24$. + +\Item{10.} $0$. + +\Item{11.} $12$. + +\Item{12.} $100$. + +\Item{13.} $80$. + +\Item{14.} $71$. + +\Item{15.} $139$. + +\Item{16.} $17$. + +\Item{17.} $8$. + +\Item{18.} $5$. + +\Item{19.} $3$. + +\Item{20.} $6$. + +\Item{21.} $5$. + +\Item{22.} $1$. + +\Item{23.} $2$. + +\Item{24.} $2$. + + +\AnsTo{Exercise}{5.} % Page 14. + +\Item{1.} $a$~plus~$b$; $a$~minus~$b$; $a$~times~$b$; $a$~divided by~$b$. + +\Item{3.} $a + b$. + +\Item{5.} $a - b$. + +\Item{7.} $x - y$. + +\Item{9.} $4x$; $x^{4}$. + +\Item{11.} $35 - x$. + +\Item{12.} $x - a$. + +\Item{14.} $14 - x$. + +\Item{15.} $a - x$. + +\Item{17.} $xy$. + +\Item{18.} $\dfrac{x}{y}$. +%% -----File: 172.png---Folio 166------- + + +\AnsTo[2]{Exercise}{6.} % Page 15. + +\Item{2.} $\dfrac{a}{b}$. + +\Item{4.} $(x - 3)$~yr.; $(x + 7)$~yr. + +\Item{6.} $7(2x - y)$. + +\Item{8.} $x + 1$; $x - 1$. + +\Item{9.} $20 - d$. + +\Item{11.} $x + 8$. + +\Item{13.} $x - 10$. + +\Item{14.} $10$. + + +\AnsTo[2]{Exercise}{7.} % Page 16. + +\Item{1.} $(40 - x)$~yr. + +\Item{2.} $(a + y)$~yr. + +\Item{3.} $4$. + +\Item{5.} $ab$. + +\Item{6.} $5x - 3x$. + +\Item{8.} $2x - 3 - (x + 1)$. + +\Item{9.} $40$. + +\Item{10.} $12$. + +\Item{11.} $100a + 25b + 10c$. + +\Item{12.} $100 - x - y$. + +\Item{14.} $xy + c$. + + +\AnsTo[2]{Exercise}{8.} % Page 17. + +\Item{2.} $xy - a^{2}$. + +\Item{3.} $\dfrac{ph}{gk}$. + +\Item{4.} $6m^{2} + 5c(d + b - a)$. + +\Item{5.} $5(2n + 1) - 6(c - a + b)$. + +\Item{6.} \$$100 - \text{\$}(a + b + 2c)$. + +\Item{8.} $\dfrac{1}{x}$. + +\Item{9.} $\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}$. + +\ResetCols[1] +\Item{10.} $n + (n + 1) + (n + 2)$; $n(n + 1)(n + 2)$. + +\ResetCols[4] +\Item{11.} $\dfrac{36}{x}$. + +\Item{12.} $qd$. + +\Item{13.} $qd + r$. + +\Item{14.} $2x$. + +\Item{15.} $\dfrac{10x}{12}$. + +\Item{16.} $\dfrac{6a}{b}$. + +\Item{17.} $4m$. + +\Item{18.} $x + 17$. + +\Item{19.} $x + y + c$. + +\Item{20.} $36x$. + +\Item{21.} $\dfrac{c}{4}$. + +\Item{22.} $x - 1$, $x$, $x + 1$. + +\Item{23.} $2n + 3$. + + +\AnsTo[5]{Exercise}{9.} % Page 24. + +\Item{1.} $4$. + +\Item{2.} $5$. + +\Item{3.} $3$. + +\Item{4.} $1$. + +\Item{5.} $13$. + +\Item{6.} $0$. + +\Item{7.} $5$. + +\Item{8.} $5$. + +\Item{9.} $7$. + +\Item{10.} $4$. + +\Item{11.} $4$. + +\Item{12.} $1$. + +\Item{13.} $25$. + +\Item{14.} $32\frac{1}{2}$. + +\Item{15.} $6$. + +\Item{16.} $10$. + +\Item{17.} $3$. + +\Item{18.} $1$. + +\Item{19.} $7$. + +\Item{20.} $13$. + +\Item{21.} $12$. + +\Item{22.} $14$. + +\Item{23.} $7$. + +\Item{24.} $5$. + +\Item{25.} $0$. + +\Item{26.} $15$. + +\Item{27.} $10\frac{1}{2}$. + +\Item{28.} $4$. + +\Item{29.} $3$. + +\Item{30.} $4$. + +\Item{31.} $\frac{2}{5}$. + +\Item{32.} $6$. + +\Item{33.} $9$. + +\Item{34.} $7$. + +\Item{35.} $5$. + +\Item{36.} $6$. + +\Item{37.} $6$. + + +\AnsTo[2]{Exercise}{10.} % Page 27. + +\Item{1.} $30$. + +\Item{2.} Son, $10$~yr.; father, $50$~yr. + +\Item{3.} $78$, $13$. + +\Item{4.} $80$~ft.\ broken off; $10$~ft.\ standing. + +\Item{5.} $23$, $30$. + +\Item{6.} $36$, $48$. + +\Item{7.} $15$, $20$. + +\Item{8.} $20$. + +\Item{9.} $12$. + +\Item{10.} $10$, $40$. + +\Item{11.} $26$, $10$. + +\Item{12.} $25$, $15$. + +\Item{13.} $10$, $20$. + +\Item{14.} $7$, $20$. + +\Item{15.} $12$, $20$. +%% -----File: 173.png---Folio 167------- + + +\AnsTo{Exercise}{11.} % Page 29. + +\Item{1.} Cow, \$$42$; horse, \$$168$. + +\Item{2.} $81$. + +\Item{3.} $2$. + +\Item{4.} $30$, $40$. + +\Item{5.} $25$, $26$, $27$. + +\Item{6.} $5$, $6$, $7$, $8$, $9$. + +\Item{7.} A, $30$~yr.; B, $10$~yr. + +\Item{8.} Father, $40$~yr., son, $10$~yr. + +\Item{9.} $40$. + +\Item{10.} $10$. + +\Item{11.} \$$40$. + +\Item{12.} $9$. + + +\AnsTo[2]{Exercise}{12.} % Page 30. + +\Item{1.} $15$~men; $30$~women; $45$~children. + +\Item{2.} $50$. + +\Item{3.} $16$. + +\Item{4.} $7$. + +\Item{5.} $35$. + +\Item{6.} $24$. + +\Item{7.} $24$. + +\Item{8.} $20$. + +\Item{9.} $970$; $1074$. + + +\AnsTo[2]{Exercise}{13.} % Page 31. + +\Item{1.} $24$. + +\Item{2.} A, \$$60$; B, \$$30$. + +\Item{3.} \$$3$~quarters; $6$~bills. + +\Item{4.} $14$. + +\Item{5.} \$$4$~quarters; $20$~half-dollars. + +\Item{6.} \$$7$~ten-dollar bills; $21$~one-dollar bills. + +\Item{7.} Father, $32$~yr.; son, $8$~yr. + +\Item{8.} $20$. + + +\AnsTo[2]{Exercise}{14.} % Page 32. + +\Item{1.} $9$~miles. + +\Item{2.} \$$60$. + +\Item{3.} \$$20$~lb.\ at $65$~cts.; + $60$~lb.\ at $45$~cts. + +\Item{4.} $40$. + +\Item{5.} $15,000$. + +\Item{6.} $15$~in.; $21$~in. + +\Item{7.} \$$2$~doz.\ at $25$~cts.; + $5$~doz.\ at $20$~cts. + +\Item{8.} \$$6$~quarters, $18$~ten-cent pieces. + + +\AnsTo[4]{Exercise}{15.} % Page 39. + +\Item{1.} $40c$. + +\Item{2.} $24a$. + +\Item{3.} $39x$. + +\Item{4.} $51y$. + +\Item{5.} $-26a$. + +\Item{6.} $-40x$. + +\Item{7.} $-17b$. + +\Item{8.} $-66z$. + +\Item{9.} $-20m$. + +\Item{10.} $2d$. + +\Item{11.} $0$. + +\Item{12.} $-18g$. + +\Item{13.} $a^{2}$. + +\Item{14.} $-21x^{3}$. + +\Item{15.} $0$. + +\Item{16.} $3mn$. + +\Item{17.} $0$. + +\Item{18.} $-3a^{3}b^{3}c^{3}$. + +\Item{19.} $-9abcd$. + +\Item{20.} $1$. + +\Item{21.} $12$. + +\Item{22.} $4$. + +\Item{23.} $-18$. + +\Item{24.} $10$. +%% -----File: 174.png---Folio 168------- + + +\AnsTo{Exercise}{16.} % Page 42. + +\Item{1.} $30a^{5}$. + +\Item{2.} $40a^{4}b^{3}$. + +\Item{3.} $63x^{2}y^{2}$. + +\Item{4.} $2a^{5}b^{5}c^{2}$. + +\Item{5.} $9a^{7}b^{9}c^{9}$. + +\Item{6.} $-10a^{2}$. + +\Item{7.} $12ab$. + +\Item{8.} $-a^{4}b^{3}$. + +\Item{9.} $10a^{5}b^{5}c$. + +\Item{10.} $12x^{7}y^{5}z^{2}$. + +\Item{11.} $105a^{7}b^{5}$. + +\Item{12.} $6a^{6}b^{5}c^{7}$. + +\Item{13.} $-12a^{5}b^{5}c^{5}x^{5}$. + +\Item{14.} $24a^{7}b^{6}c^{5}$. + +\Item{15.} $-42a^{6}m^{5}x^{7}$. + +\Item{16.} $30x^{8}y^{4}z^{6}$. + +\Item{17.} $-46$. + +\Item{18.} $-3$. + +\Item{19.} $-8$. + +\Item{20.} $-17$. + +\Item{21.} $9$. + +\Item{22.} $12$. + +\Item{23.} $-102$. + +\Item{24.} $-41$. + +\Item{25.} $174$. + +\Item{26.} $6$. + +\Item{27.} $30$. + +\Item{28.} $372$. + + +\AnsTo[4]{Exercise}{17.} % Page 45. + +\Item{1.} $x^{2}$. + +\Item{2.} $3x^{2}$. + +\Item{3.} $-7$. + +\Item{4.} $-7$. + +\Item{5.} $7x^{4}$. + +\Item{6.} $9x$. + +\Item{7.} $-4a$. + +\Item{8.} $4x^{2}y^{2}$. + +\Item{9.} $-9x$. + +\Item{10.} $5x$. + +\Item{11.} $3y^{2}$. + +\Item{12.} $-4ab^{2}$. + +\Item{13.} $12xy^{4}$. + +\Item{14.} $\dfrac{3x^{3}y^{3}}{5}$. + +\Item{15.} $-\dfrac{3a}{2}$. + +\Item{16.} $-bd$. + +\Item{17.} $acd^{2}$. + +\Item{18.} $-\dfrac{2xy}{3}$. + +\Item{19.} $5ab$. + +\Item{20.} $4mn$. + +\Item{21.} $\dfrac{yz^{3}}{3}$. + +\Item{22.} $-17cd$. + +\Item{23.} $2n^{2}p$. + +\Item{24.} $\dfrac{3r^{2}}{p}$. + +\Item{25.} $-13agt$. + +\Item{26.} $\dfrac{1}{abc}$. + +\Item{27.} $\dfrac{3}{2xy^{2}z^{3}}$. + +\Item{28.} $\dfrac{2}{mnp}$. + +\Item{29.} $-\dfrac{a}{3b^{2}}$. + +\Item{30.} $-\dfrac{g}{3mt}$. + + +\AnsTo[2]{Exercise}{18.} % Page 48. + +\Item{1.} $2a^{2} + 2b^{2}$. + +\Item{2.} $9a^{2} - 2a + 6$. + +\Item{3.} $0$. + +\Item{4.} $4x + 4y + 4z$. + +\Item{5.} $2b + 2c$. + +\Item{6.} $4a + 4b + 4c$. + +\Item{7.} $3a^{2} + 5a - 2$. + +\Item{8.} $8ab + 3ac$. + +\Item{9.} $6x^{3}$. + +\Item{10.} $5x^{2} + 3xy - 2y^{2}$. + +\Item{11.} $a^{2} - 2b^{2}$. + +\Item{12.} $4a^{3} + 6a + 2$. + +\Item{13.} $3m^{3} + 7m^{2} + 2$. + +\Item{14.} $2x^{3} - 2x^{2} + 4x + y$. + +\Item{15.} $7x^{3} + 7x^{2} + 2$. + +\Item{16.} $a^{3} + 3a^{2}b - 5ab^{2} - b^{3}$. + +\Item{17.} $-a^{3} - a^{2}b - 2ab^{2} - 2b^{3}$. + +\Item{18.} $2x^{3} + 2x^{2}y - xy^{2} + 6y^{3}$. +%% -----File: 175.png---Folio 169------- + + +\AnsTo[2]{Exercise}{19.} % Page 50. + +\Item{1.} $a - b + c$. + +\Item{2.} $2a - 2b + 6c$. + +\Item{3.} $2a + 3y -8z$. + +\Item{4.} $x + 4y + 5z$. + +\Item{5.} $2ac + 2bc$. + +\Item{6.} $2ab - 3ac + 4bc$. + +\Item{7.} $x^{3} + 3x^{2} + 2x - 8$. + +\Item{8.} $x^{3} - 7x^{2} + 4x$. + +\Item{9.} $2b^{3} + 18abc - 15c^{3}$. + +\Item{10.} $2 - x - 2x^{2} + 2x^{3}$. + +\Item{11.} $-3b^{3} - 4c^{3} + 6abc$. + +\Item{12.} $-x^{4} - 2x^{3} + 4x^{2} - 7x + 5$. + +\Item{13.} $2x + x^{2} + x^{3} + x^{5}$. + +\Item{14.} $2b^{3} - 4a^{2}b + 2ab^{2}$. + +\Item{15.} $2b^{4} - 2a^{3}b^{3} - ab^{2}$. + +\Item{16.} $4x^{3} - 3x^{2}y - 4xy^{2} + 7y^{3}$. + + +\AnsTo{Exercise}{20.} % Page 52. + +\Item{1.} $c$. + +\Item{2.} $y - b$. + +\Item{3.} $x - 3y - 7c$. + +\Item{4.} $7a + 2b-2$. + +\Item{5.} $2a + x$. + +\Item{6.} $13x - 15y + 13z$. + +\Item{7.} $2a - 2b + 2c$. + +\Item{8.} $5a + b - 4c$. + +\Item{9.} $4x - 5y + 2z$. + +\Item{10.} $3a - b$. + +\Item{11.} $2x + 3y + z$. + +\Item{12.} $-8x + y$. + +\Item{13.} $6z - 2y-z$. +\ResetCols + +\Item{15.} $(a + c)x - (a - b)y + (a - c)z$. + +\Item{16.} $(2a + 5c)x - (3a + 4b)y - (6b + 7c)z$. + +\Item{17.} $(ac - an)x - (bm + cn)y + (a + 3c)z$. + +\Item{18.} $(mn - 1)x - (mn + 1)y + (mn + 1)z$. + + +\AnsTo[2]{Exercise}{21.} % Page 53. + +\Item{1.} $x^{2} + 7x$. + +\Item{2.} $8x^{2} - 12xy$. + +\Item{3.} $14xy - 21y^{2}$. + +\Item{4.} $2ax - 4a^{2}$. + +\Item{5.} $bx - 3b^{2}$. + +\Item{6.} $-6a^{3} + 9a^{2}b$. + +\Item{7.} $10x^{2}z + 15xz^{2}$. + +\Item{8.} $5a^{3}b - 25a^{2}b^{2}$. + +\Item{9.} $-x^{2}y^{2} + 3xy^{3}$. + +\Item{10.} $4x^{5} - 6x^{4}$. + +\Item{11.} $4x^{2}y - 12y^{3}$. + +\Item{12.} $-x^{4} + 3x^{2}y^{2}$. + +\Item{13.} $-a^{3}b^{3} + a^{5}b^{2}$. + +\Item{14.} $a^{4}b^{2} + a^{5}$. + +\Item{15.} $4x^{5} - 6x^{4} + 2x^{3}$. + +\Item{16.} $5a^{3}b - 25a^{2}b^{2} - 5ab^{3}$. + +\Item{17.} $a^{5} + 2a^{4}b + 2a^{3}b^{2}$. + +\Item{18.} $a^{3}b^{3} + 2a^{2}b^{4} + 2ab^{5}$. + +\Item{19.} $8x^{3} - 12x^{2}y - 18xy^{2}$. + +\Item{20.} $x^{2}y + 2xy^{2} - y^{3}$. + +\Item{21.} $a^{5} + a^{4}b^{2} + a^{2}b^{3}$. + +\Item{22.} $x^{2}y^{2} - 2xy^{3} + y^{4}$. + +\Item{23.} $15a^{4}b^{4} - 20a^{3}b^{5} + 5a^{5}b^{3}$. + +\Item{24.} $3a^{2}x^{2}y^{2} - 9a^{2}xy^{4} + 3a^{2}y^{6}$. + +\Item{25.} $x^{15}y^{2} - x^{13}y^{5} - x^{6}y^{12}$. + +\Item{26.} $4x^{5}y^{3} - 6x^{4}y^{5} + 4x^{3}y^{6}$. + +\Item{27.} $a^{10}x^{5}y^{10} - a^{9}x^{4}y^{9} - a^{8}x^{3}y^{8}$. + +\Item{28.} $15a^{4}b^{5} - 10a^{3}b^{6} + 25a^{5}b^{4}$. +%% -----File: 176.png---Folio 170------- + + +\AnsTo[2]{Exercise}{22.} % Page 56. + +\Item{1.} $x^{2} + 13x + 42$. + +\Item{2.} $x^{2} - x - 42$. + +\Item{3.} $x^{2} + x - 42$. + +\Item{4.} $x^{2} - 13x + 42$. + +\Item{5.} $x^{2} + 3x - 40$. + +\Item{6.} $4x^{2} + 12x + 9$. + +\Item{7.} $4x^{2} - 12x + 9$. + +\Item{8.} $4x^{2} - 9$. + +\Item{9.} $-9x^{2} + 12x - 4$. + +\Item{10.} $20x^{2} - 47x + 21$. + +\Item{11.} $a^{2} + ab - 6b^{2}$. + +\Item{12.} $a^{2} - 12ab + 35b^{2}$. + +\Item{13.} $25x^{2} - 30xy + 9y^{2}$. + +\Item{14.} $x^{2} - bx - cx + bc$. + +\Item{15.} $8m^{2} - 10mp + 3p^{2}$. + +\Item{16.} $a^{2} + ab - bc - c^{2}$. + +\Item{17.} $a^{4} - a^{3}b + 2a^{2}b^{2} - ab^{3} + b^{4}$. + +\Item{18.} $x^{5} - 3x^{4} - x^{3} + 16x^{2} - 21$. + +\Item{19.} $a^{3} - b^{3}$. + +\Item{20.} $a^{3} + b^{3}$. + +\Item{21.} $2x^{4} + 13x^{3} - 9x^{2} - 50x + 40$. + +\Item{22.} $9x^{5} - 7x^{3} + 6x^{2} - 2x$. + +\Item{23.} $x^{5} - 4x^{2}y^{3} + 3xy^{4}$. + +\Item{24.} $-a^{4} + 4a^{3}b - 7ab^{3} - 2b^{4}$. +\ResetCols[1] + +\Item{25.} $-25a^{5}b^{3} + 20a^{4}b^{4} + 12a^{3}b^{5} - 5a^{2}b^{6} - 2ab^{7}$. + +\Item{26.} $a^{4} - 2a^{2}b^{2} + b^{4}$. + +\Item{27.} $a^{2}b^{2} + 2abcd - a^{2}c^{2} + c^{2}d^{2}$. + +\Item{28.} $-2x^{5}y^{3} + x^{4}y^{4} + 10x^{3}y^{5} - 8x^{2}y^{6} - 3xy^{7}$. + +\Item{29.} $x^{4} - 4x^{2}y^{2} + 4xy^{3} - y^{4}$. + +\Item{30.} $3x^{4} - 5x^{3}y - 12x^{2}y^{2} - xy^{3} + 3y^{4}$. + +\Item{31.} $-a^{4} + 6a^{2}b^{2} - b^{4}$. + +\Item{32.} $a^{4} - a^{3}c + ab^{2}c - b^{4} - 2b^{2}c^{2} + ac^{3} - c^{4}$. + +\Item{33.} $a^{4} - 16a^{2}b^{2}x^{2} + 32a^{3}b^{3}x^{3} - 16a^{4}b^{4}x^{4}$. + +\Item{34.} $6a^{4} + 5a^{3}bx - 10a^{2}b^{2}x^{2} + 7ab^{3}x^{3} - 2b^{4}x^{4}$. + +\Item{35.} $10x^{6}y^{2} + 14x^{5}y^{3} - 48x^{4}y^{4} + 32x^{3}y^{5} - 16x^{2}y^{6}$. + + +\AnsTo{Exercise}{23.} % Page 58. + +\Item{1.} $2a^{2} - a$. + +\Item{2.} $7a^{4} - a$. + +\Item{3.} $7x^{2} + 1$. + +\Item{4.} $5m^{4} - p^{2}$. + +\Item{5.} $3x^{3} - 5x^{2}$. + +\Item{6.} $-3x^{3} + 1$. + +\Item{7.} $2x^{2} - 3x$. + +\Item{8.} $-x^{2} + 2$. + +\Item{9.} $a + 2c$. + +\Item{10.} $5x - y$. + +\Item{11.} $ax - 1$. + +\Item{12.} $x + xy$. + +\Item{13.} $-3a + 4b - 2c$. + +\Item{14.} $ab - b^{4} - a^{2}b$. + +\Item{15.} $x^{2} - 2xy - 3y^{2}$. + +\Item{16.} $xy - x^{2} - y^{2}$. + +\Item{17.} $-a^{2} + ab + b^{2}$. + +\Item{18.} $-a + 1 - b$. + +\Item{19.} $-1 + xy - x^{2}y^{2}$. + +\Item{20.} $x^{2} + 2x + 1$. + +\Item{21.} $a - b - c$. + +\Item{22.} $x^{3} - x^{2}y - y^{2}$. + +\Item{23.} $ab - 2 - 3b^{2}$. + +\Item{24.} $a^{2}c^{2} + a - c$. +%% -----File: 177.png---Folio 171------- + + +\AnsTo{Exercise}{24.} % Page 62. + +\Item{1.} $x + 8$. + +\Item{2.} $x - 8$. + +\Item{3.} $x + 8$. + +\Item{4.} $x - 8$. + +\Item{5.} $a + 5$. + +\Item{6.} $3a + 1$. + +\Item{7.} $a + 5$. + +\Item{8.} $-3a - 2$. + +\Item{9.} $x^{2} - x + 1$. + +\Item{10.} $x^{4} + x^{2} + 1$. + +\Item{11.} $1 + ab + a^{2}b^{2}$. + +\Item{12.} $x^{2} + 3x + 1$. + +\Item{13.} $a - b + c$. + +\Item{14.} $a + b - c$. + +\Item{15.} $x + y - z$. + +\Item{16.} $c^{2} + c + 2$. + +\Item{17.} $x - 2y - z$. + +\Item{18.} $x - a$. + +\Item{19.} $a - 2b + 3c$. + +\Item{20.} $a^{2} + 5a + 6$. + +\Item{21.} $q^{2} + 3q + 2$. + +\Item{22.} $9a^{2} + 6ab + 4b^{2}$. + +\Item{23.} $-65$. + +\Item{24.} $10$. + +\Item{25.} $7a - 45$. + +\Item{26.} $2x^{4}$. + + +\AnsTo[1]{Exercise}{25.} % Page 63. + +\Item{1.} $2a^{2}$. + +\Item{2.} $-3a^{4} + 2a^{3}b - 2ab^{3} + 4b^{4}$. + +\Item{3.} $x$. + +\Item{4.} $a^{4} + 2a^{2}b^{2} + b^{4} - c^{4} + 2c^{2}d^{2} - d^{4}$. + +\Item{5.} $10y^{4} + 8y^{3} + 6y^{2} + 4y + 2$. + +\Item{6.} $0$. + +\Item{7.} $2z - 7y$. + +\Item{8.} $a^{3} - 3abc + b^{3} + c^{3}$. + +\Item{9.} $4y^{2} - 3xy + 2x^{2}$. + +\Item{10.} $5a^{3}b - b^{4}$. + +\Item{11.} $3x^{3} - 2x^{2} + 1$. + +\Item{12.} $3c^{2} + 24c - 12$. + +\Item{13.} $2b^{4}$. + +\Item{14.} $10 - 16x - 39x^{2} + 2x^{3} + 15x^{4}$. + +\Item{15.} $a^{4} - ax^{3} + x^{4}$. + +\Item{16.} $(a - b)x^{3} + (b + c)x^{2} - (c + 1)x$. + +\Item{17.} $(a + b)x^{4} - (a - b)x^{3} - (c + 2)x$. + +\Item{18.} $(a + 1)x^{3} - (b + c)x^{2} + (b - c)x$. + + +\AnsTo[2]{Exercise}{26.} % Page 65. + +\Item{1.} $m^{2} + 2mn + n^{2}$. + +\Item{2.} $c^{2} - 2ac + a^{2}$. + +\Item{3.} $a^{2} + 4ac + 4c^{2}$ + +\Item{4.} $9a^{2} - 12ab + 4b^{2}$. + +\Item{5.} $4a^{2} + 12ab + 9b^{2}$. + +\Item{6.} $a^{2} - 6ab + 9b^{2}$. + +\Item{7.} $4x^{2} - 4xy + y^{2}$. + +\Item{8.} $y^{2} - 4xy + 4x^{2}$. + +\Item{9.} $a^{2} + 10ab + 25b^{2}$. + +\Item{10.} $4a2 - 20ac + 25c^{2}$. + +\Item{11.} $x^{2} - y^{2}$. + +\Item{12.} $16a^{2} - b^{2}$. + +\Item{13.} $4b^{2} - 9c^{2}$. + +\Item{14.} $x^{2} + 10bx + 25b^{2}$. + +\Item{15.} $y^{2} - 4yz + 4z^{2}$. + +\Item{16.} $y^{2} - 9z^{2}$. + +\Item{17.} $4a^{2} - 9b^{2}$. + +\Item{18.} $4a^{2} - 12ab + 9b^{2}$. + +\Item{19.} $4a^{2} + 12ab + 9b^{2}$. + +\Item{20.} $25x^{2} - 9a^{2}$. +%% -----File: 178.png---Folio 172------- + + +\AnsTo[2]{Exercise}{27.} % Page 67. + +\Item{1.} $x^{2} + 11x + 28$. + +\Item{2.} $x^{2} + 4x - 21$. + +\Item{3.} $x^{2} - 6x + 8$. + +\Item{4.} $x^{2} - 16x + 60$. + +\Item{5.} $x^{2} + 3x - 28$. + +\Item{6.} $x^{2} - ax - 2a^{2}$. + +\Item{7.} $x^{2} + 2ax - 3a^{2}$. + +\Item{8.} $a^{2} + 6ac + 9c^{2}$. + +\Item{9.} $a^{2} - 2ax - 8x^{2}$. + +\Item{10.} $a^{2} - 7ab + 12b^{2}$. + +\Item{11.} $a^{4} + a^{2}c - 2c^{2}$. + +\Item{12.} $x^{2} - 20x + 51$. + +\Item{13.} $x^{2} + xy - 30y^{2}$. + +\Item{14.} $9 + 3x - 2x^{2}$. + +\Item{15.} $5 - 8x - 4x^{2}$. + +\Item{16.} $a^{2} + ab - 6b^{2}$. + +\Item{17.} $a^{4}b^{4} - 6a^{2}b^{2}x^{2} + 5x^{4}$. + +\Item{18.} $a^{6}b^{2} + 4a^{4}b^{4} - 5a^{2}b^{6}$. + +\Item{19.} $x^{4}y^{2} - 4x^{3}y^{3} + 3x^{2}y^{4}$. + +\Item{20.} $x^{4}y^{2} + 2x^{3}y^{3} + x^{2}y^{4}$. + +\Item{21.} $x^{2} + (a + b)x + ab$. + +\Item{22.} $x^{2} + (a - b)x - ab$. + +\Item{23.} $x^{2} - (a - b)x - ab$. + +\Item{24.} $x^{2} - (a + b)x + ab$. + +\Item{25.} $x^{2} + (2a + 2b)x + 4ab$. + +\Item{26.} $x^{2} - (2a - 2b)x - 4ab$. + +\Item{27.} $x^{2} + (2a - 2b)x - 4ab$. + +\Item{28.} $x^{2} - (2a + 2b)x + 4ab$. + +\Item{29.} $x^{2} + 2ax - 3a^{2}$. + +\Item{30.} $x^{2} + ax - 6a^{2}$. + + +\AnsTo{Exercise}{28.} % Page 68. + +\Item{1.} $x + 2$. + +\Item{2.} $x - 2$. + +\Item{3.} $a + 3$. + +\Item{4.} $a - 3$. + +\Item{5.} $c + 5$. + +\Item{6.} $c - 5$. + +\Item{7.} $7x + y$. + +\Item{8.} $7x - y$. + +\Item{9.} $3b + 1$. + +\Item{10.} $3b - 1$. + +\Item{11.} $4x^{2} + 5a$. + +\Item{12.} $4x^{2} - 5a$. + +\Item{13.} $3x + 5y$. + +\Item{14.} $a + b - c$. + +\Item{15.} $a - b + c$. + +\Item{16.} $a + 2b - c$. + +\Item{17.} $5a - 7b + 1$. + +\Item{18.} $5a - 7b - 1$. + +\Item{19.} $z + x - y$. + +\Item{20.} $z - x + y$. + +\Item{21.} $a - 2b + c$. + +\Item{22.} $x + 3y + z$. + +\Item{23.} $x + 3y - z$. + +\Item{24.} $a + 2b + 2c$. + +\Item{25.} $a + 2b - 2c$. + +\Item{26.} $1 - 3x + 2y$. + + +\AnsTo[2]{Exercise}{29.} % Page 69. + +\Item{1.} $1 + x + x^{2}$. + +\Item{2.} $1 + 2a + 4a^{2}$. + +\Item{3.} $1 + 3c + 9c^{2}$. + +\Item{4.} $4a^{2} + 2ab + b^{2}$. + +\Item{5.} $16b^{2} + 12bc + 9c^{2}$. + +\Item{6.} $9x^{2} + 6xy + 4y^{2}$. + +\Item{7.} $x^{2}y^{2} + xyz + z^{2}$. + +\Item{8.} $a^{2}b^{2} + 2ab + 4$. + +\Item{9.} $25a^{2} + 5ab + b^{2}$. + +\Item{10.} $a^{2} + 2ab + 4b^{2}$. + +\Item{11.} $a^{2} + 4a + 16$. + +\Item{12.} $a^{6} + 3a^{3} + 9$. + +\Item{13.} $a^{8} + a^{4}x^{2}y^{2} + x^{4}y^{4}$. + +\Item{14.} $x^{10} + x^{5}a^{3}b^{3} + a^{6}b^{6}$. + +\Item{15.} $9x^{2}y^{2} + 3xyz^{4} + z^{8}$. + +\Item{16.} $x^{2}y^{2}z^{2} + xyz + 1$. + +\Item{17.} $4a^{2}b^{2}c^{2} - 6abc + 9$. + +\Item{18.} $1 + 4xyz + 16x^{2}y^{2}z^{2}$. +%% -----File: 179.png---Folio 173------- + + +\AnsTo[2]{Exercise}{30.} % Page 70. + +\Item{1.} $1 - x + x^{2}$. + +\Item{2.} $1 - 2a + 4a^{2}$. + +\Item{3.} $1 - 3c + 9c^{2}$. + +\Item{4.} $4a^{2} - 2ab + b^{2}$. + +\Item{5.} $16b^{2} - 12bc + 9c^{2}$. + +\Item{6.} $9x^{2} - 6xy + 4y^{2}$. + +\Item{7.} $4x^{2} - 10xy + 25y^{2}$. + +\Item{8.} $x^{2}y^{2} - xyz + z^{2}$. + +\Item{9.} $a^{2}b^{2} - 2ab + 4$. + +\Item{10.} $25a^{2} - 5ab + b^{2}$. + +\Item{11.} $a^{2} - 2ab + 4b^{2}$. + +\Item{12.} $a^{4} - 4a^{2} + 16$. + +\Item{13.} $a^{6} - 3a^{3} + 9$. + +\Item{14.} $4a^{4} - 2a^{2}b + b^{2}$. + +\Item{15.} $a^{8} - a^{4}x^{2}y^{2} + x^{4}y^{4}$. + +\Item{16.} $x^{10} - x^{5}a^{3}b^{3} + a^{6}b^{6}$. + +\Item{17.} $9x^{2}y^{2} - 3xyz^{4} + z^{8}$. + +\Item{18.} $x^{2}y^{2}z^{2} - xyz + 1$. + +\Item{19.} $4a^{2}b^{2}c^{2} - 6abc + 9$. + +\Item{20.} $1 - 4xyz + 16x^{2}y^{2}z^{2}$. + +\Item{21.} $1 - 3a^{2}bc + 9a^{4}b^{2}c^{2}$. + +\Item{22.} $x^{3} + x^{2}y + xy^{2} + y^{3}$. + +\Item{23.} $x^{3} - x^{2}y + xy^{2} - y^{3}$. + +\Item{24.} $x^{4} + x^{3}y + x^{2}y^{2} + xy^{3} + y^{4}$. + +\Item{25.} $x^{4} - x^{3}y + x^{2}y^{2} - xy^{3} + y^{4}$. + +\Item{26.} $x^{5} + x^{4}y + x^{3}y^{2} + x^{2}y^{3} + xy^{4} + y^{5}$. + +\Item{27.} $x^{5} - x^{4}y + x^{3}y^{2} - x^{2}y^{3} + xy^{4} - y^{5}$. + + +\AnsTo[2]{Exercise}{31.} % Page 72. + +\Item{1.} $2x(x - 2)$. + +\Item{2.} $3a(a^{2} - 2)$. + +\Item{3.} $5a^{2}b^{2}(1 - 2ab)$. + +\Item{4.} $xy(3x + 4y)$. + +\Item{5.} $4a^{2}b^{2}(2a + b)$. + +\Item{6.} $3a^{2}(a^{2} - 4 - 2a)$. + +\Item{7.} $4x^{2}(1 - 2x^{2} - 3x^{3})$. + +\Item{8.} $5(1 - 2x^{2}y^{2} + 3x^{2}y)$. + +\Item{9.} $7a(a + 2 - 3a^{2})$. + +\Item{10.} $3x^{2}y^{2}(xy - 2x^{2}y^{2} - 3)$. + + +\AnsTo{Exercise}{32.} % Page 73. + +\Item{1.} $(x^{2} + 1)(x + 1)$. + +\Item{2.} $(x^{2} + 1)(x - 1)$. + +\Item{3.} $(x + y)(x + z)$. + +\Item{4.} $(x - y)(a - b)$. + +\Item{5.} $(a + b)(a - c)$. + +\Item{6.} $(x + 3)(x - b)$. + +\Item{7.} $(x^{2} + 2)(2x - 1)$. + +\Item{8.} $(a - b)(a - 3)$. + +\Item{9.} $(2a - c)(3a + b)$. + +\Item{10.} $(xy + c)(ab + c)$. + +\Item{11.} $(a - b - c)(x - y)$. + +\Item{12.} $(a - b - 2c)(a - b)$. + + +\AnsTo[2]{Exercise}{33.} % Page 74. + +\Item{1.} $(2 + x)(2 - x)$. + +\Item{2.} $(3 + x)(3 - x)$. + +\Item{3.} $(3a + x)(3a - x)$. + +\Item{4.} $(5 + x)(5 - x)$. + +\Item{5.} $(5x + a)(5x - a)$. + +\Item{6.} $(4a^{2} + 11)(4a^{2} - 11)$. + +\Item{7.} $(11a^{2} + 4)(11a^{2} - 4)$. + +\Item{8.} $(2ab + cd)(2ab - cd)$. + +\Item{9.} $(1 + xy)(1 - xy)$. + +\Item{10.} $(9xy + 1)(9xy - 1)$. + +\Item{11.} $(7ab + 2)(7ab - 2)$. + +\Item{12.} $(5a^{2}b^{2} + 3)(5a^{2}b^{2} - 3)$. + +\Item{13.} $(3a^{4}b^{3} + 4x^{5})(3a^{4}b^{3} - 4x^{5})$. + +\Item{14.} $(12xy + 1)(12xy - 1)$. + +\Item{15.} $(10x^{3}yz^{2} + 1)(10x^{3}yz^{2} - 1)$. + +\Item{16.} $(1 + 11a^{2}b^{4}c^{6})(1 - 11a^{2}b^{4}c^{6})$. + +\Item{17.} $(5a + 8x^{3}y^{3})(5a - 8x^{3}y^{3})$. + +\Item{18.} $(4x^{8} + 5y^{9})(4x^{8} - 5y^{9})$. + +\Item{19.} $90,000$. + +\Item{20.} $22,760$. + +\Item{21.} $732,200$. + +\Item{22.} $400$. + +\Item{23.} $28,972$. + +\Item{24.} $14,248,000$. +%% -----File: 180.png---Folio 174------- + + +\AnsTo[2]{Exercise}{34.} % Page 75. + +\Item{1.} $(x + y + z)(x + y - z)$. + +\Item{2.} $(x - y + z)(x - y - z)$. + +\Item{3.} $(z + x + y)(z - x - y)$. + +\Item{4.} $(z + x - y)(z - x + y)$. + +\Item{5.} $(x + y + 2z)(x + y - 2z)$. + +\Item{6.} $(2z + x - y)(2z - x + y)$. + +\Item{7.} $(a + 2b + c)(a + 2b - c)$. + +\Item{8.} $(a - 2b + c)(a - 2b - c)$. + +\Item{9.} $(c + a - 2b)(c - a + 2b)$. + +\Item{10.} $(2a + 5c + 1)(2a + 5c - 1)$. + +\Item{11.} $(1 + 2a - 5c)(1 - 2a + 5c)$. + +\Item{12.} $(a + 3b + 4c)(a + 3b - 4c)$. + +\Item{13.} $(a - 5b + 3c)(a - 5b - 3c)$. + +\Item{14.} $(4c + a - 5b)(4c - a + 5b)$. + +\Item{15.} $(2a + x + y)(2a - x - y)$. + +\Item{16.} $(b + a - 2x)(b - a + 2x)$. + +\Item{17.} $(2z + x + 3y)(2z - x - 3y)$. + +\Item{18.} $(3 + 3a - 7b)(3 - 3a + 7b)$. + +\Item{19.} $(4a + 2b + 5c)(4a - 2b - 5c)$. + +\Item{20.} $(5c + 3a - 2x)(4c - 3a + 2x)$. + +\Item{21.} $(3a + 3b -5c)(3a - 3b + 5c)$. + +\Item{22.} $(4y + a - 3c)(4y - a + 3c)$. + +\Item{23.} $(7m + p + 2q)(7m - p - 2q)$. + +\Item{24.} $(6n + d - 2c)(6n - d + 2c)$. + +\Item{25.} $(x + y + a + b)(x + y - a - b)$. + +\Item{26.} $(x - y + a - b)(x - y - a + b)$. + +\Item{27.} $(2x + 3 + 2a + b)(2x + 3 - 2a - b)$. + +\Item{28.} $(b - c + a - 2x)(b - c - a + 2x)$. + +\Item{29.} $(3x - y + 2a - b)(3x - y - 2a + b)$. + +\Item{30.} $(x - 3y + a + 2b)(x - 3y - a - 2b)$. + +\Item{31.} $(x + 2y + a + 3b)(x + 2y - a - 3b)$. + +\Item{32.} $(x + y + a - z)(x + y - a + z)$. + + +\AnsTo[2]{Exercise}{35.} % Page 77. + +\Item{1.} $(2x - y)(4x^{2} + 2xy + y^{2})$. + +\Item{2.} $(x - 1)(x^{2} + x + 1)$. + +\Item{3.} $(xy - z)(x^{2}y^{2} + xyz + z^{2})$. + +\Item{4.} $(x - 4)(x^{2} + 4x + 16)$. + +\Item{5.} $(5a - b)(25a^{2} + 5ab + b^{2})$. + +\Item{6.} $(a - 7)(a^{2} + 7a + 49)$. + +\Item{7.} $(ab - 3c)(a^{2}b^{2} + 3abc + 9c^{2})$. + +\Item{8.} $(xyz - 2)(x^{2}y^{2}z^{2} + 2xyz + 4)$. + +\Item{9.} $(2ab - 3y^{2})(4a^{2}b^{2} + 6aby^{2} + 9y^{4})$. + +\Item{10.} $(4x - y^{3})(16x^{2} + 4xy^{3} + y^{6})$. + +\Item{11.} $(3a - 4c^{2})(9a^{2} + 12ac^{2} + 16c^{4})$. + +\Item{12.} $(xy - 6z)(x^{2}y^{2} + 6xyz + 36z^{2})$. + +\Item{13.} $(4x - 9y)(16x^{2} + 36xy + 81y^{2})$. + +\Item{14.} $(3a - 8c)(9a^{2} + 24ac + 64c^{2})$. + +\Item{15.} $(2x^{2} - 5y)(4x^{4} + 10x^{2}y + 25y^{2})$. + +\Item{16.} $(4x^{2} - 3y^{5})(16x^{8} + 12x^{4}y^{5} + 9y^{10})$. + +\Item{17.} $(6 - 2a)(36 + 12a + 4a^{2})$. + +\Item{18.} $(7 - 3y)(49 + 21y + 9y^{2})$. +%% -----File: 181.png---Folio 175------- + + +\AnsTo[2]{Exercise}{36.} % Page 78. + +\Item{1.} $(x + 1)(a^{2} - x + 1)$. + +\Item{2.} $(2x + y)(4x^{2} - 2xy + y^{2})$. + +\Item{3.} $(x + 5)(x^{2} - 5x + 25)$. + +\Item{4.} $(4a + 3)(16a^{2} - 12a + 9)$. + +\Item{5.} $(xy + z)(x^{2}y^{2} - xyz + z^{2})$. + +\Item{6.} $(a + 4)(a^{2} - 4a + 16)$. + +\Item{7.} $(2a^{2} + b)(4a^{4} - 2a^{2}b + b^{2})$. + +\Item{8.} $(x + 7)(x^{2} - 7x + 49)$. + +\Item{9.} $(2 + xyz)(4 - 2xyz + x^{2}y^{2}z^{2})$. + +\Item{10.} $(y^{3} + 4x)(y^{6} - 4y^{3}x + 16x^{2})$. + +\Item{11.} $(ab + 3x)(a^{2}b^{2} - 3abx + 9x^{2})$. + +\Item{12.} $(2yz + x^{2})(4y^{2}z^{2} - 2yzx^{2} + x^{4})$. + +\Item{13.} $(y^{3} + 4x^{2})(y^{6} - 4y^{3}x^{2} + 16x^{4})$. + +\Item{14.} $(4a^{4} + x^{5})(16a^{8} - 4a^{4}x^{5} + x^{10})$. + +\Item{15.} $(3x^{5} + 2a^{2})(9x^{10} - 6x^{5}a^{2} + 4a^{4})$. + +\Item{16.} $(3x^{3} + 8)(9x^{6} - 24x^{3} + 64)$. + +\Item{17.} $(7 + 4x)(49 - 28x + 16x^{2})$. + +\Item{18.} $(5 + 3y)(25 - 15y + 9y^{2})$. + + +\AnsTo{Exercise}{37.} % Page 80. + +\Item{1.} $(2x + y)(2x + y)$. + +\Item{2.} $(x + 3y)(x + 3y)$. + +\Item{3.} $(x + 8)(x + 8)$. + +\Item{4.} $(x + 5a)(x + 5a)$. + +\Item{5.} $(a - 8)(a - 8)$. + +\Item{6.} $(a - 5b)(a - 5b)$. + +\Item{7.} $(c - 3d)(c - 3d)$. + +\Item{8.} $(2x - 1)(2x - 1)$. + +\Item{9.} $(2a - 3b)(2a - 3b)$. + +\Item{10.} $(3a - 4b)(3a - 4b)$. + +\Item{11.} $(x + 4y)(x + 4y)$. + +\Item{12.} $(x - 4y)(x - 4y)$. + +\Item{13.} $(2x - 5y)(2x - 5y)$. + +\Item{14.} $(1 + 10a)(1 + 10a)$. + +\Item{15.} $(7a - 2)(7a - 2)$. + +\Item{16.} $(6a + 5b)(6a + 5b)$. + +\Item{17.} $(9x - 2b)(9x - 2b)$. + +\Item{18.} $(mn + 7x^{2})(mn + 7x^{2})$. + + +\AnsTo{Exercise}{38.} % Page 82. + +\Item{1.} $(a + 2)(a + 3)$. + +\Item{2.} $(a - 2)(a - 3)$. + +\Item{3.} $(a + 1)(a + 5)$. + +\Item{4.} $(a - 1)(a - 5)$. + +\Item{5.} $(a - 1)(a + 5)$. + +\Item{6.} $(a + 1)(a - 5)$. + +\Item{7.} $(c - 3)(c - 6)$. + +\Item{8.} $(c + 3)(c + 6)$. + +\Item{9.} $(c - 3)(c + 6)$. + +\Item{10.} $(c + 3)(c - 6)$. + +\Item{11.} $(x + 2)(x + 7)$. + +\Item{12.} $(x - 2)(x - 7)$. + +\Item{13.} $(x + 2)(x - 7)$. + +\Item{14.} $(x - 4)(x - 5)$. + +\Item{15.} $(x + 4)(x - 5)$. + +\Item{16.} $(x - 4)(x + 5)$. + +\Item{17.} $(x - 3)(x - 7)$. + +\Item{18.} $(x + 3)(x - 7)$. + +\Item{19.} $(x - 3)(x + 7)$. + +\Item{20.} $(x - 7)(x - 8)$. + +\Item{21.} $(x + 7)(x - 8)$. + +\Item{22.} $(x - 1)(x - 9)$. + +\Item{23.} $(x + 3)(x + 10)$. + +\Item{24.} $(x - 3)(x + 10)$. + +\Item{25.} $(x + 3)(x - 10)$. + +\Item{26.} $(a - 2b)(a + 3b)$. + +\Item{27.} $(a + 2b)(a - 3b)$. + +\Item{28.} $(a - b)(a + 4b)$. + +\Item{29.} $(a + b)(a - 4b)$. + +\Item{30.} $(ax + 7)(ax - 9)$. + +\Item{31.} $(a - 7x)(a + 9x)$. + +\ResetCols[2] +\Item{32.} $(a - 4b)(a - 5b)$. + +\Item{33.} $(xy - 3z)(xy - 16z)$. + +\Item{34.} $(ab + 4c)(ab + 11c)$. + +\Item{35.} $(x - 4y)(x - 9y)$. + +\Item{36.} $(x + 7y)(x + 12y)$. + +\Item{37.} $(ax - 6y)(ax - 17y)$. + +\Item{38.} $(x + 2y)(x - 2y)(x^{2} - 5y^{2})$. + +\Item{39.} $(a^{2}x^{2} - 11y^{2})(a^{2}x^{2} - 13y^{2})$. + +\Item{40.} $(a^{3}b^{3} - 11c^{2})(a^{3}b^{3} - 12c^{2})$. + +\Item{41.} $(a + 4bc)(a - 24bc)$. + +\Item{42.} $(a + 8bc)(a - 12bc)$. + +\Item{43.} $(a + 6bc)(a - 16bc)$. + +\Item{44.} $(a - 3bc)(a + 32bc)$. + +\Item{45.} $(a + 2bc)(a - 48bc)$. + +\Item{46.} $(a + bc)(a + 48bc)$. + +\Item{47.} $(x + 9yz)(x - 27yz)$. + +\Item{48.} $(xy + 13z)(xy - 14z)$. +%% -----File: 182.png---Folio 176------- + + +\AnsTo[2]{Exercise}{39.} % Page 83. + +\Item{1.} $a(a^{2} - 7)$. + +\Item{2.} $ab(3ab - 2a^{2} + 3b^{2})$. + +\Item{3.} $(a - b + 1)(a - b)$. + +\Item{4.} $(a + b + 1)(a + b - 1)$. + +\Item{5.} $(a + 2b)(a^{2} - 2ab + 4b^{2})$. + +\Item{6.} $(x + 2y + 1)(x - 2y)$. + +\Item{7.} $(a - b)(a^{2} + ab + b^{2} + 1)$. + +\Item{8.} $(a - 3b)(a - 3b)$. + +\Item{9.} $(x + 1)(x - 2)$. + +\Item{10.} $(x + 1)(x - 3)$. + +\Item{11.} $(x - 3)(x + 7)$. + +\Item{12.} $(a + 2)(a - 13)$. + +\Item{13.} $(x^{2} + 3)(a + b)$. + +\Item{14.} $(x - 3)(x - y)$. + +\Item{15.} $(x - 3)(x - 4)$. + +\Item{16.} $(a + 2b)(a + 3b)$. + +\Item{17.} $(x^{2} + 5)(x^{2} + 5)$. + +\Item{18.} $(x - 9)(x - 9)$. + +\Item{19.} $(x - 10)(x - 11)$. + +\Item{20.} $(x + 8)(x + 11)$. + +\Item{21.} $(x - 8)(x + 11)$. + +\Item{22.} $(x^{2} + 1)(x - 1)$. + +\Item{23.} $x^{2}(3x + 1)(3x - 1)$. + +\Item{24.} $(1 + a - b)(1 - a + b)$. + +\Item{25.} $(a + b)(a^{2} - ab + b^{2} + 1)$. + +\Item{26.} $(m + n)(m - n)(x + y)$. + +\Item{27.} $(x - y + z)(x - y - z)$. + +\Item{28.} $(z + x - y)(z - x + y)$. + +\Item{29.} $(2a^{2} + 3a - 1)(2a^{2} - 3a + 1)$. + +\Item{30.} $(2x - y)(4x^{2} + 2xy + y^{2})$. + +\Item{31.} $x^{2}(x - 3y)$. + +\Item{32.} $(x - 3y)(x^{2} + 3xy + 9y^{2})$. + +\Item{33.} $(x - 5)(x + 8)$. + +\Item{34.} $(x - 2y)(x + 5y)$. + +\Item{35.} $(1 + 4x)(1 - 4x)$. + +\Item{36.} $a^{2}(a^{2} + 3b^{2})(a^{2} - 3b^{2})$. + +\Item{37.} $x(x + y)(x + 2y)$. + +\Item{38.} $x^{2}(x + y)(x + 3y)$. + +\Item{39.} $(x - 2y^{2})(x - 2y^{2})$. + +\Item{40.} $(4x^{2} + 1)(4x^{2} + 1)$. + +\Item{41.} $a^{2}(3a + 2c)(3a - 2c)$. + +\Item{42.} $ab(a + b)(a - 2b)$. + +\Item{43.} $(x + 2)(x^{2} - 2x + 4)(x - 1)$. + +\Item{44.} $(a + y)(a^{2} - ay + y^{2})(a - x)$. + + +\AnsTo{Exercise}{40.} % Page 86. + +\Item{1.} $6$. + +\Item{2.} $5x^{3}$. + +\Item{3.} $6ax$. + +\Item{4.} $7ab^{2}$. + +\Item{5.} $7$. + +\Item{6.} $2a^{2}b^{2}$. + +\Item{7.} $x + 3y$. + +\Item{8.} $x + 3$. + +\Item{9.} $2a + 1$. + +\Item{10.} $x + y$. + +\Item{11.} $a + x$. + +\Item{12.} $a + 2b$. + +\Item{13.} $x - 1$. + +\Item{14.} $x + 3$. + +\Item{15.} $x - 6$. + +\Item{16.} $x^{2} - x + 1$. + +\Item{17.} $x - 1$. + +\Item{18.} $x - y$. + +\Item{19.} $x - 5$. + +\Item{20.} $a - b - c$. + +\Item{21.} $x + 2y$. + +\Item{22.} $x + 4y$. + +\Item{23.} $x^{2} + 2xy + 4y^{2}$. + +\Item{24.} $x - 2$. + +\Item{25.} $1 - 3a$. + +\Item{26.} $x - 7y$. + +\Item{27.} $2a + b$. + +\Item{28.} $x + y - z$. + + +\AnsTo[2]{Exercise}{41.} % Page 88. + +\Item{1.} $18x^{2}y^{3}$. + +\Item{2.} $6a^{2}bc^{3}$. + +\Item{3.} $20a^{3}b^{3}$. + +\Item{4.} $30a^{3}b^{4}$. + +\Item{5.} $189x^{3}y^{5}$. + +\Item{6.} $x^{2}y^{3}z^{3}$. + +\Item{7.} $a^{2}(a + 1)$. + +\Item{8.} $x^{2}(x - 3)$. + +\Item{9.} $x(x + 1)(x - 1)$. + +\Item{10.} $x(x + 1)(x - 1)$. + +\Item{11.} $xy(x + y)$. + +\Item{12.} $x(x + 2)^{2}$. +%% -----File: 183.png---Folio 177------- + +\Item{13.} $(a + 2)(a + 2)(a + 3)$. + +\Item{14.} $(c - 4)(c + 5)(c - 6)$. + +\Item{15.} $(b - 5)(b - 6)(b + 7)$. + +\Item{16.} $(y - 4)(y + 5)(y - 6)$. + +\Item{17.} $(z - 5)(z - 6)(z + 7)$. + +\Item{18.} $(x - 4)(x + 8)(x - 8)(x^{2} + 4x + 16)$. + +\Item{19.} $(a + b)(a + b)(a - b)(a - b)$. + +\Item{20.} $4a^{2}b(a + b)(a + b)(a - b)$. + +\Item{21.} $(y + 2)(y + 3)(y + 4)$. + +\Item{22.} $(x + 1)(x - 1)(x^{2} + 1)$. + +\Item{23.} $(1 + x)(1 - x)(1 + x + x^{2})$. + +\Item{24.} $(x + y)(x + y)(x - y)(x - y)$. + +\Item{25.} $x(x - 3)(x + 5)(x^{2} + 3x + 9)$. + +\Item{26.} $(a + b + c)(a + b + c)(a + b - c)$. + +\Item{27.} $(x - a)(x - b)(x - c)$. + +\Item{28.} $a(a + b + c)(a + b - c)$. + + +\AnsTo[4]{Exercise}{42.} % Page 90. + +\Item{1.} $\dfrac{1}{3b}$. + +\Item{2.} $\dfrac{4m}{5n}$. + +\Item{3.} $\dfrac{3m}{4p^{2}}$. + +\Item{4.} $\dfrac{x^{2}}{2yz}$. + +\Item{5.} $\dfrac{a^{3}b^{3}}{3c^{2}}$. + +\Item{6.} $\dfrac{2xy}{3}$. + +\Item{7.} $\dfrac{2m}{3p}$. + +\Item{8.} $\dfrac{3b^{2}c}{4a^{3}}$. + +\Item{9.} $\dfrac{2y^{2}z^{4}}{3}$. + +\Item{10.} $\dfrac{b}{c}$. + +\Item{11.} $\dfrac{2a - 3b}{2a}$. + +\Item{12.} $\dfrac{3a}{a + 2}$. + +\Item{13.} $\dfrac{x}{x - 1}$. + +\Item{14.} $\dfrac{y}{x^{2} + 3xy + 9y^{2}}$. + +\Item{15.} $\dfrac{x + 1}{x - 5}$. + +\Item{16.} $\dfrac{x + 1}{x - 2}$. + +\Item{17.} $\dfrac{a + b + c}{a}$. + +\Item{18.} $\dfrac{x + 5}{x + 3}$. + +\Item{19.} $\dfrac{x + 1}{x + 3}$. + + +\AnsTo[2]{Exercise}{43.} % Page 91. + +\Item{1.} $a + b + \dfrac{2}{a - b}$. + +\Item{2.} $a - b - \dfrac{2}{a + b}$. + +\Item{3.} $a - 1 + \dfrac{2a}{a^{2} - a - 1}$. + +\Item{4.} $2x - 4 + \dfrac{5}{x + 1}$. + +\Item{5.} $4x^{2} - 2x + 1 - \dfrac{1}{2x + 1}$. + +\Item{6.} $5x + 4 + \dfrac{x + 7}{x^{2} + x - 1}$. + +\Item{7.} $a + \dfrac{5a - 2}{a^{2} + a + 2}$. + +\Item{8.} $y^{2} - yx + x^{2}$. + +\Item{9.} $x^{2} - 4x + 3 + \dfrac{2x - 4}{x^{2} + x + 1}$. + +\Item{10.} $x^{3} + x + 1 + \dfrac{2x + 2}{x^{2} - x - 1}$. +%% -----File: 184.png---Folio 178------- + + +\AnsTo[2]{Exercise}{44.} % Page 92. + +\Item{1.} $\dfrac{x^{2} + y^{2}}{x - y}$. + +\Item{2.} $\dfrac{x^{2} + y^{2}}{x + y}$. + +\Item{3.} $\dfrac{2y}{x + y}$. + +\Item{4.} $-\dfrac{2ax}{a - x}$. + +\Item{5.} $-\dfrac{x + 2}{x - 3}$. + +\Item{6.} $-\dfrac{2x^{2} - 6x + 5}{x - 2}$. + +\Item{7.} $\dfrac{x^{3} + x^{2} - 2x + 1}{x + 2}$. + +\Item{8.} $\dfrac{2a^{2} - 11a + 6}{a - 3}$. + +\Item{9.} $\dfrac{-2a^{3} + a^{2} + 2a-3}{a - 1}$. + +\Item{10.} $\dfrac{3a^{4} + 6a^{3} - 14a^{2} - 4}{3a^{2} + 1}$. + + +\AnsTo[1]{Exercise}{45.} % Page 94. + +\Item{1.} $\dfrac{x(x + a)}{(x + a)(x - a)}$, $\dfrac{x^{2}}{(x + a)(x - a)}$. + +\Item{2.} $\dfrac{a(a - b)}{(a + b)(a - b)}$, $\dfrac{a^{2}}{(a + b)(a - b)}$. + +\Item{3.} $\dfrac{1 - 2a}{(1 + 2a)(1 - 2a)}$, $\dfrac{1}{(1 + 2a)(1 - 2a)}$. + +\Item{4.} $\dfrac{9}{(4 + x)(4 - x)}$, $\dfrac{(4 - x)^{2}}{(4 + x)(4 - x)}$. + +\Item{5.} $\dfrac{a^{2}}{(3 - a)(9 + 3a + a^{2})}$, $\dfrac{a(9 + 3a + a^{2})}{(3 - a)(9 + 3a + a^{2})}$. + +\Item{6.} $\dfrac{x + 2}{(x + 2)(x - 2)(x - 3)}$, $\dfrac{x - 2}{(x + 2)(x - 2)(x - 3)}$. + + +\AnsTo{Exercise}{46.} % Page 95. + +\Item{1.} $\dfrac{4x + 2}{5}$. + +\Item{2.} $\dfrac{13x + 3}{12}$. + +\Item{3.} $\dfrac{5(9x - 13)}{42}$. + +\Item{4.} $\dfrac{51x + 31}{36}$. + +\Item{5.} $\dfrac{x - 5}{3}$. + +\Item{6.} $\dfrac{5(x - y)}{8x}$. + +\Item{7.} $\dfrac{22x - 97}{30}$. + +\Item{8.} $\dfrac{3x - 4}{15x}$. + +\Item{9.} $\dfrac{a^{3} - b^{3} + c^{3} - abc}{abc}$. + + +\AnsTo{Exercise}{47.} % Page 96}. + +\Item{1.} $\dfrac{2x + 1}{(x + 3)(x - 2)}$. + +\Item{2.} $\dfrac{2x}{x^{2} - 1}$. + +\Item{3.} $\dfrac{3x + 16}{(x - 8)(x + 2)}$. + +\Item{4.} $\dfrac{4ax}{a^{3} - x^{2}}$. + +\Item{5.} $\dfrac{ax}{x^{2} - a^{2}}$. + +\Item{6.} $-\dfrac{4ab}{4a^{2} - b^{2}}$. + +\Item{7.} $\dfrac{1}{9 - a^{2}}$. + +\Item{8.} $\dfrac{b}{a^{2} - b^{2}}$. + +\Item{9.} $\dfrac{5x + 8}{x^{2} - 4}$. + +\Item{10.} $\dfrac{1 + x}{1 - 9x^{2}}$. + +\Item{11.} $\dfrac{3(a^{2} + 4a + 1)}{a(a + 1)(a + 3)}$. + +\Item{12.} $\dfrac{2}{x^{2} - 1}$. + +\Item{13.} $\dfrac{2x^{2}}{(x + 2)(x - 3)}$. + +\Item{14.} $0$. +%% -----File: 185.png---Folio 179------- + + +\AnsTo{Exercise}{48.} % Page 97. + +\Item{1.} $0$. + +\Item{2.} $\dfrac{2a}{a + b}$. + +\Item{3.} $\dfrac{7a}{1 - a^{2}}$. + +\Item{4.} $\dfrac{x - 10y}{4x^{2} - 25y^{2}}$. + +\Item{5.} $\dfrac{2}{x + 4y}$. + +\Item{6.} $\dfrac{2(x + 6)}{4x^{2} - 9}$. + + +\AnsTo{Exercise}{49.} % Page 100. + +\Item{1.} $\dfrac{20}{3bc}$. + +\Item{2.} $\dfrac{2ay}{3}$. + +\Item{3.} $\dfrac{7p^{2}}{4xz}$. + +\Item{4.} $\dfrac{2a^{2}cm}{7}$. + +\Item{5.} $\dfrac{30}{abc}$. + +\Item{6.} $abc$. + +\Item{7.} $b^{2}$. + +\Item{8.} $\dfrac{x + a}{x - 2a}$. + +\Item{9.} $\dfrac{xy}{2c - 1}$. + +\Item{10.} $\dfrac{a + 10}{a + 3}$. + +\Item{11.} $\dfrac{3x + 2y}{x - 2}$. + +\Item{12.} $\dfrac{5a + b}{4a + 3b}$. + +\Item{13.} $\dfrac{x - 7}{a + b + c}$. + +\Item{14.} $\dfrac{x(x + 1)}{x - 5}$. + +\Item{15.} $\dfrac{a + 1}{a + 5}$. + +\Item{16.} $1$. + +\Item{17.} $\dfrac{x(x + y)}{x + 1}$. + +\Item{18.} $\dfrac{b}{a - b}$. + +\Item{19.} $abc$. + +\Item{20.} $\dfrac{(x + 2y)(x + 1)}{(x + y)(x + 2)}$. + + +\AnsTo{Exercise}{50.} % Page 102. + +\Item{1.} $\dfrac{x + y}{z}$. + +\Item{2.} $\dfrac{12x + 3y}{12x - 4y}$. + +\Item{3.} $\dfrac{abd - 21d^{2}}{21cd - 7ab}$. + +\Item{4.} $\dfrac{x^{2} + x - 2}{x^{2} - x - 2}$. + +\Item{5.} $1$. + +\Item{6.} $\dfrac{x + y}{x^{2} - 2xy + y^{2}}$. + +\Item{7.} $\dfrac{a + b}{a - b}$. + +\Item{8.} $4(3a + 8)$. + +\Item{9.} $\dfrac{y + x}{y - x}$. + +\Item{10.} $x$. + +\Item{11.} $\dfrac{1}{x}$. + +\Item{12.} $\dfrac{x^{2}(x - 3)}{x - 2}$. + +\Item{13.} $a - 1$. + +\Item{14.} $\dfrac{4a}{a - x}$. + + +\AnsTo[5]{Exercise}{51.} % Page 105. + +\Item{1.} $5$. + +\Item{2.} $7$. + +\Item{3.} $2\frac{1}{2}$. + +\Item{4.} $120$. + +\Item{5.} $12$. + +\Item{6.} $2\frac{1}{3}$. + +\Item{7.} $17$. + +\Item{8.} $4$. + +\Item{9.} $4$. + +\Item{10.} $1$. + +\Item{11.} $-16$. + +\Item{12.} $11$. + +\Item{13.} $-4$. + +\Item{14.} $-2$. + +\Item{15.} $-2$. + +\Item{16.} $5$. + +\Item{17.} $9$. + +\Item{18.} $-1$. +%% -----File: 186.png---Folio 180------- + + +\AnsTo[5]{Exercise}{52.} % Page 106. + +\Item{1.} $2$. + +\Item{2.} $2$. + +\Item{3.} $-33$. + +\Item{4.} $1$. + +\Item{5.} $\frac{2}{3}$. + +\Item{6.} $1\frac{1}{2}$. + +\Item{7.} $2$. + +\Item{8.} $8$. + +\Item{9.} $5$. + +\Item{10.} $\frac{3}{7}$. + +\Item{11.} $2$. + +\Item{12.} $1$. + +\Item{13.} $3$. + + +\AnsTo[6]{Exercise}{53.} % Page 107. + +\Item{1.} $33$. + +\Item{2.} $2$. + +\Item{3.} $3\frac{1}{2}$. + +\Item{4.} $1\frac{5}{37}$. + +\Item{5.} $7$. + +\Item{6.} $3$. + + +\AnsTo{Exercise}{54.} % Page 108. + +\Item{1.} $a + b$. + +\Item{2.} $\dfrac{a}{2}$. + +\Item{3.} $\dfrac{b}{2}$. + +\Item{4.} $2a$. + +\Item{5.} $b - a$. + +\Item{6.} $\dfrac{ab}{a + b + c}$. + +\Item{7.} $\dfrac{a^{2} - b^{2}}{2a}$. + +\Item{8.} $\dfrac{2b}{a}$. + +\Item{9.} $\dfrac{2b^{2} - a^{2}}{4b - 3a}$. + +\Item{10.} $1$. + +\Item{11.} $3(a - b)$. + + +\AnsTo[4]{Exercise}{55.} % Page 109. + +\Item{1.} $35$. + +\Item{2.} $70$. + +\Item{3.} $36$. + +\Item{4.} $57$, $58$. + + +\AnsTo[5]{Exercise}{56.} % Page 110. + +\Item{1.} $81$, $19$. + +\Item{2.} $100$, $24$. + +\Item{3.} $64$, $15$. + +\Item{4.} $103$, $12$. + +\Item{5.} $295$, $25$. + + +\AnsTo[4]{Exercise}{57.} % Page 111. + +\Item{1.} $12$~yr. + +\Item{2.} A, $60$~yr.; B, $10$~yr. + +\Item{3.} A, $25$~yr.; B, $5$~yr. + +\Item{4.} $17\frac{1}{2}$~yr. + +\Item{5.} $35$~yr. + +\Item{6.} Son, $12$~yr.; father, $36$~yr. + +\Item{7.} $25$~yr. + +\Item{8.} A, $30$~yr.; B, $15$~yr. + +\Item{9.} Son, $12$~yr.; father, $68$~yr. + + +\AnsTo[6]{Exercise}{58.} % Page 112. + +\Item{1.} $1\frac{3}{7}$~dy. + +\Item{2.} $1\frac{13}{47}$~dy. + +\Item{3.} $1\frac{1}{20}$~dy. + +\Item{4.} $15$~dy. + +\Item{5.} $12$~hr. + +\Item{6.} $10$~dy. + + +\AnsTo[5]{Exercise}{59.} % Page 113. + +\Item{1.} $7\frac{5}{13}$~hr. + +\Item{2.} $2\frac{2}{5}$~hr. + +\Item{3.} $\frac{10}{11}$~hr. + +\Item{4.} $1\frac{1}{13}$~hr. + +\Item{5.} $30$~hr. + + +\AnsTo[4]{Exercise}{60.} % Page 114. + +\Item{1.} $36$~mi. + +\Item{2.} $26$~hr. + +\Item{3.} $8$~mi. + +\Item{4.} $240$~mi. + + +\AnsTo{Exercise}{61.} % Page 115. + +\Item{1.} $600$. + +\Item{2.} $700$. + +\Item{3.} Dog, $1440$; rabbit, $1800$. +%% -----File: 187.png---Folio 181------- + +\AnsTo[2]{Exercise}{62.} % Page 116. + +\Item{1.} $27\frac{3}{11}$~min.\ past 5~o'clock. + +\Item{2.} $27\frac{3}{11}$~min.\ past 2~o'clock. + +\Item{3.} $43\frac{7}{11}$~min.\ past 2~o'clock. + +\Item{4.} $21\frac{9}{11}$~min.\ past 1~o'clock. + +\Item{5.} $38\frac{2}{11}$~min.\ past 1~o'clock. + +\Item{6.} $38\frac{2}{11}$~min.\ past 7~o'clock. + + + +\AnsTo{Exercise}{63.} % Page 117. + +\Item{1.} $1764$~sq.~ft. + +\Item{2.} $18$~ft.\ by $23$~ft. + +\Item{3.} $14$~ft.\ by $20$~ft. + +\Item{4.} $12$~ft.\ by $15$~ft. + +\Item{5.} $30$~ft.\ by $40$~ft. + + +\AnsTo[4]{Exercise}{64.} % Page 121. + +\Item{1.} $90°\,0'\,30''$; $30°\,30'$. + +\Item{2.} \$$133\frac{1}{3}$. + +\Item{3.} \$$2000$. + +\Item{4.} \$$4000$. + +\Item{5.} \$$3000$. + +\Item{6.} \$$500$. + +\Item{7.} $5$\%. + +\Item{8.} $6$\%. + +\Item{9.} $3$~yr. + +\Item{10.} $9\frac{3}{8}$~yr. + +\Item{11.} \$$25,000$. + +\Item{12.} \$$20,000$. + +\Item{13.} \$$6$\%. + +\Item{14.} $20$~yr. + + +\AnsTo{Exercise}{65.} % Page 124. + +\Item{1.} $x = 2$, $y = 1$. + +\Item{2.} $x = 3$, $y = 2$. + +\Item{3.} $x = 5$, $y = 1$. + +\Item{4.} $x = 2$, $y = 1$. + +\Item{5.} $x = 1$, $y = 2$. + +\Item{6.} $x = 6$, $y = 1$. + +\Item{7.} $x = 3$, $y = 21$. + +\Item{8.} $x = 7$, $y = 7$. + +\Item{9.} $x = 23$, $y = -1$. + +\Item{10.} $x = 2$, $y = 1$. + +\Item{11.} $x = 35$, $y = 20$. + +\Item{12.} $x = 2$, $y = 1$. + +\Item{13.} $x = 1$, $y = 2$. + +\Item{14.} $x = 3$, $y = 2$. + +\Item{15.} $x = 1$, $y = 2$. + +\Item{16.} $x = 4$, $y = 3$. + +\Item{17.} $x = 12$, $y = 4$. + +\Item{18.} $x = 12$, $y = 21$. + +\Item{19.} $x = 5$, $y = 7$. + +\Item{20.} $x = 5$, $y = 2$. + +\Item{21.} $x = 18$, $y = 6$. + +\Item{22.} $x = 3$, $y = 2$. + +\Item{23.} $x = 3$, $y = 2$. + +\Item{24.} $x = 7$, $y = 8$. + +\Item{25.} $x = 8$, $y = -2$. + +\Item{26.} $x = \dfrac{a}{(a - b)}$, $y = \dfrac{b}{(a + b)}$. + + +\AnsTo[2]{Exercise}{66.} % Page 127. + +\Item{1.} A, \$$520$; B, \$$440$. + +\Item{2.} $23$~and~$17$. + +\Item{3.} $20$~and~$16$. + +\Item{4.} Velvet, \$$6$; silk, \$$3$. + +\Item{5.} Wheat, \$$1$; rye, \$$\frac{4}{5}$. + +\Item{6.} Tea, \$$\frac{1}{2}$; coffee, \$$\frac{1}{4}$. + +\Item{7.} Horse, \$$92$; cow, \$$64$. +%% -----File: 188.png---Folio 182------- + + +\AnsTo[5]{Exercise}{67.} % Page 128. + +\Item{1.} $\frac{3}{7}$. + +\Item{2.} $\frac{13}{25}$. + +\Item{3.} $\frac{3}{20}$. + +\Item{4.} $\frac{5}{21}$. + +\Item{5.} $\frac{7}{22}$. + + +\AnsTo[4]{Exercise}{68.} % Page 129. + +\Item{1.} $45$. + +\Item{2.} $72$. + +\Item{3.} $75$~and~$57$. + +\Item{4.} $54$. + + +\AnsTo{Exercise}{69.} % Page 130. + +\Item{1.} \$$2500$ at $4$\%. + +\Item{2.} \$$1600$ at $6$\%. + +\Item{3.} \$$6000$ at $4$\%; \$$4000$ at $5$\%. + + +\AnsTo[4]{Exercise}{70.} % Page 131. + +\Item{1.} $22$~and~$18$. + +\Item{2.} $60$~and~$8$. + +\Item{3.} $\frac{14}{17}$. + +\Item{4.} Wheat, \$$1$; barley, \$$\frac{4}{5}$. + +\Item{5.} A, \$$235$; B, \$$65$. + +\Item{6.} A, \$$70$; B, \$$30$. + +\Item{7.} Lemon, $2$~cts.; orange, $3$~cts. + +\Item{8.} A, $30$~apples; B, $10$~apples. + + +\AnsTo[4]{Exercise}{71.} % Page 134. + +\Item{1.} $±2$. + +\Item{2.} $±3$. + +\Item{3.} $±5$. + +\Item{4.} $±8$. + +\Item{5.} $±7$. + +\Item{6.} $±5$. + +\Item{7.} $±5$. + +\Item{8.} $±3$. + +\Item{9.} $±3$. + +\Item{10.} $±3$. + +\Item{11.} $12$~and~$16$. + +\Item{12.} $12$~oranges at $3$~cts. + +\Item{13.} $3$~rods. + +\Item{14.} Width, $12$~rd.; length, $48$~rd. + + +\AnsTo{Exercise}{72.} % Page 137. + +\Item{1.} $9$~or~$3$. + +\Item{2.} $4$~or~$2$. + +\Item{3.} $3$~or~$1$. + +\Item{4.} $1$~or~$-\frac{1}{5}$. + +\Item{5.} $1$~or~$-3$. + +\Item{6.} $\frac{4}{3}$~or~$\frac{4}{3}$. + +\Item{7.} $1$~or~$-\frac{1}{6}$. + +\Item{8.} $3$~or~$-1$. + +\Item{9.} $\frac{3}{4}$~or~$\frac{1}{4}$. + +\Item{10.} $3$~or~$\frac{1}{3}$. + +\Item{11.} $17$~or~$-3$. + +\Item{12.} $25$~or~$9$. + +\Item{13.} $4$~or~$-5$. + +\Item{14.} $4$~or~$-3$. + +\Item{15.} $5$~or~$1$. + +\Item{16.} $2$~or~$-6$. + +\Item{17.} $2$~or~$2$. + +\Item{18.} $5$~or~$-11$. + +\Item{19.} $2$~or~$-5\frac{1}{3}$. + +\Item{20.} $4\frac{1}{3}$~or~$-3\frac{2}{3}$. + +\Item{21.} $2$~or~$\frac{1}{3}$. + +\Item{22.} $4$~or~$-\frac{2}{5}$. + +\Item{23.} $2$~or~$-3$. + +\Item{24.} $10$~or~$2$. + +\Item{25.} $3$~or~$-2\frac{1}{3}$. + +\Item{26.} $2$~or~$2$. + +\Item{27.} $\frac{1}{2}$~or~$-3$. + +\Item{28.} $5$~or~$\frac{1}{2}$. + +\Item{29.} $7$~or~$2$. + +\Item{30.} $4$~or~$-\frac{2}{3}$. + +\Item{31.} $8$~or~$2$. + +\Item{32.} $4$~or~$-7$. + +\Item{33.} $0$~or~$3$. + +\Item{34.} $0$~or~$7$. + +\Item{35.} $5$~or~$2$. + +\Item{36.} $4$~or~$-1$. +%% -----File: 189.png---Folio 183------- + + +\AnsTo{Exercise}{73.} % Page 140. + +\Item{1.} $6$~and~$5$. + +\Item{2.} $5$~and~$15$. + +\Item{3.} Son, $8$~yr.; father, $40$~yr. + +\Item{4.} $9$. + +\Item{6.} $5$~rd.\ by $7$~rd. + +\Item{7.} $12$~ft. + +\Item{8.} $20$~ft.\ by $18$~ft. + +\Item{9.} $10$~rd.\ by $12$~rd. + +\Item{10.} Son, $10$~yr.; father, $54$~yr. + + +\AnsTo{Exercise}{74.} % Page 141. + +\Item{1.} $6$~miles an hour. + +\Item{2.} $7$. + +\Item{3.} $5$. + +\Item{4.} $8$. + +\Item{5.} $36$. + + +\AnsTo[4]{Exercise}{75.} % Page 144. + +\Item{1.} $75$. + +\Item{2.} $38$. + +\Item{3.} $4\frac{1}{7}$. + +\Item{4.} $-8\frac{3}{4}$. + +\Item{5.} $23$. + +\Item{6.} $0$. + +\Item{7.} $156$. + +\Item{8.} $20$th. + +\Item{9.} $101$st. + +\Item{10.} $26$. + +\Item{11.} $a$. + +\Item{12.} $21$, $22$,~etc. + + +\AnsTo[4]{Exercise}{76.} % Page 146. + +\Item{1.} $440$. + +\Item{2.} $201$. + +\Item{3.} $4frac{1}{6}$. + +\Item{4.} $128$. + +\Item{5.} $-378$. + +\Item{6.} $187\frac{1}{2}$. + +\Item{7.} $1$, $3$, $5$. + +\Item{8.} $156$. + +\Item{9.} $300$. + +\Item{10.} $2550$~yd. + +\Item{11.} $5812.1$~ft. + +\Item{12.} $144.9$~ft. + +\Item{13.} $11$, $15$. + +\Item{14.} $7$, $9$, $11$. + +\Item{15.} $12$~miles. + +\Item{16.} $5$, $12$, $19$. + + +\AnsTo[4]{Exercise}{77.} % Page 151. + +\Item{1.} $243$. + +\Item{2.} $192$. + +\Item{3.} $\frac{3}{64}$. + +\Item{4.} $256$. + +\Item{5.} $±4$. + +\Item{6.} $±4$. + +\Item{7.} $1092$. + +\Item{8.} $765$. + +\Item{9.} $11\frac{29}{32}$. + +\Item{10.} $15\frac{15}{16}$. + +\Item{11.} $127\frac{3}{4}$. + +\Item{12.} $44$. + +\Item{13.} $1\frac{11}{54}$. + +\Item{14.} \$$1.27$. + +\Item{15.} \$$81.90$. + +\Item{16.} $14,641$. + + +\AnsTo{Exercise}{78.} % Page 153. + +\Item{1.} $a + b + c$. + +\Item{2.} $x^{2} + x + 1$. + +\Item{3.} $x^{2} - 2xy + y^{2}$. + +\Item{4.} $2a^{2} - 3ab + 5b^{2}$. + +\Item{5.} $4x^{3} + 3x^{2}y - 2y^{3}$. + +\Item{6.} $2x^{3} - xy^{2} + 3y^{3}$. + + +\AnsTo[4]{Exercise}{79.} % Page 156. + +\Item{1.} $18$. + +\Item{2.} $21$. + +\Item{3.} $23$. + +\Item{4.} $31$. + +\Item{5.} $3.2$. + +\Item{6.} $7.3$. + +\Item{7.} $232$. + +\Item{8.} $785$. + +\Item{9.} $1225$. + +\Item{10.} $589$. + +\Item{11.} $5601$. + +\Item{12.} $1234$. + +\Item{13.} $1.4142\dots$ + +\Item{14.} $1.7320\dots$ + +\Item{15.} $2.2360\dots$ + +\Item{16.} $2.4494\dots$ + +\Item{17.} $0.7071\dots$ + +\Item{18.} $0.9486\dots$ + +\Item{19.} $0.8164\dots$ + +\Item{20.} $0.8660\dots$ + +\Item{21.} $0.8944\dots$ + +\Item{22.} $0.7905\dots$ +%% -----File: 190.png---Folio 184------- + + +\AnsTo{Exercise}{80.} % Page 159. + +\Item{1.} $x + y$. + +\Item{2.} $2x - 1$. + +\Item{3.} $2x - 3y$. + +\Item{4.} $4a - 3x$. + +\Item{5.} $1 + x + x^{2}$. + +\Item{6.} $x^{2} - x + 1$. + + +\AnsTo[4]{Exercise}{81.} % Page 163. + +\Item{1.} $36$. + +\Item{2.} $35$. + +\Item{3.} $45$. + +\Item{4.} $65$. + +\Item{5.} $48$. + +\Item{6.} $637$. + +\Item{7.} $478$. + +\Item{8.} $638$. + +\Item{9.} $503$. + +\Item{10.} $728$. + +\Item{11.} $12.34$. + +\Item{12.} $12.25$. + +\Item{13.} $0.2154\dots$ + +\Item{14.} $0.3684\dots$ + +\Item{15.} $0.5848\dots$ + +\Item{16.} $1.5874\dots$ + +\Item{17.} $2.1544\dots$ + +\Item{18.} $4.4310\dots$ + +\Item{19.} $1.3572\dots$ + +\Item{20.} $1.2703\dots$ + +\Item{21.} $1.4454\dots$ + +\Item{22.} $0.8735\dots$ + +\Item{23.} $0.9085\dots$ + +\Item{24.} $0.9352\dots$ +%[** TN: Environment opened by \AnsTo] +\end{multicols} + + +%%%%%%%%%%%%%%%%%%%%%%%%% GUTENBERG LICENSE %%%%%%%%%%%%%%%%%%%%%%%%%% + +\BackMatter +\begin{PGtext} +End of the Project Gutenberg EBook of The First Steps in Algebra, by +G. A. (George Albert) Wentworth + +*** END OF THIS PROJECT GUTENBERG EBOOK THE FIRST STEPS IN ALGEBRA *** + +***** This file should be named 36670-t.tex or 36670-t.zip ***** +This and all associated files of various formats will be found in: + http://www.gutenberg.org/3/6/6/7/36670/ + +Produced by Andrew D. Hwang, Peter Vachuska, Chuck Greif +and the Online Distributed Proofreading Team at +http://www.pgdp.net. + + +Updated editions will replace the previous one--the old editions +will be renamed. + +Creating the works from public domain print editions means that no +one owns a United States copyright in these works, so the Foundation +(and you!) can copy and distribute it in the United States without +permission and without paying copyright royalties. Special rules, +set forth in the General Terms of Use part of this license, apply to +copying and distributing Project Gutenberg-tm electronic works to +protect the PROJECT GUTENBERG-tm concept and trademark. 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When a bi-no-mial is the dif-fer-ence of two squares. + [] + +[95] [96] [97] [98] +Underfull \hbox (badness 10000) in paragraph at lines 4959--4963 + + [] + +[99] [100] [101] [102] [103] [104] +Overfull \hbox (1.29138pt too wide) in paragraph at lines 5268--5270 +[]\OT1/cmr/m/n/10.95 Two num-bers whose prod-uct is $30$ are $1$ and $30$, $2$ +and $15$, $3$ and $10$, + [] + +[105] [106] [107] [108] [109] [110 + +] [111] [112] [113] [114] [115] [116] [117 + +] [118] [119] [120] [121] [122] [123] +Overfull \hbox (0.88312pt too wide) in paragraph at lines 6065--6066 +[][] $[]$\OT1/cmr/m/n/12 , $[]$. + [] + +[124] +Overfull \hbox (3.79523pt too wide) in paragraph at lines 6118--6118 +[] + [] + +[125] [126] [127] [128] [129] [130] [131] [132] [133] [134] [135] +Overfull \hbox (12.6967pt too wide) in paragraph at lines 6449--6449 +[] + [] + +[136 + +] +Overfull \hbox (13.73837pt too wide) in paragraph at lines 6508--6508 +[] + [] + +[137] [138] [139] [140] +Overfull \hbox (127.43008pt too wide) in paragraph at lines 6644--6644 +[] + [] + +[141] +Overfull \hbox (82.71672pt too wide) in paragraph at lines 6685--6685 +[] + [] + +[142] [143] +Overfull \hbox (50.27081pt too wide) in paragraph at lines 6733--6733 +[] + [] + + +Overfull \hbox (2.05074pt too wide) in paragraph at lines 6749--6749 +[] + [] + +[144] [145] +Overfull \hbox (60.99086pt too wide) in paragraph at lines 6827--6827 +[] + [] + +[146] [147] +Overfull \hbox (41.2357pt too wide) in paragraph at lines 6885--6885 +[] + [] + +[148] +Overfull \hbox (33.08694pt too wide) in paragraph at lines 6936--6936 +[] + [] + +[149] [150] +Overfull \hbox (30.04593pt too wide) in paragraph at lines 6991--6991 +[] + [] + +[151] +Overfull \hbox (57.59785pt too wide) in paragraph at lines 7039--7039 +[] + [] + +[152] +Overfull \hbox (24.40863pt too wide) in paragraph at lines 7052--7052 +[] + [] + +[153] +Overfull \hbox (54.40073pt too wide) in paragraph at lines 7088--7088 +[] + [] + +[154] +Overfull \hbox (55.29538pt too 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paragraph at lines 8136--8136 +[] + [] + +[179] +Overfull \hbox (75.6592pt too wide) in paragraph at lines 8158--8158 +[] + [] + + +Overfull \hbox (81.72191pt too wide) in paragraph at lines 8180--8180 +[] + [] + +[180] [181] [182] [183] +Overfull \hbox (14.36713pt too wide) in paragraph at lines 8349--8349 +[] + [] + + +Overfull \hbox (39.82089pt too wide) in paragraph at lines 8367--8367 +[] + [] + +[184] +Overfull \hbox (48.35448pt too wide) in paragraph at lines 8374--8374 +[] + [] + + +Overfull \hbox (14.62827pt too wide) in paragraph at lines 8395--8395 +[] + [] + +[185] [186] +Overfull \hbox (36.34747pt too wide) in paragraph at lines 8465--8465 +[] + [] + +[187] [188] [189 + +] [190] +Overfull \hbox (41.95776pt too wide) in paragraph at lines 8610--8610 +[] + [] + +[191] [192] [193] [194] [195] [196] [197 + +] +Overfull \hbox (5.98813pt too wide) in paragraph at lines 8846--8859 +[]\OT1/cmr/m/n/10.95 The 5th and 7th terms are $\OML/cmm/m/it/10.95 ar[]$ \OT1/ +cmr/m/n/10.95 and $\OML/cmm/m/it/10.95 ar[]$\OT1/cmr/m/n/10.95 , re-spec-tively +. + [] + +[198] +Overfull \hbox (1.00482pt too wide) in paragraph at lines 8878--8880 +[]\OT1/cmr/m/n/12 Hence\OT1/cmr/m/it/12 , the ge-o-met-ri-cal mean of any two n +um-bers is the square + [] + + +Overfull \hbox (84.25482pt too wide) in paragraph at lines 8894--8894 +[] + [] + +[199] +Overfull \hbox (35.57372pt too wide) in paragraph at lines 8921--8921 +[] + [] + + +Overfull \hbox (36.34883pt too wide) in paragraph at lines 8935--8935 +[] + [] + +[200] +Overfull \hbox (2.11494pt too wide) in paragraph at lines 8950--8952 +[][] \OT1/cmr/m/n/12 Find the com-mon ra-tio if the 1st and 3d terms are $2$ an +d $32$. + [] + +[201] [202] [203 + +] [204] [205] [206] [207] [208] [209] [210] [211] [212] [213] [214] [215] [216] +[217] [218] [219] [220] +Overfull \hbox (2.5148pt too wide) in paragraph at lines 9785--9786 +[][]$\OT1/cmr/m/n/10 14$. + [] + + +Overfull \hbox (2.5148pt too wide) in paragraph at lines 9787--9788 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Cartesian bounding box: [-4,4] x [0,1] +%% Actual size: 5 x 0.5in +%% Figure offset: left by 0in, down by 0in +%% +%% usepackages tikz +%% +\xdefinecolor{rgb_000000}{rgb}{0,0,0}% +\begin{tikzpicture} +\pgfsetlinewidth{0.4pt} +\useasboundingbox (0in,0in) rectangle (5in,0.5in); +\pgfsetlinewidth{0.8pt} +\draw (-0.371746in,0in)--(2.5in,0in)--(5.37175in,0in); +\draw (0in,0in)--(2.5in,0in)--(5in,0in); +\pgftext[at={\pgfpoint{0in}{0in}}] {\makebox(0,0)[b]{\hbox{\color{rgb_000000}$\rule{0.5pt}{6pt}$}}} +\pgftext[at={\pgfpoint{0.625in}{0in}}] {\makebox(0,0)[b]{\hbox{\color{rgb_000000}$\rule{0.5pt}{6pt}$}}} +\pgftext[at={\pgfpoint{1.25in}{0in}}] {\makebox(0,0)[b]{\hbox{\color{rgb_000000}$\rule{0.5pt}{6pt}$}}} +\pgftext[at={\pgfpoint{1.875in}{0in}}] {\makebox(0,0)[b]{\hbox{\color{rgb_000000}$\rule{0.5pt}{6pt}$}}} +\pgftext[at={\pgfpoint{2.5in}{0in}}] {\makebox(0,0)[b]{\hbox{\color{rgb_000000}$\rule{0.5pt}{6pt}$}}} +\pgftext[at={\pgfpoint{3.125in}{0in}}] 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+{ + picture(P(-4,0), P(4,1), "5 x 0.5in"); + + begin(); + bold(); + axis Ax(P(xmin(),0), P(xmax(),0), xsize(), P(3,10), tl); + Ax.align(t); + + line(P(xmin(),0), P(xmax(),0), 20); + Ax.label_rep(f).draw(); + + label(P(xmin() - 0.5,0), P(0, 14), "$\\cdots$", l); + label(P(xmax() + 0.5,0), P(0, 14), "$\\cdots$", r); + + tikz_format(); + end(); +} diff --git a/36670-t/images/src/fig3.eepic b/36670-t/images/src/fig3.eepic new file mode 100644 index 0000000..faf251d --- /dev/null +++ b/36670-t/images/src/fig3.eepic @@ -0,0 +1,43 @@ +%% Generated from fig3.xp on Thu Jul 7 13:03:27 EDT 2011 by +%% ePiX-1.2.4 +%% +%% Cartesian bounding box: [-5,6] x [0,1] +%% Actual size: 5 x 0.5in +%% Figure offset: left by 0in, down by 0in +%% +%% usepackages tikz +%% +\xdefinecolor{rgb_000000}{rgb}{0,0,0}% +\begin{tikzpicture} +\pgfsetlinewidth{0.4pt} +\useasboundingbox (0in,0in) rectangle (5in,0.5in); +\pgfsetlinewidth{0.8pt} +\draw (-0.371746in,0in)--(2.5in,0in)--(5.37175in,0in); +\draw 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{\makebox(0,0)[br]{\hbox{\color{rgb_000000}$+1$}}} +\pgftext[at={\pgfpoint{3.21983in}{0.139372in}}] {\makebox(0,0)[br]{\hbox{\color{rgb_000000}$+2$}}} +\pgftext[at={\pgfpoint{3.67437in}{0.139372in}}] {\makebox(0,0)[br]{\hbox{\color{rgb_000000}$+3$}}} +\pgftext[at={\pgfpoint{4.12892in}{0.139372in}}] {\makebox(0,0)[br]{\hbox{\color{rgb_000000}$+4$}}} +\pgftext[at={\pgfpoint{4.58347in}{0.139372in}}] {\makebox(0,0)[br]{\hbox{\color{rgb_000000}$+5$}}} +\pgftext[at={\pgfpoint{5.03801in}{0.139372in}}] {\makebox(0,0)[br]{\hbox{\color{rgb_000000}$+6$}}} +\pgftext[at={\pgfpoint{-0.227273in}{0.193718in}}] {\makebox(0,0)[r]{\hbox{\color{rgb_000000}$\cdots$}}} +\pgftext[at={\pgfpoint{5.22727in}{0.193718in}}] {\makebox(0,0)[l]{\hbox{\color{rgb_000000}$\cdots$}}} +\end{tikzpicture} diff --git a/36670-t/images/src/fig3.xp b/36670-t/images/src/fig3.xp new file mode 100644 index 0000000..d6a7ced --- /dev/null +++ b/36670-t/images/src/fig3.xp @@ -0,0 +1,36 @@ +/* -*-ePiX-*- */ +#include "epix.h" +using namespace ePiX; + +std::string f(double t, unsigned int prec, unsigned int base) +{ + std::ostringstream buf; + buf << "$"; + if (t == 0) + buf << "\\phantom{-}"; + + else if(0 < t) + buf << "+"; + + buf << t << "$"; + return buf.str(); +} + +int main() +{ + picture(P(-5,0), P(6,1), "5 x 0.5in"); + + begin(); + bold(); + axis Ax(P(xmin(),0), P(xmax(),0), xsize(), P(3,10), tl); + Ax.align(t); + + line(P(xmin(),0), P(xmax(),0), 20); + Ax.label_rep(f).draw(); + + label(P(xmin() - 0.5,0), P(0, 14), "$\\cdots$", l); + label(P(xmax() + 0.5,0), P(0, 14), "$\\cdots$", r); + + tikz_format(); + end(); +} diff --git a/36670-t/old/36670-t.tex b/36670-t/old/36670-t.tex new file mode 100644 index 0000000..667128f --- /dev/null +++ b/36670-t/old/36670-t.tex @@ -0,0 +1,14160 @@ +% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % +% % +% The Project Gutenberg EBook of The First Steps in Algebra, by % +% G. A. (George Albert) Wentworth % +% % +% This eBook is for the use of anyone anywhere at no cost and with % +% almost no restrictions whatsoever. You may copy it, give it away or % +% re-use it under the terms of the Project Gutenberg License included % +% with this eBook or online at www.gutenberg.net % +% % +% % +% Title: The First Steps in Algebra % +% % +% Author: G. A. (George Albert) Wentworth % +% % +% Release Date: July 9, 2011 [EBook #36670] % +% % +% Language: English % +% % +% Character set encoding: ISO-8859-1 % +% % +% *** START OF THIS PROJECT GUTENBERG EBOOK THE FIRST STEPS IN ALGEBRA ***% +% % +% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % + +\def\ebook{36670} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%% %% +%% Packages and substitutions: %% +%% %% +%% book: Required. %% +%% inputenc: Standard DP encoding. Required. %% +%% %% +%% fix-cm: For larger title page fonts. Optional. %% +%% ifthen: Logical conditionals. Required. %% +%% %% +%% amsmath: AMS mathematics enhancements. Required. %% +%% amssymb: Additional mathematical symbols. Required. %% +%% %% +%% alltt: Fixed-width font environment. Required. %% +%% array: Enhanced tabular features. Required. %% +%% %% +%% indentfirst: Indent first word of each sectional unit. Optional. %% +%% textcase: Apply \MakeUppercase (et al.) only to text, not math. %% +%% Required. %% +%% %% +%% calc: Length calculations. Required. %% +%% soul: Spaced text. Optional. %% +%% %% +%% fancyhdr: Enhanced running headers and footers. Required. %% +%% %% +%% graphicx: Standard interface for graphics inclusion. Required. %% +%% wrapfig: Illustrations surrounded by text. Required. %% +%% %% +%% geometry: Enhanced page layout package. Required. %% +%% hyperref: Hypertext embellishments for pdf output. Required. %% +%% %% +%% %% +%% Producer's Comments: %% +%% %% +%% Changes are noted in this file in two ways. %% +%% 1. \DPtypo{}{} for typographical corrections, showing original %% +%% and replacement text side-by-side. %% +%% 2. [** TN: Note]s for lengthier or stylistic comments. %% +%% %% +%% Compilation Flags: %% +%% %% +%% The following behavior may be controlled by boolean flags. %% +%% %% +%% ForPrinting (false by default): %% +%% Compile a screen-optimized PDF file. Set to true for print- %% +%% optimized file (large text block, two-sided layout, black %% +%% hyperlinks). %% +%% %% +%% Both print and screen layout are relatively loose, and contain %% +%% hard-coded page breaks (\PrintBreak, \ScreenBreak, \newpage). %% +%% %% +%% %% +%% PDF pages: 269 (if ForPrinting set to false) %% +%% PDF page size: 4.75 x 7" (non-standard) %% +%% PDF bookmarks: created, point to ToC entries %% +%% PDF document info: filled in %% +%% Images: 3 pdf diagrams %% +%% %% +%% Summary of log file: %% +%% * Large numbers of visually-harmless over-full hboxes and vboxes %% +%% from DPalign* and DPgather* environments. %% +%% %% +%% %% +%% Compile History: %% +%% %% +%% July, 2011: adhere (Andrew D. Hwang) %% +%% texlive2007, GNU/Linux %% +%% %% +%% Command block: %% +%% %% +%% pdflatex x2 %% +%% %% +%% %% +%% July 2011: pglatex. %% +%% Compile this project with: %% +%% pdflatex 36670-t.tex ..... TWO times %% +%% %% +%% pdfTeXk, Version 3.141592-1.40.3 (Web2C 7.5.6) %% +%% %% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\listfiles +\documentclass[12pt]{book}[2005/09/16] + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%% PACKAGES %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\usepackage[latin1]{inputenc}[2006/05/05] + +\usepackage{ifthen}[2001/05/26] %% Logical conditionals + +\usepackage{amsmath}[2000/07/18] %% Displayed equations +\usepackage{amssymb}[2002/01/22] %% and additional symbols + +\usepackage{alltt}[1997/06/16] %% boilerplate, credits, license + +\usepackage{array}[2005/08/23] %% extended array/tabular features + +\usepackage{multicol} + +\usepackage{graphicx}[1999/02/16]%% For diagrams + +\usepackage{indentfirst}[1995/11/23] +\usepackage{textcase}[2004/10/07] + +\usepackage{calc}[2005/08/06] + +% for running heads +\usepackage{fancyhdr} + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%% Interlude: Set up PRINTING (default) or SCREEN VIEWING %%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + +% ForPrinting=true (default) false +% Asymmetric margins Symmetric margins +% Black hyperlinks Blue hyperlinks +% Start Preface, ToC, etc. recto No blank verso pages +% +% Chapter-like ``Sections'' start both recto and verso in the scanned +% book. 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A. (George Albert) Wentworth + +This eBook is for the use of anyone anywhere at no cost and with +almost no restrictions whatsoever. You may copy it, give it away or +re-use it under the terms of the Project Gutenberg License included +with this eBook or online at www.gutenberg.net + + +Title: The First Steps in Algebra + +Author: G. A. (George Albert) Wentworth + +Release Date: July 9, 2011 [EBook #36670] + +Language: English + +Character set encoding: ISO-8859-1 + +*** START OF THIS PROJECT GUTENBERG EBOOK THE FIRST STEPS IN ALGEBRA *** +\end{PGtext} +\end{minipage} +\end{center} +\clearpage +%%%% Credits and transcriber's note %%%% +\begin{center} +\begin{minipage}{\textwidth} +\begin{PGtext} +Produced by Andrew D. Hwang, Peter Vachuska, Chuck Greif +and the Online Distributed Proofreading Team at +http://www.pgdp.net. +\end{PGtext} +\end{minipage} +\end{center} +\vfill + +\begin{minipage}{0.85\textwidth} +\small +\BookMark{0}{Transcriber's Note} +\subsection*{\centering\normalfont\scshape +\normalsize\MakeLowercase{\TransNote}} + +\raggedright +\TransNoteText +\end{minipage} +%%%%%%%%%%%%%%%%%%%%%%%%%%% FRONT MATTER %%%%%%%%%%%%%%%%%%%%%%%%%% +\FrontMatter +%% -----File: 001.png---Folio i------- +\begin{center} +\Large THE +\vfill + +\textbf{\LARGE FIRST STEPS IN ALGEBRA.} +\vfill + +\normalsize BY \\[12pt] +\Large G. A. WENTWORTH, A.M. \\[12pt] +\footnotesize AUTHOR OF A SERIES OF TEXT-BOOKS IN MATHEMATICS. +\vfill\vfill + +\large BOSTON, U.S.A.: \\ +PUBLISHED BY GINN \& COMPANY. \\ +1904. +\end{center} +%% -----File: 002.png---Folio ii------- +\newpage +\null\vfill +\begin{center} +\footnotesize +Entered according to Act of Congress, in the year 1894, by \\[4pt] +G. A. WENTWORTH, \\[4pt] +in the Office of the Librarian of Congress, at Washington. \\[4pt] +\tb +\textsc{All Rights Reserved.} +\vfill + +\textsc{Typography by J. S. Cushing \& Co., Boston, U.S.A.} \\ +\tb[2.5in] +\textsc{Presswork by Ginn \& Co., Boston, U.S.A.} +\end{center} +%% -----File: 003.png---Folio iii------- + + +\Preface + +\First{This} book is written for pupils in the upper grades of +grammar schools and the lower grades of high schools. +The introduction of the simple elements of Algebra into +these grades will, it is thought, so stimulate the mental +activity of the pupils, that they will make considerable +progress in Algebra without detriment to their progress +in Arithmetic, even if no more time is allowed for the +two studies than is usually given to Arithmetic alone. + +The great danger in preparing an Algebra for very +young pupils is that the author, in endeavoring to smooth +the path of the learner, will sacrifice much of the educational +value of the study. To avoid this real and serious +danger, and at the same time to gain the required simplicity, +great care has been given to the explanations of +the fundamental operations and rules, the arrangement +of topics, the model solutions of examples, and the making +of easy examples for the pupils to solve. + +Nearly all the examples throughout the book are new, +and made expressly for beginners. + +The first chapter clears the way for quite a full treatment +of simple integral equations with one unknown number. +In the first two chapters only \emph{positive} numbers are +%% -----File: 004.png---Folio iv------- +involved, and the learner is led to see the practical advantages +of Algebra in its most interesting applications before +he faces the difficulties of negative numbers. + +The third chapter contains a simple explanation of negative +numbers. The recognition of the facts that the real +nature of subtraction is counting backwards, and that the +real nature of multiplication is forming the product from +the multiplicand precisely as the multiplier is formed from +unity, makes an easy road to the laws of addition and subtraction +of algebraic numbers, and to the law of signs in +multiplication and division. All the principles and rules +of this chapter are illustrated and enforced by numerous +examples involving \emph{simple} algebraic expressions only. + +The ordinary processes with \emph{compound} expressions, including +simple cases of resolution into factors, and the +treatment of fractions, naturally follow the third chapter. +The immediate succession of topics that require similar +work is of the highest importance to the beginner, and it +is hoped that the half-dozen chapters on algebraic expressions +will prove interesting, and give sufficient readiness +in the use of symbols. + +A chapter on fractional equations with one unknown +number, a chapter on simultaneous equations with two +unknown numbers, and a chapter on quadratics follow in +order. Only one method of elimination is given in simultaneous +equations and one method of completing the +square in quadratics. Moreover, the solution of the examples +in quadratics requires the square roots of only small +numbers such as every pupil knows who has learned the +%% -----File: 005.png---Folio v------- +multiplication table. In each of these three chapters a +considerable number of problems is given to \emph{state} and solve. +By this means the learner is led to exercise his reasoning +faculty, and to realize that the methods of Algebra require +a strictly logical process. These problems, however, are +divided into classes, and a model solution of an example +of each class is given as a guide to the solution of other +examples of that class. + +The course may end with the chapter on quadratics, but +the simple questions of arithmetical progression and of +geometrical progression are so interesting in themselves, +and show so clearly the power of Algebra, that it will +be a great loss not to take the short chapters on these +series. + +The last chapter is on square and cube roots. It is +expected that pupils who use this book will learn how to +extract the square and cube roots by the simple formulas +of Algebra, and be spared the necessity of committing to +memory the long and tedious rules given in Arithmetic, +rules that are generally forgotten in less time than they +are learned. + +Any corrections or suggestions will be thankfully received +by the author. + +A teachers' edition is in press, containing solutions of +examples, and such suggestions as experience with beginners +has shown to be valuable. + +\Signature{G\Add{.} A. WENTWORTH.} +{\textsc{Exeter}, NH, April, 1894} +%% -----File: 006.png---Folio vi------- +\TableofContents +\iffalse +CONTENTS. + +Chapter Page + +I. Introduction.............. 1 + +II. Simple Equations............. 19 + +III. Positive and Negative Numbers....... 33 + +IV. Addition and Subtraction......... 46 + +V. Multiplication and Division........ 53 + +VI. Special Rules in Multiplication and Division . 64 + +VII. Factors............... 71 + +VIII. Common Factors and Multiples....... 84 + +IX. Fractions................ 89 + +X. Fractional Equations........... 103 + +XI. Simultaneous Equations of the First Degree . 122 + +XII. Quadratic Equations........... 132 + +XIII. Arithmetical Progression......... 142 + +XIV. Geometrical Progression.......... 148 + +XV. Square and Cube Roots.......... 152 + + Answers................ 165 +\fi +%% -----File: 007.png---Folio 1------- +\MainMatter +% FIRST STEPS IN ALGEBRA. +% [** TN: Chapter macro prints preceding line] + +\Chapter{I.}{Introduction.} + +\begin{Remark}[\First{Note}\Add{.}] +The principal definitions are put at the beginning of the +book for convenient reference. They are not to be committed to +memory. It is a good plan to have definitions and explanations +read aloud in the class, and to encourage pupils to make comments +upon them, and ask questions about them. +\end{Remark} + +\Paragraph{1. Algebra.} Algebra, like Arithmetic, treats of numbers. + +\Paragraph{2. Units.} In counting separate objects or in measuring +magnitudes, the \emph{standards} by which we count or measure +are called \Defn{units}. + +\begin{Remark} +Thus, in counting the boys in a school, the unit is a boy; in selling +eggs by the dozen, the unit is a dozen eggs; in selling bricks by +the thousand, the unit is a thousand bricks; in measuring short distances, +the unit is an inch, a foot, or a yard; in measuring long +distances, the unit is a rod or a mile. +\end{Remark} + +\Paragraph{3. Numbers.} \emph{Repetitions of the unit} are expressed by +numbers. + +\Paragraph{4. Quantities.} A number of specified units of any kind +is called a quantity; as, $4$~pounds, $5$~oranges. + +\Paragraph{5. Number-Symbols in Arithmetic.} Arithmetic employs +the arbitrary symbols, $1$,~$2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$,~$0$, called +\Defn{figures}, to represent numbers. +%% -----File: 008.png---Folio 2------- + +\Paragraph{6. Number-Symbols in Algebra.} Algebra employs \emph{the +letters of the alphabet} in addition to the figures of Arithmetic +to represent numbers. Letters are used as \emph{general} +symbols of numbers to which \emph{any particular values} may be +assigned. + + +\Section{PRINCIPAL SIGNS OF OPERATIONS.} + +\Paragraph{7.} The signs of the fundamental operations are the same +in Algebra as in Arithmetic. + +\Paragraph{8. The Sign of Addition,~$+$.} The sign~$+$ is read \emph{plus}. + +\begin{Remark} +Thus, $4 + 3$, read $4$~plus~$3$, indicates that the number~$3$ is to be +added to the number~$4$, $a + b$, read $a$~plus~$b$, indicates that the number~$b$ +is to be added to the number~$a$. +\end{Remark} + +\Paragraph{9. The Sign of Subtraction,~$-$\Add{.}} The sign~$-$ is read \emph{minus}. + +\begin{Remark} +Thus, $4 - 3$, read $4$~minus~$3$, indicates that the number~$3$ is to be +subtracted from the number~$4$, $a - b$, read $a$~minus~$b$, indicates that +the number~$b$ is to be subtracted from the number~$a$. +\end{Remark} + +\Paragraph{10. The Sign of Multiplication,~$×$\Add{.}} The sign~$×$ is read +\emph{times}. + +\begin{Remark} +Thus, $4 × 3$, read $4$~times~$3$, indicates that the number~$3$ is to be +multiplied by~$4$, $a × b$, read $a$~times~$b$, indicates that the number~$b$ +is to be multiplied by the number~$a$. +\end{Remark} + +A dot is sometimes used for the sign of multiplication. +Thus $2 · 3 · 4 · 5$ means the same as $2 × 3 × 4 × 5$. Either +sign is read \emph{multiplied by} when followed by the multiplier. +\$$a × b$, or \$$a · b$, is read $a$~dollars multiplied by~$b$. + +\Paragraph{11. The Sign of Division,~$÷$.} The sign~$÷$ is read \emph{divided by}. + +\begin{Remark} +Thus, $4 ÷ 2$, read $4$~divided by~$2$, indicates that the number~$4$ is +to be divided by~$2$, $a ÷ b$, read $a$~divided by~$b$, indicates that the +number~$a$ is to be divided by the number~$b$. +\end{Remark} +%% -----File: 009.png---Folio 3------- + +Division is also indicated by writing the dividend above +the divisor with a horizontal line between them. + +\begin{Remark} +Thus, $\dfrac{4}{2}$ means the same as~$4 ÷ 2$; $\dfrac{a}{b}$~means the same as~$a ÷ b$. +\end{Remark} + + +\Section{OTHER SIGNS USED IN ALGEBRA.} + +\Paragraph{12. The Sign of Equality,~$=$.} The sign~$=$ is read \emph{is equal +to}, when placed between two numbers and indicates that +these two numbers are equal. + +\begin{Remark} +Thus, $8 + 4 = 12$ means that $8 + 4$ and~$12$ stand for \emph{equal} numbers; +$x + y = 20$ means that $x + y$ and~$20$ stand for equal numbers. +\end{Remark} + +\Paragraph{13. The Sign of Inequality, $>$~or~$<$.} The sign $>$~or~$<$ is +read \emph{is greater than} and \emph{is less than} respectively, and when +placed between two numbers indicates that these two numbers +are unequal, and that the number toward which the +sign opens is the greater. + +\begin{Remark} +Thus, $9 + 6 > 12$ means that $9 + 6$ is greater than~$12$; and +$9 + 6 < 16$ means that $9 + 6$ is less than~$16$. +\end{Remark} + +%[** TN: [sic] no punctuation after \therefore] +\Paragraph{14. The Sign of Deduction,~$\therefore$}\quad The sign~$\therefore$ is read \emph{hence} +or \emph{therefore}. + +\Paragraph{15. The Sign of Continuation,~$\dots$.} The sign~$\dots$ is read +\emph{and so on}. + +\Paragraph{16. The Signs of Aggregation.} The signs of aggregation +are the bar~$|$, the vinculum~$\overline{\phantom{a+x}}$, the parenthesis~$(\ )$, the +bracket~$[\ ]$, and the brace~$\{\ \}$. + +\begin{Remark} +Thus, each of the expressions $\VSum{a}{b}$, $\Vinc{a + b}$, $(a + b)$, $[a + b]$, $\{a + b\}$, +signifies that $a + b$ is to be treated as a single number. +\end{Remark} +%% -----File: 010.png---Folio 4------- + + +\Section{FACTORS. COEFFICIENTS. POWERS.} + +\Paragraph{17. Factors.} When a number consists of the product of +two or more numbers, each of these numbers is called a +\Defn{factor} of the product. + +The sign~$×$ is generally omitted between a figure and a +letter, or between letters; thus, instead of $63 × a × b$, we +write~$63ab$; instead of $a × b × c$, we write~$abc$. + +The expression~$abc$ must not be confounded with $a + b + c$. +$abc$~is a product; $a + b + c$ is a sum. +\begin{DPalign*} +\lintertext{\indent If} +a = 2,\quad b &= 3,\quad c = 4, \\ +\lintertext{then} +abc &= 2 × 3 × 4 = 24; \\ +\lintertext{but} +a + b + c &= 2 + 3 + 4 = 9. +\end{DPalign*} + +\begin{Remark}[Note.] +When a sign of operation is omitted in the notation of +Arithmetic, it is always the \emph{sign of addition}; but when a sign of +operation is omitted in the notation of Algebra, it is always the +\emph{sign of multiplication}. Thus, $456$~means $400 + 50 + 6$, but $4ab$ +means $4 × a × b$. +\end{Remark} + +\Paragraph{18.} Factors expressed by letters are called \Defn{literal} factors; +factors expressed by figures are called \Defn{numerical} factors. + +\Paragraph{19.} If one factor of a product is equal to~$0$, the product +is equal to~$0$, whatever the values of the other factors. +Such a factor is called a \Defn{zero factor}. + +\Paragraph{20. Coefficients.} A known factor of a product which is +prefixed to another factor, to show the number of times that +factor is taken, is called a \Defn{coefficient}. + +\begin{Remark} +Thus, in~$7c$, $7$~is the coefficient of~$c$; in~$7ax$, $7$~is the +coefficient of~$ax$, +or, if $a$~is known, $7a$~is the coefficient of~$x$. +\end{Remark} +%% -----File: 011.png---Folio 5------- + +By coefficient, we generally mean the \textbf{numerical coefficient +with its sign}. If no numerical coefficient is written, $1$~is +understood. Thus, $ax$~means the same as~$1ax$. + +\Paragraph{21. Powers and Roots.} A product consisting of two or +more \textbf{equal factors} is called a \Defn{power} of that factor, and one +of the equal factors is called a \Defn{root} of the number. + +\begin{Remark} +Thus, $9 = 3 × 3$; that is, $9$~is a power of~$3$, and $3$~is a root of~$9$. +\end{Remark} + +\Paragraph{22. Indices or Exponents.} An index or exponent is a +number-symbol written at the right of, and a little above, +a number. + +If the index is a \emph{whole number}, it shows the number +of times the given number is taken as a factor. + +\begin{Remark} +Thus, $a^{1}$, or simply~$a$, denotes that $a$~is taken \emph{once} as a +factor; $a^{2}$~denotes +that $a$~is taken \emph{twice} as a factor; $a^{3}$~denotes that $a$~is taken +\emph{three times} as a factor; and $a^{4}$~denotes that $a$~is taken \emph{four times} as a +factor; and so on. These are read: the first power of~$a$; the second +power of~$a$; the third power of~$a$; the fourth power of~$a$; and so on. + +$a^{3}$~is written instead of~$aaa$. + +$a^{4}$~is written instead of~$aaaa$. +\end{Remark} + +\Paragraph{23.} The meaning of coefficient and exponent must be +carefully distinguished. Thus, +\begin{DPalign*} +4a &= a + a + a + a; \\ +a^{4} &= a× a× a× a. \displaybreak[1] \\ +\lintertext{\indent If $a = 3$,} +4a &= 3 + 3 + 3 + 3 = 12. \\ +a^{4} &= 3 × 3 × 3 × 3 = 81. +\end{DPalign*} + +\begin{Remark} +The second power of a number is generally called the \emph{square} of +that number; thus, $a^{2}$~is called the \emph{square} of~$a$, because if $a$~denotes +the number of units of length in the side of a square, $a^{2}$~denotes the +number of units of surface in the square. The third power of a number +is generally called the \emph{cube} of that number; thus, $a^{3}$~is called the +\emph{cube} of~$a$, because if $a$~denotes the number of units of length in the +edge of a cube, $a^{3}$~denotes the number of units of volume in the +cube. +\end{Remark} +%% -----File: 012.png---Folio 6------- + + +\Section{ALGEBRAIC EXPRESSIONS.} + +\Paragraph{24. An Algebraic Expression.} An algebraic expression is +a number written with algebraic symbols. An algebraic +expression may consist of one symbol, or of several symbols +connected by signs. + +\begin{Remark} +Thus, $a$, $3abc$, $5a + 2b - 3c$, are algebraic expressions. +\end{Remark} + +\Paragraph{25. Terms.} A \Defn{term} is an algebraic expression, the parts +of which are not separated by the sign $+$~or~$-$. + +\begin{Remark} +Thus, $a$, $5xy$, $2ab × 4cd$, $\dfrac{3ab}{4cd}$ are algebraic expressions of one +term each. A term may be separated into parts by the sign $×$~or~$÷$. +\end{Remark} + +\Paragraph{26. Simple Expressions.} An algebraic expression of \emph{one +term} is called a \Defn{simple expression} or \Defn{monomial}. + +\begin{Remark} +Thus, $5xy$, $7a × 2b$, $7a ÷ 2b$, are simple expressions. +\end{Remark} + +\Paragraph{27. Compound Expressions.} An algebraic expression of +\emph{two or more terms} is called a \Defn{compound expression} or \Defn{polynomial}. + +\begin{Remark} +Thus, $5xy + 7a$, $2x - y - 3z$, $4a - 3b + 2c - 3d$ are compound +expressions. +\end{Remark} + +\Paragraph{28.} A polynomial of two terms is called a \Defn{binomial}; of +three terms, a \Defn{trinomial}. + +\begin{Remark} +Thus, $3a - b$ is a binomial; and $3a - b + c$ is a trinomial. +\end{Remark} + +\Paragraph{29. Positive and Negative Terms.} The terms of a compound +expression preceded by the sign~$+$ are called \Defn{positive +terms}, and the terms preceded by the sign~$-$ are called +\Defn{negative terms}. The sign~$+$ before the first term is omitted. + +\Paragraph{30.} A positive and a negative term of the same numerical +value cancel each other when combined. +%% -----File: 013.png---Folio 7------- + +\Paragraph{31. Like Terms.} Terms which have the same combination +of \emph{letters} are called \Defn{like} or \Defn{similar} terms; terms which +do not have the same combination of letters are called +\Defn{unlike} or \Defn{dissimilar} terms. + +\begin{Remark} +Thus, $5a^{2}bc$, $-7a^{2}bc$, $a^{2}bc$, are like terms; but $5a^{2}bc$, $5ab^{2}c$, +$5abc^{2}$, are unlike terms. +\end{Remark} + +\Paragraph{32. Degree of a Term.} A term that is the product of +three letters is said to be of the \emph{third degree}; a term of +four letters is of the \emph{fourth degree}; and so on. + +\begin{Remark} +Thus, $5abc$~is of the third degree; $2a^{2}b^{2}c^{2}$, that is, $2aabbcc$, is of +the sixth degree. +\end{Remark} + +\Paragraph{33. Degree of a Compound Expression.} The degree of a +compound expression is the degree of that term of the +expression which is of the \emph{highest degree}. + +\begin{Remark} +Thus, $a^{2}x^{2} + bx + c$ is of the fourth degree, since $a^{2}x^{2}$~is of the +fourth degree. +\end{Remark} + +\Paragraph{34. Dominant Letter.} It often happens that there is one +letter in an expression of more importance than the rest, +and this is, therefore, called the \Defn{dominant letter}. In such +cases the degree of the expression is generally called by +the degree of the \emph{dominant letter}. + +\begin{Remark} +Thus, $a^{2}x^{2} + bx + c$ is of the \emph{second degree in~$x$}. +\end{Remark} + +\Paragraph{35. Arrangement of a Compound Expression.} A compound +expression is said to be \emph{arranged} according to the powers +of some letter when the exponents of that letter, reckoning +from left to right, either descend or ascend in \emph{the order of +magnitude}. + +\begin{Remark} +Thus, $3ax^{3} - 4bx^{2} - 6ax + 8b$ is arranged according to the descending +powers of~$x$, and $8b - 6ax - 4bx^{2} + 3ax^{3}$ is arranged +according to the ascending powers of~$x$. +\end{Remark} +%% -----File: 014.png---Folio 8------- + + +\Section{PARENTHESES.} + +\Paragraph{36.} If a compound expression is to be treated as a whole, +it is enclosed in a parenthesis. + +\begin{Remark} +Thus, $2 × (10 + 5)$ means that we are to add $5$~to~$10$ and multiply +the result by~$2$; if we were to omit the parenthesis and write +$2 × 10 + 5$, the meaning would be that we were to multiply $10$~by~$2$ +and add~$5$ to the result. +\end{Remark} + +Like the parenthesis, we use with the same meaning any +other sign of aggregation. + +\begin{Remark} +Thus, $(5 + 2)$, $[5 + 2]$, $\{5 + 2\}$, $\Vinc{5 + 2}$, $\VSum{5}{2}$, all mean that the +expression $5 + 2$ is to be treated as the single symbol~$7$. +\end{Remark} + +\Paragraph{37. Parentheses preceded by~$+$.} If a man has $10$~dollars +and afterwards collects $3$~dollars and then $2$~dollars, it +makes no difference whether he adds the $3$~dollars to his +$10$~dollars, and then the $2$~dollars, or puts the $3$~and~$2$ +dollars together and adds their sum to his $10$~dollars. + +The first process is represented by $10 + 3 + 2$. + +The second process is represented by $10 + (3 + 2)$. + +\begin{DPgather*} +\lintertext{\indent Hence,} +10 + (3 + 2) = 10 + 3 + 2. +\Tag{(1)} +\end{DPgather*} + +If a man has $10$~dollars and afterwards collects $3$~dollars +and then pays a bill of $2$~dollars, it makes no difference +whether he adds the $3$~dollars collected to his $10$~dollars +and pays out of this sum his bill of $2$~dollars, or pays the +$2$~dollars from the $3$~dollars collected and adds the remainder +to his $10$~dollars. + +The first process is represented by $10 + 3 - 2$. + +The second process is represented by $10 + (3 - 2)$. + +\begin{DPgather*} +\lintertext{\indent Hence,} +10 + (3 - 2) = 10 + 3 - 2. +\Tag{(2)} +\end{DPgather*} +%% -----File: 015.png---Folio 9------- + +From (1)~and~(2) it follows that + +\begin{Theorem} +If an expression within a parenthesis is preceded by the +sign~$+$, the parenthesis can be removed without making any +change in the signs of the expression. +\end{Theorem} + +\begin{Theorem}[\textsc{Conversely.}] Any part of an expression can lie enclosed +within a parenthesis and the sign~$+$ prefixed, without making +any change in the signs of the terms thus enclosed. +\end{Theorem} + +\Paragraph{38. Parentheses preceded by~$-$.} If a man has $10$~dollars +and has to pay two bills, one of $3$~dollars and one of $2$~dollars, +it makes no difference whether he takes $3$~dollars +and $2$~dollars in succession, or takes the $3$~and~$2$ dollars at +one time, from his $10$~dollars. + +The first process is represented by $10 - 3 - 2$. + +The second process is represented by $10 - (3 + 2)$. + +\begin{DPgather*} +\lintertext{\indent Hence,} +10 - (3 + 2) = 10 - 3 - 2. +\Tag{(3)} +\end{DPgather*} + +If a man has $10$~dollars consisting of $2$~five-dollar bills, +and has a debt of $3$~dollars to pay, he can pay his debt by +giving a five-dollar bill and receiving $2$~dollars. + +This process is represented by $10 - 5 + 2$. + +Since the debt paid is $3$~dollars, that is, $(5 - 2)$~dollars, +the number of dollars he has left can evidently be +expressed by +\begin{DPalign*} +10 &- (5 - 2). \\ +\lintertext{\indent Hence,} +10 &- (5 - 2) = 10 - 5 + 2. +\Tag{(4)} +\end{DPalign*} + +From (3)~and~(4) it follows that + +\begin{Theorem} +If an expression within a parenthesis is preceded by the +sign~$-$, the parenthesis can be removed, provided the sign +before each term within the parenthesis is changed, the +sign~$+$ to~$-$, and the sign~$-$ to~$+$. +\end{Theorem} +%% -----File: 016.png---Folio 10------- + +\begin{Theorem}[\textsc{Conversely.}] Any part of an expression can be enclosed +within a parenthesis and the sign~$-$ prefixed, provided the +sign of each term enclosed is changed, the sign~$+$ to~$-$, and +the sign~$-$ to~$+$. +\end{Theorem} + + +\Exercise{1.} + +Remove the parentheses, and combine: +\begin{multicols}{2} +\Item{1.} $9 + (3 + 2)$. + +\Item{2.} $9 + (3 - 2)$. + +\Item{3.} $7 + (5 + 1)$. + +\Item{4.} $7 + (5 - 1)$. + +\Item{5.} $6 + (4 + 3)$. + +\Item{6.} $6 + (4 - 3)$. + +\Item{7.} $3 + (8 - 2)$. + +\Item{8.} $9 - (8 - 6)$. + +\Item{9.} $10 - (9 - 5)$. + +\Item{10.} $9 - (6 + 1)$. + +\Item{11.} $8 - (3 + 2)$. + +\Item{12.} $7 - (3 - 2)$. + +\Item{13.} $9 - (4 + 3)$. + +\Item{14.} $9 - (4 - 3)$. + +\Item{15.} $7 - (5 - 2)$. + +\Item{16.} $7 - (7 - 3)$. + +\Item{17.} $(8 - 6) - 1$. + +\Item{18.} $(3 - 2) - (1 - 1)$. + +\Item{19.} $(7 - 3) - (3 - 2)$. + +\Item{20.} $(8 - 2) - (5 - 3)$. + +\Item{21.} $15 - (10 - 3 - 2)$. +\end{multicols} + +\Paragraph{39. Multiplying a Compound Expression.} The expression +$4(5 + 3)$ means that we are to take the sum of the numbers +$5$~and~$3$ four times. The process can be represented by +placing five dots in a line, and a little to the right three +more dots in the same line, and then placing a second, +third, and fourth line of dots underneath the first line and +exactly similar to it. +\[ +\begin{array}{*{9}{>{\ }r}} +\DOT & \DOT & \DOT & \DOT & \DOT & \quad & \DOT & \DOT & \DOT \\[-4pt] +\DOT & \DOT & \DOT & \DOT & \DOT & & \DOT & \DOT & \DOT \\[-4pt] +\DOT & \DOT & \DOT & \DOT & \DOT & & \DOT & \DOT & \DOT \\[-4pt] +\DOT & \DOT & \DOT & \DOT & \DOT & & \DOT & \DOT & \DOT \\[-4pt] +\end{array} +\] + +There are $(5 + 3)$ dots in each line, and $4$~lines. The +total number of dots, therefore, is $4 × (5 + 3)$. + +We see that in the left-hand group there are $4 × 5$ dots, +and in the right-hand group $4 × 3$ dots. The sum of these +%% -----File: 017.png---Folio 11------- +two numbers $(4 × 5) + (4 × 3)$ must be equal to the total +number; that is, +\begin{align*} +4(5 + 3) &= (4 × 5) + (4 × 3) \\ + &= 20 + 12. +\end{align*} + +Again, the expression $4(8 - 3)$ means that we are to +take the difference of the numbers $8$~and~$3$ four times. +The process can be represented by placing eight dots in a +line and crossing the last three, and then placing a second, +third, and fourth line of dots underneath the first line and +exactly similar to it. +\[ +%[** TN: Added gap between dot groups.] +\begin{array}{*{9}{>{\ }r}} +\DOT & \DOT & \DOT & \DOT & \DOT & \quad & \NOT & \NOT & \NOT \\[-4pt] +\DOT & \DOT & \DOT & \DOT & \DOT & & \NOT & \NOT & \NOT \\[-4pt] +\DOT & \DOT & \DOT & \DOT & \DOT & & \NOT & \NOT & \NOT \\[-4pt] +\DOT & \DOT & \DOT & \DOT & \DOT & & \NOT & \NOT & \NOT \\[-4pt] +\end{array} +\] + +The whole number of dots not crossed in each line is +evidently $(8 - 3)$, and the whole number of lines is~$4$. +Therefore the total number of dots not crossed is +\[ +4 × (8 - 3). +\] + +The total number of dots (crossed and not crossed) is +$(4 × 8)$, and the total number of dots crossed is~$(4 × 3)$. +Therefore the total number of dots not crossed is +\begin{DPalign*} +(4 × 8) &- (4 × 3); \\ +\lintertext{that is,} +4(8 - 3) &= (4 × 8) - (4 × 3) \\ + &= 32 - 12. \displaybreak[1] \\ +\intertext{\indent If $a$, $b$, and~$c$ stand for any three numbers, we have} +a (b + c) &= ab + ac, \\ +\lintertext{and} +a(b - c) &= ab - ac. +\EqText{Therefore,} +\end{DPalign*} + +\Dictum{To multiply a compound expression by a simple one}, +\begin{Theorem} +Multiply each term by the multiplier, and write the successive +products with the same signs as those of the original +terms. +\end{Theorem} +%% -----File: 018.png---Folio 12------- + +\Exercise{2.} + +Multiply and remove parentheses: +\begin{multicols}{3} +\Item{1.} $7(8 + 5)$. + +\Item{2.} $7(8 - 5)$. + +\Item{3.} $6(7 + 3)$. + +\Item{4.} $6(7 - 3)$. + +\Item{5.} $8(7 + 5)$. + +\Item{6.} $8(7 - 5)$. + +\Item{7.} $9(6 - 2)$. + +\Item{8.} $4(a + b)$. + +\Item{9.} $4(a - b)$. + +\Item{10.} $2(a^{2} + b^{2})$. + +\Item{11.} $2(a^{2} - b^{2})$. + +\Item{12.} $3(ab + c)$. + +\Item{13.} $3(ab - c)$. + +\Item{14.} $3(c - ab)$. + +\Item{15.} $a(b + c)$. + +\Item{16.} $a(b - c)$. + +\Item{17.} $3a(b + c)$. + +\Item{18.} $3a(b - c)$. + +\Item{19.} $5a(b^{2} + c)$. + +\Item{20.} $5a(b^{2} - c^{2})$. + +\Item{21.} $5a^{2}(b^{2} - c)$. +\end{multicols} + +\Paragraph{40.} The numerical value of an algebraic expression is the +number obtained by putting for the letters involved the +numbers for which these letters stand, and then performing +the operations required by the signs. + +\Item{1.} If $b = 4$, find the value of~$3b^{2}$. +\begin{Soln} +\begin{DPgather*} +\lintertext{\indent Here} +3b^{2} = 3 × 4^{2} = 3 × 16 = 48. +\end{DPgather*} +\end{Soln} + +\Item{2.} If $a = 7$, $b = 2$, $c = 3$, find the value of~$5ab^{2}c^{3}$. +\begin{Soln} +\begin{DPgather*} +\lintertext{\indent Here} +5ab^{2}c^{3} = 5 × 7 × 2^{2} × 3^{3} = 3780. +\end{DPgather*} +\end{Soln} + +\Exercise{3.} + +If $a = 7$, $b = 5$, $c = 3$, find the value of +\begin{multicols}{3} +\Item{1.} $9a$. + +\Item{2.} $8ab$. + +\Item{3.} $4b^{2}c$. + +\Item{4.} $2a^{2}$. + +\Item{5.} $3c^{3}$. + +\Item{6.} $2b^{4}$. + +\Item{7.} $5ac$. + +\Item{8.} $abc$. + +\Item{9.} $abc^{2}$. + +\Item{10.} $\frac{1}{3}abc$. + +\Item{11.} $\frac{1}{5}ab^{2}c$. + +\Item{12.} $\frac{1}{7}a^{2}bc$. +\end{multicols} + +If $a = 5$, $b = 2$, $c = 0$, $x = 1$, $y = 3$, find the value of +\begin{multicols}{3} +\Item{13.} $4acy^{2}$. + +\Item{14.} $3ax^{5}y^{2}$. + +\Item{15.} $2ab^{2}y$. + +\Item{16.} $2a^{2}b^{2}c^{2}y^{2}$. + +\Item{17.} $2a^{2}b^{2}x^{2}y^{2}$. + +\Item{18.} $2abx^{3}y^{3}$. + +\Item{19.} $3abcxy$. + +\Item{20.} $3abx^{3}y^{2}$. + +\Item{21.} $3ab^{2}xy^{2}$. +\end{multicols} +%% -----File: 019.png---Folio 13------- + +\Paragraph{41. The Numerical Value of a Compound Expression.} + +If $a$~stands for~$10$, $b$~for~$4$, and $c$~for~$3$, find the value of +the expression $5ab - 10c^{2} - 5b^{2}$. + +Find the value of each term, and combine the results. +\begin{Soln} +\begin{gather*} +\begin{alignedat}{3} + 5ab &\text{ stands for } &5 × 10 &× 4 &&= 200; \\ +10c^{2} &\text{ stands for } & 10 &× 3^{2} &&= \Z90; \\ +5b^{2} &\text{ stands for } & 5 &× 4^{2} &&= \Z80. +\end{alignedat} \\ +\begin{aligned} +&\therefore 5ab - 10c^{2} - 5b^{2} \\ +&= 200 - 90 - 80 \\ +&= 30. +\end{aligned} +\end{gather*} +\end{Soln} + +\Paragraph{42.} In finding the value of a compound expression the +operations indicated \emph{for each term} must be performed \emph{before} +the operation indicated by the sign prefixed to the term. + +When there is no sign expressed between single symbols +or between \emph{simple} and \emph{compound expressions}, it must be +remembered that the sign understood is the \emph{sign of multiplication}. +Thus $2(a - b)$ has the same meaning as $2 × (a - b)$. + +\Exercise{4.} + +If $a = 5$, $b = 4$, $c = 3$, find the value of +\begin{multicols}{2} +\Item{1.} $9a - 2bc$. + +\Item{2.} $ab + 2c$. + +\Item{3.} $abc + bc$. + +\Item{4.} $5ac + 2a$. + +\Item{5.} $2abc - 2ac^{2}$. + +\Item{6.} $ab + bc - ac$. + +\Item{7.} $ac - (b + c)$. + +\Item{8.} $a^{2} + (b^{2} + c^{2})$\Add{.} + +\Item{9.} $2a + (2b + 2c)$. + +\Item{10.} $a^{2} - b^{2} - c^{2}$. + +\Item{11.} $3(a - b + c)$. + +\Item{12.} $6ab - (bc + 8)$. + +\Item{13.} $7bc - c^{2} + a$. + +\Item{14.} $5ac - b^{2} + 3b$. + +\Item{15.} $4b^{2}c - 5c^{2} - 2b$. + +\Item{16.} $2a + (b + c)$. + +\Item{17.} $b + 2(a - c)$. + +\Item{18.} $c + 2(a - b)$. + +\Item{19.} $2a - (b + c)$. + +\Item{20.} $2b - (a - c)$. + +\Item{21.} $2c - (a - b)$. + +\Item{22.} $2c - 5(a - b)$. + +\Item{23.} $2b - 3(a - c)$. + +\Item{24.} $2c - b(a - b)$. +\end{multicols} +%% -----File: 020.png---Folio 14------- + + +\Section{ALGEBRAIC NOTATION.} + +\Exercise{5.} + +\Item{1.} Read $a + b$; $a - b$; $ab$; $a÷b$. + +\Item{2.} Write six increased by four. \Ans{$6 + 4$.} + +\Item{3.} Write $a$ increased by~$b$. + +\Item{4.} Write six diminished by four. \Ans{$6 - 4$.} + +\Item{5.} Write $a$ diminished by~$b$. + +\Item{6.} By how much does twenty-five exceed sixteen? +\Ans{$25 - 16$.} + +\Item{7.} By how much does $x$ exceed~$y$? + +\Item{8.} Write four times three; the fourth power of three. +\Ans{$4 × 3$; $3^{4}$.} + +\Item{9.} Write four times~$x$; the fourth power of~$x$. + +\Item{10.} If one part of twenty-five is fifteen, what is the +other part? +\Ans{$25 - 15$.} + +\Item{11.} If one part of~$35$ is~$x$, what is the other part? + +\Item{12.} If one part of~$x$ is~$a$, what is the other part? + +\Item{13.} How much does ten lack of being twelve? +\Ans{$12 - 10$.} + +\Item{14.} How much does $x$ lack of being fourteen? + +\Item{15.} How much does $x$ lack of being~$a$? + +\Item{16.} If a man walks four miles an hour, how many miles +will he walk in three hours? +\Ans{$3 × 4$.} + +\Item{17.} If a man walks $y$~miles an hour, how many miles +will he walk in $x$~hours? + +\Item{18.} If a man walks $y$~miles an hour, how many hours +will it take him to walk $x$~miles? +%% -----File: 021.png---Folio 15------- + +\Exercise{6.} + +\Item{1.} If the dividend is twenty and the quotient five, +what is the divisor? +\Ans{$\frac{20}{5}$.} + +\Item{2.} If the dividend is~$a$ and the quotient~$b$, what is the +divisor? + +\Item{3.} If John is twenty years old to-day, how old was he +four years ago? How old will he be five years hence? +\Ans{$20 - 4$; $20 + 5$.} + +\Item{4.} If James is $x$~years old to-day, how old was he three +years ago? How old will he be seven years hence? + +\Item{5.} Write four times the expression seven minus five. +\Ans{$4(7 - 5)$.} + +\Item{6.} Write seven times the expression $2x$~minus~$y$. + +\Item{7.} Write the next integral number above four. +\Ans{$4 + 1$.} + +\Item{8.} If $x$~is an integral number, write the next integral +number above it; the next integral number below it. + +\Item{9.} What number is less than $20$ by~$d$? + +\Item{10.} If the difference of two numbers is five, and the +smaller number is fifteen, what is the greater number? +\Ans{$15 + 5$.} + +\Item{11.} If the difference of two numbers is eight, and the +smaller number is~$x$, what is the greater number? + +\Item{12.} If the sum of two numbers is~$30$, and one of them +is~$20$, what is the other? +\Ans{$30 - 20$.} + +\Item{13.} If the sum of two numbers is~$x$, and one of them is~$10$, +what is the other? + +\Item{14.} If $100$ contains~$x$ ten times, what is the value of~$x$? +%% -----File: 022.png---Folio 16------- + +\Exercise{7.} + +\Item{1.} In $x$~years a man will be $40$~years old; what is his +present age? + +\Item{2.} How old will a man be in $y$~years, if his present age +is $a$~years? + +\Item{3.} What is the value of~$x$ if $7x$~equals~$28$? + +\Item{4.} If it takes $3$~men $4$~days to reap a field, how many +days will it take one man to reap it? +\Ans{$3 × 4$.} + +\Item{5.} If it takes $a$~men $b$~days to reap a field, how many +days will it take one man to reap it? + +\Item{6.} What is the excess of~$5x$ over~$3x$? + +\Item{7.} By how much does $20 - 3$ exceed $(10 + 1)$? +\Ans{$20 - 3 - (10 + 1)$.} + +\Item{8.} By how much does $2x - 3$ exceed $(x + 1)$? + +\Item{9.} If $x$~stands for~$10$, find the value of~$4(3x - 20)$. + +\Item{10.} If $a$~stands for~$10$, and $b$~for~$2$, find the value of +$2(a - 2b)$. + +\Item{11.} How many cents in $a$~dollars, $b$~quarters, and $c$~dimes? + +\Item{12.} A book-shelf contains French, Latin, and Greek +books. There are $100$~books in all, and there are $x$~Latin +and $y$~Greek books. How many French books are there? + +\Item{13.} A regiment of men is drawn up in $10$~ranks of $80$~men +each, and there are $15$~men over. How many men +are there in the regiment? +\Ans{$10 × 80 + 15$.} + +\Item{14.} A regiment of men is drawn up in $x$~ranks of $y$~men +each, and there are $c$~men over. How many men are there +in the regiment? +%% -----File: 023.png---Folio 17------- + +\Exercise{8.} + +\Item{1.} A room is $10$~yards long and $8$~yards wide. In the +middle there is a carpet $6$~yards square. How many +square yards of oilcloth will be required to cover the rest +of the floor? +\Ans{$10 × 8 - 6^{2}$.} + +\Item{2.} A room is $x$~yards long and $y$~yards wide. In the +middle there is a carpet $a$~yards square. How many +square yards of oilcloth will be required to cover the rest +of the floor? + +\Item{3.} How many rolls of paper $g$~feet long and $k$~feet +wide will be required to paper a room, the perimeter of +which, after proper allowance is made for doors and windows, +is $p$~feet and the height $h$~feet? + +\Item{4.} Write six times the square of~$m$, plus five~$c$ times +the expression $d$~plus $b$~minus~$a$. + +\Item{5.} Write five times the expression two~$n$ plus one, +diminished by six times the expression $c$~minus $a$~plus~$b$. + +\Item{6.} A lady bought a dress for $a$~dollars, a cloak for $b$~dollars, +two pairs of gloves for $c$~dollars a pair. She gave +a hundred-dollar bill in payment. How much money +should be returned to her? + +\Item{7.} If a man can perform a piece of work in $4$~days, how +much of it can he do in one day? +\Ans{$\frac{1}{4}$.} + +\Item{8.} If a man can perform a piece of work in $x$~days, +how much of it can he do in one day? + +\Item{9.} If A~can do a piece of work in $x$~days, B~in $y$~days, +C~in $z$~days, how much of it can they all do in one day, +working together? + +\Item{10.} Write an expression for the sum, and also for the +product, of three consecutive numbers of which the least is~$n$. +%% -----File: 024.png---Folio 18------- + +\Item{11.} The product of two factors is~$36$; if one of the +factors is~$x$, what is the other factor? + +\Item{12.} If $d$~is the divisor and $q$~the quotient, what is the +dividend? + +\Item{13.} If $d$~is the divisor, $q$~the quotient, and $r$~the remainder, +what is the dividend? + +\Item{14.} If $x$~oranges can be bought for $50$~cents, how many +oranges can be bought for $100$~cents? + +\Item{15.} What is the price in cents of $x$~apples, if they are +ten cents a dozen? + +\Item{16.} If $b$~oranges cost $6$~cents, what will $a$~oranges cost? + +\Item{17.} How many miles between two places, if a train +travelling $m$~miles an hour requires $4$~hours to make the +journey? + +\Item{18.} If a man was $x$~years old $10$~years ago, how many +years old will he be $7$~years hence? + +\Item{19.} If a man was $x$~years old $y$~years ago, how many +years old will he be $c$~years hence? + +\Item{20.} If a floor is $3x$~yards long and $12$~yards wide, how +many square yards does the floor contain? + +\Item{21.} How many hours will it take to walk $c$~miles, at +the rate of one mile in $15$~minutes? + +\Item{22.} Write three consecutive numbers of which $x$~is the +middle number. + +\Item{23.} If an odd number is represented by~$2n + 1$, what +will represent the next odd number? +%% -----File: 025.png---Folio 19------- + + +\Chapter{II.}{Simple Equations.} + +\Paragraph{43. Equations.} An equation is a statement in symbols +that two expressions stand for the same number. + +\begin{Remark} +Thus, the equation $3x + 2 = 8$ states that $3x + 2$ and~$8$ stand for +the same number. +\end{Remark} + +\Paragraph{44.} That part of the equation which precedes the sign +of equality is called the \Defn{first member}, or \Defn{left side}, and that +which follows the sign of equality is called the \Defn{second member}, +or \Defn{right side}. + +\Paragraph{45.} The statement of equality between two algebraic +expressions, if true for all values of the letters involved, is +called an \Defn{identical equation}; but if true only for certain +particular values of the letters involved, it is called an +\Defn{equation of condition}. + +\begin{Remark} +Thus, $a + b = b + a$, which is true for \emph{all values} of $a$~and~$b$, is an +\emph{identical equation}, and $3x + 2 = 8$, which is true only when $x$~stands +for~$2$, is an \emph{equation of condition}\Add{.} +\end{Remark} + +For brevity, an identical equation is called an \Defn{identity}, +and an equation of condition is called simply an \Defn{equation}. + +\Paragraph{46.} We often employ an equation to discover an \emph{unknown +number} from its relation to known numbers. We usually +represent the unknown number by one of the \emph{last} letters +of the alphabet, as $x$,~$y$,~$z$; and by way of distinction, we +use the \emph{first} letters, $a$,~$b$, $c$,~etc., to represent numbers that +%% -----File: 026.png---Folio 20------- +are supposed to be known, though not expressed in the +number-symbols of Arithmetic. + +\begin{Remark} +Thus, in the equation $ax + b = c$, $x$~is supposed to represent an +unknown number, and $a$,~$b$, and~$c$ are supposed to represent known +numbers. +\end{Remark} + +\Paragraph{47. Simple Equations.} An equation which contains the +first power of~$x$, the symbol for the unknown number, and +no higher power, is called a \Defn{simple equation}, or an \Defn{equation +of the first degree}. + +\begin{Remark} +Thus, $ax + b = c$ is a simple equation, or an equation of the first +degree \emph{in~$x$}. +\end{Remark} + +\Paragraph{48. Solution of an Equation.} To solve an equation is to +find the unknown number; that is, the number which, when +substituted for its symbol in the given equation, renders the +equation an identity. This number is said to \emph{satisfy} the +equation, and is called the \Defn{root} of the equation. + +\Paragraph{49. Axioms.} In solving an equation, we make use of the +following axioms: + +\Ax{1.} If equal numbers be added to equal numbers, +the sums will be equal. + +\Ax{2.} If equal numbers be subtracted from equal numbers, +the remainders will be equal. + +\Ax{3.} If equal numbers be multiplied by equal numbers, +the products will be equal. + +\Ax{4.} If equal numbers be divided by equal numbers, +the quotients will be equal. + +\begin{Theorem} +If, therefore, the two sides of an equation be increased by, +diminished by, multiplied by, or divided by equal numbers, +the results will be equal. +\end{Theorem} + +%[** TN: Next paragraph set in normal-size type in the original] +\begin{Remark} +Thus, if $8x = 24$, then $8x + 4 = 24 + 4$, $8x - 4 = 24 - 4$, +$4 × 8x = 4 × 24$, and $8x ÷ 4 = 24 ÷ 4$. +\end{Remark} +%% -----File: 027.png---Folio 21------- + +\Paragraph{50. Transposition of Terms.} It becomes necessary in solving +an equation to bring all the terms that contain the +symbol for the unknown number to one side of the equation, +and all the other terms to the other side. This is +called \Defn{transposing the terms}. We will illustrate by examples: + +\Item{1.} Find the number for which $x$~stands when +\[ +14x - 11 = 5x + 70. +\] + +The first object to be attained is to get all the terms +which contain~$x$ on the left side of the equation, and all the +other terms on the right side. This can be done by first +subtracting~$5x$ from both sides (Ax.~2), which gives +\[ +9x - 11 = 70, +\] +and then adding~$11$ to these equals (Ax.~1), which gives +\begin{DPalign*} +9x + 11 - 11 &= 70 + 11. \\ +\lintertext{\indent Combine,} +9x &= 81. \\ +\lintertext{\indent Divide by~$9$,} +x &= 9. +\end{DPalign*} + +\Item{2.} Find the number for which $x$~stands when $x + b = a$. +\begin{DPalign*}[m] +\lintertext{\indent The equation is} +x + b &= a. \\ +\lintertext{\indent Subtract~$b$ from each side,} +x + b - b &= a - b. +\rintertext{(Ax.~2)} +\end{DPalign*} + +Since $+b$~and~$-b$ in the left side cancel each other +(§~30), we have $x = a - b$. + +\Item{3.} Find the number for which $x$~stands when $x - b = a$. +\begin{DPalign*} +\lintertext{\indent The equation is} +x - b &= a. \\ +\lintertext{Add $+b$ to each side,} +x + b - b &= a + b. +\rintertext{(Ax.~1)} +\end{DPalign*} + +Since $+b$~and~$-b$ in the left side cancel each other +(§~30), we have $x = a + b$. +%% -----File: 028.png---Folio 22------- + +\Paragraph{51.} The effect of the operation in the preceding equations, +when Axioms (1)~and~(2) are used, is to take a term +from one side and put it on the other side with its sign +changed. We can proceed in a like manner in any other +case. Hence the general rule: + +\Paragraph{52.} \begin{Theorem}[nopar] Any term may be transposed from one side of an +equation to the other, provided its sign is changed. +\end{Theorem} + +\Paragraph{53.} Any term, therefore, which occurs on both sides +with \emph{the same sign} may be removed from both without +affecting the equality; and the sign of every term of an +equation may be changed without affecting the equality. + +\Paragraph{54. Verification.} When the root is substituted for its +symbol in the given equation, and the equation reduces to +an \emph{identity}, the root is said to be \Defn{verified}. We will illustrate +by examples: + +\Item{1.} What number added to twice itself gives~$24$? + +Let $x$~stand for the number; \\ +then $2x$~will stand for twice the number, \\ +and the number added to twice itself will be $x + 2x$. + +But the number added to twice itself is~$24$. +\begin{DPalign*} +\therefore x + 2x &= 24. \\ +\lintertext{\indent Combine $x$~and~$2x$,} +3x &= 24. \\ +\intertext{\indent Divide by~$3$, the coefficient of~$x$,} +x &= 8. +\rintertext{(Ax.~4)} +\end{DPalign*} + +Therefore the required number is~$8$. + +\begin{DPalign*} +\lintertext{\indent\textsc{Verification.}} +x + 2x &= 24, \\ +8 + 2 × 8 &= 24, \\ +8 + 16 &= 24, \\ +24 &= 24. +\end{DPalign*} +%% -----File: 029.png---Folio 23------- + +\ScreenBreak +\Item{2.} If $4x - 5$ stands for~$19$, for what number does $x$~stand? + +We have the equation +\begin{DPalign*} +4x - 5 &= 19. \\ +\lintertext{\indent Transpose $-5$,} +4x &= 19 + 5. \\ +\lintertext{\indent Combine,} +4x &= 24. \\ +\lintertext{\indent Divide by~$4$,} +x &= 6. +\rintertext{(Ax.~4)} \displaybreak[1] \\ +\lintertext{\indent \textsc{Verification.}} +4x - 5 &= 19, \\ +4 × 6 - 5 &= 19, \\ +24 - 5 &= 19, \\ +19 &= 19. +\end{DPalign*} + +\Item{3.} If $3x - 7$ stands for the same number as $14 - 4x$, +what number does $x$~stand for? + +We have the equation +\[ +3x - 7 = 14 - 4x. +\] + +Transpose $-4x$ to the left side, and $-7$ to the right side, +\begin{DPalign*} +3x + 4x &= 14 + 7. \\ +\lintertext{\indent Combine,} +7x &= 21. \\ +\lintertext{\indent Divide by~$7$,} +x &= 3. \displaybreak[1] \\ +\lintertext{\indent \textsc{Verification.}} +3x - 7 &= 14 - 4x, \\ +3 × 3 - 7 &= 14 - 4 × 3, \\ +2 &= 2. +\end{DPalign*} + +\Item{4.} Solve the equation +\[ +7(x - 1) - 30 = 4(x - 4). +\] +We have the equation +\[ +7(x - 1) - 30 = 4(x - 4). +\] +%% -----File: 030.png---Folio 24------- +%[** TN: Equation repeated at page break in the original] +% 7(x - 1) - 30 = 4(x - 4). + +Remove the parentheses, +\begin{DPalign*} +7x - 7 - 30 &= 4x - 16. \\ +\lintertext{\indent Then} +7x - 4x &= 7 + 30 - 16. \\ +\lintertext{\indent Combine,} +3x &= 21. \\ +\lintertext{\indent Divide by~$3$,} +x &= 7. \displaybreak[1] \\ +\lintertext{\indent \textsc{Verification.}} +7(7 - 1) - 30 &= 4(7 - 4), \\ +7 × 6 - 30 &= 4 x 3, \\ +42 - 30 &= 12, \\ +12 &= 12. +\end{DPalign*} + +\Exercise{9.} + +Find the number that $x$~stands for, if: +\begin{multicols}{2} +\Item{1.} $3x = x + 8$. + +\Item{2.} $3x = 2x + 5$. + +\Item{3.} $3x + 4 = x + 10$. + +\Item{4.} $4x + 6 = x + 9$. + +\Item{5.} $7x - 19 = 5x + 7$. + +\Item{6.} $3(x - 2) = 2(x - 3)$. + +\Item{7.} $8x + 7 = 4x + 27$. + +\Item{8.} $3x + 10 = x + 20$. + +\Item{9.} $5(x - 2) = 3x + 4$. + +\Item{10.} $3(x - 2) = 2(x - 1)$. + +\Item{11.} $2x + 3 = 16 - (2x - 3)$. + +\Item{12.} $19x - 3 = 2(7 + x)$. + +\Item{13.} $7x - 70 = 5x - 20$. + +\Item{14.} $2x - 22 = 108 - 2x$. + +\Item{15.} $2(x + 5) + 5(x - 4) = 32$. + +\Item{16.} $2(3x - 25) = 10$. + +\Item{17.} $33x - 70 = 3x + 20$. + +\Item{18.} $4(1 + x) + 3(2 + x) = 17$. + +\Item{19.} $8x - (x + 2) = 47$. + +\Item{20.} $3(x - 2) = 50 - (2x - 9)$. +\end{multicols} +%% -----File: 031.png---Folio 25------- + +\Item{21.} $2x - (3 + 4x - 3x + 5) = 4$. + +\Item{22.} $5(2 - x) + 7x - 21 = x + 3$. + +\Item{23.} $3(x - 2) + 2(x - 3) + (x - 4) = 3x + 5$. + +\Item{24.} $x + 1 + x + 2 + x + 4 = 2x + 12$. + +\Item{25.} $(2x - 5) - (x - 4) + (x - 3) = x - 4$. + +\Item{26.} $4 - 5x - (1 - 8x) = 63 - x$. + +\Item{27.} $3x - (x + 10) - (x - 3) = 14 - x$. + +\Item{28.} $x^{2} - 2x - 3 = x^{2} - 3x + 1$. + +\Item{29.} $(x^{2} - 9) - (x^{2} - 16) + x = 10$. + +\Item{30.} $x^{2} + 8x - (x^{2} - x - 2) = 5(x + 3) + 3$. + +\Item{31.} $x^{2} + x - 2 + x^{2} + 2x - 3 = 2x^{2} - 7x - 1$. + +\Item{32.} $10x - (x - 5) = 2x + 47$. + +\Item{33.} $7x - 5 - (6 - 8x) + 2 = 3x - 7 + 106$. + +\Item{34.} $6x + 3 - (3x + 2) = (2x - 1) + 9$. + +\Item{35.} $3(x + 10) + 4(x + 20) + 5x - 170 = 15 - 3x$. + +\Item{36.} $20 - x + 4(x - 1) - (x - 2) = 30$. + +\Item{37.} $5x + 3 - (2x - 2) + (1 - x) = 6(9 - x)$. + + +\Paragraph{55. Statement and Solution of Problems.} The difficulties +which the beginner usually meets in stating problems will +be quickly overcome if he will observe the following directions: + +Study the problem until you clearly understand its meaning +and just what is required to be found. + +Remember that $x$~must not be put for money, length, +time, weight,~etc., but for the \textbf{required number} of \emph{specified +units} of money, length, time, weight,~etc. + +Express each statement carefully in algebraic language, +and write out in full just what each expression stands for. +%% -----File: 032.png---Folio 26------- + +Do not attempt to form the equation until all the statements +are made in symbols. + +We will illustrate by examples: + +\Item{1.} John has three times as many oranges as James, and +they together have~$32$. How many has each? +\begin{Soln} +\begin{DPgather*} +\lintertext{\indent Let} +\text{$x$~stand for the \emph{number} of oranges James has;} \\ +\lintertext{then} +\text{$3x$~is the number of oranges John has;} \\ +\lintertext{and} +\text{$x + 3x$ is the number of oranges they together have.} +\end{DPgather*} + +But $32$~is the number of oranges they together have. +\begin{DPalign*} +\therefore x + 3x &= 32; \\ +\lintertext{or,} +4x &= 32, \\ +\lintertext{and} +x &= 8. \\ +\lintertext{\indent Since $x = 8$,} +3x &= 24. +\end{DPalign*} +\end{Soln} + +Therefore James has $8$~oranges, and John has $24$~oranges. + +\begin{Remark}[Note.] Beginners in stating the preceding problem generally write: +\[ +\text{Let $x = {}$\emph{what} James had.} +\] + +Now, we know \emph{what} James had. He had oranges, and we are to +discover simply the \emph{number} of oranges he had. +\end{Remark} + +\Item{2.} James and John together have~\$$24$, and James has +\$$8$~more than John. How many dollars has each? +\begin{Soln} +\begin{DPgather*} +\lintertext{\indent Let} +\text{$x$~stand for the number of dollars John has;} \\ +\lintertext{then} +\text{$x + 8$ is the number of dollars James has;} \\ +\lintertext{and} +\text{$x + (x + 8)$ is the number of dollars they both have.} +\end{DPgather*} + +But $24$~is the number of dollars they both have. +\[ +\therefore x + (x + 8) = 24. +\] + +Removing the parenthesis, +\[ +x + x + 8 = 24. +\] +\begin{DPalign*} +\therefore 2x &= 16. \\ +\lintertext{\indent Dividing by~$2$,} +x &= 8. \\ +\lintertext{\indent Since $x = 8$,} +x + 8 &= 16. +\end{DPalign*} +\end{Soln} + +Therefore John has~\$$8$, and James has~\$$16$. +%% -----File: 033.png---Folio 27------- + +\begin{Remark}[Note\Add{.}] The beginner must avoid the mistake of writing +\[ +\text{Let $x = {}$John's money\Add{.}} +\] + +We are required to find the \emph{number} of dollars John has, and therefore +$x$~must represent this required number. +\end{Remark} + +\Item{3.} The sum of two numbers is~$18$, and three times the +greater number exceeds four times the less by~$5$. Find the +numbers. + +\begin{Soln} +Let $x = {}$the greater number. + +Then, since $18$~is the sum and $x$~is one of the numbers, the other +number must be the sum minus~$x$. Hence +\[ +18 - x = \text{the smaller number}\Add{.} +\] + +Now, three times the greater number is~$3x$, and four times the less +number is~$4(18 - x)$\Add{.} +\begin{DPalign*} +\lintertext{\indent Hence,} +&3x - 4(18 - x) = \text{the excess}\Add{.} \\ +\lintertext{\indent But} +&5 = \text{the excess}\Add{,} \\ +\therefore\ &3x - 4(18 - x) = 5 \\ +\therefore\ &3x - (72 - 4x) = 5, \\ +\lintertext{or} +&3x - 72 + 4x = 5. \\ +\therefore\ &7x = 77, \\ +\lintertext{and} +&\Z x = 11\Add{.} +\end{DPalign*} +\end{Soln} + +Therefore the numbers are $11$~and~$7$. + +\Exercise{10.} + +\Item{1.} If a number is multiplied by~$9$, the product is~$270$. +Find the number. + +\Item{2.} If the sum of the ages of a father and son is $60$~years, +and the father is $5$~times as old as the son, what is the +age of each? + +\Item{3.} The sum of two numbers is~$91$, and the greater is $6$~times +the less. Find the numbers. +%% -----File: 034.png---Folio 28------- + +\Item{4.} A tree $90$~feet high was broken so that the part +broken off was $8$~times the length of the part left standing. +Find the length of each part. + +\Item{5.} The difference of two numbers is~$7$, and their sum is~$53$. +Find the numbers. + +\Item{6.} The difference of two numbers is~$12$, and their sum is~$84$. +Find the numbers. + +\Item{7.} Divide $35$ into two parts so that one part shall be +greater by~$5$ than the other part. + +\Item{8.} Three times a given number is equal to the number +increased by~$40$. Find the number. + +\Item{9.} Three times a given number diminished by~$24$ is +equal to the given number. Find the number. + +\Item{10.} One number is $4$~times another, and their difference +is~$30$. Find the numbers. + +\Item{11.} The sum of two numbers is~$36$, and one of them +exceeds twice the other by~$6$. Find the numbers. + +\begin{Remark}[Hint.] Let $x$~equal the greater number: then $36 - x$ will equal the +smaller. +\end{Remark} + +\Item{12.} The sum of two numbers is~$40$, and $5$~times the +smaller exceeds $2$~times the greater by~$25$. Find the +numbers. + +\Item{13.} The number $30$ is divided into two parts such that +$4$~times the greater part exceeds $5$~times the smaller part +by~$30$. Find the parts. + +\Item{14.} The sum of two numbers is~$27$, and twice the greater +number increased by $3$~times the less is~$61$. Find the +numbers. + +\Item{15.} The sum of two numbers is~$32$, and five times the +smaller is $3$~times the greater number. Find the numbers. +%% -----File: 035.png---Folio 29------- + +\Exercise{11.} + +\Item{1.} A farmer sold a horse and a cow for~\$$210$. He +sold the horse for four times as much as the cow. How +much did he get for each? + +\Item{2.} Three times the excess of a certain number over~$6$ +is equal to the number plus~$144$. Find the number. + +\Item{3.} Thirty-one times a certain number is as much above~$40$ +as nine times the number is below~$40$. Find the number. + +\Item{4.} Two numbers differ by~$10$, and their sum is equal to +seven times their difference. Find the numbers. + +\Item{5.} Find three consecutive numbers, $x$, $x + 1$, and~$x + 2$, +whose sum is~$78$. + +\Item{6.} Find five consecutive numbers whose sum is~$35$. + +\Item{7.} The sum of the ages of A~and~B is $40$~years, and $10$~years +hence A~will be twice as old as~B\@. Find their +present ages. + +\Item{8.} A father is four times as old as his son, and in $5$~years +he will be only three times as old. Find their present ages. + +\Item{9.} One man is $60$~years old, and another man is $50$~years. +How many years ago was the first man twice as +old as the second? + +\Item{10.} A man $50$~years old has a son $10$~years old. In +how many years will the father be three times as old as +the son? + +\Item{11.} A~has~\$$100$, and B~has~\$$20$. How much must A +give B in order that they may each have the same sum? + +\Item{12.} A banker paid \$$63$ in $5$-dollar bills and $2$-dollar +bills. He paid just as many $5$-dollar bills as $2$-dollar bills. +How many bills of each kind did he pay? +%% -----File: 036.png---Folio 30------- + +\Exercise{12.} + +\Item{1.} In a company of $90$~persons, composed of men, +women, and children, there are three times as many children +as men, and twice as many women as men. How many +are there of each? + +\Item{2.} Find the number whose double exceeds~$70$ by as +much as the number itself is less than~$80$. + +\Item{3.} A farmer employed two men to build $112$~rods of +wall. One of them built on the average $4$~rods a day, and +the other $3$~rods a day. How many days did they work? + +\Item{4.} Two men travel in \emph{opposite} directions, one $30$~miles +a day, and the other $20$~miles a day. In how many days +will they be $350$~miles apart? + +\Item{5.} Two men travel in the same direction, one $30$~miles +a day, and the other $20$~miles a day. In how many days +will they be $350$~miles apart? + +\Item{6.} A man bought $3$~equal lots of hay for~\$$408$. For +the first lot he gave \$$17$~a ton, for the second~\$$16$, for the +third~\$$18$. How many tons did he buy in all? + +\Item{7.} A farmer sold a quantity of wood for~\$$84$, one half +of it at \$$3$~a cord, and the other half at \$$4$~a cord. How +many cords did he sell? + +\begin{Remark}[Hint.] +Let $2x$~equal the number of cords. +\end{Remark} + +\Item{8.} If $2x - 3$ stands for~$29$, for what number will +$4 + x$ stand? + +\Item{9.} At an election two opposing candidates received +together $2044$~votes, and one received $104$~more votes than +the other. How many votes did each candidate receive? +%% -----File: 037.png---Folio 31------- + +\Exercise{13.} + +\Item{1.} A man walks $4$~miles an hour for $x$~hours, and +another man walks $3$~miles an hour for $x + 2$~hours. If +they each walk the same distance, how many miles does +each walk? + +\Item{2.} A has twice as much money as~B; but if A gives~B +\$$30$, it will take twice as much as A has left to equal~B's. +How much money has each? + +\Item{3.} I have \$$12.75$ in two-dollar bills and twenty-five +cent pieces, and I have twice as many bills as twenty-five +cent pieces. How many have I of each? + +\Item{4.} I have in mind a certain number. If this number +is diminished by~$8$ and the remainder multiplied by~$8$, the +result is the same as if the number was diminished by~$6$ +and the remainder multiplied by~$6$. What is the number? + +\Item{5.} I have five times as many half-dollars as quarters, and +the half-dollars and quarters amount to~\$$11$. How many +of each have I? + +\Item{6.} A man pays a debt of~\$$91$ with ten-dollar bills and +one-dollar bills, paying three times as many one-dollar bills +as ten-dollar bills. How many bills of each kind does he +pay? + +\Item{7.} A father is four times as old as his son, but $4$~years +hence he will be only three times as old as his son. How +old is each? + +\Item{8.} A workman was employed for $24$~days. For every +day he worked he was to receive~\$$1.50$, and for every day +he was idle he was to pay \$$0.50$ for his board. At the end +of the time he received~\$$28$. How many days did he +work? +%% -----File: 038.png---Folio 32------- + +\Exercise{14.} + +\Item{1.} A boy has $4$~hours at his disposal. How far can he +ride into the country at the rate of $9$~miles an hour and +walk back at the rate of $3$~miles an hour, if he returns just +on time? + +\begin{Remark}[Hint.] Let $x = {}$the number of hours he rides. + +Then $4 - x = {}$the number he walks. +\end{Remark} + +\Item{2.} A has~\$$180$, and B has~\$$80$. How much must A +give B in order that six times B's money shall be equal to +$7$~times~A's? + +\Item{3.} A grocer has two kinds of tea, one kind worth $45$~cents +a pound, and the other worth $65$~cents a pound. +How many pounds of each kind must he take to make $80$~pounds, +worth $50$~cents a pound? + +\Item{4.} A tank holding $1200$~gallons has three pipes. The +first lets in $8$~gallons a minute, the second $10$~gallons, and +the third $12$~gallons a minute. In how many minutes will +the tank be filled? + +\Item{5.} The fore and hind wheels of a carriage are $10$~feet +and $12$~feet respectively in circumference. How many +feet will the carriage have passed over when the fore wheel +has made $250$~revolutions more than the hind wheel? + +\Item{6.} Divide a yard of tape into two parts so that one part +shall be $6$~inches longer than the other part. + +\Item{7.} A boy bought $7$~dozen oranges for~\$$1.50$. For a part +he paid $20$~cents a dozen; and for the remainder, $25$~cents +a dozen. How many dozen of each kind did he buy? + +\Item{8.} How can a bill of~\$$3.30$ be paid in quarters and ten-cent +pieces so as to pay three times as many ten-cent +pieces as quarters? +%% -----File: 039.png---Folio 33------- + + +\Chapter{III.}{Positive and Negative Numbers.} + +\Paragraph{56. Quantities Opposite in Kind.} If a person is engaged +in trade, his capital will be \emph{increased} by his \emph{gains}, and +\emph{diminished} by his \emph{losses}. + +\emph{Increase} in temperature is measured by the number of +degrees the mercury rises in a thermometer, and \emph{decrease} +in temperature by the number of degrees the mercury \emph{falls}. + +In considering any quantity whatever, a quantity that +\emph{increases} the quantity considered is called a \emph{positive quantity}; +and a quantity that \emph{decreases} the quantity considered +is called a \emph{negative quantity}. + +\Paragraph{57. Positive and Negative Numbers.} If from a given +point, marked~$0$, we draw a straight line to the right, and +beginning from the \emph{zero} point lay off units of length on this +line, the successive repetitions of the unit will be expressed +by the \emph{natural series of numbers}, $1$,~$2$, $3$, $4$,~etc. Thus: +\Graphic{1} + +If we wish to \emph{add} $2$~to~$5$, we begin at~$5$, count $2$~units +\emph{forwards}, and arrive at~$7$, the sum required. If we wish +to \emph{subtract} $2$~from~$5$, we begin at~$5$, count $2$~units \emph{backwards}, +and arrive at~$3$, the difference required. If we wish +to subtract $5$~from~$5$, we count $5$~units backwards, and arrive +at~$0$. If we wish to subtract $5$~from~$2$, we cannot do it, +because when we have counted backwards from~$2$ as far as~$0$, +\emph{the natural series of numbers comes to an end}. +%% -----File: 040.png---Folio 34------- + +In order to subtract a greater number from a smaller, it +is necessary to \emph{assume} a new series of numbers, beginning +at zero and extending to the left of zero. The series to the +left of zero must proceed from zero by \emph{the repetitions of the +unit}, precisely like the natural series to the right of zero; +and the \emph{opposition} between the right-hand series and the +left-hand series must be clearly marked. This opposition +is indicated by calling every number in the right-hand +series a \Defn{positive number}, and prefixing to it, when written, +the sign~$+$; and by calling every number in the left-hand +series a \Defn{negative number}, and prefixing to it the sign~$-$. +The two series of numbers may be called the \Defn{algebraic series +of numbers}, and written thus: +\Graphic{2} + +If, now, we wish to subtract $7$~from~$4$, we begin at~$4$ in +the positive series, count $7$~units in the \emph{negative direction} +(to the left), and arrive at~$-3$ in the negative series; that +is, $4 - 7 = -3$. + +The result obtained by subtracting a greater number from +a less, when both are positive, is \emph{always a negative number}. + +In general, if $a$~and~$b$ represent any two numbers of the +positive series, the expression $a - b$ will be a positive number +when $a$~is greater than~$b$; will be zero when $a$~is equal +to~$b$; will be a negative number when $a$~is less than~$b$. + +In counting from left to right in the algebraic series, numbers +\emph{increase} in magnitude; in counting from right to left, +numbers \emph{decrease} in magnitude. Thus $-3$,~$-1$, $0$, $+2$,~$+4$, +are arranged in \emph{ascending} order of magnitude. + +\Paragraph{58.} Every algebraic number, as $+4$~or~$-4$, consists of a +\emph{sign} $+$~or~$-$ and the \emph{absolute value} of the number. The +sign shows whether the number belongs to the positive or +%% -----File: 041.png---Folio 35------- +negative series of numbers; the absolute value shows the +place the number has in the positive or negative series. + +When no sign stands before a number, the sign~$+$ is +always understood. Thus $4$~means the same as~$+4$, $a$~means +the same as~$+a$. But \emph{the sign~$-$ is never omitted}. + +\Paragraph{59.} Two algebraic numbers which have, one the sign~$+$, +and the other the sign~$-$, are said to have \emph{unlike signs}. + +Two algebraic numbers which have the same absolute +values, but unlike signs, always cancel each other when +combined. Thus $+4 - 4 = 0$; $+a - a = 0$. + +\Paragraph{60. Double Meanings of the Signs $+$~and~$-$.} The use of +the signs $+$~and~$-$ to indicate addition and subtraction +must be carefully distinguished from the use of the signs $+$~and~$-$ +to indicate in which series, the positive or the negative, +a given number belongs. In the first sense they are +signs of \emph{operations}, and are common to Arithmetic and +Algebra; in the second sense they are signs of \emph{opposition}, +and are employed in Algebra alone. + +\begin{Remark}[Note.] In Arithmetic, if the things counted are \emph{whole units}, the +numbers which count them are called \Defn{whole numbers}, \Defn{integral numbers}, +or \Defn{integers}, where the adjective is transferred from the things counted +to the numbers which count them. But if the things counted are +only \emph{parts of units}, the numbers which count them are called \Defn{fractional +numbers}, or simply \Defn{fractions}, where again the adjective is transferred +from the things counted to the numbers which count them. + +Likewise in Algebra, if the units counted are \emph{negative}, the numbers +which count them are called \Defn{negative numbers}, where the adjective +which defines the nature of the units counted is transferred to the +numbers that count them. + +A whole number means a number of whole units, a fractional number +means a number of parts of units, and a negative number means +a number of negative units. +\end{Remark} +%% -----File: 042.png---Folio 36------- + +\Paragraph{61. Addition and Subtraction of Algebraic Numbers.} An +algebraic number which is to be added or subtracted is +often \DPtypo{inclosed}{enclosed} in a parenthesis, in order that the signs $+$~and~$-$, +which are used to distinguish positive and negative +numbers, may not be confounded with the $+$~and~$-$ signs +that denote the operations of addition and subtraction. +Thus $+4 + (-3)$ expresses the sum, and $+4 - (-3)$ expresses +the difference, of the numbers $+4$~and~$-3$. + +\Paragraph{62. Addition.} In order to add two algebraic numbers, we +begin at the place in the series which the first number occupies, +and count, \emph{in the direction indicated by the sign of the +second number}, as many units as there are in the absolute +value of the second number. +\Graphic{3} + +Thus the sum of $+4 + (+3)$ is found by counting from +$+4$ three units in \emph{the positive direction}; that is, to the +right, and is, therefore,~$+7$. + +The sum of $+4 + (-3)$ is found by counting from $+4$ +three units in \emph{the negative direction}; that is, to the left, and +is, therefore,~$+1$. + +The sum of $-4 + (+3)$ is found by counting from $-4$ +three units in the positive direction, and is, therefore,~$-1$. + +The sum of $-4 + (-3)$ is found by counting from $-4$ +three units in the negative direction, and is, therefore,~$-7$. + +\Paragraph{63. Subtraction.} In order to subtract one algebraic number +from another, we begin at the place in the series which +the minuend occupies, and count, \emph{in the direction opposite to +that indicated by the sign of the subtrahend}, as many units +as there are in the absolute value of the subtrahend. + +Thus the result of subtracting $+3$~from~$+4$ is found by +%% -----File: 043.png---Folio 37------- +counting from $+4$ three units in the \emph{negative direction}; +that is, in the direction \emph{opposite to that indicated by the sign~$+$ +before~$3$}, and is, therefore,~$+1$. + +The result of subtracting $-3$~from~$+4$ is found by counting +from $+4$ three units in the \emph{positive direction}, and is, +therefore,~$+7$. + +The result of subtracting $+3$~from~$-4$ is found by counting +from $-4$ three units in the \emph{negative direction}, and is, +therefore,~$-7$. + +The result of subtracting $-3$~from~$-4$ is found by counting +from $-4$ three units in the \emph{positive direction}, and is, +therefore,~$-1$. + +\Paragraph{64.} Collecting the results obtained in addition and subtraction, +we have: +\[ +\begin{array}{c>{\quad}c} +\textsc{Addition.} & \textsc{Subtraction.} \\ + +4 + (-3) = +4 - 3 = +1. & +4 - (+3) = +4 - 3 = +1. \\ + +4 + (+3) = +4 + 3 = +7. & +4 - (-3) = +4 + 3 = +7. \\ +- 4 + (-3) = -4 - 3 = -7. & -4 - (+3) = -4 - 3 = -7. \\ +- 4 + (+3) = -4 + 3 = -1. & -4 - (-3) = -4 + 3 = -1. \\ +\end{array} +\] + +\Paragraph{65.} From these four cases of addition, therefore, + +% [**** TN: Book uses commas elsewhere] +\Dictum{To Add Algebraic Numbers}\DPtypo{:}{,} +\begin{Theorem}[I.] If the numbers have like signs, find the sum of their +absolute values, and prefix the common sign to the result. +\end{Theorem} + +\begin{Theorem}[II.] If the numbers have unlike signs, find the difference +of their absolute values, and prefix the sign of the greater +number to the result. +\end{Theorem} + +\begin{Theorem}[III.] If there are more than two numbers, find the sum +of the positive numbers and the sum of the negative numbers, +%% -----File: 044.png---Folio 38------- +take the difference between the absolute values of these two +sums, and prefix the sign of the greater sum to the result. +\end{Theorem} + +\begin{Remark}[Note.] Since the order in which numbers are added is immaterial, +we may add any two of the numbers, and then this sum to +any third number, and so on. +\end{Remark} + +\Paragraph{66.} The result is generally called the \Defn{algebraic sum}, in +distinction from the arithmetical sum; that is, the sum of +the absolute values of the numbers. + +\Paragraph{67.} From the four cases of subtraction in §~64, we see +that \textit{subtracting a positive number is equivalent to adding +an equal negative number, and subtracting a negative number +is equivalent to adding an equal positive number}. + + +\Dictum{To Subtract One Algebraic Number from Another}, +\begin{Theorem} +Change the sign of the subtrahend, and add the subtrahend +to the minuend. +\end{Theorem} + +\Paragraph{68. Examples.} + +\Item{1.} Find the sum of $3a$, $2a$, $a$, $5a$, $7a$. + +The sum of the coefficients is $3 + 2 + 1 + 5 + 7 = 18$. + +Hence the sum of the numbers is~$18a$. + +\Item{2.} Find the sum of $-5c$, $-c$, $-3c$, $-4c$, $-2c$. + +The sum of the coefficients is $-5 - 1 - 3 - 4 - 2 = -15$. + +Hence the sum of the numbers is~$-15c$. + +\Item{3.} Find the sum of $8x$, $-9x$, $-x$, $3x$, $4x$, $-12x$, $x$. + +The sum of the positive coefficients is $8 + 3 + 4 + 1 = 16$. + +The sum of the negative coefficients is $-9 - 1 - 12 = -22$. + +The difference between $16$~and~$22$ is~$6$, and the sign of +the greater is negative. + +Hence the required sum is~$-6x$. +%% -----File: 045.png---Folio 39------- + +\PrintBreak +\Exercise{15.} + +Find the sum of: +\begin{multicols}{2} +\Item{1.} $5c$, $23c$, $c$, $11c$. + +\Item{2.} $4a$, $3a$, $7a$, $10a$. + +\Item{3.} $7x$, $12x$, $11x$, $9x$. + +\Item{4.} $6y$, $8y$, $2y$, $35y$. + +\Item{5.} $-3a$, $-5a$, $-18a$. + +\Item{6.} $-5x$, $-6x$, $-18x$, $-11x$. + +\Item{7.} $-3b$, $-b$, $-9b$, $-4b$. + +\Item{8.} $-z$, $-2z$, $-10z$, $-53z$. + +\Item{9.} $-11m$, $-3m$, $-5m$, $-m$. + +\Item{10.} $5d$, $-d$, $-4d$, $2d$. +\end{multicols} + +\Item{11.} $13n$, $13n$, $-11n$, $-6n$, $-9n$, $n$, $2n$, $-3n$. + +\Item{12.} $5g$, $-3g$, $-6g$, $-4g$, $20g$, $-5g$, $-11g$, $-14g$. + +\Item{13.} $-9a^{2}$, $5a^{2}$, $6a^{2}$, $a^{2}$, $2a^{2}$, $-a^{2}$, $-3a^{2}$. + +\Item{14.} $3x^{3}$, $-2x^{3}$, $-5x^{3}$, $-7x^{3}$, $-x^{3}$, $2x^{3}$, $-10x^{3}$, $-x^{3}$. + +\Item{15.} $4a^{2}b^{2}$, $-a^{2}b^{2}$, $-6a^{2}b^{2}$, $4a^{2}b^{2}$, $-2a^{2}b^{2}$, $a^{2}b^{2}$. + +\Item{16.} $6mn$, $-5mn$, $mn$, $-3mn$, $4mn$. + +\Item{17.} $3xyz$, $-2xyz$, $5xyz$, $-7xyz$, $xyz$. + +\Item{18.} $5a^{3}b^{3}c^{3}$, $-7a^{3}b^{3}c^{3}$, $-3a^{3}b^{3}c^{3}$, $2a^{3}b^{3}c^{3}$. + +\Item{19.} $11abcd$, $-10abcd$, $-9abcd$, $-abcd$. + +\Item{20.} Subtract $-a$ from $-b$, and find the value of the +result if $a = -4$, $b = -5$. + +When $a = 4$, $b = -2$, $c = -3$, find the difference in +the values of: + +\Item{21.} $a - b + c$ and $-a + b + c$. + +\Item{22.} $a + (-b) + c$ and $a - (-b) + c$. + +\Item{23.} $-a - (-b) + c$ and $-(-a) + (-b) - c$. + +\Item{24.} $a - b + (-c)$ and $a - (-b) - (-c)$. +%% -----File: 046.png---Folio 40------- + + +\Section{MULTIPLICATION AND DIVISION OF ALGEBRAIC +NUMBERS} + +\Paragraph{69. Multiplication.} Multiplication is generally defined +in Arithmetic as the process of finding the result when one +number (the multiplicand) is taken as many times as there +are units in another number (the multiplier). This definition +fails when the \emph{multiplier is a fraction}. In multiplying +by a fraction, we divide the multiplicand into as many +equal parts as there are units in the denominator, and take +as many of these parts as there are units in the numerator. + +If, for example, we multiply $6$~by~$\frac{2}{3}$, we divide $6$ into +\emph{three} equal parts and take \emph{two} of these parts, obtaining $4$ +for the product. The multiplier,~$\frac{2}{3}$, is~$\frac{2}{3}$ of~$1$, and the +product,~$4$, is~$\frac{2}{3}$ of~$6$, in other words, \emph{the product is obtained +from the multiplicand precisely as the multiplier is obtained +from~$1$}. + +This statement is also true when the multiplier is a whole +number. Thus in $5 × 7 = 35$, the multiplier,~$5$, is equal to +\[ +1 + 1 + 1 + 1 + 1, +\] +and the product,~$35$, is equal to +\[ +7 + 7 + 7 + 7 + 7. +\] + +\Paragraph{70.} \Dictum{Multiplication may be defined}, therefore, + +As the operation of finding from two given numbers, +called \emph{multiplicand} and \emph{multiplier}, a third number called +\emph{product}, which is \emph{formed from the multiplicand as the multiplier +is formed from unity}. + +\Paragraph{71.} According to this definition of multiplication, +\begin{DPalign*}[m] +\lintertext{since} ++3 &= + 1 + 1 + 1, \\ +3 × (+8) &= +8 + 8 + 8 +\Tag{(1)} \\ +&= +24, \displaybreak[1] \\ +%% -----File: 047.png---Folio 41------- +\lintertext{and} +3 × (-8) &= -8 - 8 - 8 +\Tag{(2)} \\ +&= -24. \displaybreak[1] \\ +\lintertext{\indent Again, since} +-3 &= -1 - 1 - 1; \\ +(-3) × 8 &= -8 - 8 - 8 +\Tag{(3)} \\ +&= -24, \displaybreak[1] \\ +\lintertext{and} +(-3) × (-8) &= -(-8) - (-8) - (-8) +\Tag{(4)} \\ +&= +8 + 8 + 8 \\ +&= +24. +\end{DPalign*} + +\Paragraph{72.} From these four cases it follows that in finding +the product of two algebraic numbers, if the two numbers +have \emph{like} signs, the product will have the \emph{plus} sign, and if +\emph{unlike} signs, the product will have the \emph{minus} sign. + +Hence the \Defn{Law of Signs in Multiplication} is: +\begin{Theorem} +Like signs give~$+$, and unlike signs give~$-$. +\end{Theorem} + +If $a$~and~$b$ stand for any two numbers, we have +\begin{align*} +(+a) × (+b) &= +ab, \\ +(+a) × (-b) &= -ab, \\ +(-a) × (+b) &= -ab, \\ +(-a) × (-b) &= +ab. +\end{align*} + +\Paragraph{73. The Index Law in Multiplication.} +\begin{DPalign*} +\lintertext{\indent Since} +a^{2} &= aa, \quad\text{and}\quad a^{3} = aaa, \\ +a^{2} × a^{3} &= aa × aaa = aaaaa = a^{5} = a^{2 + 3}; \\ +a^{4} × a &= aaaa × a = aaaaa = a^{5} = a^{4 + 1}. +\end{DPalign*} + +If $a$~stands for any number, and $m$~and~$n$ for any integers, +\[ +a^{m} × a^{n} = a^{m + n}. \EqText{Hence,} +\] +\begin{Theorem} +The index of the product of two powers of the same number +is equal to the sum of the indices of the factors. +\end{Theorem} +%% -----File: 048.png---Folio 42------- + +\Paragraph{74. Examples.} + +\Item{1.} Find the product of $6a^{2}b^{2}$ and $7ab^{2}c^{3}$. + +Since the order of the factors is immaterial, +\begin{align*} +6a^{2}b^{3} × 7ab^{2}c^{3} + &= 6 × 7 × a^{2} × a × b^{3} × b^{2} × c^{3} \\ + &= 42a^{3}b^{5}c^{3}. +\end{align*} + +\Item{2.} Find the product of $-3ab$ and $7ab^{3}$. +\begin{align*} +-3ab × 7ab^{3} + &= -3 × 7 × a × a × b × b^{3} \\ + &= -21a^{2}b^{4}. +\end{align*} + +\Paragraph{75. To Find the Product of Simple Expressions}, therefore, +\begin{Theorem} +Take the product of the coefficients and the sum of the +indices of the like letters. +\end{Theorem} + +\Exercise{16.} + +Find the product of: +\begin{multicols}{2} +\Item{1.} $5a^{2}$ and $6a^{3}$. + +\Item{2.} $8ab$ and $5a^{3}b^{2}$. + +\Item{3.} $9xy$ and $7xy$. + +\Item{4.} $2a^{2}b$ and $a^{3}b^{4}c^{2}$. + +\Item{5.} $3a^{3}b^{7}c^{8}$ and $3a^{4}b^{2}c$. + +\Item{6.} $2a$ and $-5a$. + +\Item{7.} $-3a$ and $-4b$. + +\Item{8.} $-ab$ and $a^{3}b^{2}$. + +\Item{9.} $-2ab^{4}$ and $-5a^{4}bc$. + +\Item{10.} $-2x^{6}y^{3}z$ and $-6xy^{2}z$. +\end{multicols} + +\Item{11.} $3a^{2}b$, $-5ab^{2}$, and $-7a^{4}b^{2}$. + +\Item{12.} $2a^{2}bc^{3}$, $-3a^{3}b^{2}c$, and $-ab^{2}c^{3}$. + +\Item{13.} $2b^{2}c^{2}x^{2}$, $2a^{2}b^{2}c^{3}$, and $-3a^{3}bx^{3}$. + +\Item{14.} $2a^{3}b^{2}c$, $-3a^{2}b^{3}c$, and $-4a^{2}bc^{3}$. + +\Item{15.} $7am^{2}x^{3}$, $3a^{4}m^{2}x^{3}$, and $-2amx$. + +\Item{16.} $-3x^{2}y^{2}z^{2}$, $2x^{2}yz^{3}$, and $-5x^{4}yz$. +%% -----File: 049.png---Folio 43------- + +If $a = -2$, $b = 3$, and $c = -1$, find the value of: +\begin{multicols}{2} +\Item{17.} $2ab^{2} - 3bc^{2} + c$. + +\Item{18.} $4a^{2} - 2b^{2} - c^{2}$. + +\Item{19.} $5a + 2b - 4c^{4}$. + +\Item{20.} $2a^{3} - 3b + 8c^{2}$. + +\Item{21.} $-a + 3b - 2c^{2}$. + +\Item{22.} $-a^{3} - 2b - 10c$. + +\Item{23.} $3a^{3} - 3b^{3} - 3c^{3}$. + +\Item{24.} $2ab^{2} - 3bc^{2} + 2ac$. + +\Item{25.} $3abc + 5a^{2}b^{2} - 2a^{2}b$. + +\Item{26.} $ab^{2}c^{2} + 2abc^{2} + a^{2}b^{2}c^{2}$. + +\Item{27.} $2a^{2}bc + 3abc + a^{2}b^{2}c^{2}$. + +\Item{28.} $6a^{2} + 8a^{2}b^{2} - 5a^{2}bc$. +\end{multicols} + +\Paragraph{76. Division.} To divide $48$~by~$8$ is to find the number +of times it is necessary to take $8$ to make~$48$. Here the +\emph{product} and \emph{one factor} are given, and \emph{the other factor} is +required. We may, therefore, take for the definition of +division\Add{:} + +The operation by which when \emph{the product} and \emph{one factor} +are given, \emph{the other factor is found}. + +With reference to this operation the product is called +the \Defn{dividend}, the given factor the \Defn{divisor}, and the required +factor the \Defn{quotient}. + +\Paragraph{77. Law of Signs in Division.} +\begin{alignat*}{2} +&\text{Since } (+a) × (+b) = +ab,\quad && \therefore +ab ÷ (+a) = +b. \\ +&\text{Since } (+a) × (-b) = -ab, && \therefore -ab ÷ (+a) = -b. \\ +&\text{Since } (-a) × (+b) = -ab, && \therefore -ab ÷ (-a) = +b. \\ +&\text{Since } (-a) × (-b) = +ab, && \therefore +ab ÷ (-a) = -b. +\end{alignat*} + +That is, if the dividend and divisor have like signs, the +quotient has the plus sign; and if they have unlike signs, +the quotient has the minus sign. Hence, in division, +\begin{Theorem} +Like signs give~$+$, and unlike signs give~$-$. +\end{Theorem} +%% -----File: 050.png---Folio 44------- + +\Paragraph{78. Index Law in Division.} + +The dividend contains all the factors of the divisor and +of the quotient, and therefore the quotient contains the +factors of the dividend that are not found in the divisor. + +Thus, $\dfrac{abc}{bc} = a$, $\dfrac{aabx}{ab} = ax$, $\dfrac{124abc}{-4ab} = -31c$. + +Divide $a^{5}$~by~$a^{2}$, $a^{6}$~by~$a^{4}$, $a^{4}$~by~$a$\DPtypo{,}{.} +\begin{alignat*}{4} +\frac{a^{5}}{a^{2}} + &= \frac{aaaaa}{aa} &&= aaa &&= a^{3} &&= a^{5-2}; \\ +\frac{a^{6}}{a^{4}} + &= \frac{aaaaaa}{aaaa} &&= aa &&= a^{2} &&= a^{6-4}\DPtypo{,}{;} \\ +\frac{a^{4}}{a} + &= \frac{aaaa}{a} &&= aaa &&= a^{3} &&= a^{4-1}\DPtypo{,}{.} +\end{alignat*} + +If $m$~and~$n$ stand for any integers, and $m$~is greater than~$n$\Add{,} +\[ +a^{m} - a^{n} = a^{m - n}\Add{.} +\] + +\begin{Theorem} +The index of the quotient of two powers of the same letter +is equal to the index of the letter in the dividend diminished +by the index of the letter in the divisor. +\end{Theorem} + +\Paragraph{79. Examples.} + +\Item{1.} Divide $15xy$~by~$5x$\Add{.} +\[ +\frac{15xy}{5x} = \frac{3 × 5xy}{5x} = 3y. +\] + +Here we cancel the factors $5$~and~$x$, which are common +to the dividend and divisor\Add{.} + +\Item{2.} Divide $-21a^{2}b^{3}$~by~$3ab^{2}$\Add{.} +\[ +\frac{-21a^{2}b^{3}}{3ab^{2}} = -7ab. +\] +%% -----File: 051.png---Folio 45------- + +\Item{3.} Divide $54a^{5}b^{3}c$~by~$-6ab^{2}c$. +\[ +\frac{54a^{5}b^{3}c}{-6ab^{2}c} = -9a^{4}b. +\] + +\Item{4.} Divide $-45x^{4}y^{5}z^{7}$~by~$-15x^{4}y^{5}z^{5}$. +\[ +\frac{-45x^{4}y^{5}z^{7}}{-15x^{4}y^{5}z^{5}} = 3z^{2}. +\] + +\Item{5.} Divide $-15a^{3}b^{2}c^{3}$~by~$-60a^{2}bc^{3}$. +\[ +\frac{-15a^{3}b^{2}c^{3}}{-60a^{2}bc^{3}} = \frac{ab}{4}. +\] + +\Exercise{17.} + +Divide: +\begin{multicols}{2} +\Item{1.} $x^{3}$~by~$x$. + +\Item{2.} $21x^{5}$~by~$7x^{3}$. + +\Item{3.} $35x^{2}$~by~$-5x^{2}$. + +\Item{4.} $-42x^{2}$~by~$6x^{2}$. + +\Item{5.} $-63x^{5}$~by~$-9x$. + +\Item{6.} $-72x^{3}$~by~$-8x^{2}$. + +\Item{7.} $-32a^{2}b^{2}$~by~$8ab^{2}$. + +\Item{8.} $-16x^{3}y^{3}$~by~$-4xy$. + +\Item{9.} $18x^{2}y$~by~$-2xy$. + +\Item{10.} $-25x^{4}y^{2}$~by~$-5x^{3}y^{2}$. + +\Item{11.} $-51x^{2}y^{3}$~by~$-17x^{2}y$. + +\Item{12.} $-28a^{4}b^{3}$~by~$7a^{3}b$. + +\Item{13.} $-36x^{2}y^{6}$~by~$-3xy^{2}$. + +\Item{14.} $-3x^{4}y^{6}$~by~$-5xy^{3}$. + +\Item{15.} $-12a^{2}b^{3}$~by~$8ab^{3}$. + +\Item{16.} $-abcd$~by~$ac$. + +\Item{17.} $-a^{2}b^{3}c^{4}d^{5}$~by~$-ab^{3}c^{3}d^{3}$. + +\Item{18.} $2x^{2}y^{2}z^{3}$~by~$-3xyz^{3}$. + +\Item{19.} $-5a^{5}b^{3}c^{7}$~by~$-a^{4}b^{2}c^{7}$. + +\Item{20.} $52a^{2}m^{3}n^{4}$~by~$13a^{2}m^{2}n^{3}$. + +\Item{21.} $13xy^{2}z^{4}$~by~$39xyz$. + +\Item{22.} $68xc^{2}d^{3}$~by~$-4xcd^{2}$. + +\Item{23.} $-8m^{5}n^{3}p^{2}$~by~$-4m^{5}np$. + +\Item{24.} $-6pqr^{3}$~by~$-2p^{2}qr$. + +\Item{25.} $26a^{2}g^{2}t^{5}$~by~$-2agt^{4}$. + +\Item{26.} $-a^{4}b^{2}c^{3}$~by~$-a^{5}b^{3}c^{4}$. + +\Item{27.} $-3x^{2}y^{2}z^{2}$~by~$-2x^{3}y^{4}z^{5}$. + +\Item{28.} $-6mnp$~by~$-3m^{2}n^{2}p^{2}$. + +\Item{29.} $-17a^{2}b^{3}c^{4}$~by~$51ab^{5}c^{4}$. + +\Item{30.} $-19mg^{2}t^{3}$~by~$57m2^gt^{4}$. +\end{multicols} +%% -----File: 052.png---Folio 46------- + + +\Chapter{IV.}{Addition and Subtraction.} + +\Section{Integral Compound Expressions.} + +\Paragraph{80.} If an algebraic expression contains only \emph{integral +forms}, that is, contains \emph{no letter in the denominator of +any of its terms}, it is called an \Defn{integral expression}. + +Thus, $x^{3} + 7cx^{2} - c^{3} - 5c^{2}x$, is an integral expression. + +Integral and fractional expressions are so named on +account of the \emph{form of the expressions}, and with no reference +whatever to the \emph{numerical value} of the expressions +when definite numbers are put in place of the letters. + +\Paragraph{81. Addition of Integral Compound Expressions.} The addition +of two algebraic expressions can be represented by +connecting the second expression with the first by the sign~$+$. +If there are no like terms in the two expressions, the +operation is \emph{algebraically complete} when the two expressions +are thus connected. + +If, for example, it is required to add $m + n - p$ to +$a + b + c$, the result will be $a + b + c + (m + n - p)$; or, +removing the parenthesis (§~37), $a + b + c + m + n - p$. + +\Paragraph{82.} If there are like terms in the expressions, the like +terms can be \emph{collected}; that is, every set of like terms can +be replaced by a single term with a coefficient equal to +the algebraic sum of the coefficients of the like terms. +%% -----File: 053.png---Folio 47------- + +\Item{1.} Add $6x^{2} + 5x + 4$ to $x^{2} - 4x - 5$. +\begin{DPalign*} +\lintertext{\indent The sum} +&= x^{2} - 4x - 5 + (6x^{2} + 5x + 4) \\ +&= x^{2} - 4x - 5 + 6x^{2} + 5x + 4 +\rintertext{§~37} \\ +&= x^{2} + 6x^{2} - 4x + 5x - 5 + 4 \\ +&= 7x^{2} + x - 1. +\end{DPalign*} + +This process is more conveniently represented by arranging +the terms in columns, so that like terms shall stand in +the same column, as follows: +\[ +\begin{array}{r*{2}{cr}} + x^{2} &-& 4x &-& 5 \\ +6x^{2} &+& 5x &+& 4 \\ +\hline +7x^{2} &+& x &-& 1 \\ +\end{array} +\] + +The coefficient of~$x^{2}$ in the result will be $6 + 1$, or~$7$; the +coefficient of~$x$ will be $-4 + 5$, or~$1$; and the last term is +$-5 + 4$, or~$-1$. + +\begin{Remark}[Note.] When the coefficient of a term is~$1$, it is not written, but +understood. +\end{Remark} + +\Item{2.} Add $2c^{3} - 5c^{2}d + 6cd^{2} + d^{3}$; $c^{3} + 6c^{2}d - 5cd^{2} - 2d^{3}$; +and $3c^{3} - c^{2}d - 7cd^{2} - 3d^{3}$. +\[ +\begin{array}{r*{3}{cr}} +2c^{3} &-& 5c^{2}d &+& 6cd^{2} &+& d^{3} \\ + c^{3} &+& 6c^{2}d &-& 5cd^{2} &-& 2d^{3} \\ +3c^{3} &-& c^{2}d &-& 7cd^{2} &-& 3d^{3} \\ +\hline +6c^{3} & & &-& 6cd^{2} &-& 4d^{3} \\ +\end{array} +\] + +The coefficient of~$c^{3}$ in the result will be $2 + 1 + 3$, or~$6$; +the coefficient of~$c^{2}d$ will be $-5 + 6 - 1$, or~$0$; therefore +$c^{2}d$ will not appear in the result; the coefficient of~$cd^{2}$ will +be $6 - 5 - 7$, or~$-6$; and the coefficient of~$d^{3}$ will be +$1 - 2 - 3$, or~$-4$. +%% -----File: 054.png---Folio 48------- + +\Exercise{18.} + +Find the sum of: + +\Item{1.} $a^{2} - ab + b^{2}$; $a^{2} + ab + b^{2}$. + +\Item{2.} $3a^{2} + 5a-7$; $6a^{2} - 7a + 13$. + +\Item{3.} $x + 2y - 3z$; $-3x + y + 2z$; $2x - 3y + z$. + +\Item{4.} $3x + 2y - z$; $-x + 3y + 2z$; $2x - y + 3z$. + +\Item{5.} $-3a + 2b + c$; $a - 3b + 2c$; $2a + 3b - c$. + +\Item{6.} $-a + 3b + 4c$; $3a - b + 2c$; $2a + 2b - 2c$. + +\Item{7.} $4a^{2} + 3a + 5$; $-2a^{2} + 3a - 8$; $a^{2} - a + 1$. + +\Item{8.} $5ab + 6bc - 7ac$; $3ab - 9bc + 4ac$; $3bc + 6ac$. + +\Item{9.} $x^{3} + x^{2} + x$; $2x^{3} + 3x^{2} - 2x$; $3x^{3} - 4x^{2} + x$. + +\Item{10.} $3y^{2} - x^{2} - 3xy$; $5x^{2} + 6xy - 7y^{2}$; $x^{2} + 2y^{2}$. + +\Item{11.} $2a^{2} - 2ab + 3b^{2}$; $4b^{2} + 5ab - 2a^{2}$; $a^{2} - 3ab - 9b^{2}$. + +\Item{12.} $a^{3} - a^{2} + a - 1$; $a^{2} - 2a + 2$; $3a^{3} + 7a + 1$. + +\Item{13.} $2m^{3} - m^{2} - m$; $4m^{3} + 8m^{2} - 7$; $-3m^{3} + m + 9$. + +\Item{14.} $x^{3} - 3x + 6y$; $x^{2} + 2x - 5y$; $x^{3} - 3x^{2} + 5x$. + +\Item{15.} $6x^{3} - 5x + 1$; $x^{3} + 3x + 4$; $7x^{2} + 2x - 3$. + +\Item{16.} $a^{3} + 3a^{2}b - 3ab^{2}$; $-3a^{2}b - 6ab^{2} - b^{3}$; $3a^{2}b + 4ab^{2}$. + +\Item{17.} $a^{3} - 2a^{2}b - 2ab^{2}$; $a^{2}b - 3ab^{2} - b^{3}$; $3ab^{2} - 2a^{3} - b^{3}$. + +\Item{18.} $7x^{3} - 2x^{2}y + 9xy^{2} + 13y^{3}$; $5x^{2}y - 4xy^{2} - 2x^{3} - 3y^{3}$; +$y^{3} - x^{3} - 3x^{2}y - 5xy^{2}$; $2x^{2}y - 5y^{3} - 2x^{3} - xy^{2}$. + +\Item{19.} Show that $x + y + z = 0$, if $x = a - b - c$, +$y = 2b + 2c - 3a$, and $z = 2a - b - c$. + +\Item{20.} Show that $x + y = 3z$, if $x = 3a^{2} - 6a + 12$, +$y = 9a^{2} + 12a - 21$, and $z = 4a^{2} + 2a - 3$. +%% -----File: 055.png---Folio 49------- + +\Paragraph{83. Subtraction of Integral Compound Expressions.} The +subtraction of one expression from another, if none of the +terms are alike, can be represented only by connecting the +subtrahend with the minuend by means of the sign~$-$. + +If, for example, it is required to subtract $a + b + c$ from +$m + n - p$, the result will be represented by +\[ +m + n - p - (a + b + c); +\] +or, removing the parenthesis (§~38), +\[ +m + n - p - a - b - c. +\] + +If, however, some of the terms in the two expressions are +alike, we can replace two like terms by a single term. + +Thus, suppose it is required to subtract $a^{3} + 2a^{2} + 3a - 5$ +from $2a^{3} - 3a^{2} + 2a - 1$; the result may be expressed as +follows: +\[ +2a^{3} - 3a^{2} + 2a - 1 - (a^{3} + 2a^{2} + 3a - 5); +\] +or, removing the parenthesis (§~38), +\begin{align*} +&2a^{3} - 3a^{2} + 2a - 1 - a^{3} - 2a^{2} - 3a + 5 \\ +&\quad= 2a^{3} - a^{3} - 3a^{2} - 2a^{2} + 2a - 3a - 1 + 5 \\ +&\quad= a^{3} - 5a^{2} - a + 4. +\end{align*} + +This process is more easily performed by writing the subtrahend +below the minuend, \emph{mentally} changing the sign of +each term in the subtrahend, and adding. +\[ +\begin{array}{r*{3}{cr}} +2a^{3} &-& 3a^{2} &+& 2a &-& 1 \\ + a^{3} &+& 2a^{2} &+& 3a &-& 5 \\ +\hline + a^{3} &-& 5a^{2} &-& a &+& 4 \\ +\end{array} +\] + +By changing the sign of each term in the subtrahend, +the coefficient of~$a^{3}$ will be $2 - 1$, or~$1$; the coefficient of~$a^{2}$ +will be $-3 - 2$, or~$-5$; the coefficient of~$a$ will be~$2 - 3$, +or~$-1$; the last term will be $-1 + 5$, or~$4$. +%% -----File: 056.png---Folio 50------- + +Again, suppose it is required to subtract $x^{5} - 2ax^{4} - +3a^{2} x^{3} + 4a^{3} x^{2}$ from $4a^{3} x^{2} - 2a^{2} x^{3} - 5ax^{4}$. Here terms +which are alike can be written in columns, as before: +\[ +\begin{array}{r*{3}{cr}} + &-& 5ax^{4} &-& 2a^{2} x^{3} &+& 4a^{3} x^{2} \\ + x^{5} &-& 2ax^{4} &-& 3a^{2} x^{3} &+& 4a^{3} x^{2} \\ +\hline +-x^{5} &-& 3ax^{4} &+& a^{2} x^{3} & & +\end{array} +\] + +There is no term of~$x^{5}$ in the minuend, hence the coefficient +of~$x^{5}$ in the result will be~$0 - 1$, or~$-1$; the coefficient of~$ax^{4}$ +will be $-5 + 2$, or~$-3$; the coefficient\DPtypo{,}{} of~$a^{2}x^{3}$ will be +$-2 + 3$, or~$+1$; the coefficient of~$a^{3}x^{2}$ will be $-4 + 4$, or~$0$, +and therefore the term~$a^{3}x^{2}$ will not appear in the result. + +\Exercise{19.} + +Subtract: + +\Item{1.} $a - 2b + 3c$ from $2a - 3b + 4c$. + +\Item{2.} $a - 3b - 5c$ from $3a - 5b + c$. + +\Item{3.} $2x - 4y + 6z$ from $4x - y - 2z$. + +\Item{4.} $5x - 11y - 3z$ from $6x - 7y + 2z$. + +\Item{5.} $ab - ac - bc + bd$ from $ab + ac + bc + bd$. + +\Item{6.} $3ab + 2ac - 3bc + bd$ from $5ab - ac + bc + bd$. + +\Item{7.} $2x^{3} - x^{2} - 5x + 3$ from $3x^{3} + 2x^{2} - 3x - 5$. + +\Item{8.} $7x^{2} - 5x + 1 - a$ from $x^{3} - x + 1 - a$. + +\Item{9.} $7b^{3} + 8c^{3} - 15abc$ from $9b^{3} + 3abc - 7c^{3}$. + +\Item{10.} $x^{4} + x - 5x^{3} + 5$ from $7 - 2x^{2} - 3x^{3} + x^{4}$. + +\Item{11.} $a^{3} + b^{3} + c^{3} - 3abc$ from $3abc + a^{3} - 2b^{3} - 3c^{3}$. + +\Item{12.} $2x^{4} - 5x^{2} + 7x - 3$ from $x^{4} + 2 - 2x^{3} - x^{2}$. + +\Item{13.} $1 - x^{5} - x + x^{4} - x^{3}$ from $x^{4} + 1 + x + x^{2}$. + +\Item{14.} $a^{3} - b^{3} + 3a^{2}b - 3ab^{2}$ from $a^{3} + b^{3} - a^{2}b - ab^{2}$. + +\Item{15.} $a^{2} b - ab^{2} - 3a^{3} b^{3} - b^{4}$ from $b^{4} - 5a^{3} b^{3} - 2ab^{2} + a^{2} b$. + +\Item{16.} $-x^{3} + 7x^{2} y - 2y^{3} + 3xy^{2}$ from $3x^{3} + 5y^{3} - xy^{2} + 4x^{2}y$. +%% -----File: 057.png---Folio 51------- + +\Paragraph{84. Parentheses or Brackets.} We have for positive numbers +(§§~37,~38): +\begin{alignat*}{2} +a + (b + c) &= a + b + c,\qquad & \therefore a + b + c &= a + (b + c); \\ +a + (b - c) &= a + b - c, & \therefore a + b - c &= a + (b - c); \\ +a - (b + c) &= a - b - c, & \therefore a - b - c &= a - (b + c); \\ +a - (b - c) &= a - b + c, & \therefore a - b + c &= a - (b - c). +\end{alignat*} + +That is, a parenthesis preceded by~$+$ may be removed +\emph{without changing the sign of any term within the parenthesis}; +and any number of terms may be enclosed within a parenthesis +preceded by the sign~$+$, \emph{without changing the sign +of any term}. + +A parenthesis preceded by the sign~$-$ may be removed, +\emph{provided the sign of every term within the parenthesis is +changed}, namely, $+$~to~$-$, and $-$~to~$+$; and any number +of terms may be enclosed within a parenthesis preceded +by the sign~$-$, \emph{provided the sign of every term enclosed is +changed}. + +The same laws hold for \emph{negative numbers}. + +\Paragraph{85.} Expressions may occur having a parenthesis within +a parenthesis. In such cases parentheses of different shapes +are used, and the beginner when he meets with a branch +of a parenthesis~$($, or bracket~$[$, or brace~$\{$, must look carefully +for the other part, whatever may intervene; and all +that is included between the two parts of each parenthesis +must be treated as the sign before it directs, without regard +to other parentheses. It is best to remove each parenthesis +in succession, \emph{beginning with the innermost}. +\begin{align*} +a - &\bigl\{b - [c - (d - e) + f]\bigr\} \\ + &= a - \bigl\{b - [c - d + e + f]\bigr\} \\ + &= a - \bigl\{b - c + d - e - f\bigr\} \\ + &= a - b + c - d + e + f. +\end{align*} +%% -----File: 058.png---Folio 52------- + +\Exercise{20.} + +Remove the brackets and collect the like terms: + +\Item{1.} $a - b - (b - c) - a + 2b$. + +\Item{2.} $x - [x - (a - b) + a - y]$. + +\Item{3.} $3x - \bigl\{2y - [-7c - 2x] + y\bigr\}$. + +\Item{4.} $5a - [7 - (2b + 5) - 2a]$. + +\Item{5.} $x - [2x + (3a - 2x) - 5a]$. + +\Item{6.} $x - [15y - (13z + 12x)]$. + +\Item{7.} $2a - b + [4c - (b + 2c)]$. + +\Item{8.} $5a - \bigl\{b + [3c - (2b - c)]\bigr\}$. + +\Item{9.} $7x - \bigl\{5y - [3z - (3x + z)]\bigr\}$. + +\Item{10.} $(a - b + c) - (b - a - c) + (a + b - 2c)$. + +\Item{11.} $3x - [-2y - (2y - 3x) + z] + [x - (y - 2z - x)]$. + +\Item{12.} $x - [2x + (x - 2y) + 2y] - 3x - \bigl\{4x - [(x + 2y) - y]\bigr\}$. + +\Item{13.} $x - [y + z - x - (x + y) - z] + (3 x - \Vinc{2y + z})$. + +\begin{Remark}[Note.] +The expression $-\Vinc{2y + z}$ is equivalent to~$-(2y + z)$. +\end{Remark} + +Consider \emph{all the factors} that precede $x$,~$y$, and~$z$, respectively, +as the \emph{coefficients} of these letters, and collect in +brackets the coefficients of each of these letters: + +\Item{14.} $ax + by + cz - ay + az - bx += (a - b)x - (a - b)y + (a + c)z$. + +\Item{15.} $ax + az + by - cz - ay + cx$. + +\Item{16.} $2ax - 3ay - 4by + 5cx - 6bz - 7cz$. + +\Item{17.} $az - bmy + 3 cz - anx - cny + acx$. + +\Item{18.} $mnx - x - mny - y + mnz + z$. +%% -----File: 059.png---Folio 53------- + + +\Chapter{V.}{Multiplication and Division.} + +\Section{Compound Integral Expressions.} + +\Paragraph{86. Multiplication. Polynomials by Monomials.} + +We have for positive numbers (§~39), +\begin{align*} +a(b + c) &= ab + ac, \\ +a(b - c) &= ab - ac. +\end{align*} + +The same law holds for negative numbers. + +\Dictum{To multiply a polynomial by a monomial}, therefore, +\begin{Theorem} +Multiply each term of the polynomial by the monomial, +and add the partial products. +\end{Theorem} + +\Item{1.} Find the product of $ab + ac - bc$ and~$abc$. +\[ +\begin{array}{rcr} +ab + ac - bc && \\ + abc && \\ +\hline +a^{2}b^{2}c + a^{2}bc^{2} &-& ab^{2}c^{2} +\end{array} +\] + +\begin{Remark}[Note.] +We multiply~$ab$, the first term of the multiplicand, by~$abc$, +and work to the right. +\end{Remark} + +\Exercise{21.} + +Find the product of: +\begin{multicols}{2} +\Item{1.} $x + 7$ and $x$. + +\Item{2.} $2x - 3y$ and $4x$. + +\Item{3.} $2x - 3y$ and $7y$. + +\Item{4.} $x - 2a$ and $2a$. + +\Item{5.} $-x + 3b$ and $-b$. + +\Item{6.} $2a^{2} - 3ab$ and $-3a$. + +\Item{7.} $2x^{2} + 3xz$ and $5z$. + +\Item{8.} $a^{2} - 5ab$ and $5ab$. +%% -----File: 060.png---Folio 54------- + +\Item{9.} $x^{2} - 3 xy$ and $-y^{2}$. + +\Item{10.} $2 x^{3} - 3x^{2}$ and $2x^{2}$. + +\Item{11.} $x^{2} - 3y^{2}$ and $4y$. + +\Item{12.} $x^{2} - 3 y^{2}$ and $-x^{2}$. + +\Item{13.} $b^{3} - a^{2}b^{2}$ and $-a^{3}$. + +\Item{14.} $-a^{2}b^{2} - a^{3}$ and $-a^{2}$. + +\Item{15.} $2x^{3} - 3x^{2} + x$ and $2x^{2}$. + +\Item{16.} $a^{2} - 5ab - b^{2}$ and $5ab$. +\end{multicols} + +\Item{17.} $a^{3} + 2a^{2}b + 2ab^{2}$ and $a^{2}$. + +\Item{18.} $a^{3} + 2a^{2}b + 2ab^{2}$ and $b^{3}$. + +\Item{19.} $4x^{2} - 6xy - 9y^{2}$ and $2x$. + +\Item{20.} $-x^{2} - 2xy + y^{2}$ and $-y$. + +\Item{21.} $-a^{3} - a^{2}b^{2} - b^{3}$ and $-a^{2}$. + +\Item{22.} $-x^{2} + 2xy - y^{2}$ and $-y^{2}$. + +\Item{23.} $3 a^{2}b^{2} - 4 ab^{3} + a^{3}b$ and $5 a^{2}b^{2}$. + +\Item{24.} $-ax^{2} + 3axy^{2} - ay^{4}$ and $-3ay^{2}$. + +\Item{25.} $x^{12} - x^{10}y^{3} - x^{3}y^{10}$ and $x^{3}y^{2}$. + +\Item{26.} $-2x^{3} + 3x^{2}y^{2} - 2xy^{3}$ and $-2x^{2}y^{3}$. + +\Item{27.} $a^{3}x^{2}y^{5} - a^{2}xy^{4} - ay^{3}$ and $a^{7}x^{3}y^{5}$. + +\Item{28.} $3a^{2}b^{2} - 2ab^{3} + 5a^{3}b$ and $5a^{2}b^{3}$. + +\Paragraph{87. Multiplication. Polynomials by Polynomials.} + +If we have $m + n +p$ to be multiplied by $a + b + c$, we +may substitute~$M$ for the multiplier $a + b + c$. Then +\[ +M(m + n + p) = Mm + Mn + Mp. +\] + +If now we substitute $a + b + c$ for~$M$, we shall have +\begin{align*} +&(a + b + c) m + (a + b + c) n + (a + b + c) p \\ +&= am + bm + cm + an + bn + cn + ap + bp + cp\DPtypo{.}{} \\ +&= am + an + ap + bm + bn + bp + cm + cn + cp. +\end{align*} + +\Dictum{To find the product of two polynomials}, therefore, +\begin{Theorem} +Multiply every term of the multiplicand by each term of +the multiplier, and add the partial products. +\end{Theorem} +%% -----File: 061.png---Folio 55------- + +\Paragraph{88.} In multiplying polynomials, it is a convenient +arrangement to write the multiplier under the multiplicand, +and place like terms of the partial products in +columns. + +\Item{1.} Multiply $2x - 3y$ by $5x - 4y$. +\[ +\begin{array}{ccrrcr} +2x &-& 3&y && \\ +5x &-& 4&y && \\ +\cline{1-4} +10x^{2} &-& 15&xy & & \\ + &-& 8&xy &+& 12y^{2} \\ +\hline +10x^{2} &-& 23&xy &+& 12y^{2} \\ +\end{array} +\] + +We multiply~$2x$, the first term of the multiplicand, by~$5x$, +the first term of the multiplier, and obtain~$10x^{2}$, then~$-3y$, +the second term of the multiplicand, by~$5x$, and obtain~$-15xy$. +The first line of partial products is $10x^{2} - 15xy$. +In multiplying by~$-4y$, we obtain for a second line of partial +products $-8xy + 12y^{2}$, which is put one place to the +right, so that the like terms $-15xy$~and~$-8xy$ may stand +in the same column. We then add the coefficients of the +like terms, and obtain the complete product in its simplest +form. + +\Item{2.} Multiply $2a + 3 - 4a^{2}$ by $3 - 2a^{2} - 3a$. + +Arrange both multiplicand and multiplier according to +the \emph{ascending} powers of~$a$. +\[ +\begin{array}{r*{4}{cr}} +3 &+& 2a &-& 4a^{2} & & && \\ +3 &-& 3a &-& 2a^{2} & & && \\ +\cline{1-5} +9 &+& 6a &-&12a^{2} & & && \\ + &-& 9a &-& 6a^{2} &+&12a^{3} & & \\ + & & &-& 6a^{2} &-& 4a^{3} &+& 8a^{4} \\ +\hline +9 &-& 3a &-&24a^{2} &+& 8a^{3} &+& 8a^{4} \\ +\end{array} +\] +%% -----File: 062.png---Folio 56------- + +\Item{3.} Multiply $3x + x^{4} - 2x^{2}$ by $x^{3} - 2 - x$. + +Arrange according to the \emph{descending} powers of~$x$. +\[ +\begin{array}{r*{5}{cr}} +x^{4} &-& 2x^{2} &+& \PadTo[l]{3x^4}{3x} && && && \\ +x^{3} &-& \PadTo[c]{2x^2}{x} &-& \PadTo[l]{3x^4}{2}&& && && \\ +\cline{1-5} +x^{7} &-& 2x^{5} &+& 3x^{4} && && && \\ + &-& x^{5} & & &+& 2x^{3} &-& 3x^{2} && \\ + & & &-& 2x^{4} & & &+& 4x^{2} &-& 6x \\ +\hline +x^{7} &-& 3x^{5} &+& x^{4} &+& 2x^{3} &+& x^{2} &-& 6x \\ +\end{array} +\] + +\Item{4.} Multiply $a^{2} + b^{2} + c^{2} - ab - bc - ac$ by $a + b + c$. + +Arrange according to descending powers of~$a$. +\[ +\begin{array}{l*{8}{cr}} +a^{2} &-& ab &-& ac &+& b^{2} &-& bc &+& c^{2} \\ +a &+& b &+& c \\ +\cline{1-11} +a^{3} &-& a^{2}b &-& a^{2}c &+& ab^{2} &-& abc &+& ac^{2} \\ + &+& a^{2}b & & &-& ab^{2} &-& abc & & &+& b^{3} &-& b^{2}c + bc^{2} \\ + & & &+& a^{2}c & & &-& abc &-& ac^{2} & & &+& b^{2}c - bc^{2} &+& c^{3} \\ +\hline +a^{3} & & & & & & &-&3abc & & &+& b^{3} & & &+& c^{3} \\ +\end{array} +\] + +\begin{Remark}[Note.] +The pupil should observe that, with a view to bringing +like terms of the partial products in columns, the terms of the multiplicand +and multiplier are arranged in the \emph{same order}. +\end{Remark} + +\ScreenBreak +\Exercise{22.} + +Find the product of: +\begin{multicols}{2} +\Item{1.} $x + 7$ and $x + 6$. + +\Item{2.} $x - 7$ and $x + 6$. + +\Item{3.} $x + 7$ and $x - 6$. + +\Item{4.} $x - 7$ and $x - 6$. + +\Item{5.} $x + 8$ and $x - 5$. + +\Item{6.} $2x + 3$ and $2x + 3$. + +\Item{7.} $2x - 3$ and $2x - 3$. + +\Item{8.} $2x + 3$ and $2x - 3$. + +\Item{9.} $3x - 2$ and $2 - 3x$. + +\Item{10.} $5x - 3$ and $4x - 7$. + +\Item{11.} $a - 2b$ and $a + 3b$. + +\Item{12.} $a - 7b$ and $a - 5b$. +%% -----File: 063.png---Folio 57------- + +\Item{13.} $5x - 3y$ and $5x - 3y$. + +\Item{14.} $x - b$ and $x - c$. + +\Item{15.} $2m - p$ and $4m - 3p$. + +\Item{16.} $a + b + c$ and $a - c$. + +\Item{17.} $a^{2} - ab + b^{2}$ and $a^{2} + b^{2}$. + +\Item{18.} $x^{3} - 3x^{2} + 7$ and $x^{2} - 3$. + +\Item{19.} $a^{2} + ab + b^{2}$ and $a - b$. + +\Item{20.} $a^{2} - ab + b^{2}$ and $a + b$. +\end{multicols} + +\Item{21.} $x^{2} + 5x - 10$ and $2x^{2} + 3x - 4$. + +\Item{22.} $3x^{3} - 2x^{2} + x$ and $3x^{2} + 2x - 2$. + +\Item{23.} $x^{3} + 2x^{2}y + 3xy^{2}$ and $x^{2} - 2xy + y^{2}$. + +\Item{24.} $a^{2} - 3ab - b^{2}$ and $-a^{2} + ab + 2b^{2}$. + +\Item{25.} $3a^{2}b^{2} + 2 ab^{3} - 5a^{3}b$ and $5a^{2}b^{2} - ab^{3} - b^{4}$. + +\Item{26.} $a^{2} - 2ab + b^{2}$ and $a^{2} + 2ab + b^{2}$. + +\Item{27.} $ab + ac + cd$ and $ab - ac + cd$. + +\Item{28.} $3x^{2}y^{2} + xy^{3} - 2x^{3}y$ and $x^{2}y^{2} + xy^{3} - 3y^{4}$. + +\Item{29.} $x^{2} + 2xy - y^{2}$ and $x^{2} - 2xy + y^{2}$. + +\Item{30.} $3x^{2} + xy - y^{2}$ and $x^{2} - 2xy - 3y^{2}$. + +\Item{31.} $a^{2} - 2ab - b^{2}$ and $b^{2} - 2ab - a^{2}$. + +\Item{32.} $a^{2} + b^{2} + c^{2} - ac$ and $a^{2} - b^{2} - c^{2}$. + +\Item{33.} $a^{2} + 4abx - 4a^{2}b^{2}x^{2}$ and $a^{2} - 4abx + 4a^{2}b^{2}x^{2}$. + +\Item{34.} $3 a^{2} - 2abx + b^{2}x^{2}$ and $2a^{2} + 3abx - 2b^{2}x^{2}$. + +\Item{35.} $2x^{3}y + 4x^{2}y^{2} - 8xy^{3}$ and $2xy^{3} - 3x^{2}y^{2} + 5x^{3}y$. + +\Paragraph{89. Division. Polynomials by Monomials.} +\begin{DPalign*} +\lintertext{\indent Since} +a(b + c - d) &= ab + ac - ad, \\ +\therefore \frac{ab + ac - ad}{a} + &= \frac{ab}{a} + \frac{ac}{a} - \frac{ad}{a} \\ + &= b + c - d. +\end{DPalign*} +%% -----File: 064.png---Folio 58------- + +\Dictum{To divide a polynomial by a monomial}, therefore, +\begin{Theorem} +Divide each term of the dividend by the divisor, and add +the partial quotients. +\end{Theorem} + +Divide $3a^{4}b^{2}c - 9a^{3}bc^{2} - 6a^{2}c^{3}$ by $3a^{2}c$. +\begin{align*} +\frac{3a^{4}b^{2}c - 9a^{3}bc^{2} - 6a^{2}c^{3}}{3a^{2}c} + &= \frac{3a^{4}b^{2}c}{3a^{2}c} + - \frac{9a^{3}bc^{2}}{3a^{2}c} + - \frac{6a^{2}c^{3}}{3a^{2}c} \\ + &= a^{2}b^{2} - 3abc - 2c^{2}. +\end{align*} + +\Exercise{23.} + +Divide: +\begin{multicols}{2} +\Item{1.} $2a^{3} - a^{2}$ by $a$. + +\Item{2.} $42a^{5} - 6a^{2}$ by $6a$. + +\Item{3.} $21x^{4} + 3x^{2}$ by $3x^{2}$. + +\Item{4.} $35m^{4} - 7p^{2}$ by $7$. + +\Item{5.} $27x^{5} - 45x^{4}$ by $9x^{2}$. + +\Item{6.} $24x^{6} - 8x^{3}$ by $-8x^{3}$. + +\Item{7.} $34x^{3} - 51x^{2}$ by $17x$. + +\Item{8.} $5x^{5} - 10x^{3}$ by $-5x^{3}$. + +\Item{9.} $-3a^{2} - 6ac$ by $-3a$. + +\Item{10.} $-5x^{3} + x^{2}y$ by $-x^{2}$. + +\Item{11.} $2a^{5}x^{3} - 2a^{4}x^{2}$ by $2a^{4}x^{2}$. + +\Item{12.} $-x^{2}y - x^{2}y^{2}$ by $-xy$. + +\Item{13.} $9a - 12b + 6c$ by $-3$. + +\Item{14.} $a^{3}b^{2} - a^{2}b^{5} - a^{4}b^{2}$ by $a^{2}b$. + +\Item{15.} $3x^{3} - 6x^{2}y - 9xy^{2}$ by $3x$. + +\Item{16.} $x^{2}y^{2} - x^{3}y - xy^{3}$ by $xy$. + +\Item{17.} $a^{3} - a^{2}b - ab^{2}$ by $-a$. + +\Item{18.} $a^{2}b - ab + ab^{2}$ by $-ab$. + +\Item{19.} $xy - x^{2}y^{2} + x^{3}y^{3}$ by $-xy$. + +\Item{20.} $-x^{6} - 2x^{5} - x^{4}$ by $-x^{4}$. +\end{multicols} + +\Item{21.} $a^{2}x - abx - acx$ by $ax$. + +\Item{22.} $3x^{5}y^{2} - 3x^{4}y^{3} - 3x^{2}y^{4}$ by $3x^{2}y^{2}$. + +\Item{23.} $a^{2}b^{2} - 2ab - 3ab^{3}$ by $ab$. + +\Item{24.} $3a^{3}c^{3} + 3a^{2}c - 3ac^{2}$ by $3ac$. +%% -----File: 065.png---Folio 59------- + +\Paragraph{90. Division. Polynomials by Polynomials.} +\[ +\begin{array}{l*{4}{cr}} +\text{If the divisor (one factor)} + &=& & & a &+& b &+& c, \\ +\text{and the quotient (other factor)} + &=& & & n &+& p &+& q, \\ +\cline{5-9} + & & & &an &+& bn&+& cn \\ +\text{then the dividend (product)} + &=& \smash{\left\{\threelines\right.}\kern-4pt + &+&ap &+& bp&+& cp \\ + & & &+&aq &+& bq&+& cq\rlap{.} \\ +\end{array} +\] + +The first term of the dividend is~$an$; that is, the product +of~$a$, the first term of the divisor, by~$n$, the first term of the +quotient. The first term~$n$ of the quotient is therefore +found by dividing~$an$, the first term of the dividend, by~$a$, +the first term of the divisor. + +If the partial product formed by multiplying the entire +divisor by~$n$ be subtracted from the dividend, the first term +of the remainder~$ap$ is the product of~$a$, the first term of +the divisor, by~$p$, the second term of the quotient; that is, +the second term of the quotient is obtained by dividing the +first term of the remainder by the first term of the divisor. +In like manner, the third term of the quotient is obtained +by dividing the first term of the new remainder by the first +term of the divisor; and so on. + +\Dictum{To divide one polynomial by another}, therefore, +\begin{Theorem} +Arrange both the dividend and divisor in ascending or +descending powers of some common letter. + +Divide the first term of the dividend by the first term of +the divisor. + +Write the result as the first term of the quotient. + +Multiply all the terms of the divisor by the first term of +the quotient. + +Subtract the product from the dividend. + +If there is a remainder, consider it as a new dividend, +and proceed as before. +\end{Theorem} +%% -----File: 066.png---Folio 60------- + +\Paragraph{91.} It is of fundamental importance to arrange the dividend +and divisor \emph{in the same order} with respect to a common +letter, and \emph{to keep this order throughout the operation}. + +The beginner should study carefully the processes in the +following examples: + +\Item{1.} Divide $x^{2} + 18x + 77$ by $x + 7$. +\[ +\begin{array}{r*{2}{cr}|l} +x^{2} &+& 18x &+& 77 & x + 7 \\ +\cline{6-6} +x^{2} &+& 7x & & & x + 11 \\ +\cline{1-5} + & & 11x &+& \NoBar{77} \\ + & & 11x &+& \NoBar{77} \\ +\cline{3-5} +\end{array} +\] + +\begin{Remark}[Note.] +The pupil will notice that by this process we have in +effect separated the dividend into two parts, $x^{2} + 7x$ and $11x + 77$, +and divided each part by $x + 7$, and that the complete quotient is the +sum of the partial quotients $x$~and~$11$. Thus, +\begin{align*} +x^{2} + 18x + 77 + &= x^{2} + 7x + 11x + 77 = (x^{2} + 7x) + (11x + 77). \\ +\therefore \frac{x^{2} + 18x + 77}{x + 7} + &= \frac{x^{2} + 7x}{x + 7} + \frac{11x + 77}{x + 7} = x + 11. +\end{align*} +\end{Remark} + +\Item{2.} Divide $a^{2} - 2ab + b^{2}$ by $a - b$. +\[ +\begin{array}{r*{2}{cr}|l} +a^{2} &-& 2ab &+& b^{2} & a - b \\ +\cline{6-6} +a^{2} &-& ab & & & a - b \\ +\cline{1-5} + &-& ab &+& \NoBar{b^{2}} \\ + &-& ab &+& \NoBar{b^{2}} \\ +\cline{2-5} +\end{array} +\] + +\Item{3.} Divide $a^{4} - ab^{3} + b^{4} + 2a^{2}b^{2} - a^{3}b$ by $a^{2} + b^{2}$. + +Arrange according to the descending powers of~$a$. +\[ +\begin{array}{r*{4}{cr}|l} +a^{4} &-& a^{3}b &+& 2a^{2}b^{2} &-& ab^{3} &+& b^{4} & a^{2} + b^{2} \\ +\cline{10-10} +a^{4} & & &+& a^{2}b^{2} & & & & & a^{2} - ab + b^{2} \\ +\cline{1-9} + &-& a^{3}b &+& a^{2}b^{2} &-& ab^{3} &+& \NoBar{b^{4}} \\ + &-& a^{3}b & & &-& ab^{3} \\ +\cline{2-9} + & & &+& a^{2}b^{2} & & &+& \NoBar{b^{4}} \\ + & & &+& a^{2}b^{2} & & &+& \NoBar{b^{4}} \\ +\cline{4-9} +\end{array} +\] +%% -----File: 067.png---Folio 61------- + +\Item{4.} Divide $10a^{2}b^{2} - 20b^{4} - 17a^{3}b + 6a^{4} + ab^{3}$ +by $2a^{2} - 4b^{2} - 3ab$. + +Arrange according to descending powers of~$a$. +\[ +\begin{array}{r*{4}{cr}|l} +6a^{4} &-&17a^{3}b &+& 10a^{2}b^{2} &+& ab^{3} &-& 20b^{4} & 2a^{2} - 3ab - 4b^{2} \\ +\cline{10-10} +6a^{4} &-& 9a^{3}b &-& 12a^{2}b^{2} & & & & & 3a^{2} - 4ab + 5b^{2} \\ +\cline{1-9} + &-& 8a^{3}b &+& 22a^{2}b^{2} &+& ab^{3} &-& \NoBar{20b^{4}} \\ + &-& 8a^{3}b &+& 12a^{2}b^{2} &+&16ab^{3} \\ +\cline{2-9} + & & & & 10a^{2}b^{2} &-&15ab^{3} &-& \NoBar{20b^{4}} \\ + & & & & 10a^{2}b^{2} &-&15ab^{3} &-& \NoBar{20b^{4}} \\ +\cline{5-9} +\end{array} +\] + +\Item{5.} Divide $5x^{3} - 3x^{4} - 4x^{2} + 1 + x$ by $1 + 2x - 3x^{2}$. + +Arrange according to ascending powers of~$x$. +\[ +\begin{array}{r*{4}{cr}|l*{2}{cr}} +1 &+& x &-& 4x^{2} &+& 5x^{3} &-& 3x^{4} & 1 &+& 2x &-& 3x^{2} \\ +\cline{10-14} +1 &+& 2x &-& 3x^{2} & & & & & 1 &-& x &+& x^{2} \\ +\cline{1-9} + &-& x &-& x^{2} &+& 5x^{3} &-& \NoBar{3x^{4}} \\ + &-& x &-& 2x^{2} &+& 3x^{3} \\ +\cline{2-9} + & & & & x^{2} &+& 2x^{3} &-& \NoBar{3x^{4}} \\ + & & & & x^{2} &+& 2x^{3} &-& \NoBar{3x^{4}} \\ +\end{array} +\] + +\Item{6.} Divide $a^{3} + b^{3} + c^{3} - 3abc$ by $a + b + c$. + +Arrange according to descending powers of~$a$. +\[ +%[** TN: Re-formatted slightly from the original] +\begin{array}{r*{4}{cr}clcr|l} +a^{3} & & & & &-& 3abc & & &+& b^{3} &+& c^{3} & a + b + c \\ +\cline{14-14} +a^{3} &+& a^{2}b &+& a^{2}c & & & & & & & & & +\smash[b]{\begin{aligned}[t] + a^{2} &- ab - ac \\ + &+b^{2} - bc + c^{2} +\end{aligned}} \\ +\cline{1-13} + &-& a^{2}b &-& a^{2}c &-& 3abc & & &+& b^{3} &+& \NoBar{c^{3}} \\ + &-& a^{2}b &-& ab^{2} &-& abc \\ +\cline{2-13} + &-& a^{2}c &+& ab^{2} &-& 2abc & & &+& b^{3} &+& \NoBar{c^{3}} \\ + &-& a^{2}c & & &-& abc &-& ac^{2} \\ +\cline{2-13} + & & & & ab^{2} &-& abc &+& ac^{2} &+& b^{3} &+& \NoBar{c^{3}} \\ + & & & & ab^{2} & & & & &+& b^{3} &+& \NoBar{b^{2}c} \\ +\cline{4-13} + & & & & &-& abc &+& ac^{2} &-& b^{2}c&+& \NoBar{c^{3}} \\ + & & & & &-& abc & & &-& b^{2}c&-& \NoBar{bc^{2}} \\ +\cline{6-13} + & & & & & & & & ac^{2} &+& bc^{2}&+& \NoBar{c^{3}} \\ + & & & & & & & & ac^{2} &+& bc^{2}&+& \NoBar{c^{3}} \\ +\cline{9-13} +\end{array} +\] +%% -----File: 068.png---Folio 62------- + +\Exercise{24.} + +Divide: +\begin{multicols}{2} +\Item{1.} $x^{2} + 15x + 56$ by $x + 7$. + +\Item{2.} $x^{2} - 15x + 56$ by $x - 7$. + +\Item{3.} $x^{2} + x-56$ by $x - 7$. + +\Item{4.} $x^{2} - x-56$ by $x + 7$. + +\Item{5.} $2a^{2} + 11a + 5$ by $2a + 1$. + +\Item{6.} $6a^{2} - 7a-3$ by $2a - 3$. + +\Item{7.} $4a^{2} + 23a + 15$ by $4a + 3$. + +\Item{8.} $3a^{2} - 4a-4$ by $2 - a$. + +\Item{9.} $x^{4} + x^{2} + 1$ by $x^{2} + x + 1$. + +\Item{10.} $x^{8} + x^{4} + 1$ by $x^{4} - x^{2} + 1$. + +\Item{11.} $1 - a^{3}b^{3}$ by $1 - ab$. + +\Item{12.} $x^{3} - 8x-3$ by $x - 3$. +\end{multicols} + +\Item{13.} $a^{2} - 2ab + b^{2} - c^{2}$ by $a - b - c$. + +\Item{14.} $a^{2} + 2ab + b^{2} - c^{2}$ by $a + b + c$. + +\Item{15.} $x^{2} - y^{2} + 2yz - z^{2}$ by $x - y + z$. + +\Item{16.} $c^{4} + 2c^{2} - c + 2$ by $c^{2} - c + 1$. + +\Item{17.} $x^{2} - 4y^{2} - 4yz - z^{2}$ by $x + 2y + z$. + +Arrange and divide: + +\Item{18.} $x^{3} - 6a^{3} + 11a^{2}x - 6ax^{2}$ by $x^{2} + 6a^{2} - 5ax$. + +\Item{19.} $a^{2} - 4b^{2} - 9c^{2} + 12bc$ by $a - 3c + 2b$. + +\Item{20.} $2a^{3} - 8a + a^{4} + 12-7a^{2}$ by $2 + a^{2} - 3a$. + +\Item{21.} $q^{4} + 6q^{3} + 4 + 12q + 13q^{2}$ by $3q + 2 + q^{2}$. + +\Item{22.} $27a^{3} - 8b^{3}$ by $3a - 2b$. + +Find the remainder when: + +\Item{23.} $a^{4} + 9a^{2} + 15-11a - 7a^{3}$ is divided by $a - 5$. + +\Item{24.} $7 - 8c^{2} + 5c^{3} + 8c$ is divided by $5c - 3$. + +\Item{25.} $3 + 11a^{3} + 30a^{4} - 82a^{2} - 5a$ is divided by $3a^{2} - 4 + 2a$. + +\Item{26.} $2x^{3} - 16x + 10-39x^{2} + 17x^{4}$ is divided by $2 - 5x^{2} - 4x$. +%% -----File: 069.png---Folio 63------- + +\Exercise[Miscellaneous Examples.]{25.} + +\Item{1.} Add $2a^{2} - 3ac - 3ab$; $2b^{2} + 3ac + a^{2}$; $-a^{2} - 2b^{2} + 3ab$. + +\Item{2.} Subtract $3a^{4} - 2 a^{3}b + 4 a^{2}b^{2}$ from $4b^{4} - 2 ab^{3} + 4 a^{2}b^{2}$. + +\Item{3.} Simplify $x - y - \{z - x - (y - x + z)\}$. + +\Item{4.} Multiply $a^{2} + b^{2} + c^{2} - d^{2}$ by $a^{2} + b^{2} - c^{2} + d^{2}$. + +\Item{5.} Divide $10y^{6} + 2 - 12y^{5}$ by $1 + y^{2} - 2y$. + +\Item{6.} If $a = 1$, $b = 2$, and $c = -3$, find the value of + $a^{2} + b^{2} + c^{2} + 2ab + 2ac + 2bc$. + +\Item{7.} Simplify $x - (y - z) - \bigl\{4y + [2y - (z - x)]\bigr\}$. + +\Item{8.} Multiply $a^{2} + b^{2} + c^{2} - ab - ac - bc$ by $a + b + c$. + +\Item{9.} Divide $16y^{4} - 21x^{2}y^{2} + 21x^{3}y - 10x^{4}$ by $4y^{2} - 5x^{2} + 3xy$. + +\Item{10.} Add $-2a^{4} + 3a^{3}b - 4a^{2}b^{2}$; $2a^{3}b - 3a^{2}b^{2}$; $7a^{2}b^{2} + 2a^{4} - b^{4}$. + +\Item{11.} From $3x^{3} + 5x - 1$ take the sum of $x - 5 + 5x^{2}$ and +$3 + 4x -3x^{2}$. + +\Item{12.} The minuend is $9c^{2} + 11c - 5$, and the remainder is +$6c^{2} - 13c + 7$. What is the subtrahend? + +\Item{13.} Find the remainder when $a^{4} + 6b^{4}$ is divided by +$a^{2} + 2ab + 2b^{2}$. + +\Item{14.} Multiply $2 - 5x^{2} - 4x$ by $5 + 2x - 3x^{2}$. + +\Item{15.} Divide $a^{6} + a^{5}x + a^{4}x^{2} - a^{3}x^{3} + x^{6}$ by $a^{2} + ax + x^{2}$. + +Bracket the coefficients of the different powers of~$x$: + +\Item{16.} $ax^{3} - cx + bx^{2} - bx^{3} + cx^{2} - x$. + +\Item{17.} $ax^{4} - 2x + bx^{4} - cx - ax^{3} + bx^{3}$. + +\Item{18.} $x^{3} - bx^{2} - cx + bx - cx^{2} + ax^{3}$. +%% -----File: 070.png---Folio 64------- + + +\Chapter[Special Rules in Multiplication and Division.] +{VI.}{Multiplication and Division.} + +\Section{Special Rules.} + +\Paragraph{92. Special Rules of Multiplication.} Some results of multiplication +are of so great utility in shortening algebraic +work that they should be carefully noticed and remembered. +The following are important: + +\Paragraph{93. Square of the Sum of Two Numbers.} +\begin{align*} +(a + b)^{2} + &= (a + b)(a + b) \\ + &= a(a + b) + b(a + b) \\ + &= a^{2} + ab + ab + b^{2} \\ + &= a^{2} + 2ab + b^{2}. +\end{align*} + +Since $a$~and~$b$ stand for any two numbers, we have +\begin{Theorem}[\textsc{Rule 1.}] The square of the sum of two numbers is the +sum of their squares plus twice their product. +\end{Theorem} + +\Paragraph{94. Square of the Difference of Two Numbers.} +\begin{align*} +(a - b)^{2} + &= (a - b) (a - b) \\ + &= a(a - b) - b(a - b) \\ + &= a^{2} - ab - ab + b^{2} \\ + &= a^{2} - 2ab + b^{2}. +\end{align*} + +Hence we have +\begin{Theorem}[\textsc{Rule 2.}] The square of the difference of two numbers is +the sum of their squares minus twice their product. +\end{Theorem} +%% -----File: 071.png---Folio 65------- + +\Paragraph{95. Product of the Sum and Difference of Two Numbers.} +\begin{align*} +(a + b)(a - b) + &= a(a - b) + b(a - b) \\ + &= a^{2} - ab + ab - b^{2} \\ + &= a^{2} - b^{2}. +\end{align*} + +Hence, we have +\begin{Theorem}[\textsc{Rule 3.}] The product of the sum and difference of two +numbers is the difference of their squares. +\end{Theorem} + +If we put $2x$~for~$a$, and $3$~for~$b$, we have +\begin{DPalign*} +\lintertext{\indent Rule 1,} &(2x + 3)^{2} = 4x^{2} + 12x + 9. \\ +\lintertext{\indent Rule 2,} &(2x - 3)^{2} = 4x^{2} - 12x + 9. \\ +\lintertext{\indent Rule 3,} &(2x + 3)(2x - 3) = 4x^{2} - 9. +\end{DPalign*} + +\Exercise{26.} + +Write by inspection the value of: +\begin{multicols}{2} +\Item{1.} $(m + n)^{2}$. + +\Item{2.} $(c - a)^{2}$. + +\Item{3.} $(a + 2c)^{2}$. + +\Item{4.} $(3a - 2b)^{2}$. + +\Item{5.} $(2a + 3b)^{2}$. + +\Item{6.} $(a - 3b)^{2}$. + +\Item{7.} $(2x - y)^{2}$. + +\Item{8.} $(y - 2x)^{2}$. + +\Item{9.} $(a + 5b)^{2}$. + +\Item{10.} $(2a - 5c)^{2}$. + +\Item{11.} $(x + y)(x - y)$. + +\Item{12.} $(4a - b)(4a + b)$. + +\Item{13.} $(2b - 3c)(2b + 3c)$. + +\Item{14.} $(x + 5b)(x + 5b)$. + +\Item{15.} $(y - 2z)(y - 2z)$. + +\Item{16.} $(y + 3z)(y - 3z)$. + +\Item{17.} $(2a - 3b)(2a + 3b)$. + +\Item{18.} $(2a - 3b)(2a - 3b)$. + +\Item{19.} $(2a + 3b)(2a + 3b)$. + +\Item{20.} $(5x + 3a)(5x - 3a)$. +\end{multicols} +%% -----File: 072.png---Folio 66------- + +\Paragraph{96. Product of Two Binomials of the Form $x + a$, $x + b$.} +The product of two binomials which have the form $x + a$, +$x + b$, should be carefully noticed and remembered. +\begin{DPalign*} +\lintertext{\Item{1.}} +(x + 5)(x + 3) + &= x(x + 3) + 5(x + 3) \\ + &= x^{2} + 3x + 5x + 15 \\ + &= x^{2} + 8x + 15. \displaybreak[1] \\ +% +\lintertext{\Item{2.}} +(x - 5)(x - 3) + &= x(x - 3) - 5(x - 3) \\ + &= x^{2} - 3x - 5x + 15 \\ + &= x^{2} - 8x + 15. \displaybreak[1] \\ +% +\lintertext{\Item{3.}} +(x + 5)(x - 3) + &= x(x - 3) + 5(x - 3) \\ + &= x^{2} - 3x + 5x - 15 \\ + &= x^{2} + 2x - 15. \displaybreak[1] \\ +% +\lintertext{\Item{4.}} +(x - 5)(x + 3) + &= x(x + 3) - 5(x + 3) \\ + &= x^{2} + 3x - 5x - 15 \\ + &= x^{2} - 2x - 15. +\end{DPalign*} + +\Item{1.} Each of these results has three terms. + +\Item{2.} The first term of each result is the product of the first +terms of the binomials. + +\Item{3.} The last term of each result is the product of the +second terms of the binomials. + +\Item{4.} The middle term of each result has for a coefficient +the \emph{algebraic sum} of the second terms of the binomials. + +\Paragraph{97.} The intermediate step given above may be omitted, +and the products written at once by \emph{inspection}. Thus, + +\Item{1.} Multiply $x + 8$ by $x + 7$. +\begin{align*} +&8 + 7 = 15,\quad 8 × 7 = 56. \\ +\therefore\ &(x + 8)(x + 7) = x^{2} + 15x + 56. +\end{align*} +%% -----File: 073.png---Folio 67------- + +\Item{2.} Multiply $x - 8$ by $x - 7$. +\begin{align*} +&(-8) + (-7) = -15,\quad (-8)(-7) = +56. \\ +\therefore\ &(x - 8)(x - 7) = x^{2} - 15x + 56. +\end{align*} + +\Item{3.} Multiply $x - 7y$ by $x + 6y$. +\begin{align*} +&-7 + 6 = -1,\quad (-7) × 6y = -42y^{2}. \\ +\therefore\ &(x - 7y)(x + 6y) = x^{2} - xy - 42y^{2}. +\end{align*} + +\Item{4.} Multiply $x + 6y$ by $x - 5y$. +\begin{align*} +&+6 - 5 = 1,\quad 6y × (-5y) = -30y^{2}. \\ +\DPtypo{}{\therefore}\ &(x + 6y)(x - 5y) = x^{2} + xy - 30y^{2}. +\end{align*} + +\Exercise{27.} + +Write by inspection the product of: +\begin{multicols}{2} +\Item{1.} $(x + 7)(x + 4)$. + +\Item{2.} $(x - 3)(x + 7)$. + +\Item{3.} $(x - 2)(x - 4)$. + +\Item{4.} $(x - 6)(x - 10)$. + +\Item{5.} $(x + 7)(x - 4)$. + +\Item{6.} $(x + a)(x - 2a)$. + +\Item{7.} $(x + 3a)(x - a)$. + +\Item{8.} $(a + 3c)(a + 3c)$. + +\Item{9.} $(a + 2x)(a - 4x)$. + +\Item{10.} $(a - 3b)(a - 4b)$. + +\Item{11.} $(a^{2} - c)(a^{2} + 2c)$. + +\Item{12.} $(x - 17)(x - 3)$. + +\Item{13.} $(x + 6y)(x - 5y)$. + +\Item{14.} $(3 + 2x)(3 - x)$. + +\Item{15.} $(5 + 2x)(1 - 2x)$. + +\Item{16.} $(a - 2b)(a + 3b)$. + +\Item{17.} $(a^{2}b^{2} - x^{2})(a^{2}b^{2} - 5x^{2})$. + +\Item{18.} $(a^{3}b - ab^{3})(a^{3}b + 5ab^{3})$. + +\Item{19.} $(x^{2}y - xy^{2})(x^{2}y - 3xy^{2})$. + +\Item{20.} $(x^{2}y + xy^{2})(x^{2}y + xy^{2})$. + +\Item{21.} $(x + a)(x + b)$. + +\Item{22.} $(x + a)(x - b)$. + +\Item{23.} $(x - a)(x + b)$. + +\Item{24.} $(x - a)(x - b)$. + +\Item{25.} $(x + 2a)(x + 2b)$. + +\Item{26.} $(x - 2a)(x + 2b)$. + +\Item{27.} $(x + 2a)(x - 2b)$. + +\Item{28.} $(x - 2a)(x - 2b)$. + +\Item{29.} $(x - a)(x + 3a)$. + +\Item{30.} $(x - 2a)(x + 3a)$. +\end{multicols} +%% -----File: 074.png---Folio 68------- + +\Paragraph{98. Special Rules of Division.} Some results in division +are so important in abridging algebraic work that they +should be carefully noticed and remembered. + +\Paragraph{99. Difference of Two Squares.} + +Since $(a + b)(a - b) = a^{2} - b^{2}$, +\[ +\therefore +\frac{a^{2} - b^{2}}{a + b} = a - b;\quad\text{and}\quad +\frac{a^{2} - b^{2}}{a - b} = a + b. \EqText{Hence\Add{,}} +\] +\begin{Theorem}[\textsc{Rule 1.}] The difference of the squares of two numbers is +divisible by the sum, and by the difference, of the numbers. +\end{Theorem} + +\ScreenBreak +\Exercise{28.} + +Write by inspection the quotient of: +\begin{multicols}{3} +\Item{1.} $\dfrac{x^{2} - 4}{x - 2}$. + +\Item{2.} $\dfrac{x^{2} - 4}{x + 2}$. + +\Item{3.} $\dfrac{a^{2} - 9}{a - 3}$. + +\Item{4.} $\dfrac{a^{2} - 9}{a + 3}$. + +\Item{5.} $\dfrac{c^{2} - 25}{c - 5}$. + +\Item{6.} $\dfrac{c^{2} - 25}{c + 5}$. + +\Item{7.} $\dfrac{49x^{2} - y^{2}}{7x - y}$. + +\Item{8.} $\dfrac{49x^{2} - y^{2}}{7x + y}$. + +\Item{9.} $\dfrac{9b^{2} - 1}{3b - 1}$. + +\Item{10.} $\dfrac{9b^{2} - 1}{3b + 1}$. + +\Item{11.} $\dfrac{16x^{4} - 25a^{2}}{4x^{2} - 5a}$. + +\Item{12.} $\dfrac{16x^{4} - 25a^{2}}{4x^{2} + 5a}$. +\end{multicols} + +\begin{multicols}{2} +\Item{13.} $\dfrac{9x^{2} - 25y^{2}}{3x - 5y}$. + +\Item{14.} $\dfrac{a^{2}-(b - c)^{2}}{a-(b - c)}$. + +\Item{15.} $\dfrac{a^{2}-(b - c)^{2}}{a + (b - c)}$. + +\Item{16.} $\dfrac{a^{2}-(2b - c)^{2}}{a-(2b - c)}$. + +\Item{17.} $\dfrac{(5a - 7b)^{2} - 1}{(5a - 7b) - 1}$. + +\Item{18.} $\dfrac{(5a - 7b)^{2} - 1}{(5a - 7b) + 1}$. + +\Item{19.} $\dfrac{z^{2}-(x - y)^{2}}{z-(x - y)}$. + +\Item{20.} $\dfrac{z^{2}-(x - y)^{2}}{z + (x - y)}$. +%% -----File: 075.png---Folio 69------- + +\Item{21.} $\dfrac{a^{2}-(2b - c)^{2}}{a + (2b - c)}$. + +\Item{22.} $\dfrac{(x + 3y)^{2} - z^{2}}{(x + 3y) - z}$. + +\Item{23.} $\dfrac{(x + 3y)^{2} - z^{2}}{x + 3y + z}$. + +\Item{24.} $\dfrac{(a + 2b)^{2} - 4c^{2}}{(a + 2b) - 2c}$. + +\Item{25.} $\dfrac{(a + 2b)^{2} - 4c^{2}}{(a + 2b) + 2c}$. + +\Item{26.} $\dfrac{1 - (3x - 2y)^{2}}{1 + (3x - 2y)}$. +\end{multicols} + +\Paragraph{100. Difference of Two Cubes.} By performing the division +we have +\[ +\frac{a^{3} - b^{3}}{a - b} = a^{2} + ab + b^{2}. \EqText{Hence,} +\] +\begin{Theorem}[\textsc{Rule 2.}] The difference of the cubes of two numbers is +divisible by the difference of the numbers, and the quotient +is the sum of the squares of the numbers plus their product. +\end{Theorem} + +\Exercise{29.} + +Write by inspection the quotient of: +\begin{multicols}{3} +\Item{1.} $\dfrac{1 - x^{3}}{1 - x}$. + +\Item{2.} $\dfrac{1 - 8a^{3}}{1 - 2a}$. + +\Item{3.} $\dfrac{1 - 27c^{3}}{1 - 2c}$. + +\Item{4.} $\dfrac{8a^{3} - b^{3}}{2a - b}$. + +\Item{5.} $\dfrac{64b^{3} - 27c^{3}}{4b - 3c}$. + +\Item{6.} $\dfrac{27x^{3} - 8y^{3}}{3x - 2y}$. + +\Item{7.} $\dfrac{x^{3}y^{3} - z^{3}}{xy - z}$. + +\Item{8.} $\dfrac{a^{3}b^{3} - 8}{ab - 2}$. + +\Item{9.} $\dfrac{125a^{3} - b^{3}}{5a - b}$. + +\Item{10.} $\dfrac{a^{3} - 8b^{3}}{a - 2b}$. + +\Item{11.} $\dfrac{a^{3} - 64}{a - 4}$. + +\Item{12.} $\dfrac{a^{9} - 27}{a^{3} - 3}$. + +\Item{13.} $\dfrac{a^{12} - x^{6}y^{6}}{a^{4} - x^{2}y^{2}}$. + +\Item{14.} $\dfrac{x^{15} - a^{9}b^{9}}{x^{5} - a^{3}b^{3}}$. + +\Item{15.} $\dfrac{27x^{3}y^{3} - z^{12}}{3xy - z^{4}}$. + +\Item{16.} $\dfrac{x^{3}y^{3}z^{3} - 1}{xyz - 1}$. + +\Item{17.} $\dfrac{8a^{3}b^{3}c^{3} - 27}{2abc - 3}$. + +\Item{18.} $\dfrac{1 - 64x^{3}y^{3}z^{3}}{1 - 4xyz}$. +\end{multicols} +%% -----File: 076.png---Folio 70------- + +\Paragraph{101. Sum of Two Cubes.} By performing the division, +we find that +\[ +\frac{a^{3} + b^{3}}{a + b} = a^{2} - ab + b^{2}. \EqText{Hence,} +\] +\begin{Theorem}[\textsc{Rule 3.}] The sum of the cubes of two numbers is divisible +by the sum of the numbers, and the quotient is the sum +of the squares of the numbers minus their product. +\end{Theorem} + +\Exercise{30.} + +Write by inspection the quotient of: +\begin{multicols}{3} +\Item{1.} $\dfrac{1 + x^{3}}{1 + x}$. + +\Item{2.} $\dfrac{1 + 8a^{3}}{1 + 2a}$. + +\Item{3.} $\dfrac{1 + 27c^{3}}{1 + 3c}$. + +\Item{4.} $\dfrac{8a^{3} + b^{3}}{2a + b}$. + +\Item{5.} $\dfrac{64b^{3} + 27c^{3}}{4b + 3c}$. + +\Item{6.} $\dfrac{27x^{3} + 8y^{3}}{3x + 2y}$. + +\Item{7.} $\dfrac{8x^{3} + 125y^{3}}{2x + 5y}$. + +\Item{8.} $\dfrac{x^{3}y^{3} + z^{3}}{xy + z}$. + +\Item{9.} $\dfrac{a^{3}b^{3} + 8}{ab + 2}$. + +\Item{10.} $\dfrac{125a^{3} + b^{3}}{5a + b}$. + +\Item{11.} $\dfrac{a^{3} + 8b^{3}}{a + 2b}$. + +\Item{12.} $\dfrac{a^{6} + 64}{a^{2} + 4}$. + +\Item{13.} $\dfrac{a^{9} + 27}{a^{3} + 3}$. + +\Item{14.} $\dfrac{8a^{6} + b^{3}}{2a^{2} + b}$. + +\Item{15.} $\dfrac{a^{12} + x^{6}y^{6}}{a^{4} + x^{2}y^{2}}$. + +\Item{16.} $\dfrac{x^{15} + a^{9}b^{9}}{x^{5} + a^{3}b^{3}}$. + +\Item{17.} $\dfrac{27x^{3}y^{3} + z^{12}}{3xy + z^{4}}$. + +\Item{18.} $\dfrac{x^{3}y^{3}z^{3} + 1}{xyz + 1}$. + +\Item{19.} $\dfrac{8a^{3}b^{3}c^{3} + 27}{2abc + 3}$. + +\Item{20.} $\dfrac{1 + 64x^{3}y^{3}z^{3}}{1 + 4xyz}$. + +\Item{21.} $\dfrac{1 + 27a^{6}b^{3}c^{3}}{1 + 3a^{2}bc}$. +\end{multicols} + +Find by division the quotient of: +\begin{multicols}{3} +\Item{22.} $\dfrac{x^{4} - y^{4}}{x - y}$. + +\Item{23.} $\dfrac{x^{4} - y^{4}}{x + y}$. + +\Item{24.} $\dfrac{x^{5} - y^{5}}{x - y}$. + +\Item{25.} $\dfrac{x^{5} + y^{5}}{x + y}$. + +\Item{26.} $\dfrac{x^{6} - y^{6}}{x - y}$. + +\Item{27.} $\dfrac{x^{6} - y^{6}}{x + y}$. +\end{multicols} +%% -----File: 077.png---Folio 71------- + + +\Chapter{VII.}{Factors.} + +\Paragraph{102. Rational Expressions.} An expression is \emph{rational} when +none of its terms contain square or other roots. + +\Paragraph{103. Factors of Rational and Integral Expressions.} By factors +of a given integral number in arithmetic we mean +integral numbers that will divide the given number without +remainder. Likewise by factors of a rational and integral +expression in algebra we mean rational and integral +expressions that will divide the given expression without +remainder. + +\Paragraph{104. Factors of Monomials.} The factors of a monomial +may be found by inspection. Thus, the factors of~$21a^{2}b$ +are $3$, $7$, $a$, $a$, and~$b$. + +\Paragraph{105. Factors of Polynomials.} The form of a polynomial +that can be resolved into factors often suggests the process +of finding the factors. + + +\Section{Case I.} + +\Paragraph{106. When all the terms have a common factor.} + +\Item{1.} Resolve into factors $3a^{2} - 6ab$. + +Since $3a$~is seen to be a factor of each term, we have +\begin{align*} +\frac{3a^{2} - 6ab}{3a} &= \frac{3a^{2}}{3a} - \frac{6ab}{3a} = a - 2b. \\ +\therefore\ 3a^{2} - 6ab &= 3a(a - 2b). +\end{align*} + +Hence, the required factors are $3a$~and~$a - 2b$. +%% -----File: 078.png---Folio 72------- + +\Item{2.} Resolve into factors $4x^{3} + 12x^{2} - 8x$. + +Since $4x$ is seen to be a factor of each term, we have +\begin{align*} +\frac{4x^{3} + 12x^{2} - 8x}{4x} + &= \frac{4x^{3}}{4x} + \frac{12x^{2}}{4x} - \frac{8x}{4x} \\ + &= x^{2} + 3x - 2. \\ +\therefore\ +4x^{3} + 12x^{2} - 8x &= 4x(x^{2} + 3x - 2). +\end{align*} + +Hence the required factors are $4x$~and~$x^{2} + 3x - 2$. + +\Exercise{31.} + +Resolve into two factors: +\begin{multicols}{2} +\Item{1.} $2x^{2} - 4x$. + +\Item{2.} $3a^{3} - 6a$. + +\Item{3.} $5a^{2}b^{2} - 10a^{3}b^{3}$. + +\Item{4.} $3x^{2}y + 4xy^{2}$. + +\Item{5.} $8a^{3}b^{2} + 4a^{2}b^{3}$. + +\Item{6.} $3a^{4} - 12a^{2} - 6a^{3}$. + +\Item{7.} $4x^{2} - 8x^{4} - 12x^{5}$. + +\Item{8.} $5 - 10x^{2}y^{2} + 15x^{2}y$. + +\Item{9.} $7a^{2} + 14a - 21a^{3}$. + +\Item{10.} $3x^{3}y^{3} - 6x^{4}y^{4} - 9x^{2}y^{2}$. +\end{multicols} + +\Section{Case II.} + +\Paragraph{107. When the terms can be grouped so as to show a common +factor in each group.} + +\Item{1.} Resolve into factors $ac + ad + bc + bd$. +\begin{align*} +ac + ad + bc + bd + &= (ac + ad) + (bc + bd) + \Tag{(1)} \\ + &= a(c + d) + b(c + d) + \Tag{(2)} \\ + &= (a + b)(c + d). + \Tag{(3)} +\end{align*} + +\begin{Remark}[Note.] The first two terms of $ac + ad + bc + bd$ are seen to +have the common factor~$a$, and the last two terms, the common factor~$b$. +Hence we bracket the first two terms and also the last two +terms. Then we take out the factor~$a$ from $(ac + ad)$ and $b$~from +$(bc + bd)$, and get equation~(2). Since one factor is seen in~(2) to be +$c + d$, dividing by $c + d$, we obtain the other factor, $a + b$. +\end{Remark} +%% -----File: 079.png---Folio 73------- + +\Item{2.} Find the factors of $ac + ad - bc - bd$. +\begin{align*} +ac + ad - bc - bd + &= (ac + ad) - (bc + bd) \\ + &= a(c + d) - b(c + d) \\ + &= (a - b)(c + d). +\end{align*} + +\begin{Remark}[Note.] Here the last two terms, $-bc - bd$, being put within a +parenthesis preceded by the sign~$-$, have their signs changed to~$+$. +\end{Remark} + +\Item{3.} Resolve into factors $2x^{3} - 3x^{2} - 4x + 6$. +\begin{align*} +2x^{3} - 3x^{2} - 4x + 6 + &= (2x^{3} - 3x^{2}) - (4x - 6) \\ + &= x^{2}(2x - 3) - 2(2x - 3) \\ + &= (x^{2} - 2)(2x - 3). +\end{align*} + +\Item{4.} Resolve into factors $x^{3} + x^{2} - ax - a$. +\begin{align*} +x^{3} + x^{2} - ax - a + &= (x^{3} + x^{2}) - (ax + a) \\ + &= x^{2}(x + 1) - a(x + 1) \\ + &= (x^{2} - a)(x + 1). +\end{align*} + +\Item{5.} Resolve into factors $x^{3} + 3ax^{2} + x + 3a$. +\begin{align*} +x^{3} + 3ax^{2} + x + 3a + &= (x^{3} + 3ax^{2}) + (x + 3a) \\ + &= x^{2}(x + 3a) + 1(x + 3a) \\ + &= (x^{2} + 1)(x + 3a). +\end{align*} + +\Exercise{32.} + +Resolve into factors: +\begin{multicols}{2} +\Item{1.} $x^{3} + x^{2} + x + 1$. + +\Item{2.} $x^{3} - x^{2} + x - 1$. + +\Item{3.} $x^{2} + xy + xz + yz$. + +\Item{4.} $ax - bx - ay + by$. + +\Item{5.} $a^{2} - ac + ab - bc$. + +\Item{6.} $x^{2} - bx + 3x - 3b$. + +\Item{7.} $2x^{3} - x^{2} + 4x - 2$. + +\Item{8.} $a^{2} - 3a - ab + 3b$. + +\Item{9.} $6a^{2} + 2ab - 3ac - bc$. + +\Item{10.} $abxy + cxy + abc + c^{2}$. + +\Item{11.} $ax - ay - bx + cy - cx + by$. + +\Item{12.} $(a - b)^{2} - 2c(a - b)$. +\end{multicols} +%% -----File: 080.png---Folio 74------- + + +\Section{Case III.} + +\Paragraph{108. When a binomial is the difference of two squares.} + +\Item{1.} Resolve into factors $x^{2} - y^{2}$. +\begin{DPgather*} +\lintertext{\indent Since,} +(x + y)(x - y) = x^{2} - y^{2}, +\end{DPgather*} +the factors of $x^{2} - y^{2}$ are $x + y$ and~$x - y$. + +\Dictum{To find the factors of a binomial when it is the difference of +two squares}, therefore, +\begin{Theorem} +Take the square root of the first term and the square root +of the second term. + +The sum of these roots will form the first factor; + +The difference of these roots will form the second factor. +\end{Theorem} + +\Paragraph{109.} The \Defn{square root} of a \emph{monomial} is one of the \textbf{two equal +factors} of the monomial. + +Thus $9x^{8}y^{2} = 3x^{4}y × 3x^{4}y$; and $3x^{4}y$ is the square root +of~$9x^{8}y^{2}$. + +The rule for extracting the square root of a \emph{monomial, +when a perfect square}, is as follows: +\begin{Theorem} +Extract the square root of the coefficient, and divide the +index of each letter by~$2$. +\end{Theorem} + +\Exercise{33.} + +Resolve into factors: +\begin{multicols}{3} +\Item{1.} $4 - x^{2}$. + +\Item{2.} $9 - x^{2}$. + +\Item{3.} $9a^{2} - x^{2}$. + +\Item{4.} $25 - x^{2}$. + +\Item{5.} $25x^{2} - a^{2}$. + +\Item{6.} $16a^{4} - 121$. + +\Item{7.} $121a^{4} - 16$. + +\Item{8.} $4a^{2}b^{2} - c^{2}d^{2}$. + +\Item{9.} $1 - x^{2}y^{2}$. + +\Item{10.} $81x^{2}y^{2} - 1$. + +\Item{11.} $49a^{2}b^{2} - 4$. + +\Item{12.} $25a^{4}b^{4} - 9$. +\end{multicols} +%% -----File: 081.png---Folio 75------- + +\begin{multicols}{2} +\Item{13.} $9a^{8}b^{6} - 16x^{10}$. + +\Item{14.} $144x^{2}y^{2} - 1$. + +\Item{15.} $100x^{6}y^{2}z^{4} - 1$. + +\Item{16.} $1 - 121a^{4}b^{8}c^{12}$. + +\Item{17.} $25a^{2} - 64x^{6}y^{6}$. + +\Item{18.} $16x^{16}-25y^{18}$. +\end{multicols} + +Find, by resolving into factors, the value of: +\begin{multicols}{2} +\Item{19.} $(375)^{2} - (225)^{2}$. + +\Item{20.} $(579)^{2} - (559)^{2}$. + +\Item{21.} $(873)^{2} - (173)^{2}$. + +\Item{22.} $(101)^{2} - (99)^{2}$. + +\Item{23.} $(7244)^{2} - (7242)^{2}$. + +\Item{24.} $(3781)^{2} - (219)^{2}$. +\end{multicols} + +\Paragraph{110.} If the squares are compound expressions, the same +method may be employed. + +\Item{1.} Resolve into factors $(x + 3y)^{2} - 16a^{2}$. +\begin{Soln} +The square root of the first term is~$x + 3y$. + +The square root of the second term is~$4a$. + +The sum of these roots is~$x + 3y - 4a$. + +The difference of these roots is $x + 3y - 4a$. + +Therefore $(x + 3y)^{2} - 16a^{2} = (x + 3y + 4a)(x + 3y - 4a)$. +\end{Soln} + +\Item{2.} Resolve into factors $a^{2} - (3b - 5c)^{2}$. +\begin{Soln} +The square roots of the terms are $a$~and~$(3b - 5c)$. + +The sum of these roots is $a + (3b - 5c)$, or $a + 3b - 5c$. + +The difference of these roots is $a - (3b - 5c)$, or $a - 3b + 5c$. + +Therefore $a^{2} - (3b - 5c)^{2} = (a + 3b - 5c)(a - 3b + 5c)$. +\end{Soln} + +\Exercise{34.} + +\DPtypo{}{Resolve into factors:} +\begin{multicols}{2} +\Item{1.} $(x + y)^{2} - z^{2}$. + +\Item{2.} $(x - y)^{2} - z^{2}$. + +\Item{3.} $z^{2} - (x + y)^{2}$. + +\Item{4.} $z^{2} - (x - y)^{2}$. + +\Item{5.} $(x + y)^{2} - 4z^{2}$. + +\Item{6.} $4z^{2} - (x - y)^{2}$. + +\Item{7.} $(a + 2b)^{2} - c^{2}$. + +\Item{8.} $(a - 2b)^{2} - c^{2}$. + +\Item{9.} $c^{2} - (a - 2b)^{2}$. + +\Item{10.} $(2a + 5c)^{2} - 1$. +%% -----File: 082.png---Folio 76------- + +\Item{11.} $1 - (2a - 5c)^{2}$. + +\Item{12.} $(a + 3b)^{2} - 16c^{2}$. + +\Item{13.} $(a - 5b)^{2} - 9c^{2}$. + +\Item{14.} $16c^{2} - (a - 5b)^{2}$. + +\Item{15.} $4a^{2} - (x + y)^{2}$. + +\Item{16.} $b^{2} - (a - 2x)^{2}$. + +\Item{17.} $4z^{2} - (x + 3y)^{2}$. + +\Item{18.} $9 - (3a - 7b)^{2}$. + +\Item{19.} $16a^{2} - (2b + 5c)^{2}$. + +\Item{20.} $25c^{2} - (3a - 2x)^{2}$. + +\Item{21.} $9a^{2} - (3b - 5c)^{2}$. + +\Item{22.} $16y^{2} - (a - 3c)^{2}$. + +\Item{23.} $49m^{2} - (p + 2q)^{2}$. + +\Item{24.} $36n^{2} - (d - 2c)^{2}$. + +\Item{25.} $(x + y)^{2} - (a + b)^{2}$. + +\Item{26.} $(x - y)^{2} - (a - b)^{2}$. + +\Item{27.} $(2x + 3)^{2} - (2a + b)^{2}$. + +\Item{28.} $(b - c)^{2} - (a - 2x)^{2}$. + +\Item{29.} $(3x - y)^{2} - (2a - b)^{2}$. + +\Item{30.} $(x - 3y)^{2} - (a + 2b)^{2}$. + +\Item{31.} $(x + 2y)^{2} - (a + 3b)^{2}$. + +\Item{32.} $(x + y)^{2} - (a - z)^{2}$. +\end{multicols} + + +\Section{Case IV.} + +\Paragraph{111. When a binomial is the difference of two cubes.} +\begin{DPgather*} +\lintertext{\indent Since} +\frac{a^{3} - b^{3}}{a - b} = a^{2} + ab + b^{2}, +\end{DPgather*} +the factors of $a^{3} - b^{3}$ are $a - b$ and $a^{2} + ab + b^{2}$. + +In like manner we can resolve into factors any expression +which can be written as the difference of two cubes. + +\Paragraph{112.} The rule for extracting the cube root of a \emph{monomial, +when the monomial is a perfect cube}, is, +\begin{Theorem} +Extract the cube root of the coefficient, and divide the index +of each letter by~$3$. +\end{Theorem} + +\ScreenBreak +\Item{1.} Resolve into factors $8a^{3} - 27b^{6}$. + +Since $8a^{3} = (2a)^{3}$, and $27b^{6} = (3b^{2})^{3}$, we can write +$8a^{3} - 27b^{6}$ as $(2a)^{3} - (3b^{2})^{3}$. +%% -----File: 083.png---Folio 77------- +\begin{DPgather*} +\lintertext{\indent Since} +a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2}), +\end{DPgather*} +we have, by putting $2a$ for~$a$ and $3b^{2}$ for~$b$, +\begin{align*} +(2a)^{3} - (3b^{2})^{3} + &= (2a - 3b^{2})[(2a)^{2} + 2a × 3b^{2} + (3b^{2})^{2}] \\ + &= (2a - 3b^{2})(4a^{2} + 6ab^{2} + 9b^{4}). +\end{align*} + +\Item{2.} Resolve into factors $64x^{3} - 1$. +\begin{align*} +64x^{3} - 1 + &= (4x)^{3} - 1 \\ + &= (4x - 1)[(4x)^{2} + 4x +1] \\ + &= (4x - 1)(16x^{2} + 4x + 1). +\end{align*} + +\Dictum{To find the factors of a binomial when it is the difference of +two cubes}, therefore, +\begin{Theorem} +Take the difference of the cube roots of the terms for one +factor, and the sum of the squares of the cube roots of the +terms plus their product for the other factor. +\end{Theorem} + +\ScreenBreak +\Exercise{35.} + +Resolve into factors: +\begin{multicols}{3} +\Item{1.} $8x^{3} - y^{3}$. + +\Item{2.} $x^{3} - 1$. + +\Item{3.} $x^{3}y^{3} - z^{3}$. + +\Item{4.} $x^{3} - 64$. + +\Item{5.} $125a^{3} - b^{3}$. + +\Item{6.} $a^{3} - 343$. + +\Item{7.} $a^{3}b^{3} - 27c^{3}$. + +\Item{8.} $x^{3}y^{3}z^{3} - 8$. + +\Item{9.} $8a^{3}b^{3} - 27y^{6}$. + +\Item{10.} $64x^{3} - y^{9}$. + +\Item{11.} $27a^{3} - 64c^{6}$. + +\Item{12.} $x^{3}y^{3} - 216z^{3}$. + +\Item{13.} $64x^{3} - 729y^{3}$. + +\Item{14.} $27a^{3} - 512c^{3}$. + +\Item{15.} $8x^{6} - 125y^{3}$. + +\Item{16.} $64x^{12} - 27y^{15}$. + +\Item{17.} $216 - 8a^{3}$. + +\Item{18.} $343 - 27y^{3}$. +\end{multicols} + + +\Section{Case V.} + +\Paragraph{113. When a binomial is the sum of two cubes.} +\begin{DPgather*} +\lintertext{\indent Since} +\frac{a^{3} + b^{3}}{a + b} = a^{2} - ab + b^{2}, +\end{DPgather*} +the factors of $a^{3} + b^{3}$ are $a + b$ and $a^{2} - ab + b^{2}$. +%% -----File: 084.png---Folio 78------- + +In like manner we can resolve into factors any expression +which can be written as the sum of two cubes. + +\Item{1.} Resolve into factors $8x^{3} + 27y^{3}$. + +Since by §~112, $8x^{3} = (2x)^{3}$ and $27y^{3} = (3y)^{3}$, we can +write $8x^{3} + 27y^{3}$ as $(2x)^{3} + (3y)^{3}$. + +\begin{DPgather*} +\lintertext{\indent Since} +a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2}), +\end{DPgather*} +we have, by putting $2x$ for~$a$, and $3y$ for~$b$, +\begin{align*} +(2x)^{3} + (3y)^{3} + &= (2x + 3y)[(2x)^{2} - 2x × 3y + (3y)^{2}] \\ + &= (2x + 3y)(4x^{2} - 6xy + 9y^{2}). +\end{align*} + +\Item{2.} Resolve into factors $125a^{3} + 64x^{6}$\Add{.} +\begin{gather*} +125a^{3} = (5a)^{3},\quad 64x^{6} = (4x^{2})^{3}; \\ +\begin{aligned} +\therefore 125a^{3} + 64x^{6} + &= (5a + 4x^{2})[(5a)^{2} - 5a × 4x^{2} + (4x^{2})^{2}] \\ + &= (5a + 4x^{2})(25a^{2} - 20ax^{2} + 16x^{4}) +\end{aligned} +\end{gather*} + +\Dictum{To find the factors of a binomial when it is the sum of two +cubes}, therefore, +\begin{Theorem} +Take the sum of the cube roots of the terms for one factor, +and the sum of the squares of the cube roots of the terms +minus their product for the other factor. +\end{Theorem} + +\PrintBreak +\Exercise{36.} + +Resolve into factors: +\begin{multicols}{3} +\Item{1.} $x^{3} + 1$. + +\Item{2.} $8x^{3} + y^{3}$. + +\Item{3.} $x^{3} + 125$. + +\Item{4.} $64a^{3} + 27$. + +\Item{5.} $x^{3}y^{3} + z^{3}$. + +\Item{6.} $a^{3} + 64$. + +\Item{7.} $8a^{6} + b^{3}$. + +\Item{8.} $x^{3} + 343$. + +\Item{9.} $8 + x^{3}y^{3}z^{3}$. + +\Item{10.} $y^{9} + 64x^{3}$. + +\Item{11.} $a^{3}b^{3} + 27x^{3}$. + +\Item{12.} $8y^{3}z^{3} + x^{6}$. + +\Item{13.} $y^{9} + 64x^{6}$. + +\Item{14.} $64a^{12} + x^{15}$. + +\Item{15.} $27x^{15} + 8a^{6}$. + +\Item{16.} $27x^{9} + 512$. + +\Item{17.} $343 + 64x^{3}$. + +\Item{18.} $125 + 27y^{3}$. +\end{multicols} +%% -----File: 085.png---Folio 79------- + + +\Section{Case VI.} + +\Paragraph{114. When a trinomial is a perfect square.} +\begin{DPgather*} +\lintertext{\indent Since} +(x + y)^{2} = x^{2} + 2xy + y^{2}, +\end{DPgather*} +the factors of $x^{2} + 2xy + y^{2}$ are $x + y$ and $x + y$. + +\begin{DPgather*} +\lintertext{\indent Since} +(x - y)^{2} = x^{2} - 2xy + y^{2}, +\end{DPgather*} +the factors of $x^{2} - 2xy + y^{2}$ are $x - y$ and $x - y$. + +Therefore, a trinomial is a perfect square, if its first and +last terms are perfect squares and positive, and its middle +term is twice the product of their square roots. + +\Dictum{To find the factors of a trinomial when it is a perfect square}, +therefore, +\begin{Theorem} +Extract the square roots of the first and last terms, and +connect these square roots by the sign of the middle term. +\end{Theorem} + +Thus, if we wish to find the square root of +\[ +16a^{2} - 24ab + 9b^{2}, +\] +we take the square roots of $16a^{2}$ and $9b^{2}$, which are $4a$ +and~$3b$, respectively, and connect these square roots by +the minus sign, the sign of the middle term. The square +root is therefore +\[ +4a - 3b. +\] + +Again, if we wish to find the square root of +\[ +25x^{2} + 40xy + 16y^{2}, +\] +we take the square roots of $25x^{2}$ and $16y^{2}$ and connect these +roots by the plus sign, the sign of the middle term. The +square root is therefore +\[ +5x + 4y. +\] +%% -----File: 086.png---Folio 80------- + +\Exercise{37.} + +Resolve into factors: +\begin{multicols}{2} +\Item{1.} $4x^{2} + 4xy + y^{2}$. + +\Item{2.} $x^{2} + 6xy + 9y^{2}$. + +\Item{3.} $x^{2} + 16x + 64$. + +\Item{4.} $x^{2} + 10ax + 25a^{2}$. + +\Item{5.} $a^{2} - 16a + 64$. + +\Item{6.} $a^{2} - 10ab + 25b^{2}$. + +\Item{7.} $c^{2} - 6cd + 9d^{2}$. + +\Item{8.} $4x^{2} - 4x + 1$. + +\Item{9.} $4a^{2} - 12ab + 9b^{2}$. + +\Item{10.} $9a^{2} - 24ab + 16b^{2}$. + +\Item{11.} $x^{2} + 8xy + 16y^{2}$. + +\Item{12.} $x^{2} - 8xy + 16y^{2}$. + +\Item{13.} $4x^{2} - 20xy + 25y^{2}$. + +\Item{14.} $1 + 20a + 100a^{2}$. + +\Item{15.} $49a^{2} - 28a + 4$. + +\Item{16.} $36a^{2} + 60ab + 25b^{2}$. + +\Item{17.} $81x^{2} - 36bx + 4b^{2}$. + +\Item{18.} $m^{2}n^{2} + 14mnx^{2} + 49x^{2}$. +\end{multicols} + + +\PrintBreak +\Section{Case VII.} + +\Paragraph{115. When a trinomial has the form $x^{2} + ax + b$.} + +Where $a$ is the \emph{algebraic sum} of two numbers, and is +either positive or negative; and $b$~is the \emph{product} of these +two numbers, and is either positive or negative. +\begin{DPgather*} +\lintertext{\indent Since} +(x + 5)(x + 3) = x^{2} + 8x + 15, +\end{DPgather*} +the factors of $x^{2} + 8x + 15$ are $x + 5$ and $x + 3$. +\begin{DPgather*} +\lintertext{\indent Since} +(x + 5)(x - 3) = x^{2} + 2x- 15, +\end{DPgather*} +the factors of $x^{2} + 2x - 15$ are $(x + 5)$ and $(x - 3)$. + +Hence, if a trinomial of the form $x^{2} + ax + b$ is such an +expression that it can be resolved into two binomial factors, +it is obvious that the first term of each factor will be~$x$, +and that the second terms of the factors will be two +numbers whose product is~$b$, the last term of the trinomial, +and whose algebraic sum is~$a$, the coefficient of~$x$ in the +middle term of the trinomial. +%% -----File: 087.png---Folio 81------- + +\Item{1.} Resolve into factors $x^{2} + 11x + 30$. +\begin{Soln} +We are required to find two numbers whose product is~$30$ and +whose sum is~$11$. + +Two numbers whose product is $30$ are $1$~and~$30$, $2$~and~$15$, $3$~and~$10$, +$5$~and~$6$, and the sum of the last two numbers is~$11$. Hence, +\[ +x^{2} + 11x + 30 = (x + 5)(x + 6). +\] +%[** TN: Solutions sometimes printed in normal-size type; using smaller type] +\end{Soln} + +\Item{2.} Resolve into factors $x^{2} - 7x + 12$. +\begin{Soln} +We are required to find two numbers whose product is~$12$ and +whose algebraic sum is~$-7$. + +Since the product is~$+12$, the two numbers are \emph{both positive} or \emph{both +negative}, and since their sum is~$-7$, they must both be negative. + +Two negative numbers whose product is~$12$ are $-12$~and~$-1$, $-6$ +and~$-2$, $-4$~and~$-3$, and the sum of the last two numbers is~$-7$. +Hence, +\[ +x^{2} - 7x + 12 = (x - 4)(x - 3). +\] +\end{Soln} + +\Item{3.} Resolve into factors $x^{2} + 2x - 24$. +\begin{Soln} +We are required to find two numbers whose product is~$-24$ and +whose algebraic sum is~$2$. + +Since the product is~$-24$, one of the numbers is positive and the +other negative, and since their sum is~$+2$, the larger number is +positive. + +Two numbers whose product is~$-24$, and the larger number positive, +are $24$~and~$-1$, $12$~and~$-2$, $8$~and~$-3$, $6$~and~$-4$, and the sum +of the last two numbers is~$+2$. Hence, +\[ +x^{2} + 2x - 24 = (x + 6)(x - 4). +\] +\end{Soln} + +\Item{4.} Resolve into factors $x^{2} - 3x - 18$. +\begin{Soln} +Since the product is~$-18$, one of the numbers is positive and the +other negative, and since their sum is~$-3$, the larger number is +negative. + +Two numbers whose product is~$-18$, and the larger number negative, +are $-18$~and~$1$, $-9$~and~$2$, $-6$~and~$3$, and the sum of the last +two numbers is~$-3$. Hence, +\[ +x^{2} - 3x - 18 = (x - 6)(x + 3). +\] +\end{Soln} +%% -----File: 088.png---Folio 82------- + +\Exercise{38.} + +Resolve into factors: +\begin{multicols}{2} +\Item{1.} $a^{2} + 5a + 6$. + +\Item{2.} $a^{2} - 5a + 6$. + +\Item{3.} $a^{2} + 6a + 5$. + +\Item{4.} $a^{2} - 6a + 5$. + +\Item{5.} $a^{2} + 4a - 5$. + +\Item{6.} $a^{2} - 4a - 5$. + +\Item{7.} $c^{2} - 9c + 18$. + +\Item{8.} $c^{2} + 9c + 18$. + +\Item{9.} $c^{2} + 3c - 18$. + +\Item{10.} $c^{2} - 3c - 18$. + +\Item{11.} $x^{2} + 9x + 14$. + +\Item{12.} $x^{2} - 9x + 14$. + +\Item{13.} $x^{2} - 5x - 14$. + +\Item{14.} $x^{2} - 9x + 20$. + +\Item{15.} $x^{2} - x - 20$. + +\Item{16.} $x^{2} + x - 20$. + +\Item{17.} $x^{2} - 10x + 21$. + +\Item{18.} $x^{2} - 4x - 21$. + +\Item{19.} $x^{2} + 4x - 21$. + +\Item{20.} $x^{2} - 15x + 56$. + +\Item{21.} $x^{2} - x - 56$. + +\Item{22.} $x^{2} - 10x + 9$. + +\Item{23.} $x^{2} + 13x + 30$. + +\Item{24.} $x^{2} + 7x - 30$. + +\Item{25.} $x^{2} - 7x - 30$. + +\Item{26.} $a^{2} + ab - 6b^{2}$. + +\Item{27.} $a^{2} - ab - 6b^{2}$. + +\Item{28.} $a^{2} + 3ab - 4b^{2}$. + +\Item{29.} $a^{2} - 3ab - 4b^{2}$. + +\Item{30.} $a^{2}x^{2} - 2ax - 63$. + +\Item{31.} $a^{2} + 2ax - 63x^{2}$. + +\Item{32.} $a^{2} - 9ab + 20b^{2}$. + +\Item{33.} $x^{2}y^{2} - 19xyz + 48z^{2}$. + +\Item{34.} $a^{2}b^{2} + 15abc + 44c^{2}$. + +\Item{35.} $x^{2} - 13xy + 36y^{2}$. + +\Item{36.} $x^{2} + 19xy + 84y^{2}$. + +\Item{37.} $a^{2}x^{2} - 23axy + 102y^{2}$. + +\Item{38.} $x^{4} - 9x^{2}y^{2} + 20y^{4}$. + +\Item{39.} $a^{4}x^{4} - 24a^{2}x^{2}y^{2} + 143y^{4}$. + +\Item{40.} $a^{6}b^{6} - 23a^{3}b^{3}c^{2} + 132c^{4}$. + +\Item{41.} $a^{2} - 20abc - 96b^{2}c^{2}$. + +\Item{42.} $a^{2} - 4abc - 96b^{2}c^{2}$. + +\Item{43.} $a^{2} - 10abc - 96b^{2}c^{2}$. + +\Item{44.} $a^{2} + 29abc - 96b^{2}c^{2}$. + +\Item{45.} $a^{2} - 46abc - 96b^{2}c^{2}$. + +\Item{46.} $a^{2} + 49abc + 48b^{2}c^{2}$. + +\Item{47.} $x^{2} - 18xyz - 243y^{2}z^{2}$. + +\Item{48.} $x^{2}y^{2} - xyz - 182z^{2}$. +\end{multicols} +%% -----File: 089.png---Folio 83------- + +\Exercise{39.} + +\Section{Examples For Review.} + +Resolve into factors: +\begin{multicols}{2} +\Item{1.} $a^{3} - 7a$. + +\Item{2.} $3a^{2}b^{2} - 2a^{3}b + 3ab^{3}$. + +\Item{3.} $(a - b)^{2} + (a - b)$. + +\Item{4.} $(a + b)^{2} - 1$. + +\Item{5.} $a^{3} + 8b^{3}$. + +\Item{6.} $(x^{2} - 4y^{2}) + (x - 2y)$. + +\Item{7.} $(a^{3} - b^{3}) + (a - b)$. + +\Item{8.} $a^{2} - 6ab + 9b^{2}$. + +\Item{9.} $x^{2} - x -2$. + +\Item{10.} $x^{2} - 2x - 3$. + +\Item{11.} $x^{2} + 4x - 21$. + +\Item{12.} $a^{2} - 11a - 26$. + +\Item{13.} $ax^{2} + bx^{2} + 3a + 3b$. + +\Item{14.} $x^{2} - 3x- xy + 3y$. + +\Item{15.} $x^{2} - 7x + 12$. + +\Item{16.} $a^{2} + 5ab + 6b^{2}$. + +\Item{17.} $x^{4} + 10x^{2} + 25$. + +\Item{18.} $x^{2} - 18x + 81$. + +\Item{19.} $x^{2} - 21x + 110$. + +\Item{20.} $x^{2} + 19x + 88$. + +\Item{21.} $x^{2} - 19x + 88$. + +\Item{22.} $x^{3} - x^{2} + x - 1$. + +\Item{23.} $9x^{4} - x^{2}$. + +\Item{24.} $1 - (a - b)^{2}$. + +\Item{25.} $(a^{3} + b^{3}) + (a + b)$. + +\Item{26.} $m^{2}x - n^{2}x + m^{2}y - n^{2}y$. + +\Item{27.} $(x - y)^{2} - z^{2}$. + +\Item{28.} $z^{2} - (x - y)^{2}$. + +\Item{29.} $4a^{4} - (3a - 1)^{2}$. + +\Item{30.} $8x^{3} - y^{3}$. + +\Item{31.} $x^{3} - 3x^{2}y$. + +\Item{32.} $x^{3} - 27y^{3}$. + +\Item{33.} $x^{2} + 3x - 40$. + +\Item{34.} $x^{2} + 3xy - 10y^{2}$. + +\Item{35.} $1 - 16x^{2}$. + +\Item{36.} $a^{6} - 9a^{2}b^{4}$. + +\Item{37.} $x^{3} + 3x^{2}y + 2xy^{2}$. + +\Item{38.} $x^{4} + 4x^{3}y + 3x^{2}y^{2}$. + +\Item{39.} $x^{2} - 4xy^{2} + 4y^{4}$. + +\Item{40.} $16x^{4} + 8x^{2} + 1$. + +\Item{41.} $9a^{4} - 4a^{2}c^{2}$. + +\Item{42.} $a^{3}b - a^{2}b^{2} - 2ab^{3}$. + +\Item{43.} $x^{4} - x^{3} + 8x - 8$. + +\Item{44.} $a^{4} - a^{3}x + ay^{3} - xy^{3}$. +\end{multicols} +%% -----File: 090.png---Folio 84------- + + +\Chapter{VIII.}{Common Factors and Multiples.} + +\Paragraph{116. Common Factors.} A \Defn{common factor} of two or more +\emph{integral numbers} is an integral number which divides each +of them without a remainder. + +\Paragraph{117.} A \Defn{common factor} of two or more integral and rational +\emph{expressions} is an integral and rational expression which +divides each of them without a remainder. + +\begin{Remark} +Thus $5a$~is a common factor of $20a$~and~$25a$, $3x^{2}y^{2}$ is a common +factor of~$12x^{2}y^{2}$ and~$15x^{3}y^{3}$. +\end{Remark} + +\Paragraph{118.} Two \emph{numbers} are said to be \Defn{prime} to each other +when they have no common factor except~$1$. + +\Paragraph{119.} Two \emph{expressions} are said to be \Defn{prime} to each other +when they have no common factor except~$1$. + +\Paragraph{120.} The \Defn{highest common factor} of two or more integral +\emph{numbers} is the greatest number that will divide each of +them without a remainder. + +\Paragraph{121.} The \Defn{highest common factor} of two or more integral +and rational \emph{expressions} is an integral and rational expression +of highest degree that will divide each of them without +a remainder. + +\begin{Remark} +Thus $3a^{2}$ is the highest common factor of $3a^{2}$, $6a^{3}$, and~$12a^{4}$, +$5x^{2}y^{2}$ is the highest common factor of $10x^{3}y^{2}$ and~$15x^{2}y^{2}$. +\end{Remark} + +For brevity, we use \HCF\ for ``highest common factor.'' +%% -----File: 091.png---Folio 85------- + +\Paragraph{122. To Find the Highest Common Factor of Two or More +Algebraic Expressions.} + +\Item{1.} Find the \HCF\ of $42a^{3}b^{2}$ and~$30a^{2}b^{4}$. +\begin{alignat*}{2} +&42a^{3}b^{2} &&= 2 × 3 × 7 × aaa × bb; \\ +&30a^{2}b^{4} &&= 2 × 3 × 5 × aa × bbbb. \\ +\therefore\ &\text{the \HCF} &&= 2 × 3 × aa × bb, +\quad\text{or}\quad 6a^{2}b^{2}. +\end{alignat*} + +\Item{2.} Find the \HCF\ of $x^{2} - 9y^{2}$ and $x^{2} + 6xy + 9y^{2}$. +\begin{alignat*}{2} +&x^{2} - 9y^{2} &&= (x + 3y)(x - 3y); \\ +&x^{2} + 6xy + 9y^{2} &&= (x + 3y)(x + 3y). \\ +\therefore\ &\text{the \HCF} &&= (x + 3y). +\end{alignat*} + +\Item{3.} Find the \HCF\ of $4x^{2} - 4x - 80$, $2x^{2} - 18x + 40$. +\begin{alignat*}{3} +&4x^{2} -& 4x &- 80 &&= 4(x^{2} - x-20) \\ +&&& &&= 4(x - 5)(x + 4); \\ +&2x^{2} -& 18x &+ 40 &&= 2(x^{2} - 9x + 20) \\ +&&& &&= 2(x - 5)(x - 4). \\ +\therefore\ &\rlap{\text{the \HCF}} &&&&= 2(x - 5). +\end{alignat*} + +\Dictum{To find the \HCF\ of two or more expressions}, therefore, +\begin{Theorem} +Resolve each expression into its simplest factors. + +Find the product of all the common factors, taking each +factor the least number of times it occurs in any of the given +expressions. +\end{Theorem} + +\begin{Remark}[Note.] The \emph{highest common factor} in Algebra corresponds to the +\emph{greatest common measure}, or \emph{greatest common divisor} in Arithmetic. +We cannot apply the terms \emph{greatest} and \emph{least} to an algebraic expression +in which particular values have not been given to the letters +contained in the expression. Thus $a$~is \emph{greater} than~$a^{2}$, if $a$~stands +for~$\frac{1}{4}$. +\end{Remark} +%% -----File: 092.png---Folio 86------- + +\Exercise{40.} + +Find the \HCF\ of: +\begin{multicols}{2} +\Item{1.} $330$ and $546$. + +\Item{2.} $20x^{3}$ and $15x^{4}$. + +\Item{3.} $42ax^{2}$ and $60a^{2}x$. + +\Item{4.} $35a^{2}b^{2}$ and $49ab^{3}$. + +\Item{5.} $28x^{4}$ and $63y^{4}$. + +\Item{6.} $54a^{2}b^{2}$ and $56a^{3}b^{3}$. + +\Item{7.} $x^{3} + 3x^{2}y$ and $x^{3} + 27y^{3}$. + +\Item{8.} $x^{2} + 3x$ and $x^{2} - 9$. + +\Item{9.} $2ax^{3} + x^{3}$ and $8a^{3} + 1$. + +\Item{10.} $(x + y)^{2}$ and $x^{2} - y^{2}$. + +\Item{11.} $a^{3} + a^{2}x$ and $a^{2} - x^{2}$. + +\Item{12.} $a^{2} - 4b^{2}$ and $a^{2} + 2ab$. +\end{multicols} + +\Item{13.} $x^{2} - 1$ and $x^{2} + 2x - 3$. + +\Item{14.} $x^{2} + 5x + 6$ and $x^{2} + 4x + 3$. + +\Item{15.} $x^{2} - 9x + 18$ and $x^{2} - 10x + 24$. + +\Item{16.} $x^{3} + 1$ and $x^{2} - x + 1$. + +\Item{17.} $x^{2} - 3x + 2$ and $x^{2} - 4x + 3$. + +\Item{18.} $x^{2} - 3xy + 2y^{2}$ and $x^{2} - 2xy + y^{2}$. + +\Item{19.} $x^{2} - 4x - 5$ and $x^{2} - 25$. + +\Item{20.} $(a - b)^{2} - c^{2}$ and $ab - b^{2} - bc$. + +\Item{21.} $x^{2} + xy -2y^{2}$ and $x^{2} + 5xy + 6y^{2}$. + +\Item{22.} $x^{2} + 7xy + 12y^{2}$ and $x^{2} + 3xy - 4y^{2}$. + +\Item{23.} $x^{3} - 8y^{3}$ and $x^{2} + 2xy + 4y^{2}$. + +\Item{24.} $x^{3} - 2x^{2} - x + 2$ and $x^{2} - 4x + 4$. + +\Item{25.} $1 - 5a + 6a^{2}$ and $1 - 7a + 12a^{2}$. + +\Item{26.} $x^{2} - 8xy + 7y^{2}$ and $x^{2} - 3xy - 28y^{2}$. + +\Item{27.} $8a^{3} + b^{3}$ and $4a^{2} + 4ab + b^{2}$. + +\Item{28.} $x^{2} - (y - z)^{2}$ and $(x + y)^{2} - z^{2}$. +%% -----File: 093.png---Folio 87------- + +\Paragraph{123. Common Multiples.} A \Defn{common multiple} of two or +more integral \emph{numbers} is a number which is exactly divisible +by each, of the numbers. + +A \Defn{common multiple} of two or more \emph{expressions} is an expression +which is exactly divisible by each of the expressions. + +\Paragraph{124.} The \Defn{lowest common multiple} of two or more \emph{numbers} +is the least number that is exactly divisible by each of the +given numbers. + +The \Defn{lowest common multiple} of two or more \emph{expressions} is +the expression of lowest degree that is exactly divisible by +each of the given expressions. + +We use \LCM\ for ``lowest common multiple.'' + +\Paragraph{To find the lowest common multiple of two or more algebraic +expressions.} + +\Item{1.} Find the \LCM\ of $42a^{3}b^{2}$, $30a^{2}b^{4}$, and~$66ab^{3}$. +\begin{align*} +42a^{3}b^{2} &= 2 × 3 × 7 × a^{3} × b^{2}; \\ +30a^{2}b^{4} &= 2 × 3 × 5 × a^{2} × b^{4}; \\ +66ab^{3} &= 2 × 3 × 11 × a × b^{3}. +\end{align*} + +\begin{Soln} +The \LCM\ must evidently contain each factor the greatest number +of times that it occurs in any expression. +\begin{align*} +\therefore\ \text{\LCM} + &= 2 × 3 × 7 × 5 × 11a^{3} × b^{4}, \\ + &= 2310a^{3}b^{4}. +\end{align*} +\end{Soln} + +\Item{2.} Find the \LCM\ of $4x^{2} - 4x - 80$ and $2x^{2} - 18x + 40$. +\begin{Soln} +\begin{alignat*}{3} +&4x^{2} -& 4x &- 80 &&= 4(x^{2} - x - 20) = 4(x - 5)(x + 4); \\ +&2x^{2} -& 18x &+ 40 &&= 2(x^{2} - 9x + 20) = 2(x - 5)(x - 4). \\ +\therefore\ &\rlap{\text{\LCM}}&&& &= 4(x - 5)(x + 4)(x - 4). +\end{alignat*} +\end{Soln} + +\Dictum{To find the \LCM\ of two or more expressions}, therefore, +%% -----File: 094.png---Folio 88------- +\begin{Theorem} +Resolve each expression into its simplest factors. + +Find the product of all the different factors, taking each +factor the greatest number of times it occurs in any of the +given expressions. +\end{Theorem} + +\Exercise{41.} + +Find the \LCM\ of: +\begin{multicols}{2} +\Item{1.} $9xy^{3}$ and $6x^{2}y$. + +\Item{2.} $3abc^{2}$ and $2a^{2}bc^{3}$. + +\Item{3.} $4a^{3}b$ and $10ab^{3}$. + +\Item{4.} $6a^{3}b^{3}$ and $15a^{2}b^{4}$. + +\Item{5.} $21xy^{3}$ and $27x^{3}y^{5}$. + +\Item{6.} $xy^{3}z^{2}$ and $x^{2}y^{2}z^{3}$. + +\Item{7.} $a^{2}$ and $a^{2} + a$. + +\Item{8.} $x^{2}$ and $x^{3} - 3x^{2}$. + +\Item{9.} $x^{2} - 1$ and $x^{2} + x$. + +\Item{10.} $x^{2} - 1$ and $x^{2} - x$. + +\Item{11.} $x^{2} + xy$ and $xy + y^{2}$. + +\Item{12.} $x^{2} + 2x$ and $(x + 2)^{2}$. +\end{multicols} + +\Item{13.} $a^{2} + 4a + 4$ and $a^{2} + 5a + 6$. + +\Item{14.} $c^{2} + c - 20$ and $c^{2} - c - 30$. + +\Item{15.} $b^{2} + b - 42$ and $b^{2} - 11b + 30$. + +\Item{16.} $y^{2} - 10y + 24$ and $y^{2} + y -20$. + +\Item{17.} $z^{2} + 2z - 35$ and $z^{2} - 11z + 30$. + +\Item{18.} $x^{2} - 64; x^{3} - 64$; and $x + 8$. + +\Item{19.} $a^{2} - b^{2}; (a + b)^{2}$; and $(a - b)^{2}$. + +\Item{20.} $4ab(a + b)^{2}$ and $2a^{2}(a^{2} - b^{2})$. + +\Item{21.} $y^{2} + 7y + 12; y^{2} + 6y + 8$; and $y^{2} + 5y +6$. + +\Item{22.} $x^{2} - 1; x^{3} + x^{2} + x + 1$; and $x^{3} - x^{2} + x - 1$. + +\Item{23.} $1 - x^{2}; 1 - x^{3}$; and $1 + x$. + +\Item{24.} $x^{2} + 2xy + y^{2}; x^{2} - y^{2}$; and $x^{2} - 2xy + y^{2}$. + +\Item{25.} $x^{3} - 27; x^{2} + 2x - 15; x^{2} + 5x$. + +\Item{26.} $(a + b)^{2} - c^{2}; (a + b + c)^{2}$; and $a + b - c$. + +\Item{27.} $x^{2} - (a + b)x + ab$ and $x^{2} - (a + c)x + ac$. + +\Item{28.} $(a + b)^{2} - c^{2}$ and $a^{2} + ab + ac$. +%% -----File: 095.png---Folio 89------- + + +\Chapter{IX.}{Fractions.} + +\Paragraph{125.} An \Defn{algebraic fraction} is the indicated quotient of +two expressions, written in the form~$dfrac{a}{b}$. + +The dividend~$a$ is called the \Defn{numerator}, and the divisor~$b$ +is called the \Defn{denominator}; and the numerator and denominator +are called the \Defn{terms} of the fraction. + +\Paragraph{126.} The introduction of the same factor into the dividend +and divisor does not alter the value of the quotient, +and the rejection of the same factor from the dividend and +divisor does not alter the value of the quotient. +\begin{DPgather*} +\lintertext{\indent Thus} +\frac{12}{4} = 3;\quad +\frac{2 × 12}{2 × 4} = 3; \frac{12÷2}{4÷2} = 3. \EqText{Hence,} +\end{DPgather*} +\begin{Theorem} +The value of a fraction is not altered if the numerator and +denominator are both multiplied, or both divided, by the +same factor. +\end{Theorem} + + +\Section{Reduction of Fractions.} + +\Paragraph{127.} To reduce a fraction is to change its \emph{form} without +altering its \emph{value}. + + +\Section{Case I.} + +\Paragraph{128. To Reduce a Fraction to its Lowest Terms.} + +A fraction is in its \emph{lowest terms} when the numerator and +denominator have no common factor. We have, therefore, +the following rule: +%% -----File: 096.png---Folio 90------- +\begin{Theorem} +Resolve the numerator and denominator into their prime +factors, and cancel all the common factors. +\end{Theorem} + +Reduce the following fractions to their lowest terms: + +\Item{1.} $\dfrac{38a^{2}b^{3}c^{4}}{57a^{3}bc^{2}} = \dfrac{2 × 19a^{2}b^{3}c^{4}}{3 × 19a^{3}bc^{2}} = \dfrac{2b^{2}c^{2}}{3a}$. + +\Item{2.} $\dfrac{a^{3} - x^{3}}{a^{2} - x^{2}} = \dfrac{(a - x)(a^{2} + ax + x^{2})}{(a - x)(a + x)} = \dfrac{a^{2} + ax + x^{2}}{a + x}$. + +\Item{3.} $\dfrac{a^{2} + 7a + 10}{a^{2} + 5a + 6} = \dfrac{(a + 5)(a + 2)}{(a + 3)(a + 2)} = \dfrac{a + 5}{a + 3}$. + +\Exercise{42.} + +Reduce to lowest terms: +\begin{multicols}{3} +\Item{1.} $\dfrac{2a}{6ab}$. + +\Item{2.} $\dfrac{12m^{2}n}{15mn^{2}}$. + +\Item{3.} $\dfrac{21m^{2}p^{2}}{28mp^{4}}$. + +\Item{4.} $\dfrac{3x^{3}y^{2}z}{6xy^{3}z^{2}}$. + +\Item{5.} $\dfrac{5a^{3}b^{3}c^{3}}{15c^{5}}$. + +\Item{6.} $\dfrac{34x^{3}y^{4}z^{5}}{51x^{2}y^{3}z^{5}}$. + +\Item{7.} $\dfrac{46m^{2}np^{3}}{69mnp^{4}}$. + +\Item{8.} $\dfrac{39a^{2}b^{3}c^{4}}{52a^{5}bc^{3}}$. + +\Item{9.} $\dfrac{58xy^{4}z^{6}}{87xy^{2}z^{2}}$. +\end{multicols} + +\begin{multicols}{2} +\Item{10.} $\dfrac{abx - bx^{2}}{acx - cx^{2}}$. + +\Item{11.} $\dfrac{4a^{2} - 9b^{2}}{4a^{2} + 6ab}$. + +\Item{12.} $\dfrac{3a^{2} + 6a}{a^{2} + 4a + 4}$. + +\Item{13.} $\dfrac{x^{2} + 5x}{x^{2} + 4x-5}$. + +\Item{14.} $\dfrac{xy - 3y^{2}}{x^{3} - 27y^{3}}$. + +\Item{15.} $\dfrac{x^{2} + 5x + 4}{x^{2} - x-20}$. + +\Item{16.} $\dfrac{x^{2} + 2x + 1}{x^{2} - x-2}$. + +\Item{17.} $\dfrac{(a + b)^{2} - c^{2}}{a^{2} + ab-ac}$. + +\Item{18.} $\dfrac{x^{2} + 9x + 20}{x^{2} + 7x + 12}$. + +\Item{19.} $\dfrac{x^{2} - 14x - 15}{x^{2} - 12x - 45}$. +\end{multicols} +%% -----File: 097.png---Folio 91------- + + +\PrintBreak +\Section{Case II.} + +\Paragraph{129. To Reduce a Fraction to an Integral or Mixed Expression.} + +\Item{1.} Reduce $\dfrac{x^{3} - 1}{x - 1}$ to an integral or mixed expression. + +By division, $\dfrac{x^{3} - 1}{x - 1} = x^{2} + x + 1$. + +\Item{2.} Reduce $\dfrac{x^{3} - 1}{x + 1}$ to an integral or mixed expression. + +By division, $\dfrac{x^{3} - 1}{x + 1} = x^{2} - x + 1 - \dfrac{2}{x + 1}$. + +\ScreenBreak +\Dictum{To reduce a fraction to an integral or mixed expression}, +therefore, +\begin{Theorem} +Divide the numerator by the denominator. +\end{Theorem} + +\begin{Remark}[Note.] If there is a remainder, this remainder must be written +as the numerator of a fraction of which the divisor is the denominator, +and this fraction with its proper sign must be annexed to the integral +part of the quotient. +\end{Remark} + +\Exercise{43.} + +Reduce to integral or mixed expressions: +\begin{multicols}{2} +\Item{1.} $\dfrac{a^{2} - b^{2} + 2}{a - b}$. + +\Item{2.} $\dfrac{a^{2} - b^{2} - 2}{a + b}$. + +\Item{3.} $\dfrac{a^{3} - 2a^{2} + 2a + 1}{a^{2} - a - 1}$. + +\Item{4.} $\dfrac{2x^{2} - 2x + 1}{x + 1}$. + +\Item{5.} $\dfrac{8x^{3}}{2x + 1}$. + +\Item{6.} $\dfrac{5x^{3} + 9x^{2} + 3}{x^{2} + x - 1}$. + +\Item{7.} $\dfrac{a^{3} + a^{2} + 7a - 2}{a^{2} + a + 2}$. + +\Item{8.} $\dfrac{y^{4} + y^{2}x^{2} + x^{4}}{y^{2} + yx + x^{2}}$. + +\Item{9.} $\dfrac{x^{4} - 3x^{3} + x - 1}{x^{2} + x + 1}$. + +\Item{10.} $\dfrac{x^{5} - x^{4} + 1}{x^{2} - x - 1}$. +\end{multicols} +%% -----File: 098.png---Folio 92------- + + +\ScreenBreak +\Section{Case III.} + +\Paragraph{130. To Reduce a Mixed Expression to a Fraction.} + +The process is precisely the same as in Arithmetic. Hence, +\begin{Theorem} +Multiply the integral expression by the denominator, to +the product add the numerator, and under the result write +the denominator. +\end{Theorem} + +Reduce to a fraction $a - b - \dfrac{a^{2} - ab - b^{2}}{a + b}$. +\begin{align*} +a - b - \frac{a^{2} - ab - b^{2}}{a + b} + &= \frac{(a - b)(a + b) - (a^{2} - ab - b^{2})}{a + b} \\ + &= \frac{a^{2} - b^{2} - a^{2} + ab + b^{2}}{a + b} \\ + &= \frac{ab}{a + b}. +\end{align*} + +\begin{Remark}[Note.] The dividing line between the terms of a fraction has the +force of a vinculum affecting the numerator. If, therefore, a \emph{minus +sign} precedes the dividing line, as in the preceding Example, and +this line is removed, the numerator of the given fraction must be +enclosed in a parenthesis preceded by the minus sign, or the sign of +every term of the numerator must be changed. +\end{Remark} + +\ScreenBreak +\Exercise{44.} + +Reduce to a fraction: +\begin{multicols}{2} +\Item{1.} $x - y + \dfrac{2xy}{x - y}$. + +\Item{2.} $x + y - \dfrac{2xy}{x + y}$. + +\Item{3.} $1 - \dfrac{x - y}{x + y}$. + +\Item{4.} $a - x - \dfrac{a^{2} + x^{2}}{a - x}$. + +\Item{5.} $x + 2 - \dfrac{x^{2} - 4}{x - 3}$. + +\Item{6.} $\dfrac{x - 3}{x - 2} - 2x + 1$. + +\Item{7.} $\dfrac{x + 3}{x + 2} + x^{2} - x - 1$. + +\Item{8.} $2a - 1 + \dfrac{3 - 4a}{a - 3}$. + +\Item{9.} $1 - 2a^{2} - \dfrac{a^{2} - a + 2}{a - 1}$. + +\Item{10.} $a^{2} + 2a - 5 - \dfrac{2a - 1}{3a^{2} + 1}$. +\end{multicols} +%% -----File: 099.png---Folio 93------- + + +\Section{Case IV.} + +\Paragraph{131. To Reduce Fractions to their Lowest Common Denominator.} + +The process is the same as in Arithmetic. Hence: +\begin{Theorem} +Find the lowest common multiple of the denominators; +this will be the required denominator. Divide this denominator +by the denominator of each fraction. + +Multiply the first numerator by the first quotient, the second +numerator by the second quotient, and so on. + +The products will be the respective numerators of the +equivalent fractions. +\end{Theorem} + +\begin{Remark}[Note.] Every fraction should be in its lowest terms before the +common denominator is found. +\end{Remark} + +\Item{1.} Reduce $\dfrac{3x}{4a^{2}}$, $\dfrac{2y}{3a}$, and $\dfrac{5}{6a^{3}}$ to equivalent fractions +having the lowest common denominator. +\begin{Soln} +The \LCM\ of $4a^{2}$, $3a$, and $6a^{3} = 12a^{3}$. + +The respective quotients are $3a$, $4a^{2}$, and~$2$. + +The products are $9ax$, $8a^{2}y$, and~$10$. + +Hence, the required fractions are +\[ +\frac{9ax}{12a^{3}},\quad +\frac{8a^{2}y}{12a^{3}}, \quad\text{and}\quad +\frac{10}{12a^{3}}. +\] +\end{Soln} + +\Item{2.} Express $\dfrac{1}{x^{2} + 5x + 6}$ and $\dfrac{1}{x^{2} + 4x + 3}$ with lowest +common denominator. +\begin{Soln} +The factors of the denominators are $x + 3$, $x + 2$; and $x + 3$, $x + 1$. + +Hence the lowest common denominator (\LCD) is $(x + 3)(x + 2)(x + 1)$, +and the required numerators are $x + 1$ and $x + 2$. Hence the required +fractions are +\[ +%[** TN: Equations not displayed in the original] +\frac{x + 1}{(x + 3)(x + 2)(x + 1)} \quad\text{and}\quad +\frac{x + 2}{(x + 3)(x + 2)(x + 1)}. +\] +\end{Soln} +%% -----File: 100.png---Folio 94------- + +\ScreenBreak +\Exercise{45.} + +Express with lowest common denominator: + +\begin{multicols}{2} +\Item{1.} $\dfrac{x}{x - a}$, $\dfrac{x^{2}}{x^{2} - a^{2}}$. + +\Item{2.} $\dfrac{a}{a + b}$, $\dfrac{a^{2}}{a^{2} - b^{2}}$. + +\Item{3.} $\dfrac{1}{1 + 2a}$, $\dfrac{1}{1 - 4a^{2}}$. + +\Item{4.} $\dfrac{9}{16 - x^{2}}$, $\dfrac{4 - x}{4 + x}$. + +\Item{5.} $\dfrac{a^{2}}{27 - a^{3}}$, $\dfrac{a}{3 - a}$. + +\Item{6.} $\dfrac{1}{x^{2} - 5x + 6}$, $\dfrac{1}{x^{2} - x-6}$. +\end{multicols} + + +\Section{Addition and Subtraction of Fractions.} + +\Paragraph{132.} The algebraic sum of two or more fractions which +have the same denominator, is a fraction whose numerator +is the algebraic sum of the numerators of the given fractions, +and whose denominator is the common denominator +of the given fractions. Hence, + +\Dictum{To add fractions}, +\begin{Theorem} +Reduce the fractions to equivalent fractions having the +same denominator; and write the algebraic sum of the +numerators of these fractions over the common denominator. +\end{Theorem} + +\Paragraph{133. When the denominators are simple expressions.} + +\Item{1.} Simplify $\dfrac{3a - 4b}{4} - \dfrac{2a - b + c}{3} + \dfrac{a - 4c}{12}$. +\begin{Soln} +The $\text{\LCD} = 12$. + +The multipliers, that is, the quotients obtained by dividing $12$ by +$4$, $3$, and $12$, are $3$,~$4$, and~$1$. + +Hence the sum of the fractions equals +\begin{align*} +&\frac{9a - 12b}{12} - \frac{8a - 4b + 4c}{12} + \frac{a - 4c}{12} \\ +&\quad= \frac{9a - 12b - 8a + 4b - 4c + a - 4c}{12} \\ +&\quad= \frac{2a - 8b-8c}{12} = \frac{a - 4b - 4c}{6}. +\end{align*} +\end{Soln} +%% -----File: 101.png---Folio 95------- + +The preceding work may be arranged as follows: +\begin{Soln} +The $\text{\LCD} = 12$. + +The multipliers are $3$,~$4$, and~$1$, respectively. +\begin{gather*} +\begin{array}{l*{5}{cr}cl} +3(3a &-& 4b & & ) &=& 9a &-&12b & & &=& \text{1st numerator.} \\ +\llap{$-$} +4(2a &-& b &+& c) &=&-8a &+& 4b &-& 4c &=& \text{2d numerator.} \\ +1( a &-& & &4c) &=& a & & &-& 4c &=& \text{3d numerator.} \\ +\cline{7-11} + & & & & & & 2a &-& 8b &-& 8c \\ + & & & & &\rlap{or}& 2(a &-& 4b &-& 4c) &=& \text{the sum of the numerators.} +\end{array} \\ +\therefore\ \text{sum of fractions} = \frac{2(a - 4b - 4c)}{12} = \frac{a - 4b - 4c}{6}. +\end{gather*} +\end{Soln} + +\Exercise{46.} + +Find the sum of: + +\Item{1.} $\dfrac{x + 1}{2} + \dfrac{x - 3}{5} + \dfrac{x + 5}{10}$. + +\Item{2.} $\dfrac{2x - 1}{3} + \dfrac{x + 5}{4} + \dfrac{x - 4}{6}$. + +\Item{3.} $\dfrac{7x - 1}{6} - \dfrac{3x - 2}{7} + \dfrac{x - 5}{3}$. + +\Item{4.} $\dfrac{3x - 2}{9} - \dfrac{x - 2}{6} + \dfrac{5x + 3}{4}$. + +\Item{5.} $\dfrac{x - 1}{6} - \dfrac{x - 3}{3} + \dfrac{x - 5}{2}$. + +\Item{6.} $\dfrac{x - 2y}{2x} + \dfrac{x + 5y}{4x} - \dfrac{x + 7y}{8x}$. + +\Item{7.} $\dfrac{5x - 11}{3} - \dfrac{2x - 1}{10} - \dfrac{11x - 5}{15}$. + +\Item{8.} $\dfrac{x - 3}{3x} - \dfrac{x^{2} - 6x}{5x^{2}} - \dfrac{7x^{2} - x^{3}}{15x^{3}}$. + +\Item{9.} $\dfrac{ac - b^{2}}{ac} - \dfrac{ab - c^{2}}{ab} + \dfrac{a^{2} - bc}{bc}$. +%% -----File: 102.png---Folio 96------- + +\PrintBreak +\Paragraph{134. When the denominators have compound expressions, +arranged in the same order.} + +\Item{1.} Simplify $\dfrac{a + b}{a - b} - \dfrac{a - b}{a + b} - \dfrac{4ab}{a^{2} - b^{2}}$. +\begin{Soln} +The \LCD\ is $(a - b)(a + b)$. + +The multipliers are $a + b$, $a - b$, and~$1$, respectively. +\begin{gather*} +\begin{array}{l*{3}{cr}cl} + (a + b)(a + b) &=& a^{2} &+& 2ab &+& b^{2} &=& \text{1st numerator.} \\ +\llap{$-$} + (a - b)(a - b) &=&-a^{2} &+& 2ab &-& b^{2} &=& \text{2d numerator.} \\ +\llap{$-$} + 1(4ab) &=& &-& 4ab & & &=& \text{3d numerator.} \\ +\cline{3-7} + & & \multicolumn{5}{c}{0} &=& \text{sum of numerators.} +\end{array} \\ +\therefore\ \text{Sum of fractions${} = 0$.} +\end{gather*} +\end{Soln} + +\Exercise{47.} + +Find the sum of: +\begin{multicols}{2} +\Item{1.} $\dfrac{1}{x + 3} + \dfrac{1}{x - 2}$. + +\Item{2.} $\dfrac{1}{x + 1} + \dfrac{1}{x - 1}$. + +\Item{3.} $\dfrac{4}{x - 8} - \dfrac{1}{x + 2}$. + +\Item{4.} $\dfrac{a + x}{a - x} - \dfrac{a - x}{a + x}$. + +\Item{5.} $\dfrac{x}{x - a} - \dfrac{x^{2}}{x^{2} - a^{2}}$. + +\Item{6.} $\dfrac{4a^{2} + b^{2}}{4a^{2} - b^{2}} - \dfrac{2a + b}{2a - b}$. + +\Item{7.} $\dfrac{7}{9 - a^{2}} - \dfrac{1}{3 + a} - \dfrac{1}{3 - a}$. + +\Item{8.} $\dfrac{1}{a - b} - \dfrac{1}{a + b} - \dfrac{b}{a^{2} - b^{2}}$. +\end{multicols} +%[** TN: Adjusted layout] + +\Item{9.} $\dfrac{2}{x - 2} - \dfrac{2}{x + 2} + \dfrac{5x}{x^{2} - 4}$. + +\Item{10.} $\dfrac{3 - x}{1 - 3x} - \dfrac{3 + x}{1 + 3x} - \dfrac{15x - 1}{1 - 9x^{2}}$. + +\begin{multicols}{2} +\Item{11.} $\dfrac{1}{a} - \dfrac{1}{a + 3} + \dfrac{3}{a + 1}$. + +\Item{12.} $\dfrac{x}{x - 1} - \dfrac{1} - \dfrac{1}{x + 1}$. +\end{multicols} + +\Item{13.} $\dfrac{x + 1}{x + 2} + \dfrac{x - 2}{x - 3} + \dfrac{2x + 7}{x^{2} - x - 6}$. + +\Item{14.} $\dfrac{1}{x(x - 1)} - \dfrac{2}{x^{2} - 1} + \dfrac{1}{x(x + 1)}$. +%% -----File: 103.png---Folio 97------- + +\Exercise{48.} + +Find the sum of: + +\Item{1.} $\dfrac{1}{2x + 1} + \dfrac{1}{2x - 1} - \dfrac{4x}{4x^{2} - 1}$. + +\Item{2.} $\dfrac{a^{2} + b^{2}}{a^{2} - b^{2}} + \dfrac{a}{a + b} - \dfrac{b}{a - b}$. + +\Item{3.} $\dfrac{3a}{1 - a^{2}} + \dfrac{2}{1 - a} - \dfrac{2}{1 + a}$. + +\Item{4.} $\dfrac{1}{2x + 5y} - \dfrac{3x}{4x^{2} - 25y^{2}} + \dfrac{1}{2x + 5y}$. + +\Item{5.} $\dfrac{1}{x + 4y} - \dfrac{8y}{x^{2} - 16y^{2}} + \dfrac{1}{x - 4y}$. + +\Item{6.} $\dfrac{3}{2x - 3} - \dfrac{2}{2x + 3} - \dfrac{3}{4x^{2} - 9}$. + + +\Section{Multiplication and Division of Fractions.} + +\Paragraph{135.} Find the product of $\dfrac{a}{b} × \dfrac{c}{d}$. + +Let $\dfrac{a}{b} = x$, and $\dfrac{c}{d} = y$. + +Then $a = bx$, and $c = dy$. + +The product of these two equations is +\begin{DPalign*} +ac &= bdxy. \\ +\lintertext{\indent Divide by~$bd$,} +\frac{ac}{bd} &= xy. \displaybreak[1] \\ +\lintertext{\indent But} +\frac{a}{b} × \frac{c}{d} = xy. \displaybreak[1] \\ +\lintertext{\indent Therefore} +\frac{a}{b} × \frac{c}{d} = \frac{ac}{bd}. +\end{DPalign*} +%% -----File: 104.png---Folio 98------- + +\Dictum{To find the product of two fractions}, therefore, +\begin{Theorem} +Find the product of the numerators for the required +numerator, and the product of the denominators for the +required denominator. +\end{Theorem} + +In like manner, +\[ +\frac{a}{b} × \frac{c}{d} × \frac{e}{f} = \frac{ac}{bd} × \frac{e}{f} = \frac{ace}{bdf}. +\] + +\Paragraph{136. Reciprocals.} If the product of two numbers is equal +to~$1$, each of the numbers is called the \Defn{reciprocal} of the +other. + +The reciprocal of $\dfrac{a}{b}$ is $\dfrac{b}{a}$, for $\dfrac{b}{a} × \dfrac{a}{b} = \dfrac{ba}{ab} = 1$. + +The reciprocal of a fraction, therefore, is the fraction +inverted. + +%[** TN: Next line broken in the original] +Since $\dfrac{a}{b} ÷ \dfrac{a}{b} = 1$, and $\dfrac{b}{a} × \dfrac{a}{b} = 1$, +it follows that +\begin{Theorem} +To divide by a fraction is the same as to multiply by its +reciprocal. +\end{Theorem} + +\Paragraph{137.} \Dictum{To Divide by a Fraction}, therefore, +\begin{Theorem} +Invert the divisor and multiply. +\end{Theorem} + +\begin{Remark}[Note.] Every mixed expression should first be reduced to a fraction, +and every integral expression should be written as a fraction +having $1$~for the denominator. Both terms of each fraction should +be expressed in their prime factors, and if a factor is common to a +numerator and denominator, it should be cancelled, as the cancelling +of a common factor \emph{before} the multiplication is evidently equivalent +to cancelling it \emph{after} the multiplication. +\end{Remark} +%% -----File: 105.png---Folio 99------- + +\Item{1.} Find the product of $\dfrac{3a^{2}b}{2x^{2}y} × \dfrac{6xy^{2}}{7ab} × \dfrac{7abc}{9a^{2}by^{2}}$. +\[ +\frac{3a^{2}b}{2x^{2}y} × \frac{6xy^{2}}{7ab} × \frac{7abc}{9a^{2}by^{2}} + = \frac{3 × 6 × 7a^{3}b^{2}cxy^{2}}{2 × 7 × 9a^{3}b^{2}x^{2}y^{3}} + = \frac{c}{xy}. +\] + +\Item{2.} Find the product of $\dfrac{ab - b^{2}}{a + b} × \dfrac{ab + b^{2}}{a^{2} - b^{2}}$. +\[ +\frac{ab - b^{2}}{a + b} × \frac{ab + b^{2}}{a^{2} - b^{2}} + = \frac{b(a - b)}{(a + b)} × \frac{b(a + b)}{(a - b)(a + b)} + = \frac{b^{2}}{a + b}. +\] + +\Item{3.} Find quotient of $\dfrac{ab}{(a - b)^{2}} ÷ \dfrac{ac}{a^{2} - b^{2}}$. +\begin{align*} +\frac{ab}{(a - b)^{2}} ÷ \frac{ac}{a^{2} - b^{2}} + &= \frac{ab}{(a - b)(a - b)} × \frac{(a - b)(a + b)}{ac} \\ + &= \frac{b(a + b)}{c(a - b)}. +\end{align*} + +\Item{4.} Find the result of $\dfrac{1}{x} × \dfrac{x^{2} - 1}{x^{2} - 4x - 5} ÷ \dfrac{x^{2} + 2x - 3}{x^{2} - 25}$. +\begin{align*} +&\frac{1}{x} × \frac{x^{2} - 1}{x^{2} - 4x - 5} ÷ \frac{x^{2} + 2x - 3}{x^{2} - 25} \\ +&\qquad= \frac{1}{x} × \frac{x^{2} - 1}{x^{2} - 4x - 5} × \frac{x^{2} - 25}{x^{2} + 2x - 3} \\ +&\qquad= \frac{1}{x} × \frac{(x - 1)(x + 1)}{(x - 5)(x + 1)} × \frac{(x - 5)(x + 5)}{(x + 3)(x - 1)} \\ +&\qquad= \frac{x + 5}{x(x + 3)}. +\end{align*} +%% -----File: 106.png---Folio 100------- + +\Exercise{49.} + +Express in the simplest form: +\begin{multicols}{2} +\Item{1.} $\dfrac{15a^{2}}{7b^{2}} × \dfrac{28ab}{9a^{3}c}$. + +\Item{2.} $\dfrac{3x^{2}y^{2}z^{3}}{4a^{2}b^{2}c^{2}} × \dfrac{8a^{3}b^{2}c^{2}}{9x^{2}yz^{3}}$. + +\Item{3.} $\dfrac{5m^{2}n^{2}p^{4}}{3x^{2}yz^{3}} × \dfrac{21xyz^{2}}{20m^{2}n^{2}p^{2}}$. + +\Item{4.} $\dfrac{16a^{4}b^{2}c^{3}}{21m^{2}x^{3}y^{4}} × \dfrac{3m^{3}x^{3}y^{4}}{8a^{2}b^{2}c^{2}}$. + +\Item{5.} $\dfrac{2a}{bc} × \dfrac{3b}{ac} × \dfrac{5c}{ab}$. + +\Item{6.} $\dfrac{2a^{3}}{3bc} × \dfrac{3b^{3}}{5ac} × \dfrac{5c^{3}}{2ab}$. + +\Item{7.} $\dfrac{5abc^{3}}{3x^{2}} ÷ \dfrac{10ac^{3}}{6bx^{2}}$. + +\Item{8.} $\dfrac{x^{2} - a^{2}}{x^{2} - 4a^{2}} × \dfrac{x + 2a}{x - a}$. + +\Item{9.} $\dfrac{x^{2}y^{2} + 3xy}{4c^{2} - 1} × \dfrac{2c + 1}{xy + 3}$. + +\Item{10.} $\dfrac{a^{2} - 100}{a^{2} - 9} × \dfrac{a - 3}{a - 10}$. + +\Item{11.} $\dfrac{9x^{2} - 4y^{2}}{x^{2} - 4} × \dfrac{x + 2}{3x - 2y}$. + +\Item{12.} $\dfrac{25a^{2} - b^{2}}{16a^{2} - 9b^{2}} ÷ \dfrac{5a - b}{4a - 3b}$. +\end{multicols} +%[** TN: Moved end of two-column layout up two questions] + +\Item{13.} $\dfrac{x^{2} - 49}{(a + b)^{2} - c^{2}} ÷ \dfrac{x + 7}{(a + b) - c}$. + +\Item{14.} $\dfrac{x^{2} + 2x + 1}{x^{2} - 25} ÷ \dfrac{x + 1}{x^{2} + 5x}$. + +\Item{15.} $\dfrac{a^{2} + 3a + 2}{a^{2} + 5a + 6} × \dfrac{a^{2} + 7a + 12}{a^{2} + 9a + 20}$. + +\Item{16.} $\dfrac{y^{2} - y-30}{y^{2} - 36} × \dfrac{y^{2} - y-2}{y^{2} + 3y-10} × \dfrac{y^{2} + 6y}{y^{2} + y}$. + +\Item{17.} $\dfrac{x^{2} - 2x + 1}{x^{2} - y^{2}} × \dfrac{x^{2} + 2xy + y^{2}}{x - 1} ÷ \dfrac{x^{2} - 1}{x^{2} - xy}$. + +\Item{18.} $\dfrac{a^{2} - b^{2}}{a^{2} - 3ab + 2b^{2}} × \dfrac{ab - 2b^{2}}{a^{2} + ab} ÷ \dfrac{(a - b)^{2}}{a(a - b)}$. + +\Item{19.} $\dfrac{(a + b)^{2} - c^{2}}{a^{2} + ab-ac} × \dfrac{a^{2}b^{2}c^{2}}{a^{2} + ab + ac} ÷ \dfrac{b^{2}c^{2}}{abc}$. + +\Item{20.} $\dfrac{x^{2} + 7xy + 10y^{2}}{x^{2} + 6xy + 5y^{2}} × \dfrac{x + 1}{x^{2} + 4x + 4} ÷ \dfrac{1}{x + 2}$. +%% -----File: 107.png---Folio 101------- + +\Paragraph{138. Complex Fractions.} A complex fraction is one that +has a fraction in the numerator, or in the denominator, or +in both. + +The shortest way to reduce to its simplest form a complex +fraction is to multiply both terms of the fraction by +the \LCD\ of the fractions contained in the numerator and +denominator. + +\Item{1.} Simplify $\dfrac{3x}{x - \frac{1}{4}}$. + +Multiply both terms by~$4$, and we have +\[ +\frac{12x}{4x - 1}. +\] + +\Item{2.} Simplify $\dfrac{\dfrac{a + x}{a - x} - \dfrac{a - x}{a + x}}{\dfrac{a + x}{a - x} + \dfrac{a - x}{a + x}}$. + +The \LCD\ of the fractions in the numerator and denominator +is +\[ +(a - x)(a + x). +\] + +Multiply by $(a - x)(a + x)$, and the result is +\begin{align*} +&\frac{(a + x)^{2} - (a - x)^{2}}{(a + x)^{2} + (a - x)^{2}} \\ +&\qquad= \frac{(a^{2} + 2ax + x^{2}) - (a^{2} - 2ax + x^{2})} + {(a^{2} + 2ax + x^{2}) + (a^{2} - 2ax + x^{2})} \\ +&\qquad= \frac{a^{2} + 2ax + x^{2} - a^{2} + 2ax - x^{2}} + {a^{2} + 2ax + x^{2} + a^{2} - 2ax + x^{2}} \\ +&\qquad= \frac{4ax}{2a^{2} + 2x^{2}} \\ +&\qquad= \frac{2ax}{a^{2} + x^{2}}. +\end{align*} +%% -----File: 108.png---Folio 102------- + +\Exercise{50.} + +Reduce to the simplest form: +\begin{multicols}{2} +\Item{1.} $\dfrac{\dfrac{x}{b} + \dfrac{y}{b}}{\dfrac{z}{b}}$. + +\Item{2.} $\dfrac{x + \dfrac{y}{4}}{x - \dfrac{y}{3}}$. + +\Item{3.} $\dfrac{\dfrac{ab}{7} - 3d}{3c - \dfrac{ab}{d}}$. + +\Item{4.} $\dfrac{1 + \dfrac{1}{x + 1}}{1 - \dfrac{1}{x - 1}}$. + +\Item{5.} $\dfrac{\dfrac{2m + x}{m + x} - 1}{1 - \dfrac{x}{m + x}}$. + +\Item{6.} $\dfrac{\dfrac{x + y}{x^{2} - y^{2}}}{\dfrac{x - y}{x + y}}$. + +\Item{7.} $\dfrac{a + \dfrac{ab}{a - b}}{a - \dfrac{ab}{a + b}}$. + +\Item{8.} $\dfrac{9a^{2} - 64}{a - 1 - \dfrac{a + 4}{4}}$. + +\Item{9.} $\dfrac{\dfrac{1}{x} + \dfrac{1}{y}}{\dfrac{1}{x} - \dfrac{1}{y}}$. + +\Item{10.} $\dfrac{x + 3 + \dfrac{2}{x}}{1 + \dfrac{3}{x} + \dfrac{2}{x^{2}}}$. + +\Item{11.} $\dfrac{\dfrac{1}{x} - \dfrac{2}{x^{2}} + \dfrac{1}{x^{3}}}{\dfrac{(1 - x)^{2}}{x^{2}}}$. + +\Item{12.} $\dfrac{x^{2} - x-6}{1 - \dfrac{4}{x^{2}}}$. + +\Item{13.} $\dfrac{a^{2} - a + \dfrac{a - 1}{a + 1}}{a + \dfrac{1}{a + 1}}$. + +\Item{14.} $\dfrac{\dfrac{4a(a - x)}{a^{2} - x^{2}}}{\dfrac{a - x}{a + x}}$. +\end{multicols} +%% -----File: 109.png---Folio 103------- + + +\Chapter{X.}{Fractional Equations.} + +\Paragraph{139. To Reduce Equations containing Fractions.} + +\Item{1.} Solve $\dfrac{x}{3} - \dfrac{x - 1}{11} = x - 9$. +\begin{Soln} +Multiply by~$33$, the \LCM\ of the denominators. +\begin{DPalign*} +\lintertext{\indent Then,} +11x - 3x + 3 &= 33x - 297. \\ +\lintertext{\indent Transpose,} +11x - 3x - 33x &= - 297 - 3. \\ +\lintertext{\indent Combine,} +- 25x &= - 300. \\ +\lintertext{\indent Divide by~$-25$,} +x &= 12. +\end{DPalign*} +\end{Soln} + +\begin{Remark}[Note.] Since the minus sign precedes the second fraction, in removing +the denominator, the~$+$ (understood) before~$x$, the first term +of the numerator, is changed to~$-$, and the~$-$ before~$1$, the second +term of the numerator, is changed to~$+$. +\end{Remark} + +\Dictum{To clear an equation of fractions}, therefore, +\begin{Theorem} +Multiply each term by the \LCM\ of the denominators. +\end{Theorem} + +If a fraction is preceded by a \textbf{minus sign}, \emph{the sign of +every term of the numerator must be changed when the +denominator is removed}. + +\Item{2.} Solve $\dfrac{x + 1}{4} - \frac{1}{5}(x - 1) = 1$. +\begin{Soln} +Multiply by~$20$, the \LCD +\begin{DPalign*} +5x + 5 - 4(x - 1) &= 20. \\ +5x + 5 - 4x + 4 &= 20. \\ +\lintertext{\indent Transpose,} +5x - 4x &= 20 - 5 - 4. \\ +\lintertext{\indent Combine,} +x &= 11. +\end{DPalign*} +\end{Soln} +%% -----File: 110.png---Folio 104------- + +\Item{3.} Solve +\[ +7x - \frac{(2x - 3)(3x - 5)}{5} = \frac{153}{10} - \frac{(4x - 5)(3x - 1)}{10}. +\] +\begin{Soln} +Multiply by~$10$, the \LCD, and we have +\[ +70x - 2(2x - 3)(3x - 5) = 153 - (4x - 5) (3x - 1). +\] + +Find the products of $(2x - 3)(3x - 5)$ and $(4x - 5)(3x - 1)$. +\[ +70x - 2(6x^{2} - 19x + 15) = 153 - (12x^{2} - 19x + 5). +\] + +Remove the parentheses, +\[ +70x - 12x^{2} + 38x - 30 = 153 - 12x^{2} + 19x - 5. +\] + +Cancel the~$-12x^{2}$ on each side and transpose, +\begin{DPalign*} +70x + 38x - 19x &= 153 + 30 - 5. \\ +\lintertext{\indent Combine,} +89x &= 178. \\ +\lintertext{\indent Divide by~$89$,} +x &= 2. +\end{DPalign*} +\end{Soln} + +\Item{4.} Solve $\dfrac{2x + 1}{2x - 1} - \dfrac{2x - 1}{2x + 1} = \dfrac{8}{4x^{2} - 1}$. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Since} +4x^{2} - 1 &= (2x + 1)(2x - 1), \\ +\lintertext{the \LCD} + &= (2x + 1)(2x - 1). +\end{DPalign*} +Multiply by the \LCD, and we have, +\begin{DPalign*} +4x^{2} + 4x + 1-(4x^{2} - 4x + 1) &= 8. \\ +\therefore 4x^{2} + 4x + 1 - 4x^{2} + 4x - 1 &= 8. \\ +\therefore 8x &= 8. \\ +\therefore x &= 1. +\end{DPalign*} +\end{Soln} + +\Item{5.} Solve $\dfrac{4}{x + 1} - \dfrac{x + 1}{x - 1} + \dfrac{x^{2} - 3}{x^{2} - 1} = 0$. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Since} +x^{2} - 1 &= (x + 1)(x - 1) \\ +\lintertext{the \LCD} + &= (x + 1)(x - 1). +\end{DPalign*} + +Multiply by the \LCD, $x^{2} - 1$, and we have, +\begin{DPalign*} +4(x - 1) - (x + 1)(x + 1) + x^{2} - 3 &= 0. \\ +\therefore 4x - 4 - x^{2} - 2x - 1 + x^{2} - 3 &= 0. \\ +\therefore 2x &= 8. \\ +\therefore x &= 4. +\end{DPalign*} +\end{Soln} +%% -----File: 111.png---Folio 105------- + +%[** TN: Force page break in both print and screen layout] +\newpage +\Exercise{51.} + +Solve: +\begin{multicols}{2} +\Item{1.} $\dfrac{x - 1}{2} = \dfrac{x + 1}{3}$. + +\Item{2.} $\dfrac{3x - 1}{4} = \dfrac{2x + 1}{3}$. + +\Item{3.} $\dfrac{6x - 19}{2} = \dfrac{2x - 11}{3}$. + +\Item{4.} $\dfrac{7x - 40}{8} = \dfrac{9x - 80}{10}$. +\end{multicols} +%[** TN: Move end of two-column layout up six questions] + +\Item{5.} $\dfrac{3x - 116}{4} + \dfrac{180 - 5x}{6} = 0$. + +\Item{6.} $\dfrac{3x - 4}{2} - \dfrac{3x - 1}{16} = \dfrac{6x - 5}{8}$. + +\Item{7.} $\dfrac{x - 1}{8} - \dfrac{x + 1}{18} = 1$. + +\Item{8.} $\dfrac{60 - x}{14} - \dfrac{3x - 5}{7} = \dfrac{3x}{4}$. + +\Item{9.} $\dfrac{3x - 1}{11} - \dfrac{2 - x}{10} = \dfrac{6}{5}$. + +\Item{10.} $\dfrac{4x}{x + 1} - \dfrac{x}{x - 2} = 3$. + +\Item{11.} $\dfrac{2x + 1}{4} - \dfrac{4x - 1}{10} + 1 - \frac{1}{4} = 0$. + +\Item{12.} $\dfrac{x - 1}{5} - \dfrac{43 - 5x}{6} - \dfrac{3x - 1}{8} = 0$. + +\Item{13.} $\dfrac{1}{x + 7} = \dfrac{2}{x + 1} - \dfrac{1}{x + 3}$. + +\Item{14.} $\dfrac{1}{x + 4} + \dfrac{2}{x + 6} - \dfrac{3}{x + 5} = 0$. + +\Item{15.} $\dfrac{4}{x^{2} - 1} + \dfrac{1}{x - 1} + \dfrac{1}{x + 1} = 0$. + +\Item{16.} $\dfrac{3x + 1}{4} - \dfrac{5x - 4}{7} = 12 - 2x - \dfrac{x - 2}{3}$. + +\Item{17.} $\frac{1}{8}(5x + 3) - \frac{1}{3}(3 - 4x) + \frac{1}{6}(9 - 5x) = \frac{1}{2}(31 - x)$. + +\Item{18.} $\frac{1}{15} (34x - 56) - \frac{1}{5}(7x - 3) - \frac{1}{3}(7x - 5) = 0$. +%% -----File: 112.png---Folio 106------- + +\Exercise{52.} + +Solve: + +\Item{1.} $\frac{2}{3}(x + 1) - \frac{1}{7}(x + 5) = 1$. + +\Item{2.} $\frac{6}{7}(x - 9) - \frac{1}{3}(5 - x) + 3x + 1 = 0$. + +\Item{3.} $\frac{1}{3}(5x - 24) + \frac{1}{7}(x - 2) - 2(x - 1) = 0$. + +\Item{4.} $\dfrac{x + 3}{4} + \dfrac{7x - 2}{5} = \dfrac{5x - 1}{4} + \dfrac{5x + 4}{9}$. + +\Item{5.} $\dfrac{x + 1}{3} - \dfrac{x - 1}{4} = \dfrac{x - 2}{5} - \dfrac{x - 3}{6} + \dfrac{31}{60}$. + +\Item{6.} $\dfrac{(2x - 1)(2 - x)}{2} + x^{2} - \dfrac{1 + 3x}{2} = 0$. + +\Item{7.} $\dfrac{6x - 11}{4} - \dfrac{3 - 4x}{6} = \dfrac{4}{3} - \dfrac{x}{8}$. + +\Item{8.} $\dfrac{x + 6}{4} - \dfrac{16 - 3x}{12} = 4\frac{1}{6}$. + +\Item{9.} $x - \dfrac{x - 2}{3} = \dfrac{x + 23}{4} - \dfrac{10 + x}{5}$. + +\Item{10.} $\dfrac{5x + 3}{x - 1} + \dfrac{2x - 3}{2x - 1} = 6$. + +\Item{11.} $\dfrac{3x}{4x + 1} + 1 = 2 - \dfrac{x}{2(2x - 1)}$. + +\Item{12.} $\dfrac{8x + 7}{5x + 4} - 1 = 1 - \dfrac{2x}{5x + 1}$. + +\Item{13.} $\dfrac{x + 1}{2(x - 1)} - \dfrac{x - 1}{x + 1} = \dfrac{17 - x^{2}}{2(x^{2} - 1)}$. +%% -----File: 113.png---Folio 107------- + +\Paragraph{140.} If the denominators contain both simple and compound +expressions, it is generally best to remove the simple +expressions first, and then the compound expressions. After +each multiplication the result should be reduced to the +simplest form. + +\Item{1.} Solve $\dfrac{4x + 3}{10} - \dfrac{2x + 3}{5x - 1} = \dfrac{2x - 1}{5}$. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Multiply by~$10$,} +4x + 3 - \frac{10(2x + 3)}{5x - 1} &= 4x - 2. \\ +\lintertext{\indent Transpose,} +4x + 3 - 4x + 2 &= \frac{10(2x + 3)}{5x - 1}. \\ +\lintertext{\indent Combine,} +5 &= \frac{10(2x + 3)}{5x - 1}. \displaybreak[1] \\ +\lintertext{\indent Divide by~$5$,} +1 &= \frac{2(2x + 3)}{5x - 1}. \\ +\lintertext{\indent Multiply by $5x - 1$,} +5x - 1 &= 4x + 6. \\ +\lintertext{\indent Transpose and combine,} +x &= 7. +\end{DPalign*} +\end{Soln} + +\Exercise{53.} + +Solve: + +\Item{1.} $\dfrac{10x + 13}{18} - \dfrac{x + 2}{x - 3} = \dfrac{5x - 4}{9}$. + +\Item{2.} $\dfrac{6x + 7}{10} - \dfrac{3x + 1}{5} = \dfrac{x - 1}{3x - 4}$. + +\Item{3.} $\dfrac{11x - 12}{14} - \dfrac{11x - 7}{19x + 7} = \dfrac{22x - 36}{28}$. + +\Item{4.} $\dfrac{2x - 1}{5} + \dfrac{2x - 3}{17x - 12} = \dfrac{4x - 3}{10}$. + +\Item{5.} $\dfrac{11x - 13}{7} - \dfrac{13x + 7}{3x + 7} = \dfrac{22x - 75}{14}$. + +\Item{6.} $\dfrac{6x - 13}{2x + 3} + \dfrac{6x + 7}{9} - \dfrac{2x + 4}{3} = 0$. +%% -----File: 114.png---Folio 108------- + +\Paragraph{141. Literal Equations.} Literal equations are equations +in which some or all of the known numbers are represented +by letters; the numbers regarded as known numbers are +usually represented by the \emph{first} letters of the alphabet. + +\Item{1.} Solve $\dfrac{x + a}{x - b} + \dfrac{x + b}{x - a} = 2$. +\begin{Soln} +Multiply by $(x - a)(x - b)$, +\begin{DPalign*} +\llap{$(x + a)(x - a) + (x + b)(x - b)$} &= 2(x - a)(x - b), \\ +\lintertext{or} +x^{2} - a^{2} + x^{2} - b^{2} &= 2x^{2} - 2ax - 2bx + 2ab. \displaybreak[1] \\ +\lintertext{\indent Transpose,} +x^{2} + x^{2} - 2x^{2} + 2ax + 2bx &= a^{2} + 2ab + b^{2}\Add{.} \displaybreak[1] \\ +\lintertext{\indent Combine,} +2ax + 2bx &= a^{2} + 2ab + b^{2}, \\ +\lintertext{or} +2(a + b)x &= a^{2} + 2ab + b^{2}. \displaybreak[1] \\ +\lintertext{\indent \rlap{Divide by~$a + b$,}} +2x &= a + b\Add{,} \\ +\therefore x &= \frac{a + b}{2}. +\end{DPalign*} +\end{Soln} + +\Exercise{54.} + +Solve: + +\Item{1.} $a(x - a) = b(x - b)$. + +\Item{2.} $(a + b)x + (a - b)x = a^{2}$. + +\Item{3.} $(a + b)x - (a - b)x = b^{2}$. + +\Item{4.} $(2x - a) + (x - 2a) = 3a$. + +\Item{5.} $(x + a + b) + (x + a - b) = 2b$. + +\Item{6.} $(x - a)(x - b) = x(x + c)$. + +\Item{7.} $x^{2} + b^{2} = (a - x)(a - x)$. + +\Item{8.} $(a + b)(2 - x) = (a - b)(2 + x)$. + +\Item{9.} $(x - a)(2x - a) = 2(x - b)^{2}$. + +\Item{10.} $(a + bx)(c + d) = (a + b)(c + dx)$. + +\Item{11.} $\dfrac{x}{a - b} - \dfrac{3a}{a + b} = \dfrac{bx}{a^{2} - b^{2}}$. +%% -----File: 115.png---Folio 109------- + +\PrintBreak +\Paragraph{142. Problems involving Fractional Equations.} + +\Exercise{55.} + +Ex. The sum of the third and fifth parts of a certain +number exceeds two times the difference of the fourth and +sixth parts by~$22$. Find the number. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{the number\Add{.}} \\ +\lintertext{\indent Then} +\frac{x}{3} + \frac{x}{5} &= \text{the sum of its third and fifth parts,} \\ +\frac{x}{4} - \frac{x}{6} &= \text{the difference of its fourth and sixth parts,} \\ +2\left(\frac{x}{4} - \frac{x}{6}\right) + &= \text{$2$~times the difference of} \\ + &\qquad\text{its fourth and sixth parts,} +\end{DPalign*} +\begin{DPalign*} +\frac{x}{3} + \frac{x}{5} - 2\left(\frac{x}{4} - \frac{x}{6}\right) + &= \text{the given excess.} \\ +\lintertext{\indent But} +22 &= \text{the given excess.} \\ +\therefore \frac{x}{3} + \frac{x}{5} - 2\left(\frac{x}{4} - \frac{x}{6}\right) + &= 22. +\end{DPalign*} + +Multiply by~$60$ the \LCD\ of the fractions. +\begin{DPalign*} +20x + 12x - 30x + 20x &= 60 × 22. \\ +\lintertext{\indent Combining,} +22x &= 60 × 22\Add{,} \\ +\therefore x &= 60. +\end{DPalign*} + +The required number, therefore, is~$60$. +\end{Soln} + +\Item{1.} The difference between the fifth and seventh parts of +a certain number is~$2$. Find the number. + +\Item{2.} One-half of a certain number exceeds the sum of its +fifth and seventh parts by~$11$. Find the number. + +\Item{3.} The sum of the third and sixth parts of a certain +number exceeds the difference of its sixth and ninth parts +by~$16$. Find the number. + +\Item{4.} There are two consecutive numbers, $x$~and~$x + 1$, such +that one-half the larger exceeds one-third the smaller +number by~$10$. Find the numbers. +%% -----File: 116.png---Folio 110------- + +\Exercise{56.} + +Ex. The sum of two numbers is~$63$, and if the greater +is divided by the smaller number, the quotient is~$2$ and the +remainder~$3$. Find the numbers. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{the greater number.} \\ +\lintertext{\indent Then} +63 - x &= \text{the smaller number.} \\ +\text{Since the quotient} + &= \frac{\text{Dividend} - \text{Remainder}}{\text{Divisor}}, +\end{DPalign*} +and since, in this problem, the dividend is~$x$, the remainder is~$3$, +and the divisor is~$63 - x$, we have +\begin{DPalign*} +\frac{x - 3}{63 - x} = 2. \\ +\lintertext{\indent Solving,} +x &= 43. +\end{DPalign*} + +The two numbers, therefore, are $43$~and~$20$. +\end{Soln} + +\Item{1.} The sum of two numbers is~$100$, and if the greater is +divided by the smaller number, the quotient is~$4$ and the +remainder~$5$. Find the numbers. + +\Item{2.} The sum of two numbers is~$124$, and if the greater is +divided by the smaller number, the quotient is~$4$ and the +remainder~$4$. Find the numbers. + +\Item{3.} The difference of two numbers is~$49$, and if the greater +is divided by the smaller, the quotient is~$4$ and the remainder~$4$. +Find the numbers. + +\Item{4.} The difference of two numbers is~$91$, and if the +greater is divided by the smaller, the quotient is~$8$ and the +remainder~$7$. Find the numbers. + +\Item{5.} Divide $320$ into two parts such that the smaller part +is contained in the larger part $11$~times, with a remainder +of~$20$. +%% -----File: 117.png---Folio 111------- + +\Exercise{57.} + +Ex. Eight years ago a boy was one-fourth as old as he +will be one year hence. How old is he now? +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{the number of years old he is now.} \\ +\lintertext{\indent Then} +x - 8 &= \text{the number of years old he was eight years ago,} \\ +\lintertext{and} +x + 1 &= \text{the number of years old he will be one year hence.} +\end{DPalign*} +\begin{DPalign*} +\therefore x - 8 &= \tfrac{1}{4}(x + 1). \\ +\lintertext{\indent Solving,} +x &= 11. +\end{DPalign*} + +Therefore the boy is $11$~years old. +\end{Soln} + +\Item{1.} A son is one-fourth as old as his father. In $24$~years +he will be one-half as old. Find the age of the son. + +\Item{2.} B's~age is one-sixth of A's~age. In $15$~years B's~age +will be one-third of A's~age. Find their ages. + +\Item{3.} The sum of the ages of A~and~B is $30$~years, and $5$~years +hence B's~age will be one-third of~A's. Find their +ages. + +\Item{4.} A father is $35$~years old, and his son is one-fourth of +that age. In how many years will the son be half as old +as his father? + +\Item{5.} A is $60$~years old, and B's~age is two-thirds of~A's. +How many years ago was B's~age one-fifth of~A's? + +\Item{6.} A son is one-third as old as his father. Four years +ago he was only one-fourth as old as his father. What is +the age of each? + +\Item{7.} A is $50$~years old, and B~is half as old as~A\@. In how +many years will B be two-thirds as old as~A? + +\Item{8.} B~is one-half as old as~A\@. Ten years ago he was +one-fourth as old as~A\@. What are their present ages? + +\Item{9.} The sum of the ages of a father and his son is $80$~years. +The son's age increased by $5$~years is one-fourth of +the father's age. Find their ages. +%% -----File: 118.png---Folio 112------- + +\Exercise{58.} + +Ex. A~can do a piece of work in $2$~days, and B~can do +it in $3$~days. How long will it take both together to do +the work? +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{the number of days it will take both together.} \\ +\lintertext{\indent Then} +\frac{1}{x} &= \text{\emph{the part both together can do in one day},} \\ +\tfrac{1}{2} &= \text{the part A can do in one day,} \\ +\tfrac{1}{3} &= \text{the part B can do in one day,} \\ +\lintertext{and} +\tfrac{1}{2} + \tfrac{1}{3} + &= \text{\emph{the part both together can do in one day}.} +\end{DPalign*} +\begin{DPalign*} +\therefore \frac{1}{2} + \frac{1}{3} &= \frac{1}{x}. \\ +\lintertext{\indent Solving,} +x &= 1\tfrac{1}{5}. +\end{DPalign*} + +Therefore they together can do the work in $1\frac{1}{5}$~days. +\end{Soln} + +\Item{1.} A~can do a piece of work in $3$~days, B~in $5$~days, and +C~in $6$~days. How long will it take them to do it working +together? + +\Item{2.} A~can do a piece of work in $5$~days, B~in $4$~days, and +C~in $3$~days. How long will it take them together to do +the work? + +\Item{3.} A~can do a piece of work in $2\frac{1}{2}$~days, B~in $3\frac{1}{2}$~days, +and C~in $3\frac{3}{4}$~days. How long will it take them together to +do the work? + +\Item{4.} A~can do a piece of work in $10$~days, B~in $12$~days; +A~and~B together, with the help of~C, can do the work in +$4$~days. How long will it take C~alone to do the work? + +\Item{5.} A~and~B together can mow a field in $10$~hours, A~and~C +in $12$~hours, and A~alone in $20$~hours. In what time +can B~and~C together mow the field? + +\Item{6.} A~and~B together can build a wall in $12$~days, A~and~C +in $15$~days, B~and~C in $20$~days. In what time can they +build the wall if they all work together? + +\begin{Remark}[Hint.] By working \emph{$2$~days each} they build $\frac{1}{12} + \frac{1}{15} + \frac{1}{20}$ of it. +\end{Remark} +%% -----File: 119.png---Folio 113------- + +\Exercise{59.} + +Ex. A cistern can be filled by three pipes in $15$, $20$, and +$30$~hours, respectively. In what time will it be filled by +all the pipes together? +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{the number of hours it will take all together.} \\ +\lintertext{\indent Then} +\frac{1}{x} &= \text{the part all together can fill in one hour\Add{.}} \\ +\llap{$\tfrac{1}{15} + \tfrac{1}{20}$} + \tfrac{1}{30} + &= \text{the part all together can fill in one hour\Add{.}} +\end{DPalign*} +\begin{DPalign*} +\frac{1}{15} + \frac{1}{20} + \frac{1}{30} &= \frac{1}{x}\Add{.} \\ +\lintertext{\indent Solving,} +x &= 6 \tfrac{2}{3}\Add{.} +\end{DPalign*} + +Therefore the pipes together can fill it in $6 \frac{2}{3}$~hours. +\end{Soln} + +\Item{1.} A cistern can be filled by three pipes in $16$, $24$, and +$32$~hours, respectively. In what time will it be filled by +all the pipes together? + +\Item{2.} A tank can be filled by two pipes in $3$~hours and $4$~hours, +respectively, and can be emptied by a third pipe in +$6$~hours. In what time will the cistern be filled if the pipes +are all running together? + +\Item{3.} A tank can be filled by three pipes in $1$~hour and $40$ +minutes, $3$~hours and $20$ minutes, and $5$~hours, respectively. +In what time will the tank be filled if all three pipes are +running together? + +\Item{4.} A cistern can be filled by three pipes in $2\frac{1}{3}$~hours, $3\frac{1}{2}$~hours, +and $4\frac{2}{3}$~hours, respectively. In what time will the +cistern be filled if all the pipes are running together? + +\Item{5.} A cistern has three pipes. The first pipe will fill the +cistern in $12$~hours, the second in $20$~hours, and all three +pipes together will fill it in $6$~hours. How long will it take +the third pipe alone to fill it? +%% -----File: 120.png---Folio 114------- + +\Exercise{60.} + +Ex. A courier who travels $6$~miles an hour is followed, +after $2$~hours, by a second courier who travels $7\frac{1}{2}$~miles an +hour. In how many hours will the second courier overtake +the first? +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{the number of hours the first travels.} \\ +\lintertext{\indent Then} +x - 2 &= \text{the number of hours the second travels,} \\ +6x &= \text{the number of miles the first travels,} \\ +\lintertext{and} +(x - 2) 7\tfrac{1}{2} &= \text{the number of miles the second travels.} +\displaybreak[1] \\ +\intertext{\indent They both travel the same distance.} +\therefore 6x &= (x - 2) 7\tfrac{1}{2}, \\ +\lintertext{or} +12x &= 15x - 30. \\ +\therefore x &= 10. +\end{DPalign*} + +Therefore the second courier will overtake the first in $10 - 2$, or +$8$~hours. +\end{Soln} + +\Item{1.} A sets out from Boston and walks towards Portland +at the rate of $3$~miles an hour. Three hours afterward B +sets out from the same place and walks in the same direction +at the rate of $4$~miles an hour. How far from Boston +will B~overtake~A? + +\Item{2.} A courier who goes at the rate of $6\frac{1}{2}$~miles an hour is +followed, after $4$~hours, by another who goes at the rate of +$7\frac{1}{2}$~miles an hour. In how many hours will the second +overtake the first? + +\Item{3.} A person walks to the top of a mountain at the rate +of two miles an hour, and down the same way at the rate +of $4$~miles an hour. If he is out $6$~hours, how far is it to +the top of the mountain? + +\Item{4.} In going a certain distance, a train travelling at the +rate of $40$~miles an hour takes $2$~hours less than a train +travelling $30$~miles an hour. Find the distance. +%% -----File: 121.png---Folio 115------- + +\Exercise{61.} + +Ex. A hare takes $4$~leaps to a greyhound's~$3$; but $2$~of +the greyhound's leaps are equivalent to $3$~of the hare's. +The hare has a start of $50$~leaps. How many leaps must +the greyhound take to catch the hare? +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +3x &= \text{the number of leaps taken by the greyhound.} \\ +\lintertext{\indent Then} +4x &= \text{the number of leaps of the hare in the same time.} \\ +%[** TN: Hack to coax spacing] +\lintertext{\indent Also, l\rlap{et}} +a &= \text{the number of feet in one leap of the hare.} \displaybreak[1] \\ +\lintertext{\indent Then} +\frac{3a}{2} &= \text{the number of feet in one leap of the hound.} +\displaybreak[1] \\ +\lintertext{\indent \rlap{Therefore,}} +&\quad 3x x \frac{3a}{2} \quad\text{or}\quad +\frac{9ax}{2} = \text{the whole distance}. +\end{DPalign*} + +As the hare has a start of $50$~leaps, and takes $4x$~leaps more before +she is caught, and as each leap is $a$~feet, +\begin{DPalign*} +(50 + 4x)a &= \text{the whole distance.} \\ +\therefore \frac{9ax}{2} &= (50 + 4x)a. \\ +\lintertext{\indent Multiply by~$2$,} +9ax &= (100 + 8x)a, \\ +\lintertext{\indent Divide by~$a$,} +9x &= 100 + 8x, \\ +x &= 100, \\ +\therefore 3x &= 300. +\end{DPalign*} + +Therefore the greyhound must take $300$~leaps. +\end{Soln} + +\Item{1.} A hound makes $3$~leaps while a rabbit makes~$5$; but +$1$~of the hound's leaps is equivalent to $2$~of the rabbit's. +The rabbit has a start of $120$~leaps. How many leaps +will the rabbit take before she is caught? + +\Item{2.} A rabbit takes $6$~leaps to a dog's~$5$, and $7$~of the dog's +leaps are equivalent to $9$~of the rabbit's. The rabbit has +a start of~$60$ of her own leaps. How many leaps must the +dog take to catch the rabbit? + +\Item{3.} A dog makes $4$~leaps while a rabbit makes~$5$; but $3$~of +the dog's leaps are equivalent to $4$~of the rabbit's. The +rabbit has a start of $90$~of the \emph{dog's leaps}. How many +leaps will each take before the rabbit is caught? +%% -----File: 122.png---Folio 116------- + +\Exercise{62.} + +Ex. Find the time between $2$ and $3$~o'clock when the +hands of a clock are together. +\begin{Soln} +At 2 o'clock the hour-hand is 10 minute-spaces ahead of the +minute-hand. +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{\emph{the number of spaces the minute-hand moves over}.} \\ +\lintertext{\indent The\rlap{n}} +x - 10 &= \text{the number of spaces the hour-hand moves over.} +\displaybreak[1] \\ +\intertext{\indent Now, as the minute-hand moves $12$~times as fast as the hour-hand,} +\llap{$12(x - 10)$} &= \text{\emph{the number of spaces the minute-hand moves over}.} +\end{DPalign*} +\begin{DPalign*} +\therefore 12(x - 10) &= x, \\ +\lintertext{and} +11x &= 120. \\ +\therefore x &= 10\tfrac{10}{11}. +\end{DPalign*} + +Therefore the time is $10\frac{10}{11}$~minutes past $2$~o'clock. +\end{Soln} + +\Item{1.} Find the time between $5$ and $6$~o'clock when the +hands of a clock are together. + +\Item{2.} Find the time between $2$ and $3$~o'clock when the +hands of a clock are at right angles to each other. + +\begin{Remark}[Hint.] In this case the minute-hand is $15$~minutes ahead of the +hour-hand. +\end{Remark} + +\Item{3.} Find the time between $2$ and $3$~o'clock when the +hands of a clock point in opposite directions. + +\begin{Remark}[Hint.] In this case the minute-hand is $30$~minutes ahead of the +hour-hand. +\end{Remark} + +\Item{4.} Find the time between $1$ and $2$~o'clock when the +hands of a clock are at right angles to each other. + +\Item{5.} Find the time between $1$ and $2$~o'clock when the +hands of a clock point in opposite directions. + +\Item{6.} At what time between $7$ and $8$~o'clock are the hands +of a watch together? +%% -----File: 123.png---Folio 117------- + +\Exercise{63.} + +Ex. A rectangle has its length $6$~feet more and its width +$5$~feet less than the side of its equivalent square. Find the +dimensions of the rectangle. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{the number of feet in a side of the square.} \\ +\lintertext{\indent Then} +x + 6 &= \text{the number of feet in the length of the rectangle,} \\ +\lintertext{and} +x - 5 &= \text{the number of feet in the width of the rectangle.} +\intertext{Since the area of a rectangle is equal to the product of the number +of units of length in the length and width of the rectangle,} +\llap{$(x + 6)(x - 5)$} &= \text{the area of the rectangle in square feet,} \\ +\lintertext{and} +x × x &= \text{the area of the square in square feet.} +\end{DPalign*} + +But these areas are equal. +\begin{DPalign*} +\therefore (x + 6)(x - 5) &= x^{2}. \\ +\lintertext{\indent Solving,} +x &= 30. +\end{DPalign*} + +Therefore the dimensions of the rectangle are $36$~feet and $25$~feet. +\end{Soln} + +\Item{1.} A rectangle has its length and breadth respectively $7$~feet +longer and $6$~feet shorter than the side of the equivalent +square. Find its area. + +\Item{2.} The length of a floor exceeds the breadth by $5$~feet. +If each dimension were $1$~foot more, the area of the floor +would be $42$~sq.~ft.\ more. Find its dimensions. + +\Item{3.} A rectangle whose length is $6$~feet more than its +breadth, would have its area $35$~sq.~ft.\ more, if each dimension +were $1$~foot more. Find its dimensions. + +\Item{4.} The length of a rectangle exceeds its width by $3$~feet. +If the length is increased by $3$~feet and the width diminished +by $2$~feet, the area will not be altered. Find its +dimensions. + +\Item{5.} The length of a floor exceeds its width by $10$~feet +If each dimension were $2$~feet more, the area would be $144$~sq.~ft.\ +more. Find its dimensions. +%% -----File: 124.png---Folio 118------- + +\Paragraph{143. Formulas and Rules.} When the \emph{given} numbers of a +problem are represented by letters, the result obtained from +solving the problem is a general expression which includes +all problems of that class. Such an expression is called a +\Defn{formula}, and the translation of this formula into words is +called a \Defn{rule}. + +\Paragraph{144.} We will illustrate by examples. + +\Item{1.} The sum of two numbers is~$s$, and their difference~$d$. +Find the numbers. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{the smaller number;} \\ +\lintertext{then} +x + d &= \text{the larger number.} \displaybreak[1] \\ +\lintertext{\indent Hence} +x + x + d &= s, \\ +\lintertext{or} +2x &= s - d. \\ +\therefore x &= \frac{s - d}{2}, \displaybreak[1] \\ +\lintertext{and} +x + d &= \frac{s - d}{2} + d = \frac{s - d + 2d}{2}, \\ + &= \frac{s + d}{2}. +\end{DPalign*} + +Therefore the numbers are $\dfrac{s + d}{2}$ and $\dfrac{s - d}{2}$. +\end{Soln} + +As these formulas hold true whatever numbers $s$~and~$d$ +stand for, we have the general rule for finding two numbers +when their sum and difference are given: +\begin{Theorem} +Add the difference to the sum and take half the result for +the greater number. + +Subtract the difference from the sum and take half the +result for the smaller number. +\end{Theorem} + +\Item{2.} If A~can do a piece of work in $a$~days, and B~can +do the same work in $b$~days, in how many days can both +together do it? +%% -----File: 125.png---Folio 119------- +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{the required number of days\Add{.}} \\ +\lintertext{\indent Then,} +\frac{1}{x} &= \text{\emph{the part both together can do in one day}\Add{.}} +\displaybreak[1] \\ +\lintertext{\indent Now} +\frac{1}{a} &= \text{the part A can do in one day,} \\ +\lintertext{and} +\frac{1}{b} &= \text{the part B can do in one day\Add{,}} \\ +\lintertext{therefore} +\frac{1}{a} + \frac{1}{b} + &= \text{\emph{the part both together can do in one day}} +\end{DPalign*} +\begin{DPalign*} +\frac{1}{a} + \frac{1}{b} &= \frac{1}{x}\Add{.} \\ +\lintertext{\indent Whence} +x &= \frac{ab}{a + b}\Add{.} +\end{DPalign*} +\end{Soln} + +The translation of this formula gives the following rule +for finding the time required by two agents together to +produce a given result when the time required by each +agent separately is known. +\begin{Theorem} +Divide the product of the numbers which express the units +of time required by each to do the work by the sum of these +numbers, the quotient is the time required by both together. +\end{Theorem} + +\Paragraph{145. Interest Formulas.} The elements involved in computation +of interest are the \emph{principal}, \emph{rate}, \emph{time}, \emph{interest}, +and \emph{amount}. +\begin{DPalign*} +\lintertext{\indent Let} +p &= \text{the principal,} \\ +r &= \text{the interest of \$$1$ for $1$~year, at the given rate,} \\ +t &= \text{the time expressed in years,} \\ +i &= \text{the interest for the given time and rate,} \\ +a &= \text{the amount (sum of principal and interest).} +\end{DPalign*} + +\Paragraph{146. Given the Principal, Rate, and Time. Find the Interest.} + +Since $r$ is the interest of~\$$1$ for $1$~year, $pr$~is the interest +of~\$$p$ for $1$~year, and $prt$~is the interest of~\$$p$ for $t$~years +\begin{DPgather*} +i = prt. +\rintertext{(Formula 1.)} +\end{DPgather*} + +\begin{Theorem}[\textsc{Rule.}] Find the product of the principal, rate, and time\Add{.} +\end{Theorem} +%% -----File: 126.png---Folio 120------- + +\Paragraph{147. Given the Interest, Rate, and Time. Find the Principal.} +\begin{DPalign*} +\lintertext{\indent By formula 1,} +prt &= i. \\ +\lintertext{\indent Divide by~$rt$,} +p &= \frac{i}{rt}. +\rintertext{(Formula 2.)} +\end{DPalign*} + +\begin{Theorem}[\textsc{Rule.}] Divide the interest by the product of the rate and +time. +\end{Theorem} + +\Paragraph{148. Given the Amount, Rate, and Time. Find the Principal.} +\begin{DPalign*} +\lintertext{\indent From formula 1,} +p + prt &= a, \\ +\lintertext{or} +p(1 + rt) &= a. \\ +\lintertext{\indent Divide by $1 + rt$,} +p &= \frac{a}{1 + rt}. +\rintertext{(Formula 3.)} +\end{DPalign*} + +\begin{Theorem}[\textsc{Rule.}] Divide the amount by $1$~plus the product of the +rate and time. +\end{Theorem} + +\Paragraph{149. Given the Amount, Principal, and Rate. Find the Time.} +\begin{DPalign*} +\lintertext{\indent From formula 1,} +p + prt &= a. \\ +\lintertext{\indent Transpose~$p$,} +prt &= a - p. \\ +\lintertext{\indent Divide by~$pr$,} +t &= \frac{a - p}{pr}. +\rintertext{(Formula 4.)} +\end{DPalign*} + +\begin{Theorem}[\textsc{Rule.}] Subtract the principal from the amount, and +divide the result by the product of the principal and rate. +\end{Theorem} + +\Paragraph{150. Given the Amount, Principal, and Time. Find the Rate.} +\begin{DPalign*} +\lintertext{\indent From formula 1,} +p + prt &= a. \\ +\lintertext{\indent Transpose~$p$,} +prt &= a - p. \\ +\lintertext{\indent Divide by~$pt$,} +r &= \frac{a - p}{pt}. +\rintertext{(Formula 5.)} +\end{DPalign*} + +\begin{Theorem}[\textsc{Rule.}] Subtract the principal from the amount, and +divide the result by the product of the principal and time. +\end{Theorem} +%% -----File: 127.png---Folio 121------- + +\Exercise{64.} + +Solve the following examples by the preceding formulas: + +\Item{1.} The sum of two angles is $120°\, 30'\, 30''$ and their difference +$59°\, 30'\, 30''$. Find the angles. + +\Item{2.} Find the interest of \$$1000$ for $3$~years and $4$~months +at~$4$\%. + +\Item{3.} Find the principal that will amount to \$$2280$ in $3$~years +and $6$~months at~$4$\%. + +\Item{4.} Find the principal that will produce \$$280$ interest in +$2$~years and $4$~months at~$3$\%. + +\Item{5.} Find the principal that will produce \$$270$ interest +in $1$~year and $6$~months at~$6$\%. + +\Item{6.} Find the principal that will amount to \$$590$ in $4$~years +at~$4\frac{1}{2}$\%. + +\Item{7.} Find the rate if the amount of \$$250$ for $4$~years +is~\$$300$. + +\Item{8.} Find the rate if \$$1000$ amounts to \$$2000$ in $16$~years +and $8$~months. + +\Item{9.} Find the time required for the interest on \$$400$ to +be \$$54$ at~$4\frac{1}{2}$\%. + +\Item{10.} Find the time required for \$$160$ to amount to \$$250$ +at~$6$\%. + +\Item{11.} How much money must be invested at~$5$\% to yield +an annual income of~\$$1250$? + +\Item{12.} Find the principal that will produce \$$100$ a month +if invested at $6$\%~per annum. + +\Item{13.} Find the rate if the interest on \$$1000$ for $8$~months +is~\$$40$. + +\Item{14.} Find the time for a sum of money on interest at~$5$\% +to double itself. +%% -----File: 128.png---Folio 122------- + + +\Chapter{XI.}{Simultaneous Equations of the First +Degree.} + +\Paragraph{151.} If we have two unknown numbers and but one relation +between them, we can find an unlimited number of +pairs of values for which the given relation will hold true. +Thus, if $x$~and~$y$ are unknown, and we have given only the +one relation $x + y = 10$, we can \emph{assume} any value for~$x$, +and then from the relation $x + y = 10$ find the corresponding +value of~$y$. For from $x + y = 10$ we find $y = 10 - x$. +If $x$~stands for~$1$, $y$~stands for~$9$; if $x$~stands for~$2$, $y$~stands +for~$8$; if $x$~stands for~$-2$, $y$~stands for~$12$; and so on without +end. + +\Paragraph{152.} We may, however, have two equations that express +\emph{different} relations between the two unknown numbers. +Such equations are called \Defn{independent equations}. Thus, +$x + y = 10$ and $x - y = 2$ are independent equations, for +they evidently express \emph{different} relations between $x$~and~$y$. + +\Paragraph{153.} Independent equations involving the \emph{same} unknown +numbers are called \Defn{simultaneous equations}. + +If we have two unknown numbers, and two independent +equations involving them, there is but \emph{one} pair of values +which will hold true for both equations. Thus, if besides +the relation $x + y = 10$, we have also the relation $x - y = 2$, +the only pair of values for which both equations will hold +true is the pair $x = 6$, $y = 4$. + +Observe that in this problem $x$~stands for the same number +in \emph{both} equations; so also does~$y$. +%% -----File: 129.png---Folio 123------- + +\Paragraph{154.} Simultaneous equations are solved by combining +the equations so as to obtain a single equation with one +unknown number. + +This process is called \Defn{Elimination}. + +\Paragraph{155. Elimination by Addition or Subtraction.} + +%[** TN: Omitted large brace for grouping, horizontal bar indicating summation] +\Item{1.} Solve: +\begin{alignat*}{2} +5x &- 3y &&= 20 +\Tag{(1)} \\ +2x &+ 5y &&= 39 +\Tag{(2)} +\end{alignat*} +\begin{Soln} +Multiply (1) by~$5$, and (2) by~$3$, +\begin{DPalign*} +25x - 15y &= 100 +\Tag{(3)} \\ + 6x + 15y &= 117 +\Tag{(4)} \\ +\lintertext{\indent Add (3) and (4),} +31x \PadTo{{} + 15y}{} &= 217 \\ +\therefore x &= 7. +\intertext{\indent Substitute the value of~$x$ in~(2),} +14 + 5y &= 39. \\ +5y &= 25. \\ +\therefore y &= 5. +\end{DPalign*} + +In this solution $y$~is eliminated by \emph{addition}. +\end{Soln} + +\Item{2.} Solve: +\begin{alignat*}{2} +6x &+ 35y &&= 177 +\Tag{(1)} \\ +8x &- 21y &&= \Z33 +\Tag{(2)} +\end{alignat*} +\begin{Soln} +Multiply (1) by~$4$, and (2) by~$3$, +\begin{DPalign*} +24x + 140y &= 708 +\Tag{(3)} \\ +24x - \Z63y &= \Z99 +\Tag{(4)} \\ +\lintertext{\indent Subtract,} +203y &= 609 \\ +\therefore y &= 3. \\ +\intertext{\indent Substitute the value of~$y$ in (2),} +8x - 63 &= 33. \\ +8x &= 96. \\ +\therefore x &= 12. +\end{DPalign*} + +In this solution $x$~is eliminated by \emph{subtraction}. +\end{Soln} +%% -----File: 130.png---Folio 124------- + +\Paragraph{156.} \Dictum{To eliminate by addition or subtraction}, therefore, +\begin{Theorem} +Multiply the equations by such numbers as will make the +coefficients of one of the unknown numbers equal in the +resulting equations. + +Add the resulting equations if these equal coefficients have +unlike signs; subtract one from the other if these equal coefficients +have like signs. +\end{Theorem} + +\begin{Remark}[Note.] It is generally best to select the letter to be eliminated +which requires the smallest multipliers to make its coefficients equal; +and the smallest multiplier for each equation is found by dividing +the \LCM\ of the coefficients of this letter by the given coefficient +in that equation. Thus, in example~2, the \LCM\ of $6$~and~$8$ (the +coefficients of~$x$) is~$24$, and hence the smallest multipliers of the two +equations are $4$~and~$3$, respectively. +\end{Remark} + +Sometimes the solution is simplified by first adding the +given equations, or by subtracting one from the other. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Ex.} +x + 49y &= 51 +\Tag{(1)} \\ +49x + \PadTo{49y}{y} &= 99 +\Tag{(2)} \displaybreak[1] \\ +\lintertext{\indent Add (1) and (2),} +50x + 50y &= 150 +\Tag{(3)} \displaybreak[1] \\ +\lintertext{\indent Divide (3) by~$50$,} +x + y &= 3. +\Tag{(4)} \displaybreak[1] \\ +\lintertext{\indent \rlap{Subtract (4) from (1),}} +48y &= 48. \\ +\therefore y &= 1. \displaybreak[1] \\ +\lintertext{\indent \rlap{Subtract (4) from (2),}} +48x &= 96. \\ +\therefore x &= 2. +\end{DPalign*} +\end{Soln} + +\Exercise{65.} + +Solve by addition or subtraction: +\begin{multicols}{2} +\Item{1.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + 5x &+ 4y &&= 14 \\ +17x &- 3y &&= 31 +\end{alignedat}\right\}$} + +\Item{2.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + 3x &- 2y &&= \Z5 \\ + 2x &+ 5y &&= 16 +\end{alignedat}\right\}$} + +\Item{3.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + 2x &- 3y &&= \Z7 \\ + 5x &+ 2y &&= 27 +\end{alignedat}\right\}$} + +\Item{4.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + 7x &+ 6y &&= 20 \\ + 2x &+ 5y &&= \Z9 +\end{alignedat}\right\}$} + +\Item{5.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + x &+ 5y &&= 11 \\ + 3x &+ 2y &&= \Z7 +\end{alignedat}\right\}$} + +\Item{6.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + 3x &- 5y &&= 13 \\ + 4x &- 7y &&= 17 +\end{alignedat}\right\}$} +%% -----File: 131.png---Folio 125------- + +\Item{7.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + 8x &- \Z y &&= \Z3 \\ + 7x &+ 2y &&= 63 +\end{alignedat}\right\}$} + +\Item{8.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + 5x &- 4y &&= \Z7 \\ + 7x &+ 3y &&= 70 +\end{alignedat}\right\}$} + +\Item{9.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + x &+ 21y &&= \Z2 \\ + 2x &+ 27y &&= 19 +\end{alignedat}\right\}$} + +\Item{10.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + 6x &- 13y &&= -1 \\ + 5x &- 12y &&= -2 +\end{alignedat}\right\}$} + +\Item{11.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + 7x &+ \Z y &&= 265 \\ + 3x &- 5y &&= \Z\Z5 +\end{alignedat}\right\}$} + +\Item{12.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + 2x &+ 3y &&= \Z7 \\ + 8x &- 5y &&= 11 +\end{alignedat}\right\}$} + +\Item{13.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + 5x &+ 7y &&= 19 \\ + 7x &+ 4y &&= 15 +\end{alignedat}\right\}$} + +\Item{14.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} +11x &- 12y &&= \Z9 \\ + 4x &+ \Z5y &&= 22 +\end{alignedat}\right\}$} + +\Item{15.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + x &+ 8y &&= 17 \\ + 7x &- 3y &&= \Z1 +\end{alignedat}\right\}$} + +\Item{16.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + 4x &+ 3y &&= 25 \\ + 5x &- 4y &&= \Z8 +\end{alignedat}\right\}$} +\end{multicols} + +Clear of fractions and solve: +\begin{multicols}{2} +\Item{17.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + \frac{2x}{3} &- \frac{5y}{4} &&= 3 \\ + \frac{7x}{4} &- \frac{5y}{3} &&= \frac{43}{3} +\end{alignedat}\right\}$} + +\Item{18.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + \frac{7x}{6} &+ \frac{6y}{7} &&= 32 \\ + \frac{5x}{4} &- \frac{2y}{3} &&= 1 +\end{alignedat}\right\}$} + +\Item{19.} \raisebox{-0.5\baselineskip}{$\left.\begin{aligned} + \frac{x + y}{4} - \frac{7x - 5y}{11} &= 3 \\ + \frac{x}{5} - \frac{2y}{7} + 1 &= 0 +\end{aligned}\right\}$} + +\Item{20.} \raisebox{-0.5\baselineskip}{$\left.\begin{aligned} + \frac{ 6x + 7y}{2} &= 22 \\ + \frac{55y - 2x}{5} &= 20 +\end{aligned}\right\}$} +\end{multicols} + +\Item{21.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + \frac{x + y}{2} &- \frac{x - y}{3} &&= \Z8 \\ + \frac{x + y}{3} &+ \frac{x - y}{4} &&= 11 +\end{alignedat}\right\}$} + +\Item{22.} \raisebox{-0.5\baselineskip}{$\left.\begin{aligned} + \frac{8x - 5y}{7} + \frac{11y - 4x}{5} &= 4 \\ + \frac{17x - 13y}{5} + \frac{2x}{3} &= 7 +\end{aligned}\right\}$} +%% -----File: 132.png---Folio 126------- + +\Item{23.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + \frac{ 5x - 3y}{3} &+ \frac{7x - 5y}{11} &&= 4 \\ + \frac{15y - 3x}{7} &+ \frac{7y - 3x}{5} &&= 4 + \end{alignedat}\right\}$} + +\Item{24.} \raisebox{-0.5\baselineskip}{$\left.\begin{aligned} + \frac{2x - 3}{4} - \frac{y - 8}{5} &= \frac{y + 3}{4} \\ + \frac{x - 7}{3} + \frac{4y + 1}{11} &= 3 + \end{aligned}\right\}$} + +\Item{25.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + \frac{x - 2y}{6} &- \frac{x + 3y}{4} &&= \frac{3}{2} \\ + \frac{2x - y}{6} &- \frac{3x + y}{4} &&= \frac{5y}{4} + \end{alignedat}\right\}$} + +\Item{26.} \raisebox{-0.5\baselineskip}{$\left.\begin{alignedat}{2} + \frac{x}{a + b} &+ \frac{y}{a - b} &&= \frac{1}{a - b} \\ + \frac{x}{a + b} &- \frac{y}{a - b} &&= \frac{1}{a + b} + \end{alignedat}\right\}$} + +\begin{Remark}[Note.] To find $x$ in problem~26, add the equations; to find~$y$, +subtract one from the other. Do not clear of fractions. +\end{Remark} + +\Paragraph{157. Problems involving Two Unknown Numbers.} + +Ex. If A gives B \$$10$, B~will have three times as much +money as~A\@. If B gives A \$$10$, A~will have twice as +much money as~B\@. How much has each? +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{the number of dollars A has,} \\ +\lintertext{and} +y &= \text{the number of dollars B has.} \displaybreak[1] \\ +\intertext{\indent Then, after A gives B \$$10,$} +x - 10 &= \text{the number of dollars A has,} \\ +y + 10 &= \text{the number of dollars B has.} \displaybreak[1] \\ +\intertext{\indent Since B's money is now $3$~times A's, we have,} +y + 10 &= 3(x - 10). +\Tag{(1)} \displaybreak[1] \\ +%% -----File: 133.png---Folio 127------- +\intertext{\indent If B gives A \$$10$,} +x + 10 &= \text{the number of dollars A has,} \\ +y - 10 &= \text{the number of dollars B has.} \displaybreak[1] \\ +\intertext{\indent Since A's money is now $2$~times B's, we have} +x + 10 &= 2(y - 10). +\Tag{(2)} +\end{DPalign*} + +From the solution of equations (1) and (2), $x = 22$, and $y = 26$. + +Therefore A has \$$22$, and B has~\$$26$. +\end{Soln} + +\Exercise{66.} + +\Item{1.} If A gives B \$$200$, A~will then have half as much +money as~B; but if B gives A \$$200$, B~will have one-third +as much as A\@. How much has each? + +\Item{2.} Half the sum of two numbers is~$20$, and three times +their difference is~$18$. Find the numbers. + +\Item{3.} The sum of two numbers is~$36$, and their difference +is equal to one-eighth of the smaller number increased by~$2$. +Find the numbers. + +\Item{4.} If $4$~yards of velvet and $3$~yards of silk are sold for~\$$33$, +and $5$~yards of velvet and $6$~yards of silk for~\$$48$, +what is the price per yard of the velvet and of the silk? + +\Item{5.} If $7$~bushels of wheat and $10$~of rye are sold for~\$$15$, +and $4$~bushels of wheat and $5$~of rye are sold for~\$$8$, what +is the price per bushel of the wheat and of the rye? + +\Item{6.} If $12$~pounds of tea and $4$~pounds of coffee cost~\$$7$, +and $4$~pounds of tea and $12$~pounds of coffee cost~\$$5$, what is +the price per pound of tea and of coffee? + +\Item{7.} Six horses and $7$~cows can be bought for~\$$1000$, +and $11$~horses and $13$~cows for~\$$1844$. Find the value of +a horse and of a cow. +%% -----File: 134.png---Folio 128------- + +\Exercise{67.} + +Ex. A certain fraction becomes equal to~$\frac{1}{2}$ if $2$~is added +to its numerator, and equal to~$\frac{1}{3}$ if $3$~is added to its denominator. +Find the fraction. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +\frac{x}{y} &= \text{the required fraction.} \\ +\lintertext{\indent Then} +\frac{x + 2}{y} &= \frac{1}{2}, \\ +\lintertext{and} +\frac{x}{y + 3} &= \frac{1}{3}. +\end{DPalign*} + +The solution of these equations gives $7$ for~$x$, and $18$ for~$y$. + +Therefore the required fraction is~$\frac{7}{18}$. +\end{Soln} + +\Item{1.} If the numerator of a certain fraction is increased +by~$2$ and its denominator diminished by~$2$, its value will +be~$1$. If the numerator is increased by the denominator +and the denominator is diminished by~$5$, its value will be~$5$. +Find the fraction. + +\Item{2.} If $1$~is added to the denominator of a fraction, its +value will be~$\frac{1}{2}$. If $2$~is added to its numerator, its value +will be~$\frac{3}{5}$. Find the fraction. + +\Item{3.} If $1$~is added to the numerator of a fraction, its value +will be~$\frac{1}{5}$. If $1$~is added to its denominator, its value will +be~$\frac{1}{7}$. Find the fraction. + +\Item{4.} If the numerator of a fraction is doubled and its +denominator diminished by~$1$, its value will be~$\frac{1}{2}$. If its +denominator is doubled and its numerator increased by~$1$, +its value will be~$\frac{1}{7}$. Find the fraction. + +\Item{5.} In a certain proper fraction the difference between +the numerator and the denominator is~$15$. If the numerator +is multiplied by~$4$ and the denominator increased by~$6$, +its value will be~$1$. Find the fraction. +%% -----File: 135.png---Folio 129------- + +\Exercise{68.} + +The expression $64$ means $60 + 4$, that is, $10$~\emph{times} $6 + 4$, +and has for its \emph{digits} $6$~and~$4$. If the digits were unknown +and represented by $x$~and~$y$, the number would be represented +by~$10x + y$. + +Ex. The sum of the two digits of a number is~$10$, and if +$18$~is added to the number, the digits will be reversed. +Find the number. +\begin{Soln} +\begin{DPalign*}[m] +\lintertext{\indent Let} +x &= \text{the tens' digit,} \\ +\lintertext{and} +y &= \text{the units' digit.} \\ +\lintertext{\indent Then} +10x + y &= \text{the number.} \displaybreak[1] \\ +\lintertext{\indent Hence} +x + y &= 10, +\Tag{(1)} \\ +\lintertext{and} +10x + y + 18 &= 10y + x. +\Tag{(2)} \displaybreak[1] \\ +\lintertext{\indent From (2),} +9x - 9y &= -18, \\ +\lintertext{or} +x - y &= -2. +\Tag{(3)} \displaybreak[1] \\ +\lintertext{\indent \rlap{Add (1) and (3),}} +2x &= 8, \\ +\lintertext{and therefore} +x &= 4. \\ +\lintertext{\indent \rlap{Subtract (3) from (1),}} +2y &= 12, \\ +\lintertext{and therefore} +y &= 6. +\end{DPalign*} + +Therefore the number is~$46$. +\end{Soln} + +\Item{1.} The sum of the two digits of a number is~$9$, and if $9$ +is added to the number, the digits will be reversed. Find +the number. + +\Item{2.} A certain number of two digits is equal to eight times +the sum of its digits, and if $45$~is subtracted from the +number, the digits will be reversed. Find the number. + +\Item{3.} The sum of a certain number of two digits and the +number formed by reversing the digits is~$132$, and the +difference of these numbers is~$18$. Find the numbers. + +\Item{4.} The sum of the two digits of a number is~$9$, and if +the number is divided by the sum of its digits, the quotient +is~$6$. Find the number. +%% -----File: 136.png---Folio 130------- + +\Exercise{69.} + +Ex. A sum of money, at simple interest, amounted to +\$$2480$ in $4$~years, and to \$$2600$ in $5$~years. Find the sum +and the rate of interest. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{the number of dollars in the principal,} \\ +\lintertext{and} +y &= \text{the rate of interest.} +\end{DPalign*} + +The interest for one year is $\dfrac{y}{100}$ of the principal; that is, $\dfrac{xy}{100}$. +For $4$~years the interest is~$\dfrac{4xy}{100}$, and for $5$~years~$\dfrac{5xy}{100}$. The amount +is principal $+$~interest, +\begin{DPalign*} +\lintertext{or} +x + \frac{4xy}{100} &= 2480. \\ +x + \frac{5xy}{100} &= 2600. \\ +\lintertext{\indent Hence} +100x + 4xy &= 248,000. +\Tag{(1)} \\ +100x + 5xy &= 260,000. +\Tag{(2)} \displaybreak[1] \\ +\intertext{\indent Divide (1) by~$4$ and (2) by~$5$, and we have} +25x + xy &= 62,000 \\ +20x + xy &= 52,000 \displaybreak[1] \\ +\lintertext{\indent Subtract,} +5x \PadTo{{}+ xy}{} &= 10,000. \displaybreak[1] \\ +\lintertext{\indent Therefore} +\PadTo{5x}{x} \PadTo{{}+ xy}{} &= \PadTo{10,000}{2000}. +\end{DPalign*} + +Substitute the value of~$x$ in~(1), $y = 6$. + +Therefore the sum is \$$2000$, and the rate~$6$%. +\end{Soln} + +\Item{1.} A sum of money, at simple interest, amounted in $5$~years +to~\$$3000$, and in $6$~years to~\$$3100$. Find the sum +and the rate of interest. + +\Item{2.} A sum of money, at simple interest, amounted in $10$~months +to~\$$1680$, and in $18$~months to~\$$1744$. Find the +sum and the rate of interest. + +\Item{3.} A man has \$$10,000$ invested, a part at~$4$\%, and the +remainder at~$5$\%. The annual income from his $4$\%~investment +is \$$40$~more than from his $5$\%~investment. Find the +sum invested at~$4$\% and at~$5$\%. +%% -----File: 137.png---Folio 131------- + +\Exercise{70.} + +\Section{Miscellaneous Examples.} + +\Item{1.} Half the sum of two numbers is~$20$; and $5$~times +their difference is~$20$. Find the numbers. + +\Item{2.} A certain number when divided by a second number +gives $7$ for a quotient and $4$ for a remainder. If three +times the first number is divided by twice the second +number, the quotient is~$11$ and the remainder~$4$. Find +the numbers. + +\Item{3.} A fraction becomes $\frac{4}{5}$ in value by the addition of~$2$ to +its numerator and $3$~to its denominator. If $2$~is subtracted +from its numerator and $1$~from its denominator, the value +of the fraction is~$\frac{3}{4}$. Find the fraction. + +\Item{4.} A farmer sold $50$~bushels of wheat and $30$~of barley +for $74$~dollars; and at the same prices he sold $30$~bushels +of wheat and $50$~bushels of barley for $70$~dollars. What +was the price of the wheat and of the barley per bushel? + +\Item{5.} If A gave \$$10$ to~B, he would then have three times +as much money as~B; but if B gave \$$5$ to~A, A~would +have four times as much as~B\@. How much has each? + +\Item{6.} A and B have together \$$100$. If A~were to spend +one-half of his money, and B~one-third of his, they would +then have only \$$55$ between them. How much money +has each? + +\Item{7.} A fruit-dealer sold $6$~lemons and $3$~oranges for $21$~cents, +and $3$~lemons and $8$~oranges for $30$~cents. What +was the price of each? + +\Item{8.} If A gives me $10$~apples, he will have just twice as +many as~B\@. If he gives the $10$~apples to~$B$ instead of to +me, A~and~B will each have the same number. How +many apples has each? +%% -----File: 138.png---Folio 132------- + + +\Chapter{XII.}{Quadratic Equations.} + +\Paragraph{158.} An equation which contains the \emph{square} of the +unknown number, but no higher power, is called a \Defn{quadratic +equation}. + +\Paragraph{159.} A quadratic equation which involves but one unknown +number as~$x$, can contain only: + +1. Terms involving the square of~$x$. + +2. Terms involving the first power of~$x$. + +3. Terms which do not involve~$x$. + +Collecting similar terms, every quadratic equation can be +made to assume the form +\[ +ax^{2} + bx + c = 0, +\] +where $a$,~$b$, and~$c$ are known numbers, and $x$~the unknown +number. + +If $a$,~$b$,~$c$ are numbers expressed by figures, the equation +is a \Defn{numerical quadratic}. If $a$,~$b$,~$c$ are numbers represented +wholly or in part by letters, the equation is a \Defn{literal quadratic}. + +\Paragraph{160.} In the equation $ax^{2} + bx + c = 0$, $a$,~$b$, and~$c$ are +called the \Defn{coefficients} of the equation. The third term,~$c$, is +called the \Defn{constant term}. + +If the first power of~$x$ is wanting, the equation is a \Defn{pure +quadratic}; in this case $b = 0$. + +If the first power of~$x$ is present, the equation is an +\Defn{affected} or \Defn{complete quadratic}. +%% -----File: 139.png---Folio 133------- + + +\Section{Pure Quadratic Equations.} + +\Paragraph{161. Examples.} + +\Item{1.} Solve the equation $5x^{2} - 48 = 2x^{2}$. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent We have} +5x^{2} - 48 &= 2x^{2}. \\ +\lintertext{\indent Collect the terms,} +3x^{2} &= 48. \displaybreak[1] \\ +\lintertext{\indent Divide by~$3$,} +x^{2} &= 16. \\ +\lintertext{\indent Extract the square root,} +x &= ±4. +\end{DPalign*} + +The sign~$±$ before the~$4$, read \emph{plus or minus}, shows that the root +is either $+$~or~$-$. For $(+4) × (+4) = 16$, and $(-4) × (-4) = 16$ + +The square root of any number is positive or negative. Hitherto +we have given only the positive value. In this chapter we shall +give both values. This sign~$\surd$, called the \Defn{radical sign}, is used to +indicate that a root is to be extracted. Thus $\sqrt{4}$~means the square +root of~$4$ is required. $\sqrt[3]{4}$~means the third root of~$4$ is required; the +small figure placed over the radical sign is called the \Defn{index} of the +root, and shows the root required. +\end{Soln} + +\Item{2.} Solve the equation $3x^{2} - 15 = 0$. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent We have} +3x^{2} &= 15, \\ +\lintertext{or} +x^{2} &= 5. \\ +\lintertext{\indent Extract the square root,} +x &= ±\sqrt{5}. +\end{DPalign*} + +The roots cannot be found exactly, since the square root of~$5$ cannot +be found exactly; it can, however, be determined approximately +to any required degree of accuracy; for example, the roots lie between +$2.23606$ and $2.23607$; and between $-2.23606$ and~$-2.23607$. +\end{Soln} + +\Item{3.} Solve the equation $3x^{2} + 15 = 0$. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent We have} +3x^{2} &= -15, \\ +\lintertext{or} +x^{2} &= -5. \\ +\lintertext{\indent Extract the square root,} +x &= ±\sqrt{-5}. +\end{DPalign*} + +There is no square root of a negative number, since the square of +any number, positive or negative, is positive; $(-5) × (-5) = +25$. + +The square root of~$-5$ differs from the square root of~$+5$ in that +the latter can be found as accurately as we please, while the former +cannot be found at all. +\end{Soln} +%% -----File: 140.png---Folio 134------- + +\Paragraph{162.} A root which can be found exactly is called an \Defn{exact} +or \Defn{rational} root. Such roots are either whole numbers or +fractions. + +A root which is indicated but can be found only approximately +is called a \Defn{surd}. Such roots involve the roots of +imperfect powers. + +Rational and surd roots are together called \Defn{real} roots. + +A root which is indicated but cannot be found, either +exactly or approximately, is called an \Defn{imaginary} root. Such +roots involve the even roots of negative numbers. + +\Exercise{71.} + +Solve: +\begin{multicols}{2} +\Item{1.} $5x^{2} - 2 = 3x^{2} + 6$. + +\Item{2.} $3x^{2} + 1 = 2x^{2} + 10$. + +\Item{3.} $4x^{2} - 50 = x^{2} + 25$. + +\Item{4.} $(x - 6)(x + 6) = 28$. +\end{multicols} + +\Item{5.} $(x - 5)(x + 5) = 24$. + +\Item{6.} $3(x^{2} - 11) + 2(x^{2} - 5) = 82$. + +\Item{7.} $11(x^{2} + 5) + 6(3 - x^{2}) = 198$. + +\Item{8.} $5x^{2} + 3 - 2(17 - x^{2}) = 32$. + +\Item{9.} $4(x + 1) - 4(x - 1) = x^{2} - 1$. + +\Item{10.} $86 - 52x = 2(8 - x)(2 - 3x)$. + +\Item{11.} Find two numbers that are to each other as $3$~to~$4$, +and the difference of whose squares is~$112$. + +\begin{Remark}[Hint.] Let $3x$~stand for the smaller and $4x$~for the larger number. +\end{Remark} + +\Item{12.} A boy bought a number of oranges for $36$~cents. +The price of an orange was to the number bought as $1$~to~$4$. +How many oranges did he buy, and how many cents +did each orange cost? + +\Item{13.} A certain street contains $144$~square rods, and the +length is $16$~times the width. Find the width. +%% -----File: 141.png---Folio 135------- + +\Item{14.} Find the number of rods in the length, and in the +width of a rectangular field containing $3\frac{3}{5}$~acres, if the +length is $4$~times the width. + + +\Section{Affected Quadratic Equations.} + +\Paragraph{163.} Since +\[ +(x + b)^{2} = x^{2} + 2bx + b^{2}, \quad\text{and}\quad +(x - b)^{2} = x^{2} - 2bx + b^{2}, +\] +it is evident that the expression $x^{2} + 2bx$ or $x^{2} - 2bx$ lacks +only the \emph{third term},~$b^{2}$, of being a perfect square. + +This third term is the square of half the coefficient of~$x$. + +Every affected quadratic may be made to assume the +form $x^{2} + 2bx = c$ or $x^{2} - 2bx = c$, by dividing the equation +through by the coefficient of~$x^{2}$. + +To \textbf{solve} such an equation: + +The first step is to add to both members \emph{the square of +half the coefficient of~$x$}. This is called \emph{completing the square}. + +The second step is to \emph{extract the square root} of each member +of the resulting equation. + +The third step is to \emph{reduce} the two resulting simple +equations. + +\Item{1.} Solve the equation $x^{2} - 8x = 20$. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent We have} +x^{2} - 8x &= 20. \\ +\lintertext{\indent Complete the square,} +x^{2} - 8x + 16 &= 36. \\ +\lintertext{\indent Extract the square root,} +x - 4 &= ±6. \displaybreak[1] \\ +\lintertext{\indent Reduce, using the upper \rlap{sign,}} +x &= 4 + 6 = 10, \\ +\lintertext{or using the lower sign,} +x &= 4 - 6 = -2. +\end{DPalign*} + +The roots are $10$ and~$-2$. + +Verify by putting these numbers for~$x$ in the given equation. +\[ +\begin{array}{rcl<{\qquad}|>{\qquad}rcl} + x &=& 10, & x &=& -2, \\ +10^{2} - 8(10) &=& 20, & (-2)^{2} - 8(-2) &=& 20, \\ + 100 - 80 &=& 20. & 4 + 16 &=& 20. \\ +\end{array} +\] +\end{Soln} +%% -----File: 142.png---Folio 136------- + +\Item{2.} Solve the equation $\dfrac{x + 1}{x - 1} = \dfrac{4x - 3}{x + 9}$. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Free from fractions,} +(x + 1)(x + 9) &= (x - 1)(4x - 3). \\ +\lintertext{\indent Therefore,} +-3x^{2} + 17x &= -6. +\end{DPalign*} + +Since the square root of a negative number cannot be taken, the +coefficient of~$x^{2}$ must be changed to~$+$. +\begin{DPalign*} +\lintertext{\indent Divide by~$-3$,} +x^{2} - \tfrac{17}{3}x &= 2. \displaybreak[1] \\ +\intertext{\indent +Half the coefficient of~$x$ is $\frac{1}{2}$~of $-\frac{17}{3} = -\frac{17}{6}$, and the square of~$-\frac{17}{6}$ +is~$\frac{289}{36}$. Add the square of~$-\frac{17}{6}$ to both sides, and we have} +x^{2} - \frac{17x}{3} + \left(\frac{17}{6}\right)^{2} &= 2 + \frac{289}{36}. \displaybreak[1] \\ +\lintertext{\indent Now} +2 + \frac{289}{36} = \frac{72}{36} + \frac{289}{36} &= \frac{361}{36}, \\ +\lintertext{therefore,} +x^{2} - \tfrac{17}{3}x + \left(\frac{17}{6}\right)^{2} &= \frac{361}{36}. \displaybreak[1] \\ +\lintertext{\indent \rlap{Extract the root,}} +x - \frac{17}{6} &= ±\frac{19}{6}. \displaybreak[1] \\ +\lintertext{\indent Reduce,} +x - \frac{17}{6} &= ±\frac{19}{6}. \\ +\therefore x &= \frac{17}{6} + \frac{19}{6} = \frac{36}{6} = 6, \\ +\lintertext{or} +x &= \frac{17}{6} - \frac{19}{6} = -\frac{2}{6} = -\frac{1}{3}. +\end{DPalign*} + +The roots are $6$~and~$-\dfrac{1}{3}$. + +Verify by putting these numbers for~$x$ in the original equation. +\[ +\begin{array}[t]{rcl<{\qquad}|} +x &=& 6. \\ +\dfrac{6 + 1}{6 - 1} &=& \dfrac{24 - 3}{6 + 9}. \\ +\dfrac{7}{5} &=& \dfrac{21}{15} \\ +\dfrac{7}{5} &=& \dfrac{7}{5} \\ +\end{array} +\begin{array}[t]{>{\qquad}rcl} +x &=& -\dfrac{1}{3} \\ +\dfrac{-\dfrac{1}{3} + 1}{-\dfrac{1}{3} - 1} + &=& \dfrac{-\dfrac{4}{3} - 3}{-\dfrac{1}{3} + 9}. \\ +-\dfrac{2}{4} &=& -\dfrac{13}{26}. +\end{array} +\] +\end{Soln} +%% -----File: 143.png---Folio 137------- + +\PrintBreak +\Exercise{72.} + +Solve: +\begin{multicols}{2} +\Item{1.} $x^{2} - 12x + 27 = 0$. + +\Item{2.} $x^{2} - 6x + 8 = 0$. + +\Item{3.} $x^{2} - 4 = 4x - 7$. + +\Item{4.} $5x^{2} - 4x-1 = 0$. + +\Item{5.} $4x - 3 = 2x - x^{2}$. + +\Item{6.} $9x^{2} - 24x + 16 = 0$. + +\Item{7.} $6x^{2} - 5x-1 = 0$. + +\Item{8.} $4x + 3 = x^{2} + 2x$. + +\Item{9.} $16x^{2} - 16x + 3 = 0$. + +\Item{10.} $3x^{2} - 10x + 3 = 0$. + +\Item{11.} $x^{2} - 14x - 51 = 0$. + +\Item{12.} $34x - x^{2} - 225 = 0$. + +\Item{13.} $x^{2} + x - 20 = 0$. + +\Item{14.} $x^{2} - x - 12 = 0$. + +\Item{15.} $2x^{2} - 12x = - 10$. + +\Item{16.} $3x^{2} + 12x - 36 = 0$. + +\Item{17.} $(2x - 1)^{2} + 9 = 6(2x - 1)$. + +\Item{18.} $6(9x^{2} - x) = 55(x^{2} - 1)$. + +\Item{19.} $32 - 3x^{2} - 10x = 0$. + +\Item{20.} $9x^{2} - 6x - 143 = 0$. + +\Item{21.} $\dfrac{x}{x - 1} - \dfrac{x - 1}{x} = \dfrac{3}{2}$. + +\Item{22.} $\dfrac{1}{x - 2} + \dfrac{2}{x + 2} = \dfrac{5}{6}$. + +\Item{23.} $\dfrac{5x + 7}{x - 1} = 3x + 11$. + +\Item{24.} $\dfrac{7}{x + 4} - \dfrac{1}{4 - x} = \dfrac{2}{3}$. + +\Item{25.} $\dfrac{2}{x + 3} + \dfrac{x + 3}{2} = \dfrac{10}{3}$. + +\Item{26.} $\dfrac{2x}{x + 2} + \dfrac{x + 2}{2x} = 2$. + +\Item{27.} $\dfrac{3(x - 1)}{x + 1} - \dfrac{2(x + 1)}{x - 1} = 5$. + +\Item{28.} $\dfrac{2x + 5}{2x - 5} = \dfrac{7x - 5}{2x}$. + +\Item{29.} $\dfrac{3x - 1}{4x + 7} = \dfrac{x + 1}{x + 7}$. + +\Item{30.} $\dfrac{2x - 1}{x + 3} = \dfrac{x + 3}{2x - 1}$. + +\Item{31.} $\dfrac{x + 4}{x - 4} - \dfrac{x + 2}{x - 3} = 1$. + +\Item{32.} $\dfrac{4}{x - 1} - \dfrac{5}{x + 2} = \dfrac{1}{2}$. + +\Item{33.} $\dfrac{2}{x - 1} = \dfrac{3}{x - 2} + \dfrac{2}{x - 4}$. + +\Item{34.} $\dfrac{5}{x - 2} - \dfrac{3}{x - 1} = \dfrac{1}{2}$. + +\Item{35.} $\dfrac{x}{7 - x} + \dfrac{7 - x}{x} = \dfrac{29}{10}$. + +\Item{36.} $\dfrac{2x - 1}{x - 1} + \dfrac{1}{6} = \dfrac{2x - 3}{x - 2}$. +\end{multicols} +%% -----File: 144.png---Folio 138------- + +\ScreenBreak +\Paragraph{164. Problems involving Quadratics.} Problems which involve +quadratic equations apparently have two solutions, +since a quadratic equation has two roots. + +When both roots of the quadratic equation are positive +integers, they will, in general, both be admissible solutions. +Fractional and negative roots will in some problems give +admissible solutions; in other problems they will not give +admissible solutions. + +The reason that every root of the equation will not +always satisfy the conditions of the problem is that the +problem may have certain restrictions, expressed or implied, +that cannot be expressed in the equation. + +No difficulty will be found in selecting the result which +belongs to the particular problem we are solving. Sometimes, +by a change in the statement of the problem, we +may form a new problem which corresponds to the result +that was inapplicable to the original problem. + +Here as in simple equations x stands for an unknown +\emph{number}. + +\Item{1.} The sum of the squares of two consecutive numbers is~$41$. +Find the numbers. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{one number,} \\ +\lintertext{and} +x + 1 &= \text{the other.} \\ +\lintertext{\indent Then} +x^{2} + (x + 1)^{2} &= \text{the sum of the squares;} \\ +\lintertext{but} +41 &= \text{the sum of the squares.} +\end{DPalign*} +\begin{align*} +\therefore x^{2} + (x + 1)^{2} &= 41. \\ +x^{2} + x^{2} + 2x + 1 &= 41. \\ +2x^{2} + 2x &= 40. \\ +x^{2} + x &= 20. +\end{align*} + +The solution of this equation gives $x = 4$, or~$-5$. + +The positive root~$4$ gives for the numbers $4$~and~$5$. +\end{Soln} + +The negative root~$-5$ is inapplicable to the problem, as +\emph{consecutive numbers} are understood to be integers which +follow each other in the common scale: $1$,~$2$, $3$, $4\dots$. +%% -----File: 145.png---Folio 139------- + +\Item{2.} In a certain nest seven times the number of birds in +the nest is equal to twice the square of the number increased +by~$3$. Find the number. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{the number of birds\Add{.}} \\ +\lintertext{\indent Then} +7x &= \text{$7$~times the number} \\ +\lintertext{and} +2x^{2} + 3 &= \text{twice the square of the number plus~$3$.} +\intertext{\indent As these two expressions are equal we have} +2x^{2} + 3 &= 7x\Add{.} +\end{DPalign*} + +The solution of this \emph{equation} gives $x = 3$ or $x = \frac{1}{2}$. + +The value $\frac{1}{2}$ is not applicable to the \emph{problem} for the number of +birds must be a whole number. +\end{Soln} + +\Item{3.} A cistern has two pipes. By one of them it can be +filled $6$~hours sooner than by the other, and by both +together in $4$~hours. Find the time it will take each pipe +alone to fill it. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{the number of hours it takes the smaller pipe.} \\ +\lintertext{\indent Then} +x - 6 &= \text{the number of hours it takes the larger pipe.} +\end{DPalign*} +\begin{DPalign*} +\lintertext{\indent Therefore} +\frac{1}{x} + \frac{1}{x - 6} &= \text{the part both can fill in one hour.} \\ +\lintertext{\indent But} +\frac{1}{4} &= \text{the part both can fill in one hour.} \\ +\frac{1}{x} + \frac{1}{x - 6} &= \frac{1}{4} +\end{DPalign*} + +The solution of this \emph{equation} gives $x = 12$ or $x = 2$. + +The value~$2$ is not applicable to the \emph{problem}. + +Therefore it takes one pipe $12$~hr.\ and the other $6$~hr. +\end{Soln} + +\Item{4.} A rug is $1$~yard longer than it is broad. The number +of sq.~yds.\ in the rug is~$12$. Find its length and breadth. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{the number of yards in the breadth\Add{.}} \\ +\lintertext{\indent Then} +x + 1 &= \text{the number of yards in the length} \\ +\lintertext{and} +x(x + 1) &= \text{the number of sq yds in the rug\Add{.}} \\ +\lintertext{\indent Hence} +x(x + 1) &= 12\Add{.} +\end{DPalign*} + +The solution of this equation gives $x = 3$ or $x = -4$. + +The dimensions are therefore $3$~yards and $4$~yards. +\end{Soln} +%% -----File: 146.png---Folio 140------- + +\Exercise{73.} + +\Item{1.} Find two numbers whose sum is~$11$, and whose +product is~$30$. + +\Item{2.} Find two numbers whose difference is~$10$, and the +sum of whose squares is~$250$. + +\Item{3.} A man is five times as old as his son, and the square +of the son's age diminished by the father's age is~$24$. Find +their ages. + +\Item{4.} A number increased by its square is equal to nine +times the next higher number. Find the number. + +\Item{5.} The square of the sum of any two consecutive numbers +lacks~$1$ of being twice the sum of the squares of the +numbers. Show that this statement is true. + +\Item{6.} The length of a rectangular court exceeds its breadth +by $2$~rods. If the length and breadth were each increased +by $3$~rods, the area of the court would be $80$~square rods. +Find the dimensions of the court. + +\Item{7.} The area of a certain square will be doubled, if its +dimensions are increased by $6$~feet and $4$~feet respectively. +Find its dimensions. + +\Item{8.} The perimeter of a rectangular floor is $76$~feet and +the area of the floor is $360$~square feet. Find the dimensions +of the floor. + +\Item{9.} The length of a rectangular court exceeds its breadth +by $2$~rods, and its area is $120$~square rods. Find the +dimensions of the court. + +\Item{10.} The combined ages of a father and son amount to +$64$~years. Twice the father's age exceeds the square of the +son's age by $8$~years. Find their respective ages. +%% -----File: 147.png---Folio 141------- + +\Exercise{74.} + +Ex. A boat sails $30$~miles at a uniform rate. If the +rate had been $1$~mile an hour more, the time of the sailing +would have been $1$~hour less. Find the rate of the sailing. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Let} +x &= \text{the rate in miles per hour.} \\ +\lintertext{\indent Then} +\frac{30}{x} &= \text{the number of hours.} +\end{DPalign*} + +On the other supposition the rate would have been $x + 1$~miles +an hour and the time~$\dfrac{30}{x + 1}$. +\begin{DPalign*} +\lintertext{\indent Hence} +\frac{30}{x} - \frac{30}{x + 1} &= \text{the difference in hours for the sailing.} \\ +\lintertext{\indent But} +1 &= \text{the difference in hours for the sailing.} \\ +\frac{30}{x} - \frac{30}{x + 1} &= 1\Add{.} +\end{DPalign*} + +The solution of this equation gives $x = 5$, or $x = -6$. + +Therefore, the rate of the sailing is $5$~miles an hour. +\end{Soln} + +\Item{1.} A boat sails $30$~miles at a uniform rate. If the rate +had been $1$~mile an hour less, the time of the sailing would +have been $1$~hour more. Find the rate of the sailing. + +\Item{2.} A laborer built $35$~rods of stone wall. If he had +built $2$~rods less each day, it would have taken him $2$~days +longer. How many rods did he build a day on the +average? + +\Item{3.} A man bought flour for~\$$30$. Had he bought $1$~barrel +more for the same sum, the flour would have cost +him \$$1$~less per barrel. How many barrels did he buy? + +\Item{4.} A man bought some knives for~\$$6$. Had he bought +$2$~less for the same money, he would have paid $25$~cents +more for each knife. How many knives did he buy? + +\Item{5.} What number exceeds its square root by~$30$? + +\begin{Remark}[Hint.] Let $x^{2}$ denote the number. +\end{Remark} +%% -----File: 148.png---Folio 142------- + + +\Chapter{XIII.}{Arithmetical Progression.} + +\Paragraph{165.} A series of numbers is said to form an \Defn{Arithmetical +Progression} if the difference between any term and the preceding +term is the same throughout the series. + +\begin{Remark} +Thus $a$, $b$, $c$, $d$, etc., are in arithmetical progression if $b - a$, $c - b$, +$d - c$,~etc., are all equal. +\end{Remark} + +\Paragraph{166.} This difference is called the \Defn{common difference} of the +progression, and is represented by~$d$. If $d$~is positive, the +progression is an \emph{increasing} series; if $d$~is negative, the progression +is a \emph{decreasing} series. + +What is the common difference in each of the following +series? +\[ +\begin{array}{r*{4}{>{\quad}r}} + 1, & \Neg4, & 7, & 10, & \dots \\ + 5, & 7, & 9, & 11, & \dots \\ +10, & 9, & 8, & 7, & \dots \\ + 7, & 3, & -1, & -5, & \dots \\ +\end{array} +\] + +\Paragraph{167.} If the first term of an arithmetical progression is +represented by~$a$ and the common difference by~$d$, then +\begin{alignat*}{2} +&\text{the \emph{second} term will be } && a + d, \\ +&\text{the \emph{third} term will be } && a + 2d, \\ +&\text{the \emph{fourth} term will be } && a + 3d, +\end{alignat*} +and so on, the coefficient of~$d$ in each term being always +less by~$1$ than the \emph{number of the term}. + +Hence the $n$th~term will be $a + (n - 1)d$. + +If we represent the $n$th~term by~$l$, we have +\[ +l = a + (n - 1)d. +\Tag{Formula (1)} +\] +%% -----File: 149.png---Folio 143------- + +\Paragraph{168.} We can, therefore, find any term of an arithmetical +progression if the first term and common difference are +given, or if any \emph{two} terms are given. + +\Item{1.} Find the 10th~term of an arithmetical progression if +the 1st~term is~$3$ and the common difference is~$4$. +\begin{Soln} +By formula~(1), the 10th~term is~$3 + (10 - 1)4$, or~$39$. +\end{Soln} + +\Item{2.} If the 8th~term of an arithmetical progression is~$25$, +and the 23d~term~$70$, find the series. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent By formula (1),} +&\text{the 23d term is $a + 22d$,} \\ +\lintertext{and} +&\text{the 8th term is $a + 7d$.} +\end{DPalign*} +\begin{DPalign*} +\lintertext{\indent Therefore,} +a + 22d &= 70 \\ +\lintertext{and} +a + \Z7d &= 25\Add{.} \displaybreak[1] \\ +\lintertext{\indent Subtract,} +15d &= 45 \\ +\lintertext{and} +d &= 3, \\ +\lintertext{whence} +a &= 4. +\end{DPalign*} + +The series is therefore $4$, $7$, $10$, $13$,~etc. +\end{Soln} + +\Paragraph{169. Arithmetical Mean.} If three numbers are in arithmetical +progression, the middle number is called the arithmetical +mean of the other two numbers. + +If $a$, $A$, $b$ are in arithmetical progression, $A$~is the arithmetical +mean of $a$~and~$b$. Hence, by the definition of an +arithmetical series, +\begin{DPalign*} +A - a &= b - A, \\ +\lintertext{whence} +A &= \frac{a + b}{2}. +\rintertext{Formula (2)} +\end{DPalign*} +\begin{Theorem} +Hence, the arithmetical mean of any two numbers is found +by taking half their sum. +\end{Theorem} + +\Paragraph{170.} Sometimes it is required to insert several arithmetical +means between two numbers. +%% -----File: 150.png---Folio 144------- + +If $m = {}$the number of means, and $n = {}$the whole number +of terms, then $m + 2 = n$. If $m + 2$ is substituted for~$n$ +in formula~(1), +\begin{DPalign*} +l &= a + (n - 1)d, \\ +\lintertext{the result is} +l &= a + (m + 1)d. \displaybreak[1] \\ +\lintertext{\indent By transposing~$a$,} +l - a &= (m + 1) d. \\ +\therefore \frac{l - a}{m + 1} &= d. +\Tag{\llap{Formula (3)}} +\end{DPalign*} +\begin{Remark} +Thus, if it be required to insert six means between $3$~and~$17$, +the value of~$d$ is found to be $\dfrac{17 - 3}{6 + 1} = 2$; and the series will be $3$,~$5$, +$7$, $9$, $11$, $13$, $15$,~$17$. +\end{Remark} + +\Exercise{75.} + +\Item{1.} Find the 25th~term in the series $3$, $6$, $9$,~$\dots$. + +\Item{2.} Find the 13th~term in the series $50$, $49$, $48$,~$\dots$. + +\Item{3.} Find the 15th~term in the series $\frac{1}{7}$, $\frac{3}{7}$, $\frac{5}{7}$,~$\dots$. + +\Item{4.} Find the 19th~term in the series $\frac{1}{4}$, $-\frac{1}{4}$, $-\frac{3}{4}$,~$\dots$. + +\Item{5.} Find the 10th~term in an arithmetical progression +whose 1st~term is~$5$ and 3d~term~$9$. + +\Item{6.} Find the 11th~term in an arithmetical progression +whose 1st~term is~$10$ and whose 6th~term is~$5$. + +\Item{7.} If the 3d~term of an arithmetical progression is~$20$ +and the 13th~term is~$100$, what is the 20th~term? + +\Item{8.} Which term of the series $5$, $7$, $9$, $11$,~$\dots$, is~$43$? + +\Item{9.} Which term of the series $\frac{4}{3}$, $\frac{3}{2}$, $\frac{5}{3}$,~$\dots$, is~$18$? + +\Item{10.} What is the arithmetical mean of $20$~and~$32$? + +\Item{11.} What is the arithmetical mean of $a + b$ and $a - b$? + +\Item{12.} Insert $8$ arithmetical means between $20$~and~$29$. +%% -----File: 151.png---Folio 145------- + +\Paragraph{171. To Find the Sum of Any Number of Terms of an Arithmetical +Series.} + +If $l$ denote the last term, $a$~the first term, $n$~the number +of terms, $d$~the common difference, and $s$~the sum of the +terms, it is evident that the series beginning with the first +term will be $a$, $a + d$, $a + 2d$,~etc., and beginning with the +last term will be $l$, $l - d$, $l - 2d$,~etc. Therefore, +\begin{gather*} +\begin{array}{r*{12}{c}} +s &=& a &+& (a + d) &+& (a + 2d) &+& \Add{\dots} &+& (l - d) &+& l, \rlap{\quad\text{or}} \\ +s &=& l &+& (l - d) &+& (l - 2d) &+& \Add{\dots} &+& (a + d) &+& a \\ +\hline +2s &=& (a + l) &+& (a + l) &+& (a + l) &+& \Add{\dots} &+& (a + l) &+& (a + l) \\ +2s &=& \multicolumn{11}{l}{\text{$(a + l)$ taken as many times as there are \emph{terms},}} +\end{array} \displaybreak[1] \\ +\begin{aligned}[b] +2s &= n(a + l), \\ +\text{and } s &= \frac{n}{2}(a + l)\Add{.} +\end{aligned} +\Tag{Formula (4)} +\end{gather*} + +Putting for $l$~its value $a + (n - 1)d$, in formula~(4), we +have +\begin{align*} +s &= \frac{n}{2}\bigl\{a + a + (n - 1)d\bigr\} \\ + &= \frac{n}{2}\bigl\{2a + (n - 1)d\bigr\} +\Tag{Formula (5)} +\end{align*} + +\Item{1.} Find the sum of the first $16$~terms of the series $5$, $7$, +$9$,~$11$\DPtypo{,}{.} +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Here} +a &= 5,\quad d = 2,\quad n = 16\Add{.} \\ +\intertext{\indent Putting these values in formula~(5) we have} +s &= \tfrac{16}{2}(10 + 15 × 2) \\ + &= 320 +\end{DPalign*} +\end{Soln} + +\Item{2.} Show that the sum of any number of odd numbers, +beginning with~$1$, is a square number. +\begin{Soln} +The series of odd numbers is $1$, $3$, $5$, $7$,~$\dots$. +%% -----File: 152.png---Folio 146------- +\begin{DPalign*} +\lintertext{\indent Here} +a &= 1 \quad\text{and}\quad d = 2\Add{.} +\intertext{\indent Putting these values in formula (5) we have} +s &= \frac{n}{2} \bigl\{2 + (n - 1)2\bigr\} \\ + &= \frac{n}{2} × 2n \\ + &= n^{2}\Add{.} +\end{DPalign*} + +Therefore the sum of the first $5$~odd numbers is~$5^{2}$ or~$25$\Add{,} of the first +$8$ odd numbers is~$8^{2}$ or~$64$, and so on. +\end{Soln} + +\Item{3.} The sum of $20$~terms of an arithmetical progression +is~$420$, and the first term is~$2$. Find the common difference. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Here} +s &= 420,\quad n = 20, \quad\text{and}\quad a = 2\Add{.} \\ +\intertext{\indent Putting these values in formula~(5), we have} +420 &= \tfrac{20}{2}(4 + 19d) \\ + &= 40 + 190d \\ +190d &= 380 \\ +\Add{\therefore} d &= 2\Add{.} +\end{DPalign*} + +Therefore the common difference is~$2$. +\end{Soln} + +\Exercise{76.} + +\Item{1.} Find the sum of $3$, $5$, $7$, $\dots$, to $20$~terms. + +\Item{2.} Find the sum of $14$, $14\frac{1}{2}$, $15$, $\dots$, to $12$~terms. + +\Item{3.} Find the sum of $\frac{7}{6}$, $1$, $\frac{5}{6}$, $\dots$, to $10$~terms. + +\Item{4.} Find the sum of $-7$, $-5$, $-3$, $\dots$, to $16$~terms. + +\Item{5.} Find the sum of $12$, $9$, $6$, $\dots$, to $21$~terms. + +\Item{6.} Find the sum of $-10\frac{1}{2}$, $-9$, $-7\frac{1}{2}$, $\dots$, to $25$~terms. + +\Item{7.} The sum of three numbers in arithmetical progression +is~$9$, and the sum of their squares is~$35$. Find the numbers. + +\begin{Remark}[Hint.] Let $x - y$, $x$, $x + y$, stand for the numbers. +\end{Remark} +%% -----File: 153.png---Folio 147------- + +\Item{8.} A common clock strikes the hours from $1$ to~$12$. +How many times does it strike every $24$~hours? + +\Item{9.} The Greenwich clock strikes the hours from $1$ to~$24$. +How many times does it strike in $24$~hours? + +\Item{10.} In a potato race each man picked up $50$~potatoes +placed in line a yard apart, and the first potato one yard +from the basket, picking up one potato at a time and bringing +it to the basket. How many yards did each man run, +the start being made from the basket? + +\Item{11.} A heavy body falling from a height falls $16.1$~feet +the first second, and in each succeeding second $32.2$~feet +more than in the second next preceding. How far will a +body fall in $19$~seconds? + +\Item{12.} A stone dropped from a bridge reached the water in +just $3$~seconds. Find the height of the bridge. (See Ex.~11.) + +\Item{13.} The arithmetical mean between two numbers is~$13$, +and the mean between the double of the first and the triple +of the second is~$33\frac{1}{2}$. Find the numbers. + +\Item{14.} Find three numbers of an arithmetical series whose +sum shall be~$27$, and the sum of the first and second shall +be~$frac{4}{5}$ of the sum of the second and third. + +\Item{15.} A travels uniformly $20$~miles a day; B~travels $8$~miles +the first day, $12$~the second, and so on, in arithmetical +progression. If they start Monday morning from the same +place and travel in the same direction, how far apart will +they be Saturday night? + +\Item{16.} The sum of three terms of an arithmetical progression +is~$36$, and the square of the mean exceeds the product of +the other two terms by~$49$. Find the numbers. +%% -----File: 154.png---Folio 148------- + + +\Chapter{XIV.}{Geometrical Progression.} + +\Paragraph{172.} A series of numbers is said to be in \Defn{Geometrical Progression} +when the quotient of any term divided by the +preceding term is the same throughout the series. + +\begin{Remark} +Thus $a$, $b$, $c$, $d$, etc., are in geometrical progression if $\dfrac{b}{a} = \dfrac{c}{b} = \dfrac{d}{c}$,~etc. +\end{Remark} + +\Paragraph{173.} This quotient is called the \Defn{common ratio}, and is represented +by~$r$. + +State the common ratio of the following series: +\[ +\begin{array}{r*{4}{>{\quad}r}} +1, & 3, & \Neg9, &27, & \dots \\ +2, & 4, & 8, &16, & \dots \\ +16,& 8, & 4, &2, & \dots \\ +\frac{2}{3}, & 1, & \frac{3}{2}, & \frac{9}{4}, & \dots \\ +4, & -2, & 1, & -\frac{1}{2}, & \dots \\ +\end{array} +\] + +\Paragraph{174.} If the first term of a geometrical progression is represented +by~$a$, and the common ratio by~$r$, then +\begin{align*} +&\text{the \emph{second} term will be~$ar$,} \\ +&\text{the \emph{third} term will be~$ar^{2}$,} \\ +&\text{the \emph{fourth} term will be~$ar^{3}$,} +\end{align*} +and so on, the index of~$r$ being always less by~$1$ than the +\emph{number of the term in the series}. + +Hence the $n$th~term will be~$ar^{n - 1}$. +%% -----File: 155.png---Folio 149------- + +If we denote the $n$th~term by~$l$, we have +\[ +l = ar^{n - 1}. +\Tag{Formula (1)} +\] + +\Paragraph{175.} If the first term and common ratio are given, or if +any \emph{two terms} are given, we can find the series. + +\Item{1.} Find the 5th~term of a geometrical progression if the +first is~$3$ and the common ratio~$2$. +\begin{Soln} +In formula~(1), put $5$~for~$n$, $3$~for~$a$, and $2$~for~$r$. +\begin{DPalign*} +\lintertext{\indent Then} +l &= 3 × 2^{4} = 48. +\end{DPalign*} + +Therefore the 5th~term is~$48$. +\end{Soln} + +\Item{2.} Find the geometrical series if the 5th~term is~$48$ and +the 7th~term is~$192$. +\begin{Soln} +The 5th and 7th~terms are $ar^{4}$~and~$ar^{6}$, respectively. +\begin{DPalign*} +\lintertext{\indent Whence} +ar^{4} &= 48, +\Tag{(1)} \\ +\lintertext{and} +ar^{6} &= 192. +\Tag{(2)} \\ +\lintertext{\indent Divide (2) by~(1),} +r^{2} &= 4. \\ +\therefore r &= ±2. \\ +\lintertext{\indent From~(1),} +a = \tfrac{48}{16} &= 3. +\end{DPalign*} + +Therefore the series is $3$, $±6$, $12$, $±24$,~$\dots$. +\end{Soln} + +\Paragraph{176. Geometrical Mean.} If three numbers are in geometrical +progression, the middle number is called the \emph{geometrical +mean} of the other two numbers. Hence, if +$a$,~$G$,~$b$ are in geometrical progression, $G$~is the geometrical +mean of $a$~and~$b$. + +By the definition of a geometrical progression, +\begin{DPalign*} +\frac{G}{a} &= \frac{b}{G}. \\ +\therefore G^{2} &= ab, \\ +\lintertext{and} +G &= ± \sqrt{ab}. +\Tag{Formula (2)} +\end{DPalign*} +\begin{Theorem}[Hence], the geometrical mean of any two numbers is the +square root of their product. +\end{Theorem} +%% -----File: 156.png---Folio 150------- + +\PrintBreak +\Paragraph{177. To Find the Sum of Any Number of Terms of a Geometrical +Progression.} + +If $l$~denote the last term, $a$~the first term, $n$~the number +of terms, $r$~the common ratio, and $s$~the sum of the $n$~terms, +then +\begin{DPalign*} +s &= a + ar + ar^{2} + ar^{3} + \dots+ ar^{n - 1}. \\ +\lintertext{\indent Multiply by~$r$,} +rs &= ar + ar^{2} + ar^{3} + \dots + ar^{n - 1} + ar^{n}. +\end{DPalign*} + +Therefore, by subtracting the first equation from the +second, +\begin{DPalign*} +rs - s &= ar^{n} - a, \\ +\lintertext{or} +(r - 1)s &= a(r^{n} - 1). \\ +\therefore s &= \frac{a(r^{n} - 1)}{r - 1}. +\Tag{Formula (3)} +\end{DPalign*} + +\Paragraph{178.} When $r$~is $< 1$, this formula will be more convenient +if written +\[ +s = \frac{a(1 - r^{n})}{1 - r}. +\] + +\Item{1.} Find the sum of $8$~terms of the series +\[ +1,\quad 2,\quad 4,\quad \dots\Add{.} +\] +\begin{DPalign*} +\lintertext{\indent Here} +a &= 1,\quad r = 2,\quad n = 8. \\ +\lintertext{\indent From formula~(3),} +s &= 1(2^{8} - 1) = 255. +\end{DPalign*} + +\Item{2.} Find the sum of $6$~terms of the series +\[ +2,\quad 3,\quad \tfrac{9}{2},\quad \dots\Add{.} +\] +\begin{DPalign*} +\lintertext{\indent Here} +a &= 2,\quad r = \tfrac{3}{2},\quad n = 6. \\ +\lintertext{\indent From formula~(3),} +s &= \frac{2\bigl\{(\frac{3}{2})^{6} - 1\bigr\}}{\frac{3}{2} - 1} \\ + &= \frac{2\bigl\{\frac{729}{64} - 1\bigr\}}{\frac{1}{2}} \\ + &= \frac{4\{729 - 64\}}{64} \\ + &= 41\tfrac{9}{16}. +\end{DPalign*} +%% -----File: 157.png---Folio 151------- + +\Exercise{77.} + +\Item{1.} Find the 5th~term of $3$, $9$, $27$\Add{,}~$\dots$. + +\Item{2.} Find the 7th~term of $3$, $6$, $12$\Add{,}~$\dots$. + +\Item{3.} Find the 8th~term of $6$, $3$, $\frac{3}{2}$\Add{,}~$\dots$. + +\Item{4.} Find the 9th~term of $1$, $-2$, $4$\Add{,}~$\dots$. + +\Item{5.} Find the geometrical mean between $2$~and~$8$. + +\Item{6.} Find the common ratio if the 1st~and 3d~terms are +$2$~and~$32$. + +Find the sum of the series: + +\Item{7.} $3$, $9$, $27$, $\dots$ to $6$~terms. + +\Item{8.} $3$, $6$, $12$, $\dots$ to $8$~terms. + +\Item{9.} $6$, $3$, $\frac{3}{2}$, $\dots$ to $7$~terms. + +\Item{10.} $8$, $4$, $2$, $\dots$ to $8$~terms. + +\Item{11.} $64$, $32$, $16$, $\dots$ to $9$~terms. + +\Item{12.} $64$, $-32$, $16$, $\dots$ to $5$~terms. + +\Item{13.} $\frac{1}{2}$, $\frac{1}{3}$, $\frac{2}{9}$, $\dots$ to $4$~terms. + +\Item{14.} If a blacksmith uses seven nails in putting a shoe on +a horse's foot, and receives $1$~cent for the first nail, $2$~cents +for the second nail, and so on, what does he receive for +putting on the shoe? + +\Item{15.} If a boy receives $2$~cents for his first day's work, +$4$~cents for his second day, $8$~cents for the third day, and +so on for $12$~days, what will his wages amount to? + +\Item{16.} If the population of a city is $10,000$, and increases +$10$\%~a year for four years, what will be its population at +the end of the four years? (Here $l = ar^{4}$.) +%% -----File: 158.png---Folio 152------- + + +\Chapter{XV.}{Square and Cube Roots.} + +\Section{Square Roots of Compound Expressions.} + +\Paragraph{179.} Since the square of $a + b$ is $a^{2} + 2ab + b^{2}$, the square +root of $a^{2} + 2ab + b^{2}$ is $a + b$. + +It is required to find a method of extracting the root +$a + b$ when $a^{2} + 2ab + b^{2}$ is given. +\begin{Soln} +Ex. The first term,~$a$, of the root is obviously the square root of +the first term,~$a^{2}$, in the expression. +\[ +\begin{array}{r*{3}{cr}l} + & & a^{2} &+& 2ab &+& \TbBar{b^{2}} & a + b \\ +\cline{8-8} + & & a^{2} \\ +\cline{3-3} +2a &+& \TbBar{b} & & 2ab &+& b^{2} \\ + & & \TbBar{ } & & 2ab &+& b^{2} \\ +\cline{4-7} +\end{array} +\] + +If the $a^{2}$ is subtracted from the given +expression, the remainder is $2ab + b^{2}$\Add{.} +Therefore the second term,~$b$, of the root +is obtained when the first term of this +remainder is divided by~$2a$; that is, by +\emph{double the part of the root already found}. Also, since +\[ +2ab + b^{2} = (2a + b)b, +\] +the divisor is \emph{completed by adding to the trial-divisor the new term of +the root}\Add{.} +\end{Soln} + +\ScreenBreak +Ex. Find the square root of $25x^{2} - 20x^{3}y + 4x^{4}y^{2}$. +\begin{Soln} +\[ +\begin{array}{rcccrll} +& && & 25x^{2} & \TbBar{-20 x^{3}y + 4x^{4}y^{2}} & 5x - 2x^{2}y \\ +\cline{7-7} +\text{Here $a^{2}$} + &=& (5x)^{2} &=& 25x^{2} \\ +\cline{5-5} +2a + b &=& 10x &-& \TbBar{2x^{2}y} & -20x^{3}y + 4x^{4}y^{2} \\ + & & & & \TbBar{ } & -20x^{3}y + 4x^{4}y^{2} \\ +\cline{6-6} +\end{array} +\] + +The expression is \emph{arranged} according to the ascending powers of~$x$\Add{.} + +The square root of the first term is~$5x$, hence $5x$~is the first term +of the root. $(5x)^{2}$ or $25x^{2}$ is subtracted, and the remainder is +\[ +-20x^{3}y + 4x^{4}y^{2}. +\] + +The second term of the root, $-2x^{2}y$, is obtained by dividing +$-20x^{3}y$ by~$10x$, the double of~$5x$, and this new term of the root is +also annexed to the divisor,~$10x$, to complete the divisor. +\end{Soln} +%% -----File: 159.png---Folio 153------- + +\Paragraph{180.} The same method will apply to longer expressions, +if care be taken to obtain the \emph{trial-divisor} at each stage of +the process, \emph{by doubling the part of the root already found}, +and to obtain the \emph{complete divisor by annexing the new term +of the root to the trial-divisor}. + +\ScreenBreak +Ex. Find the square root of +\[ +1 + 10x^{2} + 25x^{4} + 16x^{6} - 24x^{5} - 20x^{3} - 4x. +\] +\begin{Soln} +\[ +%[** TN: Spacing hack] +\qquad\makebox[0pt][c]{$\begin{array}{r*{2}{cr}lll} +16x^{6} &-& 24x^{5} &+& 25x^{4} &-20x^{3} + 10x^{2} &\TbBar{- 4x + 1} & 4x^{3} - 3x^{2} + 2x - 1 \\ +\cline{8-8} +16x^{6} \\ +\cline{1-1} +\TbBar{\llap{$8x^{3} - 3x^{2}$}} + &-& 24x^{5} &+& 25x^{4} \\ +\TbBar{}&-& 24x^{5} &+& 9x^{4} \\ +\cline{2-5} +\multicolumn{4}{r|}{8x^{3} - 6x^{2} + 2x} & + 16x^{4} &-20x^{3} + 10x^{2} \\ + & & & \TbBar{ }& 16x^{4} &-12x^{3} + \Z4x^{2} \\ +\cline{5-6} +\multicolumn{5}{r|}{8x^{3} - 6x^{2} + 4x - 1} + &-\Z8x^{3} + \Z6x^{2} & -4x + 1 \\ + & & & & \TbBar{}&-\Z8x^{3} + \Z6x^{2} & -4x + 1 \\ +\cline{6-7} +\end{array}$} +\] + +The expression is arranged according to the descending powers of~$x$. + +It will be noticed that each successive trial-divisor may be obtained +by taking the preceding complete divisor with its \emph{last term doubled}. +\end{Soln} + +\Exercise{78.} + +Find the square root of: +\Item{1.} $a^{2} + 2ab + 2ac + b^{2} + 2bc + c^{2}$. + +\Item{2.} $x^{4} + 2x^{3} + 3x^{2} + 2x + 1$. + +\Item{3.} $x^{4} - 4x^{3}y + 6x^{2}y^{2} - 4xy^{3} + y^{4}$. + +\Item{4.} $4a^{4} - 12a^{3}b + 29a^{2}b^{2} - 30ab^{3} + 25b^{4}$. + +\Item{5.} $16x^{6} + 24x^{5}y + 9x^{4}y^{2} - 16x^{3}y^{3} - 12x^{2}y^{4} + 4y^{6}$. + +\Item{6.} $4x^{6} - 4x^{4}y^{2} + 12x^{3}y^{3} + x^{2}y^{4} - 6xy^{5} + 9y^{6}$. + +\Paragraph{181. Arithmetical Square Roots.} In the general method +of extracting the square root of a number expressed by +figures, the first step is to mark off the figures into \emph{groups}. +%% -----File: 160.png---Folio 154------- + +Since $1 = 1^{2}$, $100 = 10^{2}$, $10,000 = 100^{2}$, and so on, it is +evident that the square root of a number between $1$~and +$100$ lies between $1$~and~$10$; of a number between $100$ and +$10,000$ lies between $10$~and~$100$. In other words, the +square root of a number expressed by \emph{one} or \emph{two} figures is +a number of \emph{one} figure, of a number expressed by \emph{three} or +\emph{four} figures is a number of \emph{two} figures, and so on. + +If, therefore, an integral square number is divided into +groups of two figures each, from the right to the left, the +number of figures in the root will be equal to the number +of groups of figures. The last group to the left may have +only one figure. + +Ex. Find the square root of~$3249$. +%[** TN: Tabulated calculation inset in the original] +\begin{Soln} +\[ +\begin{array}{rc@{}ll} + &3&249&(57 \\ + &2&5 \\ +\cline{2-3} +107\,\rlap{)}&&749 \\ + &&749 \\ +\cline{3-3} +\end{array} +\] + +In this case, $a$~in the typical form $a^{2} + 2ab + b^{2}$ +represents $5$~tens, that is,~$50$, and $b$~represents~$7$\Add{.} The +$25$ subtracted is really~$2500$, that is,~$a^{2}$, and the complete +divisor $2a + b$ is $2 × 50 + 7 = 107$. +\end{Soln} + +\Paragraph{182.} The same method will apply to numbers of more +than two groups of figures by considering $a$~in the typical +form to represent at each step \emph{the part of the root already +found}. + +It must be observed that \emph{$a$~represents so many tens with +respect to the next figure of the root}. + +Ex. Find the square root of~$94,249$. +\begin{Soln} +\[ +\begin{array}{r*{3}{c@{\,}}l} + &9& 42&49&(307 \\ + &9 \\ +\cline{2-4} +607\rlap{)} & & \multicolumn{2}{l}{4249} \\ + & & \multicolumn{2}{l}{4249} \\ +\cline{3-4} +\end{array} +\] +\end{Soln} + +\begin{Remark}[Note.] Since the first trial divisor,~$60$, is not contained in~$42$, we +put a zero in the root, and bring down the next group,~$49$. +\end{Remark} +%% -----File: 161.png---Folio 155------- + +\Paragraph{183.} If the square root of a number has decimal places, +the number itself will have \emph{twice} as many. Thus, if $0.21$ +is the square root of some number, this number will be +$(0.21)^{2} = 0.21 × 0.21 = 0.0441$, and if $0.111$ be the root, +the number will be $(0.111)^{2} = 0.111 x 0.111 = 0.012321$. + +Therefore, the number of \emph{decimal} places in every square +decimal will be \emph{even}, and the number of decimal places in +the root will be \emph{half} as many as in the given number itself. + +Hence, if a given number contain a decimal, we divide +it into groups of two figures each, by beginning at the +decimal point and marking toward the left for the integral +number, and toward the right for the decimal. We must +have the last group on the right of the decimal point contain +\emph{two} figures, annexing a cipher when necessary. + +Ex. Find the square roots of $41.2164$ and $965.9664$. +\begin{Soln} +\[ +\begin{array}{r@{}r@{}l@{\,}ll} + & 41&.21&64& (6.42 \\ + & 36 \\ +\cline{2-3} +124\rlap{)} & 5&21 \\ + & 4&96 \\ +\cline{3-4} +\multicolumn{2}{r}{1282\rlap{)}}&\multicolumn{2}{l}{2564} \\ + & &\multicolumn{2}{l}{2564} \\ +\cline{3-4} +\end{array}\qquad\qquad\qquad +\begin{array}{rl@{}l@{\,}ll} + 9&65&.96& 64&(31.08 \\ + 9 \\ +\cline{2-2} + 61\rlap{)}&65 \\ + &61 \\ +\cline{2-3} +6208\rlap{)}&49&664 \\ + &49&664 \\ +\cline{2-3} +\end{array} +\] +\end{Soln} + +\Paragraph{184.} If a number contain an \emph{odd} number of decimal +places, or if any number give a \emph{remainder} when as many +figures in the root have been obtained as the given number +has groups, then its exact square root cannot be found. We +may, however, approximate to its exact root as near as we +please by annexing ciphers and continuing the operation. + +The square root of a common fraction whose denominator +is not a perfect square can be found approximately by +reducing the fraction to a decimal and then extracting the +root; or by reducing the fraction to an equivalent fraction +whose denominator is a perfect square, and extracting the +square root of both terms of the fraction. +%% -----File: 162.png---Folio 156------- + +%[** Force page break in both print and screen layout] +\newpage +\Item{1.} Find the square roots of $3$ and $357.357$. +\begin{Soln} +\[ +\begin{array}{rcccl} + &3.&\multicolumn{3}{l}{(\rlap{$1.732\dots$}} \\ + &1 \\ +\cline{2-3} +27\rlap{)} + &2&00 \\ + &1&89 \\ +\cline{3-4} +\multicolumn{2}{r}{343\rlap{)}} + &11&00 \\ + &&10&29 \\ +\cline{4-5} +\multicolumn{3}{r}{3462\rlap{)}} + &71&00 \\ +&& &69&24 \\ +\cline{4-5} +\end{array}\qquad\qquad\qquad +\begin{array}{rcclll} + &3&57.&35&70&\rlap{$(18.903\dots$} \\ + &1 \\ +\cline{2-3} +28\rlap{)} + &2&57 \\ + &2&24 \\ +\cline{3-4} +\multicolumn{2}{r}{369\rlap{)}} + &33&35 \\ +& &33&21 \\ +\cline{4-6} +\multicolumn{3}{r}{37803\rlap{)}} + &14&70&00 \\ +&& &11&34&09 \\ +\cline{4-6} +\end{array} +\] +\end{Soln} + +\Item{2.} Find the square root of~$\frac{5}{8}$. +\begin{Soln} +\begin{DPalign*} +\lintertext{\indent Since} +\frac{5}{8} &= 0.625, \\ +\lintertext{the square root of} +\frac{5}{8} &= \sqrt{0.625} \\ + &= 0.79057. \displaybreak[1] \\ +\lintertext{\indent Or,} +\frac{5}{8} &= \frac{10}{16}, \\ +\lintertext{and the square root of} +\frac{5}{8} &= \frac{\sqrt{10}}{\sqrt{16}} = \tfrac{1}{4}\sqrt{10} \\ + &= \tfrac{1}{4}(3.16227) \\ + &= 0.79057. +\end{DPalign*} +\end{Soln} + +\ScreenBreak +\Exercise{79.} + +Find the square root of: +\begin{multicols}{3} +\Item{1.} $324$. + +\Item{2.} $441$. + +\Item{3.} $529$. + +\Item{4.} $961$. + +\Item{5.} $10.24$. + +\Item{6.} $53.29$. + +\Item{7.} $53,824$. + +\Item{8.} $616,225$. + +\Item{9.} $1,500,625$. + +\Item{10.} $346,921$. + +\Item{11.} $31,371,201$. + +\Item{12.} $1,522,756$. +\end{multicols} + +\PrintBreak +Find to four decimal places the square root of: +\begin{multicols}{5} +\Item{13.} $2$. + +\Item{14.} $3$. + +\Item{15.} $5$. + +\Item{16.} $6$. + +\Item{17.} $0.5$. + +\Item{18.} $0.9$. + +\Item{19.} $\frac{2}{3}$. + +\Item{20.} $\frac{3}{4}$. + +\Item{21.} $\frac{4}{5}$. + +\Item{22.} $\frac{5}{8}$. +\end{multicols} +%% -----File: 163.png---Folio 157------- + + +\Section{Cube Roots of Compound Expressions.} + +\Paragraph{185.} Since the cube of $a + b$ is $a^{3} + 3a^{2}b + 3ab^{2} + b^{3}$, +the cube root of a$^{3} + 3a^{2}b + 3ab^{2} + b^{3}$ is $a + b$. + +It is required to devise a method for extracting the cube +root $a + b$ when $a^{3} + 3a^{2}b + 3ab^{2} + b^{3}$ is given. + +\Item{1.} Find the cube root of $a^{3} + 3a^{2}b + 3ab^{2} + b^{3}$. +\begin{Soln} +\[ +\begin{array}{rl<{\quad}rcll} + & & a^{3} &+& \TbBar{3a^{2}b + 3ab^{2} + b^{3}} & a + b \\ +\cline{6-6} +3a^{2} & & a^{3} \\ +\cline{3-3} + &+ 3ab + b^{2} &\TbBar{}& & 3a^{2}b + 3ab^{2} + b^{3} \\ +\cline{1-3} +3a^{2} &+ 3ab + b^{2} &\TbBar{}& & 3a^{2}b + 3ab^{2} + b^{3} \\ +\cline{4-5} +\end{array} +\] + +The first term $a$ of the root is obviously the cube root of the first +term~$a^{3}$ of the given expression. + +If $a^{3}$ be subtracted, the remainder is $3a^{2}b + 3ab^{2} + b^{3}$; therefore, +the second term~$b$ of the root is obtained by dividing the first term of +this remainder by \emph{three times the square of~$a$}. + +Also, since $3a^{2}b + 3ab^{2} + b^{3} = (3a^{2} + 3ab + b^{2})b$, the \emph{complete +divisor} is obtained by adding $3ab + b^{2}$ to the \emph{trial divisor~$3a^{2}$}. +\end{Soln} + +\Item{2.} Find the cube root of $8x^{3} + 36x^{2}y + 54xy^{2} + 27y^{3}$. +\begin{Soln} +\[ +\begin{array}{rlrlcll} + & & & 8x^{3} &+& \TbBar{36x^{2}y + 54xy^{2} + 27y^{3}} & 2x + 3y \\ +\cline{7-7} + & 12x^{2} & & 8x^{3} \\ +\cline{4-4} +(6x + 3y)3y =& & 18xy +& \TbBar{9y^{2}} & & 36x^{2}y + 54xy^{2} + 27y^{3} \\ +\cline{2-4} + & 12x^{2} &+ 18xy +& \TbBar{9y^{2}} & & 36x^{2}y + 54xy^{2} + 27y^{3} \\ +\cline{5-6} +\end{array} +\] + +The cube root of the first term is~$2x$, and this is therefore the first +term of the root. $8x^{3}$,~the cube of~$2x$, is subtracted. + +The second term of the root,~$3y$, is obtained by dividing $36x^{2}y$ by +$3(2x)^{2} = 12x^{2}$, which corresponds to $3a^{2}$ in the typical form, and the +divisor is completed by annexing to~$12x^{2}$ the expression +\[ +\bigl\{3(2x) + 3y\bigr\}3y = 18xy + 9y^{2}, +\] +which corresponds to $3ab + b^{2}$ in the typical form. +\end{Soln} +%% -----File: 164.png---Folio 158------- + +\Paragraph{186.} The same method may be applied to longer expressions +by considering~$a$ in the typical form $3a^{2} + 3ab + b^{2}$ +to represent at each stage of the process \emph{the part of the root +already found}. Thus, if the part of the root already found +is $x + y$, then $3a^{2}$~of the typical form will be represented +by $3(x + y)^{2}$; and if the third term of the root be~$+z$, \DPtypo{the}{then} +$3ab + b^{2}$ will be represented by $3(x + y)z + z^{2}$. So that +the complete divisor, $3a^{2} + 3ab + b^{2}$, will be represented +by $3(x + y)^{2} + 3(x + y)z + z^{2}$. + +Ex. Find the cube root of $x^{6} - 3x^{5} + 5x^{3} - 3x - 1$. +\begin{Soln} +%[** TN: Special spacing] +\ifthenelse{\not\boolean{ForPrinting}}{\footnotesize}{} +\[ +\ifthenelse{\not\boolean{ForPrinting}}{\quad}{} +\begin{array}{*{2}{r}@{}*{3}{r}*{4}{l}} + & & & \TbBar{}& x^{2}& \multicolumn{3}{@{}l}{{} - x - 1} \\ +\cline{5-6} + & & & & x^{6}& {} - 3x^{5} &\multicolumn{2}{@{}l}{\phantom{{}+ 3x^{4}} + 5x^{3} - 3x - 1} \\ + & 3x^{4}& & & x^{6} \\ +\cline{5-5} + (3x^{2} - x)(-x) =& &{} - 3x^{3}&+& \TbBar{x^{2}}&{} - 3x^{5}&\phantom{{}+ 3x^{4}} + 5x^{3} \\ +\cline{3-5} + & 3x^{4}&{} - 3x^{3}&+& \TbBar{x^{2}}&{} - 3x^{5}&{} + 3x^{4} - \Z x^{3} \\ +\cline{6-7} + & & & & &\TbBar{} &{} - 3x^{4} + 6x^{3} - 3x - 1 \\ + 3(x^{2} - x)^{2} =& 3x^{4}&{} - 6x^{3}&+& 3x^{2}&\TbBar{} \\ +\llap{$(3x^{2} - 3x - 1)$}(-1) =& & &-& 3x^{2}& \multicolumn{1}{l|}{{}+ 3x + 1} \\ +\cline{2-6} + & 3x^{4}&{} - 6x^{3}& & & \multicolumn{1}{l|}{{}+ 3x + 1}&{} - 3x^{4} + 6x^{3} - 3x - 1 \\ +\cline{7-7} +\end{array} +\] +\end{Soln} + +\begin{Remark}[Note.] The root is placed \emph{above} the given expression because +there is no room for it on the page at the right of the expression. + +The first term of the root,~$x^{2}$, is obtained by taking the cube root +of the first term of the given expression; and the first trial-divisor,~$3x^{4}$, +is obtained by taking three times the square of this term. + +The next term of the root is found by dividing~$-3x^{5}$, the first term +of the remainder after~$x^{6}$ is subtracted, by~$3x^{4}$, and the first complete +divisor, $3x^{4} - 3x^{3} + x^{2}$, is found by annexing to the trial divisor +$(3x^{2} - x)(-x)$, which expression corresponds to $(3a + b)b$ in the +typical form. + +\emph{The part of the root already found}~($a$) is now represented by $x^{2} - x$, +therefore $3a^{2}$~is represented by $3(x^{2} - x)^{2} = 3x^{4} - 6x^{3} + 3x^{2}$, the +second trial divisor\Add{,} and $(3a + b)b$ by $(3x^{2} - 3x - 1)(-1)$, since $b$~in +this case is found to be~$-1$, therefore, in the second complete +divisor, $3a^{2} + (3a + b)b$ is represented by +\[ +(3x^{4} - 6x^{3} + 3x^{2}) + (3x^{2} - 3x - 1)(-1) = 3x^{4} - 6x^{3} + 3x + 1. +\] +\end{Remark} +%% -----File: 165.png---Folio 159------- + +\Exercise{80.} + +Find the cube root of + +\Item{1.} $x^{3} + 3x^{2} y + 3xy^{2} + y^{3}$\Add{.} + +\Item{2.} $8x^{3} - 12x^{2} + 6x - 1$. + +\Item{3.} $8x^{3} - 36x^{2} y + 54xy^{2} - 27y^{3}$. + +\Item{4.} $64a^{3} - 144a^{2} x + 108ax^{2} - 27x^{3}$\Add{.} + +\Item{5.} $1 + 3x + 6x^{2} + 7x^{3} + 6x^{4} + 3x^{5} + x^{6}$. + +\Item{6.} $x^{6} - 3x^{5} + 6x^{4} - 7x^{3} + 6x^{2} - 3x + 1$. + +\ScreenBreak +\Paragraph{187. Arithmetical Cube Roots.} In extracting the cube root +of a number expressed by figures, the first step is to mark +it off into groups\Add{.} + +Since $1 = 1^{3}$, $1000 = 10^{3}$, $1,000,000 = 100^{3}$, and so on, it +follows that the cube root of any number between $1$~and +$1000$, that is, of any number which has \emph{one}, \emph{two}, or \emph{three} +figures, is a number of \emph{one} figure, and that the cube root +of any number between $1000$ and $1,000,000$, that is, of any +number which has \emph{four}, \emph{five}, or \emph{six} figures, is a number of +\emph{two} figures, and so on. + +If, therefore, an integral cube number be divided into +groups of three figures each, from right to left, the number +of figures in the root will be equal to the number of groups\Add{.} +The last group to the left may consist of one, two, or three +figures. + +\Paragraph{188.} If the cube root of a number have decimal places, +the number itself will have \emph{three times} as many. Thus, if +$0.11$ be the cube root of a number\Add{,} the number is $0.11 × 0.11 +x 0.11 = 0.001331$. Hence\Add{,} if a given number contain a +decimal, we divide the figures of the number into groups +of three figures each, by beginning at the decimal point +and marking toward the left for the integral number, and +%% -----File: 166.png---Folio 160------- +toward the right for the decimal. We must be careful to +have the last group on the right of the decimal point contain +\emph{three} figures, annexing ciphers when necessary. + +\ScreenBreak +Extract the cube root of~$42875$. +\begin{Soln} +\[ +\begin{array}{rrrlll} + & & & 42&875&(35 \\ + & & a^{3} =& 27 \\ +\cline{4-5} +3a^{2} =& 3 × 30^{2} =& \TbBar{2700} & 15&875 \\ + 3ab =& 3 × (30 × 5) =& \TbBar{450} \\ + b^{2} =& 5^{2} =& \TbBar{25} \\ +\cline{3-3} + & & \TbBar{3175} & 15&875 \\ +\cline{4-5} +\end{array} +\] + +Since $42875$ has two groups, the root will have two figures. + +The first group, $42$, contains the cube of the tens of the root. + +The greatest cube in~$42$ is~$27$, and the cube root of~$27$ is~$3$. Hence +$3$~is the tens' figure of the root. + +We subtract $27$ from~$42$, and bring down the next group,~$875$. +Since $a$~is $3$~tens or~$30$, $3a^{2} = 3 × 30^{2}$, or~$2700$. This trial-divisor is +contained $5$~times in~$15875$. The trial-divisor is completed by adding +$3ab + b^{2}$; that is, $450 + 25$, to the trial-divisor. +\end{Soln} + +\Paragraph{189.} The same method will apply to numbers of more +than two groups of figures, by considering in each case~$a$, +the part of the root already found, as so many tens with +respect to the next figure of the root. + +\PrintStretch{12pt} +Extract the cube root of~$57512456$. +\begin{Soln} +\[ +\begin{array}{*{3}{r}*{4}{l}} + & & & 57&512&456&(386 \\ + & & a^{3} =& 27 \\ +\cline{4-5} +3a^{2} =& 3 × 30^{2} =& \TbBar{2700} & 30&512 \\ + 3ab =& 3 × (30 × 8) =& \TbBar{720} \\ + b^{2} =& 8^{2} =& \TbBar{64} \\ +\cline{3-3} + & & \TbBar{3484} & 27&872 \\ +\cline{4-6} + & & \TbBar{} & \Z2&640&456 \\ +3a^{2} =& 3 × 380^{2} =& \TbBar{433200} \\ + 3ab =& 3 × (380 × 6) =& \TbBar{6840} \\ + b^{2} =& 6^{2} =& \TbBar{36} \\ +\cline{3-3} + & & \TbBar{440076} & \Z2&640&456 \\ +\cline{4-6} +\end{array} +\] +\end{Soln} +%% -----File: 167.png---Folio 161------- + +Extract the cube root of~$187.149248$. +\begin{Soln} +\[ +\begin{array}{*{3}{r}@{}*{4}{r}l} + & & & \multicolumn{2}{r}{187\rlap{.}}&149&248&(5.72 \\ + & & a^{3} =& \multicolumn{2}{r}{125} \\ +\cline{4-6} +3a^{2} =& 3 × 50^{2} =& \TbBar{7500}&& 62&149 \\ + 3ab =& 3 × (50 × 7) =& \TbBar{1050} \\ + b^{2} =& 7^{2} =& \TbBar{49} \\ +\cline{3-3} + & & \TbBar{8599}&& 60&193 \\ +\cline{4-7} + & & & \TbBar{}& 1&956&248 \\ +3a^{2} =& 3 × 570^{2} =& 9747&\TbBar{00} \\ + 3ab =& 3 × (570 × 2) =& 34&\TbBar{20} \\ + b^{2} =& 2^{2} =& & \TbBar{4} \\ +\cline{3-4} + & & 9781&\TbBar{24}& 1&956&248 \\ +\cline{5-7} +\end{array} +\] + +It will be seen from the groups of figures that the root will have +one integral and two decimal places. +\end{Soln} + +\Paragraph{190.} If the given number is not a perfect cube, ciphers +may be annexed, and a value of the root may be found as +near to the \emph{true} value as we please. + +Extract the cube root of~$1250.6894$. +\begin{Soln} +\[ +\begin{array}{rrrcclll} + & & & &1&\multicolumn{3}{l}{\rlap{250.689\,400\,(10.77}} \\ + & & a^{3} &=&1 \\ +\cline{4-6} +3a^{2} =& \PadTo{3 × (1070 × 7)}{3 × 10^{2}} =& \TbBar{\Z\Z300}& && 250 & \phantom{999} & \phantom{999} \\ +\end{array} +\] +Since $300$ is not contained in~$250$, the next figure of the root will +be~$0$. +\[ +\begin{array}{rrr@{}ccrll} +3a^{2} =& 3 × 100^{2} =& \TbBar{30000}& &\Z&250&689 \\ + 3ab =& 3 × (100 × 7) =& \TbBar{2100} \\ + b^{2} =& 7^{2} =& \TbBar{49} \\ +\cline{3-3} + & & \TbBar{32149}& && 225&043 \\ +\cline{4-8} + & & &\TbBar{}&& 25&646&400 \\ +3a^{2} =& 3 × 1070^{2} =& 34347&\TbBar{00} \\ + 3ab =& 3 × (1070 × 7) =& 224&\TbBar{70} \\ + b^{2} =& 7^{2} =& &\TbBar{49} \\ +\cline{3-4} + & & 34572&\TbBar{19}&& 24&200&533 \\ +\cline{5-8} + & & & && 1&445&867 \\ +\end{array} +\] +\end{Soln} +%% -----File: 168.png---Folio 162------- + +\Paragraph{191.} Notice that if $a$~denotes the first term, and $b$~the +second term of the root, the \emph{first complete divisor} is +\[ +3a^{2} + 3ab + b^{2}, +\] +and the \emph{second trial-divisor} is $3(a + b)^{2}$, that is, +\[ +3a^{2} + 6ab + 3b^{2}. +\] + +This expression may be obtained by adding to the preceding +complete divisor, $3a^{2} + 3ab + b^{2}$, \emph{its second term and +twice its third term}. Thus: +\[ +\begin{array}{rr} +3a^{2} + 3ab + & b^{2} \\ + 3ab + &2b^{2} \\ +\cline{1-2} +3a^{2} + 6ab + &3b^{2} \\ +\end{array} +\] + +This method of obtaining \emph{trial-divisors} is of great importance +for shortening numerical work, as may be seen in the +following example: + +Ex. Extract the cube root of~$5$ to five places of decimals. +\begin{Soln} +\[ +\begin{array}{rrr@{}r@{}rcrrl} + & & & &&5\rlap{.}&\multicolumn{3}{l}{000\,(1.70997} \\ + & && \multicolumn{3}{r}{a^{3} = 1} \\ +\cline{4-7} +3a^{2} =& 3 × 10^{2} =& 300&\TbBar{}&&4&000 \\ + 3ab =& 3(10 × 7) =& 210&\TbBar{} \\ + b^{2} =& 7^{2} =& 49&\TbBar{}\\ +\cline{3-3} + & & 559&\TbBar{\BB}&&3&913 \\ +\cline{5-9} + & & 259& & &\TbBar{}& 87& 000& 000 \\ +\cline{3-5} +3a^{2} =& 3 × 1700^{2} =& 867&00&00&\TbBar{} \\ + 3ab =& 3(1700 × 9) =& 4&59&00&\TbBar{} \\ + b^{2} =& 9^{2} =& & &81&\TbBar{} \\ +\cline{3-5} + & & 871&59&81&\TbBar{\BB}& 78& 443& 829 \\ +\cline{7-9} + & & 4&59&81&\TbBar{}& 8& 556& 1710 \\ +\cline{3-5} +3a^{2} =& 3 × 1709^{2} =& 876&20&43&\TbBar{}& 7& 885& 8387 \\ +\cline{7-9} + & & & & &\TbBar{}& & 670& 33230 \\ + & & & & &\TbBar{}& & 613& 34301 \\ +\cline{7-9} +\end{array} +\] +%% -----File: 169.png---Folio 163------- + +After the first two figures of the root are found, the next trial-divisor +is obtained by bringing down~$259$, the sum of the $210$ and $49$ +obtained in completing the preceding divisor, then \emph{adding the three +lines connected by the brace}, and annexing two ciphers to the result. + +This trial divisor is~$86,700$, and if we add $3ab + b^{2}$ to complete the +divisor, when $b = 1$, the complete divisor will be $86,700 + 511 = 87,211$, +and this is larger than the dividend~$87,000$. We therefore put $0$ for +the next figure of the root. We then bring down another group +and annex two more ciphers to the trial-divisor. + +The last two figures of the root are found by division. The rule +in such cases is, that two less than the number of figures already +obtained may be found without error by division, the divisor being +three times the square of the part of the root already found. +\end{Soln} + +\Paragraph{192.} The cube root of a common fraction whose denominator +is not a perfect cube can be found approximately by +reducing the fraction to a decimal, and then extracting +the root. + +\Exercise{81.} + +Find the cube root of: +\begin{multicols}{2} +\Item{1.} $46,656$. + +\Item{2.} $42,875$. + +\Item{3.} $91,125$. + +\Item{4.} $274,625$. + +\Item{5.} $110,592$. + +\Item{6.} $258,474,853$. + +\Item{7.} $109,215,352$. + +\Item{8.} $259,694,072$. + +\Item{9.} $127,263,527$. + +\Item{10.} $385,828,352$. + +\Item{11.} $1879.080904$. + +\Item{12.} $1838.265625$. +\end{multicols} + +Find to four decimal places the cube root of: +\begin{multicols}{4} +\Item{13.} $0.01$. + +\Item{14.} $0.05$. + +\Item{15.} $0.2$. + +\Item{16.} $4$. + +\Item{17.} $10$. + +\Item{18.} $87$. + +\Item{19.} $2.5$. + +\Item{20.} $2.05$. + +\Item{21.} $3.02$. + +\Item{22.} $\frac{2}{3}$. + +\Item{23.} $\frac{3}{4}$. + +\Item{24.} $\frac{9}{11}$. +\end{multicols} +%% -----File: 170.png---Folio 164------- +%[Blank Page] +%% -----File: 171.png---Folio 165------- + + +\Answers + +\AnsTo[7]{Exercise}{1.} % Page 10. + +\Item{1.} $14$. + +\Item{2.} $10$. + +\Item{3.} $13$. + +\Item{4.} $11$. + +\Item{5.} $13$. + +\Item{6.} $7$. + +\Item{7.} $9$. + +\Item{8.} $7$. + +\Item{9.} $6$. + +\Item{10.} $2$. + +\Item{11.} $3$. + +\Item{12.} $6$. + +\Item{13.} $2$. + +\Item{14.} $8$. + +\Item{15.} $4$. + +\Item{16.} $3$. + +\Item{17.} $1$. + +\Item{18.} $1$. + +\Item{19.} $3$. + +\Item{20.} $4$. + +\Item{21.} $10$. + + +\AnsTo[7]{Exercise}{2.} % Page 12. + +\Item{1.} $91$. + +\Item{2.} $21$. + +\Item{3.} $60$. + +\Item{4.} $24$. + +\Item{5.} $96$. + +\Item{6.} $16$. + +\Item{7.} $36$. + +\ResetCols[3] +\Item{8.} $4a + 4b$. + +\Item{9.} $4a - 4b$. + +\Item{10.} $2a^{2} + 2b^{2}$. + +\Item{11.} $2a^{2} - 2b^{2}$. + +\Item{12.} $3ab + 3c$. + +\Item{13.} $3ab - 3c$. + +\Item{14.} $3c - 3ab$. + +\Item{15.} $ab + ac$. + +\Item{16.} $ab - ac$. + +\Item{17.} $3ab + 3ac$. + +\Item{18.} $3ab - 3ac$. + +\Item{19.} $5ab^{2} + 5ac$. + +\Item{20.} $5ab^{2} - 5ac^{2}$. + +\Item{21.} $5a^{2}b^{2} - 5a^{2}c$. + + +\AnsTo[5]{Exercise}{3.} % Page 12. + +\Item{1.} $63$. + +\Item{2.} $280$. + +\Item{3.} $300$. + +\Item{4.} $98$. + +\Item{5.} $81$. + +\Item{6.} $1250$. + +\Item{7.} $105$. + +\Item{8.} $105$. + +\Item{9.} $315$. + +\Item{10.} $35$. + +\Item{11.} $105$. + +\Item{12.} $105$. + +\Item{13.} $0$. + +\Item{14.} $135$. + +\Item{15.} $120$. + +\Item{16.} $0$. + +\Item{17.} $1800$. + +\Item{18.} $540$. + +\Item{19.} $0$. + +\Item{20.} $270$. + +\Item{21.} $540$. + + +\AnsTo[6]{Exercise}{4.} % Page 13. + +\Item{1.} $21$. + +\Item{2.} $26$. + +\Item{3.} $72$. + +\Item{4.} $85$. + +\Item{5.} $30$. + +\Item{6.} $17$. + +\Item{7.} $8$. + +\Item{8.} $50$. + +\Item{9.} $24$. + +\Item{10.} $0$. + +\Item{11.} $12$. + +\Item{12.} $100$. + +\Item{13.} $80$. + +\Item{14.} $71$. + +\Item{15.} $139$. + +\Item{16.} $17$. + +\Item{17.} $8$. + +\Item{18.} $5$. + +\Item{19.} $3$. + +\Item{20.} $6$. + +\Item{21.} $5$. + +\Item{22.} $1$. + +\Item{23.} $2$. + +\Item{24.} $2$. + + +\AnsTo{Exercise}{5.} % Page 14. + +\Item{1.} $a$~plus~$b$; $a$~minus~$b$; $a$~times~$b$; $a$~divided by~$b$. + +\Item{3.} $a + b$. + +\Item{5.} $a - b$. + +\Item{7.} $x - y$. + +\Item{9.} $4x$; $x^{4}$. + +\Item{11.} $35 - x$. + +\Item{12.} $x - a$. + +\Item{14.} $14 - x$. + +\Item{15.} $a - x$. + +\Item{17.} $xy$. + +\Item{18.} $\dfrac{x}{y}$. +%% -----File: 172.png---Folio 166------- + + +\AnsTo[2]{Exercise}{6.} % Page 15. + +\Item{2.} $\dfrac{a}{b}$. + +\Item{4.} $(x - 3)$~yr.; $(x + 7)$~yr. + +\Item{6.} $7(2x - y)$. + +\Item{8.} $x + 1$; $x - 1$. + +\Item{9.} $20 - d$. + +\Item{11.} $x + 8$. + +\Item{13.} $x - 10$. + +\Item{14.} $10$. + + +\AnsTo[2]{Exercise}{7.} % Page 16. + +\Item{1.} $(40 - x)$~yr. + +\Item{2.} $(a + y)$~yr. + +\Item{3.} $4$. + +\Item{5.} $ab$. + +\Item{6.} $5x - 3x$. + +\Item{8.} $2x - 3 - (x + 1)$. + +\Item{9.} $40$. + +\Item{10.} $12$. + +\Item{11.} $100a + 25b + 10c$. + +\Item{12.} $100 - x - y$. + +\Item{14.} $xy + c$. + + +\AnsTo[2]{Exercise}{8.} % Page 17. + +\Item{2.} $xy - a^{2}$. + +\Item{3.} $\dfrac{ph}{gk}$. + +\Item{4.} $6m^{2} + 5c(d + b - a)$. + +\Item{5.} $5(2n + 1) - 6(c - a + b)$. + +\Item{6.} \$$100 - \text{\$}(a + b + 2c)$. + +\Item{8.} $\dfrac{1}{x}$. + +\Item{9.} $\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}$. + +\ResetCols[1] +\Item{10.} $n + (n + 1) + (n + 2)$; $n(n + 1)(n + 2)$. + +\ResetCols[4] +\Item{11.} $\dfrac{36}{x}$. + +\Item{12.} $qd$. + +\Item{13.} $qd + r$. + +\Item{14.} $2x$. + +\Item{15.} $\dfrac{10x}{12}$. + +\Item{16.} $\dfrac{6a}{b}$. + +\Item{17.} $4m$. + +\Item{18.} $x + 17$. + +\Item{19.} $x + y + c$. + +\Item{20.} $36x$. + +\Item{21.} $\dfrac{c}{4}$. + +\Item{22.} $x - 1$, $x$, $x + 1$. + +\Item{23.} $2n + 3$. + + +\AnsTo[5]{Exercise}{9.} % Page 24. + +\Item{1.} $4$. + +\Item{2.} $5$. + +\Item{3.} $3$. + +\Item{4.} $1$. + +\Item{5.} $13$. + +\Item{6.} $0$. + +\Item{7.} $5$. + +\Item{8.} $5$. + +\Item{9.} $7$. + +\Item{10.} $4$. + +\Item{11.} $4$. + +\Item{12.} $1$. + +\Item{13.} $25$. + +\Item{14.} $32\frac{1}{2}$. + +\Item{15.} $6$. + +\Item{16.} $10$. + +\Item{17.} $3$. + +\Item{18.} $1$. + +\Item{19.} $7$. + +\Item{20.} $13$. + +\Item{21.} $12$. + +\Item{22.} $14$. + +\Item{23.} $7$. + +\Item{24.} $5$. + +\Item{25.} $0$. + +\Item{26.} $15$. + +\Item{27.} $10\frac{1}{2}$. + +\Item{28.} $4$. + +\Item{29.} $3$. + +\Item{30.} $4$. + +\Item{31.} $\frac{2}{5}$. + +\Item{32.} $6$. + +\Item{33.} $9$. + +\Item{34.} $7$. + +\Item{35.} $5$. + +\Item{36.} $6$. + +\Item{37.} $6$. + + +\AnsTo[2]{Exercise}{10.} % Page 27. + +\Item{1.} $30$. + +\Item{2.} Son, $10$~yr.; father, $50$~yr. + +\Item{3.} $78$, $13$. + +\Item{4.} $80$~ft.\ broken off; $10$~ft.\ standing. + +\Item{5.} $23$, $30$. + +\Item{6.} $36$, $48$. + +\Item{7.} $15$, $20$. + +\Item{8.} $20$. + +\Item{9.} $12$. + +\Item{10.} $10$, $40$. + +\Item{11.} $26$, $10$. + +\Item{12.} $25$, $15$. + +\Item{13.} $10$, $20$. + +\Item{14.} $7$, $20$. + +\Item{15.} $12$, $20$. +%% -----File: 173.png---Folio 167------- + + +\AnsTo{Exercise}{11.} % Page 29. + +\Item{1.} Cow, \$$42$; horse, \$$168$. + +\Item{2.} $81$. + +\Item{3.} $2$. + +\Item{4.} $30$, $40$. + +\Item{5.} $25$, $26$, $27$. + +\Item{6.} $5$, $6$, $7$, $8$, $9$. + +\Item{7.} A, $30$~yr.; B, $10$~yr. + +\Item{8.} Father, $40$~yr., son, $10$~yr. + +\Item{9.} $40$. + +\Item{10.} $10$. + +\Item{11.} \$$40$. + +\Item{12.} $9$. + + +\AnsTo[2]{Exercise}{12.} % Page 30. + +\Item{1.} $15$~men; $30$~women; $45$~children. + +\Item{2.} $50$. + +\Item{3.} $16$. + +\Item{4.} $7$. + +\Item{5.} $35$. + +\Item{6.} $24$. + +\Item{7.} $24$. + +\Item{8.} $20$. + +\Item{9.} $970$; $1074$. + + +\AnsTo[2]{Exercise}{13.} % Page 31. + +\Item{1.} $24$. + +\Item{2.} A, \$$60$; B, \$$30$. + +\Item{3.} \$$3$~quarters; $6$~bills. + +\Item{4.} $14$. + +\Item{5.} \$$4$~quarters; $20$~half-dollars. + +\Item{6.} \$$7$~ten-dollar bills; $21$~one-dollar bills. + +\Item{7.} Father, $32$~yr.; son, $8$~yr. + +\Item{8.} $20$. + + +\AnsTo[2]{Exercise}{14.} % Page 32. + +\Item{1.} $9$~miles. + +\Item{2.} \$$60$. + +\Item{3.} \$$20$~lb.\ at $65$~cts.; + $60$~lb.\ at $45$~cts. + +\Item{4.} $40$. + +\Item{5.} $15,000$. + +\Item{6.} $15$~in.; $21$~in. + +\Item{7.} \$$2$~doz.\ at $25$~cts.; + $5$~doz.\ at $20$~cts. + +\Item{8.} \$$6$~quarters, $18$~ten-cent pieces. + + +\AnsTo[4]{Exercise}{15.} % Page 39. + +\Item{1.} $40c$. + +\Item{2.} $24a$. + +\Item{3.} $39x$. + +\Item{4.} $51y$. + +\Item{5.} $-26a$. + +\Item{6.} $-40x$. + +\Item{7.} $-17b$. + +\Item{8.} $-66z$. + +\Item{9.} $-20m$. + +\Item{10.} $2d$. + +\Item{11.} $0$. + +\Item{12.} $-18g$. + +\Item{13.} $a^{2}$. + +\Item{14.} $-21x^{3}$. + +\Item{15.} $0$. + +\Item{16.} $3mn$. + +\Item{17.} $0$. + +\Item{18.} $-3a^{3}b^{3}c^{3}$. + +\Item{19.} $-9abcd$. + +\Item{20.} $1$. + +\Item{21.} $12$. + +\Item{22.} $4$. + +\Item{23.} $-18$. + +\Item{24.} $10$. +%% -----File: 174.png---Folio 168------- + + +\AnsTo{Exercise}{16.} % Page 42. + +\Item{1.} $30a^{5}$. + +\Item{2.} $40a^{4}b^{3}$. + +\Item{3.} $63x^{2}y^{2}$. + +\Item{4.} $2a^{5}b^{5}c^{2}$. + +\Item{5.} $9a^{7}b^{9}c^{9}$. + +\Item{6.} $-10a^{2}$. + +\Item{7.} $12ab$. + +\Item{8.} $-a^{4}b^{3}$. + +\Item{9.} $10a^{5}b^{5}c$. + +\Item{10.} $12x^{7}y^{5}z^{2}$. + +\Item{11.} $105a^{7}b^{5}$. + +\Item{12.} $6a^{6}b^{5}c^{7}$. + +\Item{13.} $-12a^{5}b^{5}c^{5}x^{5}$. + +\Item{14.} $24a^{7}b^{6}c^{5}$. + +\Item{15.} $-42a^{6}m^{5}x^{7}$. + +\Item{16.} $30x^{8}y^{4}z^{6}$. + +\Item{17.} $-46$. + +\Item{18.} $-3$. + +\Item{19.} $-8$. + +\Item{20.} $-17$. + +\Item{21.} $9$. + +\Item{22.} $12$. + +\Item{23.} $-102$. + +\Item{24.} $-41$. + +\Item{25.} $174$. + +\Item{26.} $6$. + +\Item{27.} $30$. + +\Item{28.} $372$. + + +\AnsTo[4]{Exercise}{17.} % Page 45. + +\Item{1.} $x^{2}$. + +\Item{2.} $3x^{2}$. + +\Item{3.} $-7$. + +\Item{4.} $-7$. + +\Item{5.} $7x^{4}$. + +\Item{6.} $9x$. + +\Item{7.} $-4a$. + +\Item{8.} $4x^{2}y^{2}$. + +\Item{9.} $-9x$. + +\Item{10.} $5x$. + +\Item{11.} $3y^{2}$. + +\Item{12.} $-4ab^{2}$. + +\Item{13.} $12xy^{4}$. + +\Item{14.} $\dfrac{3x^{3}y^{3}}{5}$. + +\Item{15.} $-\dfrac{3a}{2}$. + +\Item{16.} $-bd$. + +\Item{17.} $acd^{2}$. + +\Item{18.} $-\dfrac{2xy}{3}$. + +\Item{19.} $5ab$. + +\Item{20.} $4mn$. + +\Item{21.} $\dfrac{yz^{3}}{3}$. + +\Item{22.} $-17cd$. + +\Item{23.} $2n^{2}p$. + +\Item{24.} $\dfrac{3r^{2}}{p}$. + +\Item{25.} $-13agt$. + +\Item{26.} $\dfrac{1}{abc}$. + +\Item{27.} $\dfrac{3}{2xy^{2}z^{3}}$. + +\Item{28.} $\dfrac{2}{mnp}$. + +\Item{29.} $-\dfrac{a}{3b^{2}}$. + +\Item{30.} $-\dfrac{g}{3mt}$. + + +\AnsTo[2]{Exercise}{18.} % Page 48. + +\Item{1.} $2a^{2} + 2b^{2}$. + +\Item{2.} $9a^{2} - 2a + 6$. + +\Item{3.} $0$. + +\Item{4.} $4x + 4y + 4z$. + +\Item{5.} $2b + 2c$. + +\Item{6.} $4a + 4b + 4c$. + +\Item{7.} $3a^{2} + 5a - 2$. + +\Item{8.} $8ab + 3ac$. + +\Item{9.} $6x^{3}$. + +\Item{10.} $5x^{2} + 3xy - 2y^{2}$. + +\Item{11.} $a^{2} - 2b^{2}$. + +\Item{12.} $4a^{3} + 6a + 2$. + +\Item{13.} $3m^{3} + 7m^{2} + 2$. + +\Item{14.} $2x^{3} - 2x^{2} + 4x + y$. + +\Item{15.} $7x^{3} + 7x^{2} + 2$. + +\Item{16.} $a^{3} + 3a^{2}b - 5ab^{2} - b^{3}$. + +\Item{17.} $-a^{3} - a^{2}b - 2ab^{2} - 2b^{3}$. + +\Item{18.} $2x^{3} + 2x^{2}y - xy^{2} + 6y^{3}$. +%% -----File: 175.png---Folio 169------- + + +\AnsTo[2]{Exercise}{19.} % Page 50. + +\Item{1.} $a - b + c$. + +\Item{2.} $2a - 2b + 6c$. + +\Item{3.} $2a + 3y -8z$. + +\Item{4.} $x + 4y + 5z$. + +\Item{5.} $2ac + 2bc$. + +\Item{6.} $2ab - 3ac + 4bc$. + +\Item{7.} $x^{3} + 3x^{2} + 2x - 8$. + +\Item{8.} $x^{3} - 7x^{2} + 4x$. + +\Item{9.} $2b^{3} + 18abc - 15c^{3}$. + +\Item{10.} $2 - x - 2x^{2} + 2x^{3}$. + +\Item{11.} $-3b^{3} - 4c^{3} + 6abc$. + +\Item{12.} $-x^{4} - 2x^{3} + 4x^{2} - 7x + 5$. + +\Item{13.} $2x + x^{2} + x^{3} + x^{5}$. + +\Item{14.} $2b^{3} - 4a^{2}b + 2ab^{2}$. + +\Item{15.} $2b^{4} - 2a^{3}b^{3} - ab^{2}$. + +\Item{16.} $4x^{3} - 3x^{2}y - 4xy^{2} + 7y^{3}$. + + +\AnsTo{Exercise}{20.} % Page 52. + +\Item{1.} $c$. + +\Item{2.} $y - b$. + +\Item{3.} $x - 3y - 7c$. + +\Item{4.} $7a + 2b-2$. + +\Item{5.} $2a + x$. + +\Item{6.} $13x - 15y + 13z$. + +\Item{7.} $2a - 2b + 2c$. + +\Item{8.} $5a + b - 4c$. + +\Item{9.} $4x - 5y + 2z$. + +\Item{10.} $3a - b$. + +\Item{11.} $2x + 3y + z$. + +\Item{12.} $-8x + y$. + +\Item{13.} $6z - 2y-z$. +\ResetCols + +\Item{15.} $(a + c)x - (a - b)y + (a - c)z$. + +\Item{16.} $(2a + 5c)x - (3a + 4b)y - (6b + 7c)z$. + +\Item{17.} $(ac - an)x - (bm + cn)y + (a + 3c)z$. + +\Item{18.} $(mn - 1)x - (mn + 1)y + (mn + 1)z$. + + +\AnsTo[2]{Exercise}{21.} % Page 53. + +\Item{1.} $x^{2} + 7x$. + +\Item{2.} $8x^{2} - 12xy$. + +\Item{3.} $14xy - 21y^{2}$. + +\Item{4.} $2ax - 4a^{2}$. + +\Item{5.} $bx - 3b^{2}$. + +\Item{6.} $-6a^{3} + 9a^{2}b$. + +\Item{7.} $10x^{2}z + 15xz^{2}$. + +\Item{8.} $5a^{3}b - 25a^{2}b^{2}$. + +\Item{9.} $-x^{2}y^{2} + 3xy^{3}$. + +\Item{10.} $4x^{5} - 6x^{4}$. + +\Item{11.} $4x^{2}y - 12y^{3}$. + +\Item{12.} $-x^{4} + 3x^{2}y^{2}$. + +\Item{13.} $-a^{3}b^{3} + a^{5}b^{2}$. + +\Item{14.} $a^{4}b^{2} + a^{5}$. + +\Item{15.} $4x^{5} - 6x^{4} + 2x^{3}$. + +\Item{16.} $5a^{3}b - 25a^{2}b^{2} - 5ab^{3}$. + +\Item{17.} $a^{5} + 2a^{4}b + 2a^{3}b^{2}$. + +\Item{18.} $a^{3}b^{3} + 2a^{2}b^{4} + 2ab^{5}$. + +\Item{19.} $8x^{3} - 12x^{2}y - 18xy^{2}$. + +\Item{20.} $x^{2}y + 2xy^{2} - y^{3}$. + +\Item{21.} $a^{5} + a^{4}b^{2} + a^{2}b^{3}$. + +\Item{22.} $x^{2}y^{2} - 2xy^{3} + y^{4}$. + +\Item{23.} $15a^{4}b^{4} - 20a^{3}b^{5} + 5a^{5}b^{3}$. + +\Item{24.} $3a^{2}x^{2}y^{2} - 9a^{2}xy^{4} + 3a^{2}y^{6}$. + +\Item{25.} $x^{15}y^{2} - x^{13}y^{5} - x^{6}y^{12}$. + +\Item{26.} $4x^{5}y^{3} - 6x^{4}y^{5} + 4x^{3}y^{6}$. + +\Item{27.} $a^{10}x^{5}y^{10} - a^{9}x^{4}y^{9} - a^{8}x^{3}y^{8}$. + +\Item{28.} $15a^{4}b^{5} - 10a^{3}b^{6} + 25a^{5}b^{4}$. +%% -----File: 176.png---Folio 170------- + + +\AnsTo[2]{Exercise}{22.} % Page 56. + +\Item{1.} $x^{2} + 13x + 42$. + +\Item{2.} $x^{2} - x - 42$. + +\Item{3.} $x^{2} + x - 42$. + +\Item{4.} $x^{2} - 13x + 42$. + +\Item{5.} $x^{2} + 3x - 40$. + +\Item{6.} $4x^{2} + 12x + 9$. + +\Item{7.} $4x^{2} - 12x + 9$. + +\Item{8.} $4x^{2} - 9$. + +\Item{9.} $-9x^{2} + 12x - 4$. + +\Item{10.} $20x^{2} - 47x + 21$. + +\Item{11.} $a^{2} + ab - 6b^{2}$. + +\Item{12.} $a^{2} - 12ab + 35b^{2}$. + +\Item{13.} $25x^{2} - 30xy + 9y^{2}$. + +\Item{14.} $x^{2} - bx - cx + bc$. + +\Item{15.} $8m^{2} - 10mp + 3p^{2}$. + +\Item{16.} $a^{2} + ab - bc - c^{2}$. + +\Item{17.} $a^{4} - a^{3}b + 2a^{2}b^{2} - ab^{3} + b^{4}$. + +\Item{18.} $x^{5} - 3x^{4} - x^{3} + 16x^{2} - 21$. + +\Item{19.} $a^{3} - b^{3}$. + +\Item{20.} $a^{3} + b^{3}$. + +\Item{21.} $2x^{4} + 13x^{3} - 9x^{2} - 50x + 40$. + +\Item{22.} $9x^{5} - 7x^{3} + 6x^{2} - 2x$. + +\Item{23.} $x^{5} - 4x^{2}y^{3} + 3xy^{4}$. + +\Item{24.} $-a^{4} + 4a^{3}b - 7ab^{3} - 2b^{4}$. +\ResetCols[1] + +\Item{25.} $-25a^{5}b^{3} + 20a^{4}b^{4} + 12a^{3}b^{5} - 5a^{2}b^{6} - 2ab^{7}$. + +\Item{26.} $a^{4} - 2a^{2}b^{2} + b^{4}$. + +\Item{27.} $a^{2}b^{2} + 2abcd - a^{2}c^{2} + c^{2}d^{2}$. + +\Item{28.} $-2x^{5}y^{3} + x^{4}y^{4} + 10x^{3}y^{5} - 8x^{2}y^{6} - 3xy^{7}$. + +\Item{29.} $x^{4} - 4x^{2}y^{2} + 4xy^{3} - y^{4}$. + +\Item{30.} $3x^{4} - 5x^{3}y - 12x^{2}y^{2} - xy^{3} + 3y^{4}$. + +\Item{31.} $-a^{4} + 6a^{2}b^{2} - b^{4}$. + +\Item{32.} $a^{4} - a^{3}c + ab^{2}c - b^{4} - 2b^{2}c^{2} + ac^{3} - c^{4}$. + +\Item{33.} $a^{4} - 16a^{2}b^{2}x^{2} + 32a^{3}b^{3}x^{3} - 16a^{4}b^{4}x^{4}$. + +\Item{34.} $6a^{4} + 5a^{3}bx - 10a^{2}b^{2}x^{2} + 7ab^{3}x^{3} - 2b^{4}x^{4}$. + +\Item{35.} $10x^{6}y^{2} + 14x^{5}y^{3} - 48x^{4}y^{4} + 32x^{3}y^{5} - 16x^{2}y^{6}$. + + +\AnsTo{Exercise}{23.} % Page 58. + +\Item{1.} $2a^{2} - a$. + +\Item{2.} $7a^{4} - a$. + +\Item{3.} $7x^{2} + 1$. + +\Item{4.} $5m^{4} - p^{2}$. + +\Item{5.} $3x^{3} - 5x^{2}$. + +\Item{6.} $-3x^{3} + 1$. + +\Item{7.} $2x^{2} - 3x$. + +\Item{8.} $-x^{2} + 2$. + +\Item{9.} $a + 2c$. + +\Item{10.} $5x - y$. + +\Item{11.} $ax - 1$. + +\Item{12.} $x + xy$. + +\Item{13.} $-3a + 4b - 2c$. + +\Item{14.} $ab - b^{4} - a^{2}b$. + +\Item{15.} $x^{2} - 2xy - 3y^{2}$. + +\Item{16.} $xy - x^{2} - y^{2}$. + +\Item{17.} $-a^{2} + ab + b^{2}$. + +\Item{18.} $-a + 1 - b$. + +\Item{19.} $-1 + xy - x^{2}y^{2}$. + +\Item{20.} $x^{2} + 2x + 1$. + +\Item{21.} $a - b - c$. + +\Item{22.} $x^{3} - x^{2}y - y^{2}$. + +\Item{23.} $ab - 2 - 3b^{2}$. + +\Item{24.} $a^{2}c^{2} + a - c$. +%% -----File: 177.png---Folio 171------- + + +\AnsTo{Exercise}{24.} % Page 62. + +\Item{1.} $x + 8$. + +\Item{2.} $x - 8$. + +\Item{3.} $x + 8$. + +\Item{4.} $x - 8$. + +\Item{5.} $a + 5$. + +\Item{6.} $3a + 1$. + +\Item{7.} $a + 5$. + +\Item{8.} $-3a - 2$. + +\Item{9.} $x^{2} - x + 1$. + +\Item{10.} $x^{4} + x^{2} + 1$. + +\Item{11.} $1 + ab + a^{2}b^{2}$. + +\Item{12.} $x^{2} + 3x + 1$. + +\Item{13.} $a - b + c$. + +\Item{14.} $a + b - c$. + +\Item{15.} $x + y - z$. + +\Item{16.} $c^{2} + c + 2$. + +\Item{17.} $x - 2y - z$. + +\Item{18.} $x - a$. + +\Item{19.} $a - 2b + 3c$. + +\Item{20.} $a^{2} + 5a + 6$. + +\Item{21.} $q^{2} + 3q + 2$. + +\Item{22.} $9a^{2} + 6ab + 4b^{2}$. + +\Item{23.} $-65$. + +\Item{24.} $10$. + +\Item{25.} $7a - 45$. + +\Item{26.} $2x^{4}$. + + +\AnsTo[1]{Exercise}{25.} % Page 63. + +\Item{1.} $2a^{2}$. + +\Item{2.} $-3a^{4} + 2a^{3}b - 2ab^{3} + 4b^{4}$. + +\Item{3.} $x$. + +\Item{4.} $a^{4} + 2a^{2}b^{2} + b^{4} - c^{4} + 2c^{2}d^{2} - d^{4}$. + +\Item{5.} $10y^{4} + 8y^{3} + 6y^{2} + 4y + 2$. + +\Item{6.} $0$. + +\Item{7.} $2z - 7y$. + +\Item{8.} $a^{3} - 3abc + b^{3} + c^{3}$. + +\Item{9.} $4y^{2} - 3xy + 2x^{2}$. + +\Item{10.} $5a^{3}b - b^{4}$. + +\Item{11.} $3x^{3} - 2x^{2} + 1$. + +\Item{12.} $3c^{2} + 24c - 12$. + +\Item{13.} $2b^{4}$. + +\Item{14.} $10 - 16x - 39x^{2} + 2x^{3} + 15x^{4}$. + +\Item{15.} $a^{4} - ax^{3} + x^{4}$. + +\Item{16.} $(a - b)x^{3} + (b + c)x^{2} - (c + 1)x$. + +\Item{17.} $(a + b)x^{4} - (a - b)x^{3} - (c + 2)x$. + +\Item{18.} $(a + 1)x^{3} - (b + c)x^{2} + (b - c)x$. + + +\AnsTo[2]{Exercise}{26.} % Page 65. + +\Item{1.} $m^{2} + 2mn + n^{2}$. + +\Item{2.} $c^{2} - 2ac + a^{2}$. + +\Item{3.} $a^{2} + 4ac + 4c^{2}$ + +\Item{4.} $9a^{2} - 12ab + 4b^{2}$. + +\Item{5.} $4a^{2} + 12ab + 9b^{2}$. + +\Item{6.} $a^{2} - 6ab + 9b^{2}$. + +\Item{7.} $4x^{2} - 4xy + y^{2}$. + +\Item{8.} $y^{2} - 4xy + 4x^{2}$. + +\Item{9.} $a^{2} + 10ab + 25b^{2}$. + +\Item{10.} $4a2 - 20ac + 25c^{2}$. + +\Item{11.} $x^{2} - y^{2}$. + +\Item{12.} $16a^{2} - b^{2}$. + +\Item{13.} $4b^{2} - 9c^{2}$. + +\Item{14.} $x^{2} + 10bx + 25b^{2}$. + +\Item{15.} $y^{2} - 4yz + 4z^{2}$. + +\Item{16.} $y^{2} - 9z^{2}$. + +\Item{17.} $4a^{2} - 9b^{2}$. + +\Item{18.} $4a^{2} - 12ab + 9b^{2}$. + +\Item{19.} $4a^{2} + 12ab + 9b^{2}$. + +\Item{20.} $25x^{2} - 9a^{2}$. +%% -----File: 178.png---Folio 172------- + + +\AnsTo[2]{Exercise}{27.} % Page 67. + +\Item{1.} $x^{2} + 11x + 28$. + +\Item{2.} $x^{2} + 4x - 21$. + +\Item{3.} $x^{2} - 6x + 8$. + +\Item{4.} $x^{2} - 16x + 60$. + +\Item{5.} $x^{2} + 3x - 28$. + +\Item{6.} $x^{2} - ax - 2a^{2}$. + +\Item{7.} $x^{2} + 2ax - 3a^{2}$. + +\Item{8.} $a^{2} + 6ac + 9c^{2}$. + +\Item{9.} $a^{2} - 2ax - 8x^{2}$. + +\Item{10.} $a^{2} - 7ab + 12b^{2}$. + +\Item{11.} $a^{4} + a^{2}c - 2c^{2}$. + +\Item{12.} $x^{2} - 20x + 51$. + +\Item{13.} $x^{2} + xy - 30y^{2}$. + +\Item{14.} $9 + 3x - 2x^{2}$. + +\Item{15.} $5 - 8x - 4x^{2}$. + +\Item{16.} $a^{2} + ab - 6b^{2}$. + +\Item{17.} $a^{4}b^{4} - 6a^{2}b^{2}x^{2} + 5x^{4}$. + +\Item{18.} $a^{6}b^{2} + 4a^{4}b^{4} - 5a^{2}b^{6}$. + +\Item{19.} $x^{4}y^{2} - 4x^{3}y^{3} + 3x^{2}y^{4}$. + +\Item{20.} $x^{4}y^{2} + 2x^{3}y^{3} + x^{2}y^{4}$. + +\Item{21.} $x^{2} + (a + b)x + ab$. + +\Item{22.} $x^{2} + (a - b)x - ab$. + +\Item{23.} $x^{2} - (a - b)x - ab$. + +\Item{24.} $x^{2} - (a + b)x + ab$. + +\Item{25.} $x^{2} + (2a + 2b)x + 4ab$. + +\Item{26.} $x^{2} - (2a - 2b)x - 4ab$. + +\Item{27.} $x^{2} + (2a - 2b)x - 4ab$. + +\Item{28.} $x^{2} - (2a + 2b)x + 4ab$. + +\Item{29.} $x^{2} + 2ax - 3a^{2}$. + +\Item{30.} $x^{2} + ax - 6a^{2}$. + + +\AnsTo{Exercise}{28.} % Page 68. + +\Item{1.} $x + 2$. + +\Item{2.} $x - 2$. + +\Item{3.} $a + 3$. + +\Item{4.} $a - 3$. + +\Item{5.} $c + 5$. + +\Item{6.} $c - 5$. + +\Item{7.} $7x + y$. + +\Item{8.} $7x - y$. + +\Item{9.} $3b + 1$. + +\Item{10.} $3b - 1$. + +\Item{11.} $4x^{2} + 5a$. + +\Item{12.} $4x^{2} - 5a$. + +\Item{13.} $3x + 5y$. + +\Item{14.} $a + b - c$. + +\Item{15.} $a - b + c$. + +\Item{16.} $a + 2b - c$. + +\Item{17.} $5a - 7b + 1$. + +\Item{18.} $5a - 7b - 1$. + +\Item{19.} $z + x - y$. + +\Item{20.} $z - x + y$. + +\Item{21.} $a - 2b + c$. + +\Item{22.} $x + 3y + z$. + +\Item{23.} $x + 3y - z$. + +\Item{24.} $a + 2b + 2c$. + +\Item{25.} $a + 2b - 2c$. + +\Item{26.} $1 - 3x + 2y$. + + +\AnsTo[2]{Exercise}{29.} % Page 69. + +\Item{1.} $1 + x + x^{2}$. + +\Item{2.} $1 + 2a + 4a^{2}$. + +\Item{3.} $1 + 3c + 9c^{2}$. + +\Item{4.} $4a^{2} + 2ab + b^{2}$. + +\Item{5.} $16b^{2} + 12bc + 9c^{2}$. + +\Item{6.} $9x^{2} + 6xy + 4y^{2}$. + +\Item{7.} $x^{2}y^{2} + xyz + z^{2}$. + +\Item{8.} $a^{2}b^{2} + 2ab + 4$. + +\Item{9.} $25a^{2} + 5ab + b^{2}$. + +\Item{10.} $a^{2} + 2ab + 4b^{2}$. + +\Item{11.} $a^{2} + 4a + 16$. + +\Item{12.} $a^{6} + 3a^{3} + 9$. + +\Item{13.} $a^{8} + a^{4}x^{2}y^{2} + x^{4}y^{4}$. + +\Item{14.} $x^{10} + x^{5}a^{3}b^{3} + a^{6}b^{6}$. + +\Item{15.} $9x^{2}y^{2} + 3xyz^{4} + z^{8}$. + +\Item{16.} $x^{2}y^{2}z^{2} + xyz + 1$. + +\Item{17.} $4a^{2}b^{2}c^{2} - 6abc + 9$. + +\Item{18.} $1 + 4xyz + 16x^{2}y^{2}z^{2}$. +%% -----File: 179.png---Folio 173------- + + +\AnsTo[2]{Exercise}{30.} % Page 70. + +\Item{1.} $1 - x + x^{2}$. + +\Item{2.} $1 - 2a + 4a^{2}$. + +\Item{3.} $1 - 3c + 9c^{2}$. + +\Item{4.} $4a^{2} - 2ab + b^{2}$. + +\Item{5.} $16b^{2} - 12bc + 9c^{2}$. + +\Item{6.} $9x^{2} - 6xy + 4y^{2}$. + +\Item{7.} $4x^{2} - 10xy + 25y^{2}$. + +\Item{8.} $x^{2}y^{2} - xyz + z^{2}$. + +\Item{9.} $a^{2}b^{2} - 2ab + 4$. + +\Item{10.} $25a^{2} - 5ab + b^{2}$. + +\Item{11.} $a^{2} - 2ab + 4b^{2}$. + +\Item{12.} $a^{4} - 4a^{2} + 16$. + +\Item{13.} $a^{6} - 3a^{3} + 9$. + +\Item{14.} $4a^{4} - 2a^{2}b + b^{2}$. + +\Item{15.} $a^{8} - a^{4}x^{2}y^{2} + x^{4}y^{4}$. + +\Item{16.} $x^{10} - x^{5}a^{3}b^{3} + a^{6}b^{6}$. + +\Item{17.} $9x^{2}y^{2} - 3xyz^{4} + z^{8}$. + +\Item{18.} $x^{2}y^{2}z^{2} - xyz + 1$. + +\Item{19.} $4a^{2}b^{2}c^{2} - 6abc + 9$. + +\Item{20.} $1 - 4xyz + 16x^{2}y^{2}z^{2}$. + +\Item{21.} $1 - 3a^{2}bc + 9a^{4}b^{2}c^{2}$. + +\Item{22.} $x^{3} + x^{2}y + xy^{2} + y^{3}$. + +\Item{23.} $x^{3} - x^{2}y + xy^{2} - y^{3}$. + +\Item{24.} $x^{4} + x^{3}y + x^{2}y^{2} + xy^{3} + y^{4}$. + +\Item{25.} $x^{4} - x^{3}y + x^{2}y^{2} - xy^{3} + y^{4}$. + +\Item{26.} $x^{5} + x^{4}y + x^{3}y^{2} + x^{2}y^{3} + xy^{4} + y^{5}$. + +\Item{27.} $x^{5} - x^{4}y + x^{3}y^{2} - x^{2}y^{3} + xy^{4} - y^{5}$. + + +\AnsTo[2]{Exercise}{31.} % Page 72. + +\Item{1.} $2x(x - 2)$. + +\Item{2.} $3a(a^{2} - 2)$. + +\Item{3.} $5a^{2}b^{2}(1 - 2ab)$. + +\Item{4.} $xy(3x + 4y)$. + +\Item{5.} $4a^{2}b^{2}(2a + b)$. + +\Item{6.} $3a^{2}(a^{2} - 4 - 2a)$. + +\Item{7.} $4x^{2}(1 - 2x^{2} - 3x^{3})$. + +\Item{8.} $5(1 - 2x^{2}y^{2} + 3x^{2}y)$. + +\Item{9.} $7a(a + 2 - 3a^{2})$. + +\Item{10.} $3x^{2}y^{2}(xy - 2x^{2}y^{2} - 3)$. + + +\AnsTo{Exercise}{32.} % Page 73. + +\Item{1.} $(x^{2} + 1)(x + 1)$. + +\Item{2.} $(x^{2} + 1)(x - 1)$. + +\Item{3.} $(x + y)(x + z)$. + +\Item{4.} $(x - y)(a - b)$. + +\Item{5.} $(a + b)(a - c)$. + +\Item{6.} $(x + 3)(x - b)$. + +\Item{7.} $(x^{2} + 2)(2x - 1)$. + +\Item{8.} $(a - b)(a - 3)$. + +\Item{9.} $(2a - c)(3a + b)$. + +\Item{10.} $(xy + c)(ab + c)$. + +\Item{11.} $(a - b - c)(x - y)$. + +\Item{12.} $(a - b - 2c)(a - b)$. + + +\AnsTo[2]{Exercise}{33.} % Page 74. + +\Item{1.} $(2 + x)(2 - x)$. + +\Item{2.} $(3 + x)(3 - x)$. + +\Item{3.} $(3a + x)(3a - x)$. + +\Item{4.} $(5 + x)(5 - x)$. + +\Item{5.} $(5x + a)(5x - a)$. + +\Item{6.} $(4a^{2} + 11)(4a^{2} - 11)$. + +\Item{7.} $(11a^{2} + 4)(11a^{2} - 4)$. + +\Item{8.} $(2ab + cd)(2ab - cd)$. + +\Item{9.} $(1 + xy)(1 - xy)$. + +\Item{10.} $(9xy + 1)(9xy - 1)$. + +\Item{11.} $(7ab + 2)(7ab - 2)$. + +\Item{12.} $(5a^{2}b^{2} + 3)(5a^{2}b^{2} - 3)$. + +\Item{13.} $(3a^{4}b^{3} + 4x^{5})(3a^{4}b^{3} - 4x^{5})$. + +\Item{14.} $(12xy + 1)(12xy - 1)$. + +\Item{15.} $(10x^{3}yz^{2} + 1)(10x^{3}yz^{2} - 1)$. + +\Item{16.} $(1 + 11a^{2}b^{4}c^{6})(1 - 11a^{2}b^{4}c^{6})$. + +\Item{17.} $(5a + 8x^{3}y^{3})(5a - 8x^{3}y^{3})$. + +\Item{18.} $(4x^{8} + 5y^{9})(4x^{8} - 5y^{9})$. + +\Item{19.} $90,000$. + +\Item{20.} $22,760$. + +\Item{21.} $732,200$. + +\Item{22.} $400$. + +\Item{23.} $28,972$. + +\Item{24.} $14,248,000$. +%% -----File: 180.png---Folio 174------- + + +\AnsTo[2]{Exercise}{34.} % Page 75. + +\Item{1.} $(x + y + z)(x + y - z)$. + +\Item{2.} $(x - y + z)(x - y - z)$. + +\Item{3.} $(z + x + y)(z - x - y)$. + +\Item{4.} $(z + x - y)(z - x + y)$. + +\Item{5.} $(x + y + 2z)(x + y - 2z)$. + +\Item{6.} $(2z + x - y)(2z - x + y)$. + +\Item{7.} $(a + 2b + c)(a + 2b - c)$. + +\Item{8.} $(a - 2b + c)(a - 2b - c)$. + +\Item{9.} $(c + a - 2b)(c - a + 2b)$. + +\Item{10.} $(2a + 5c + 1)(2a + 5c - 1)$. + +\Item{11.} $(1 + 2a - 5c)(1 - 2a + 5c)$. + +\Item{12.} $(a + 3b + 4c)(a + 3b - 4c)$. + +\Item{13.} $(a - 5b + 3c)(a - 5b - 3c)$. + +\Item{14.} $(4c + a - 5b)(4c - a + 5b)$. + +\Item{15.} $(2a + x + y)(2a - x - y)$. + +\Item{16.} $(b + a - 2x)(b - a + 2x)$. + +\Item{17.} $(2z + x + 3y)(2z - x - 3y)$. + +\Item{18.} $(3 + 3a - 7b)(3 - 3a + 7b)$. + +\Item{19.} $(4a + 2b + 5c)(4a - 2b - 5c)$. + +\Item{20.} $(5c + 3a - 2x)(4c - 3a + 2x)$. + +\Item{21.} $(3a + 3b -5c)(3a - 3b + 5c)$. + +\Item{22.} $(4y + a - 3c)(4y - a + 3c)$. + +\Item{23.} $(7m + p + 2q)(7m - p - 2q)$. + +\Item{24.} $(6n + d - 2c)(6n - d + 2c)$. + +\Item{25.} $(x + y + a + b)(x + y - a - b)$. + +\Item{26.} $(x - y + a - b)(x - y - a + b)$. + +\Item{27.} $(2x + 3 + 2a + b)(2x + 3 - 2a - b)$. + +\Item{28.} $(b - c + a - 2x)(b - c - a + 2x)$. + +\Item{29.} $(3x - y + 2a - b)(3x - y - 2a + b)$. + +\Item{30.} $(x - 3y + a + 2b)(x - 3y - a - 2b)$. + +\Item{31.} $(x + 2y + a + 3b)(x + 2y - a - 3b)$. + +\Item{32.} $(x + y + a - z)(x + y - a + z)$. + + +\AnsTo[2]{Exercise}{35.} % Page 77. + +\Item{1.} $(2x - y)(4x^{2} + 2xy + y^{2})$. + +\Item{2.} $(x - 1)(x^{2} + x + 1)$. + +\Item{3.} $(xy - z)(x^{2}y^{2} + xyz + z^{2})$. + +\Item{4.} $(x - 4)(x^{2} + 4x + 16)$. + +\Item{5.} $(5a - b)(25a^{2} + 5ab + b^{2})$. + +\Item{6.} $(a - 7)(a^{2} + 7a + 49)$. + +\Item{7.} $(ab - 3c)(a^{2}b^{2} + 3abc + 9c^{2})$. + +\Item{8.} $(xyz - 2)(x^{2}y^{2}z^{2} + 2xyz + 4)$. + +\Item{9.} $(2ab - 3y^{2})(4a^{2}b^{2} + 6aby^{2} + 9y^{4})$. + +\Item{10.} $(4x - y^{3})(16x^{2} + 4xy^{3} + y^{6})$. + +\Item{11.} $(3a - 4c^{2})(9a^{2} + 12ac^{2} + 16c^{4})$. + +\Item{12.} $(xy - 6z)(x^{2}y^{2} + 6xyz + 36z^{2})$. + +\Item{13.} $(4x - 9y)(16x^{2} + 36xy + 81y^{2})$. + +\Item{14.} $(3a - 8c)(9a^{2} + 24ac + 64c^{2})$. + +\Item{15.} $(2x^{2} - 5y)(4x^{4} + 10x^{2}y + 25y^{2})$. + +\Item{16.} $(4x^{2} - 3y^{5})(16x^{8} + 12x^{4}y^{5} + 9y^{10})$. + +\Item{17.} $(6 - 2a)(36 + 12a + 4a^{2})$. + +\Item{18.} $(7 - 3y)(49 + 21y + 9y^{2})$. +%% -----File: 181.png---Folio 175------- + + +\AnsTo[2]{Exercise}{36.} % Page 78. + +\Item{1.} $(x + 1)(a^{2} - x + 1)$. + +\Item{2.} $(2x + y)(4x^{2} - 2xy + y^{2})$. + +\Item{3.} $(x + 5)(x^{2} - 5x + 25)$. + +\Item{4.} $(4a + 3)(16a^{2} - 12a + 9)$. + +\Item{5.} $(xy + z)(x^{2}y^{2} - xyz + z^{2})$. + +\Item{6.} $(a + 4)(a^{2} - 4a + 16)$. + +\Item{7.} $(2a^{2} + b)(4a^{4} - 2a^{2}b + b^{2})$. + +\Item{8.} $(x + 7)(x^{2} - 7x + 49)$. + +\Item{9.} $(2 + xyz)(4 - 2xyz + x^{2}y^{2}z^{2})$. + +\Item{10.} $(y^{3} + 4x)(y^{6} - 4y^{3}x + 16x^{2})$. + +\Item{11.} $(ab + 3x)(a^{2}b^{2} - 3abx + 9x^{2})$. + +\Item{12.} $(2yz + x^{2})(4y^{2}z^{2} - 2yzx^{2} + x^{4})$. + +\Item{13.} $(y^{3} + 4x^{2})(y^{6} - 4y^{3}x^{2} + 16x^{4})$. + +\Item{14.} $(4a^{4} + x^{5})(16a^{8} - 4a^{4}x^{5} + x^{10})$. + +\Item{15.} $(3x^{5} + 2a^{2})(9x^{10} - 6x^{5}a^{2} + 4a^{4})$. + +\Item{16.} $(3x^{3} + 8)(9x^{6} - 24x^{3} + 64)$. + +\Item{17.} $(7 + 4x)(49 - 28x + 16x^{2})$. + +\Item{18.} $(5 + 3y)(25 - 15y + 9y^{2})$. + + +\AnsTo{Exercise}{37.} % Page 80. + +\Item{1.} $(2x + y)(2x + y)$. + +\Item{2.} $(x + 3y)(x + 3y)$. + +\Item{3.} $(x + 8)(x + 8)$. + +\Item{4.} $(x + 5a)(x + 5a)$. + +\Item{5.} $(a - 8)(a - 8)$. + +\Item{6.} $(a - 5b)(a - 5b)$. + +\Item{7.} $(c - 3d)(c - 3d)$. + +\Item{8.} $(2x - 1)(2x - 1)$. + +\Item{9.} $(2a - 3b)(2a - 3b)$. + +\Item{10.} $(3a - 4b)(3a - 4b)$. + +\Item{11.} $(x + 4y)(x + 4y)$. + +\Item{12.} $(x - 4y)(x - 4y)$. + +\Item{13.} $(2x - 5y)(2x - 5y)$. + +\Item{14.} $(1 + 10a)(1 + 10a)$. + +\Item{15.} $(7a - 2)(7a - 2)$. + +\Item{16.} $(6a + 5b)(6a + 5b)$. + +\Item{17.} $(9x - 2b)(9x - 2b)$. + +\Item{18.} $(mn + 7x^{2})(mn + 7x^{2})$. + + +\AnsTo{Exercise}{38.} % Page 82. + +\Item{1.} $(a + 2)(a + 3)$. + +\Item{2.} $(a - 2)(a - 3)$. + +\Item{3.} $(a + 1)(a + 5)$. + +\Item{4.} $(a - 1)(a - 5)$. + +\Item{5.} $(a - 1)(a + 5)$. + +\Item{6.} $(a + 1)(a - 5)$. + +\Item{7.} $(c - 3)(c - 6)$. + +\Item{8.} $(c + 3)(c + 6)$. + +\Item{9.} $(c - 3)(c + 6)$. + +\Item{10.} $(c + 3)(c - 6)$. + +\Item{11.} $(x + 2)(x + 7)$. + +\Item{12.} $(x - 2)(x - 7)$. + +\Item{13.} $(x + 2)(x - 7)$. + +\Item{14.} $(x - 4)(x - 5)$. + +\Item{15.} $(x + 4)(x - 5)$. + +\Item{16.} $(x - 4)(x + 5)$. + +\Item{17.} $(x - 3)(x - 7)$. + +\Item{18.} $(x + 3)(x - 7)$. + +\Item{19.} $(x - 3)(x + 7)$. + +\Item{20.} $(x - 7)(x - 8)$. + +\Item{21.} $(x + 7)(x - 8)$. + +\Item{22.} $(x - 1)(x - 9)$. + +\Item{23.} $(x + 3)(x + 10)$. + +\Item{24.} $(x - 3)(x + 10)$. + +\Item{25.} $(x + 3)(x - 10)$. + +\Item{26.} $(a - 2b)(a + 3b)$. + +\Item{27.} $(a + 2b)(a - 3b)$. + +\Item{28.} $(a - b)(a + 4b)$. + +\Item{29.} $(a + b)(a - 4b)$. + +\Item{30.} $(ax + 7)(ax - 9)$. + +\Item{31.} $(a - 7x)(a + 9x)$. + +\ResetCols[2] +\Item{32.} $(a - 4b)(a - 5b)$. + +\Item{33.} $(xy - 3z)(xy - 16z)$. + +\Item{34.} $(ab + 4c)(ab + 11c)$. + +\Item{35.} $(x - 4y)(x - 9y)$. + +\Item{36.} $(x + 7y)(x + 12y)$. + +\Item{37.} $(ax - 6y)(ax - 17y)$. + +\Item{38.} $(x + 2y)(x - 2y)(x^{2} - 5y^{2})$. + +\Item{39.} $(a^{2}x^{2} - 11y^{2})(a^{2}x^{2} - 13y^{2})$. + +\Item{40.} $(a^{3}b^{3} - 11c^{2})(a^{3}b^{3} - 12c^{2})$. + +\Item{41.} $(a + 4bc)(a - 24bc)$. + +\Item{42.} $(a + 8bc)(a - 12bc)$. + +\Item{43.} $(a + 6bc)(a - 16bc)$. + +\Item{44.} $(a - 3bc)(a + 32bc)$. + +\Item{45.} $(a + 2bc)(a - 48bc)$. + +\Item{46.} $(a + bc)(a + 48bc)$. + +\Item{47.} $(x + 9yz)(x - 27yz)$. + +\Item{48.} $(xy + 13z)(xy - 14z)$. +%% -----File: 182.png---Folio 176------- + + +\AnsTo[2]{Exercise}{39.} % Page 83. + +\Item{1.} $a(a^{2} - 7)$. + +\Item{2.} $ab(3ab - 2a^{2} + 3b^{2})$. + +\Item{3.} $(a - b + 1)(a - b)$. + +\Item{4.} $(a + b + 1)(a + b - 1)$. + +\Item{5.} $(a + 2b)(a^{2} - 2ab + 4b^{2})$. + +\Item{6.} $(x + 2y + 1)(x - 2y)$. + +\Item{7.} $(a - b)(a^{2} + ab + b^{2} + 1)$. + +\Item{8.} $(a - 3b)(a - 3b)$. + +\Item{9.} $(x + 1)(x - 2)$. + +\Item{10.} $(x + 1)(x - 3)$. + +\Item{11.} $(x - 3)(x + 7)$. + +\Item{12.} $(a + 2)(a - 13)$. + +\Item{13.} $(x^{2} + 3)(a + b)$. + +\Item{14.} $(x - 3)(x - y)$. + +\Item{15.} $(x - 3)(x - 4)$. + +\Item{16.} $(a + 2b)(a + 3b)$. + +\Item{17.} $(x^{2} + 5)(x^{2} + 5)$. + +\Item{18.} $(x - 9)(x - 9)$. + +\Item{19.} $(x - 10)(x - 11)$. + +\Item{20.} $(x + 8)(x + 11)$. + +\Item{21.} $(x - 8)(x + 11)$. + +\Item{22.} $(x^{2} + 1)(x - 1)$. + +\Item{23.} $x^{2}(3x + 1)(3x - 1)$. + +\Item{24.} $(1 + a - b)(1 - a + b)$. + +\Item{25.} $(a + b)(a^{2} - ab + b^{2} + 1)$. + +\Item{26.} $(m + n)(m - n)(x + y)$. + +\Item{27.} $(x - y + z)(x - y - z)$. + +\Item{28.} $(z + x - y)(z - x + y)$. + +\Item{29.} $(2a^{2} + 3a - 1)(2a^{2} - 3a + 1)$. + +\Item{30.} $(2x - y)(4x^{2} + 2xy + y^{2})$. + +\Item{31.} $x^{2}(x - 3y)$. + +\Item{32.} $(x - 3y)(x^{2} + 3xy + 9y^{2})$. + +\Item{33.} $(x - 5)(x + 8)$. + +\Item{34.} $(x - 2y)(x + 5y)$. + +\Item{35.} $(1 + 4x)(1 - 4x)$. + +\Item{36.} $a^{2}(a^{2} + 3b^{2})(a^{2} - 3b^{2})$. + +\Item{37.} $x(x + y)(x + 2y)$. + +\Item{38.} $x^{2}(x + y)(x + 3y)$. + +\Item{39.} $(x - 2y^{2})(x - 2y^{2})$. + +\Item{40.} $(4x^{2} + 1)(4x^{2} + 1)$. + +\Item{41.} $a^{2}(3a + 2c)(3a - 2c)$. + +\Item{42.} $ab(a + b)(a - 2b)$. + +\Item{43.} $(x + 2)(x^{2} - 2x + 4)(x - 1)$. + +\Item{44.} $(a + y)(a^{2} - ay + y^{2})(a - x)$. + + +\AnsTo{Exercise}{40.} % Page 86. + +\Item{1.} $6$. + +\Item{2.} $5x^{3}$. + +\Item{3.} $6ax$. + +\Item{4.} $7ab^{2}$. + +\Item{5.} $7$. + +\Item{6.} $2a^{2}b^{2}$. + +\Item{7.} $x + 3y$. + +\Item{8.} $x + 3$. + +\Item{9.} $2a + 1$. + +\Item{10.} $x + y$. + +\Item{11.} $a + x$. + +\Item{12.} $a + 2b$. + +\Item{13.} $x - 1$. + +\Item{14.} $x + 3$. + +\Item{15.} $x - 6$. + +\Item{16.} $x^{2} - x + 1$. + +\Item{17.} $x - 1$. + +\Item{18.} $x - y$. + +\Item{19.} $x - 5$. + +\Item{20.} $a - b - c$. + +\Item{21.} $x + 2y$. + +\Item{22.} $x + 4y$. + +\Item{23.} $x^{2} + 2xy + 4y^{2}$. + +\Item{24.} $x - 2$. + +\Item{25.} $1 - 3a$. + +\Item{26.} $x - 7y$. + +\Item{27.} $2a + b$. + +\Item{28.} $x + y - z$. + + +\AnsTo[2]{Exercise}{41.} % Page 88. + +\Item{1.} $18x^{2}y^{3}$. + +\Item{2.} $6a^{2}bc^{3}$. + +\Item{3.} $20a^{3}b^{3}$. + +\Item{4.} $30a^{3}b^{4}$. + +\Item{5.} $189x^{3}y^{5}$. + +\Item{6.} $x^{2}y^{3}z^{3}$. + +\Item{7.} $a^{2}(a + 1)$. + +\Item{8.} $x^{2}(x - 3)$. + +\Item{9.} $x(x + 1)(x - 1)$. + +\Item{10.} $x(x + 1)(x - 1)$. + +\Item{11.} $xy(x + y)$. + +\Item{12.} $x(x + 2)^{2}$. +%% -----File: 183.png---Folio 177------- + +\Item{13.} $(a + 2)(a + 2)(a + 3)$. + +\Item{14.} $(c - 4)(c + 5)(c - 6)$. + +\Item{15.} $(b - 5)(b - 6)(b + 7)$. + +\Item{16.} $(y - 4)(y + 5)(y - 6)$. + +\Item{17.} $(z - 5)(z - 6)(z + 7)$. + +\Item{18.} $(x - 4)(x + 8)(x - 8)(x^{2} + 4x + 16)$. + +\Item{19.} $(a + b)(a + b)(a - b)(a - b)$. + +\Item{20.} $4a^{2}b(a + b)(a + b)(a - b)$. + +\Item{21.} $(y + 2)(y + 3)(y + 4)$. + +\Item{22.} $(x + 1)(x - 1)(x^{2} + 1)$. + +\Item{23.} $(1 + x)(1 - x)(1 + x + x^{2})$. + +\Item{24.} $(x + y)(x + y)(x - y)(x - y)$. + +\Item{25.} $x(x - 3)(x + 5)(x^{2} + 3x + 9)$. + +\Item{26.} $(a + b + c)(a + b + c)(a + b - c)$. + +\Item{27.} $(x - a)(x - b)(x - c)$. + +\Item{28.} $a(a + b + c)(a + b - c)$. + + +\AnsTo[4]{Exercise}{42.} % Page 90. + +\Item{1.} $\dfrac{1}{3b}$. + +\Item{2.} $\dfrac{4m}{5n}$. + +\Item{3.} $\dfrac{3m}{4p^{2}}$. + +\Item{4.} $\dfrac{x^{2}}{2yz}$. + +\Item{5.} $\dfrac{a^{3}b^{3}}{3c^{2}}$. + +\Item{6.} $\dfrac{2xy}{3}$. + +\Item{7.} $\dfrac{2m}{3p}$. + +\Item{8.} $\dfrac{3b^{2}c}{4a^{3}}$. + +\Item{9.} $\dfrac{2y^{2}z^{4}}{3}$. + +\Item{10.} $\dfrac{b}{c}$. + +\Item{11.} $\dfrac{2a - 3b}{2a}$. + +\Item{12.} $\dfrac{3a}{a + 2}$. + +\Item{13.} $\dfrac{x}{x - 1}$. + +\Item{14.} $\dfrac{y}{x^{2} + 3xy + 9y^{2}}$. + +\Item{15.} $\dfrac{x + 1}{x - 5}$. + +\Item{16.} $\dfrac{x + 1}{x - 2}$. + +\Item{17.} $\dfrac{a + b + c}{a}$. + +\Item{18.} $\dfrac{x + 5}{x + 3}$. + +\Item{19.} $\dfrac{x + 1}{x + 3}$. + + +\AnsTo[2]{Exercise}{43.} % Page 91. + +\Item{1.} $a + b + \dfrac{2}{a - b}$. + +\Item{2.} $a - b - \dfrac{2}{a + b}$. + +\Item{3.} $a - 1 + \dfrac{2a}{a^{2} - a - 1}$. + +\Item{4.} $2x - 4 + \dfrac{5}{x + 1}$. + +\Item{5.} $4x^{2} - 2x + 1 - \dfrac{1}{2x + 1}$. + +\Item{6.} $5x + 4 + \dfrac{x + 7}{x^{2} + x - 1}$. + +\Item{7.} $a + \dfrac{5a - 2}{a^{2} + a + 2}$. + +\Item{8.} $y^{2} - yx + x^{2}$. + +\Item{9.} $x^{2} - 4x + 3 + \dfrac{2x - 4}{x^{2} + x + 1}$. + +\Item{10.} $x^{3} + x + 1 + \dfrac{2x + 2}{x^{2} - x - 1}$. +%% -----File: 184.png---Folio 178------- + + +\AnsTo[2]{Exercise}{44.} % Page 92. + +\Item{1.} $\dfrac{x^{2} + y^{2}}{x - y}$. + +\Item{2.} $\dfrac{x^{2} + y^{2}}{x + y}$. + +\Item{3.} $\dfrac{2y}{x + y}$. + +\Item{4.} $-\dfrac{2ax}{a - x}$. + +\Item{5.} $-\dfrac{x + 2}{x - 3}$. + +\Item{6.} $-\dfrac{2x^{2} - 6x + 5}{x - 2}$. + +\Item{7.} $\dfrac{x^{3} + x^{2} - 2x + 1}{x + 2}$. + +\Item{8.} $\dfrac{2a^{2} - 11a + 6}{a - 3}$. + +\Item{9.} $\dfrac{-2a^{3} + a^{2} + 2a-3}{a - 1}$. + +\Item{10.} $\dfrac{3a^{4} + 6a^{3} - 14a^{2} - 4}{3a^{2} + 1}$. + + +\AnsTo[1]{Exercise}{45.} % Page 94. + +\Item{1.} $\dfrac{x(x + a)}{(x + a)(x - a)}$, $\dfrac{x^{2}}{(x + a)(x - a)}$. + +\Item{2.} $\dfrac{a(a - b)}{(a + b)(a - b)}$, $\dfrac{a^{2}}{(a + b)(a - b)}$. + +\Item{3.} $\dfrac{1 - 2a}{(1 + 2a)(1 - 2a)}$, $\dfrac{1}{(1 + 2a)(1 - 2a)}$. + +\Item{4.} $\dfrac{9}{(4 + x)(4 - x)}$, $\dfrac{(4 - x)^{2}}{(4 + x)(4 - x)}$. + +\Item{5.} $\dfrac{a^{2}}{(3 - a)(9 + 3a + a^{2})}$, $\dfrac{a(9 + 3a + a^{2})}{(3 - a)(9 + 3a + a^{2})}$. + +\Item{6.} $\dfrac{x + 2}{(x + 2)(x - 2)(x - 3)}$, $\dfrac{x - 2}{(x + 2)(x - 2)(x - 3)}$. + + +\AnsTo{Exercise}{46.} % Page 95. + +\Item{1.} $\dfrac{4x + 2}{5}$. + +\Item{2.} $\dfrac{13x + 3}{12}$. + +\Item{3.} $\dfrac{5(9x - 13)}{42}$. + +\Item{4.} $\dfrac{51x + 31}{36}$. + +\Item{5.} $\dfrac{x - 5}{3}$. + +\Item{6.} $\dfrac{5(x - y)}{8x}$. + +\Item{7.} $\dfrac{22x - 97}{30}$. + +\Item{8.} $\dfrac{3x - 4}{15x}$. + +\Item{9.} $\dfrac{a^{3} - b^{3} + c^{3} - abc}{abc}$. + + +\AnsTo{Exercise}{47.} % Page 96}. + +\Item{1.} $\dfrac{2x + 1}{(x + 3)(x - 2)}$. + +\Item{2.} $\dfrac{2x}{x^{2} - 1}$. + +\Item{3.} $\dfrac{3x + 16}{(x - 8)(x + 2)}$. + +\Item{4.} $\dfrac{4ax}{a^{3} - x^{2}}$. + +\Item{5.} $\dfrac{ax}{x^{2} - a^{2}}$. + +\Item{6.} $-\dfrac{4ab}{4a^{2} - b^{2}}$. + +\Item{7.} $\dfrac{1}{9 - a^{2}}$. + +\Item{8.} $\dfrac{b}{a^{2} - b^{2}}$. + +\Item{9.} $\dfrac{5x + 8}{x^{2} - 4}$. + +\Item{10.} $\dfrac{1 + x}{1 - 9x^{2}}$. + +\Item{11.} $\dfrac{3(a^{2} + 4a + 1)}{a(a + 1)(a + 3)}$. + +\Item{12.} $\dfrac{2}{x^{2} - 1}$. + +\Item{13.} $\dfrac{2x^{2}}{(x + 2)(x - 3)}$. + +\Item{14.} $0$. +%% -----File: 185.png---Folio 179------- + + +\AnsTo{Exercise}{48.} % Page 97. + +\Item{1.} $0$. + +\Item{2.} $\dfrac{2a}{a + b}$. + +\Item{3.} $\dfrac{7a}{1 - a^{2}}$. + +\Item{4.} $\dfrac{x - 10y}{4x^{2} - 25y^{2}}$. + +\Item{5.} $\dfrac{2}{x + 4y}$. + +\Item{6.} $\dfrac{2(x + 6)}{4x^{2} - 9}$. + + +\AnsTo{Exercise}{49.} % Page 100. + +\Item{1.} $\dfrac{20}{3bc}$. + +\Item{2.} $\dfrac{2ay}{3}$. + +\Item{3.} $\dfrac{7p^{2}}{4xz}$. + +\Item{4.} $\dfrac{2a^{2}cm}{7}$. + +\Item{5.} $\dfrac{30}{abc}$. + +\Item{6.} $abc$. + +\Item{7.} $b^{2}$. + +\Item{8.} $\dfrac{x + a}{x - 2a}$. + +\Item{9.} $\dfrac{xy}{2c - 1}$. + +\Item{10.} $\dfrac{a + 10}{a + 3}$. + +\Item{11.} $\dfrac{3x + 2y}{x - 2}$. + +\Item{12.} $\dfrac{5a + b}{4a + 3b}$. + +\Item{13.} $\dfrac{x - 7}{a + b + c}$. + +\Item{14.} $\dfrac{x(x + 1)}{x - 5}$. + +\Item{15.} $\dfrac{a + 1}{a + 5}$. + +\Item{16.} $1$. + +\Item{17.} $\dfrac{x(x + y)}{x + 1}$. + +\Item{18.} $\dfrac{b}{a - b}$. + +\Item{19.} $abc$. + +\Item{20.} $\dfrac{(x + 2y)(x + 1)}{(x + y)(x + 2)}$. + + +\AnsTo{Exercise}{50.} % Page 102. + +\Item{1.} $\dfrac{x + y}{z}$. + +\Item{2.} $\dfrac{12x + 3y}{12x - 4y}$. + +\Item{3.} $\dfrac{abd - 21d^{2}}{21cd - 7ab}$. + +\Item{4.} $\dfrac{x^{2} + x - 2}{x^{2} - x - 2}$. + +\Item{5.} $1$. + +\Item{6.} $\dfrac{x + y}{x^{2} - 2xy + y^{2}}$. + +\Item{7.} $\dfrac{a + b}{a - b}$. + +\Item{8.} $4(3a + 8)$. + +\Item{9.} $\dfrac{y + x}{y - x}$. + +\Item{10.} $x$. + +\Item{11.} $\dfrac{1}{x}$. + +\Item{12.} $\dfrac{x^{2}(x - 3)}{x - 2}$. + +\Item{13.} $a - 1$. + +\Item{14.} $\dfrac{4a}{a - x}$. + + +\AnsTo[5]{Exercise}{51.} % Page 105. + +\Item{1.} $5$. + +\Item{2.} $7$. + +\Item{3.} $2\frac{1}{2}$. + +\Item{4.} $120$. + +\Item{5.} $12$. + +\Item{6.} $2\frac{1}{3}$. + +\Item{7.} $17$. + +\Item{8.} $4$. + +\Item{9.} $4$. + +\Item{10.} $1$. + +\Item{11.} $-16$. + +\Item{12.} $11$. + +\Item{13.} $-4$. + +\Item{14.} $-2$. + +\Item{15.} $-2$. + +\Item{16.} $5$. + +\Item{17.} $9$. + +\Item{18.} $-1$. +%% -----File: 186.png---Folio 180------- + + +\AnsTo[5]{Exercise}{52.} % Page 106. + +\Item{1.} $2$. + +\Item{2.} $2$. + +\Item{3.} $-33$. + +\Item{4.} $1$. + +\Item{5.} $\frac{2}{3}$. + +\Item{6.} $1\frac{1}{2}$. + +\Item{7.} $2$. + +\Item{8.} $8$. + +\Item{9.} $5$. + +\Item{10.} $\frac{3}{7}$. + +\Item{11.} $2$. + +\Item{12.} $1$. + +\Item{13.} $3$. + + +\AnsTo[6]{Exercise}{53.} % Page 107. + +\Item{1.} $33$. + +\Item{2.} $2$. + +\Item{3.} $3\frac{1}{2}$. + +\Item{4.} $1\frac{5}{37}$. + +\Item{5.} $7$. + +\Item{6.} $3$. + + +\AnsTo{Exercise}{54.} % Page 108. + +\Item{1.} $a + b$. + +\Item{2.} $\dfrac{a}{2}$. + +\Item{3.} $\dfrac{b}{2}$. + +\Item{4.} $2a$. + +\Item{5.} $b - a$. + +\Item{6.} $\dfrac{ab}{a + b + c}$. + +\Item{7.} $\dfrac{a^{2} - b^{2}}{2a}$. + +\Item{8.} $\dfrac{2b}{a}$. + +\Item{9.} $\dfrac{2b^{2} - a^{2}}{4b - 3a}$. + +\Item{10.} $1$. + +\Item{11.} $3(a - b)$. + + +\AnsTo[4]{Exercise}{55.} % Page 109. + +\Item{1.} $35$. + +\Item{2.} $70$. + +\Item{3.} $36$. + +\Item{4.} $57$, $58$. + + +\AnsTo[5]{Exercise}{56.} % Page 110. + +\Item{1.} $81$, $19$. + +\Item{2.} $100$, $24$. + +\Item{3.} $64$, $15$. + +\Item{4.} $103$, $12$. + +\Item{5.} $295$, $25$. + + +\AnsTo[4]{Exercise}{57.} % Page 111. + +\Item{1.} $12$~yr. + +\Item{2.} A, $60$~yr.; B, $10$~yr. + +\Item{3.} A, $25$~yr.; B, $5$~yr. + +\Item{4.} $17\frac{1}{2}$~yr. + +\Item{5.} $35$~yr. + +\Item{6.} Son, $12$~yr.; father, $36$~yr. + +\Item{7.} $25$~yr. + +\Item{8.} A, $30$~yr.; B, $15$~yr. + +\Item{9.} Son, $12$~yr.; father, $68$~yr. + + +\AnsTo[6]{Exercise}{58.} % Page 112. + +\Item{1.} $1\frac{3}{7}$~dy. + +\Item{2.} $1\frac{13}{47}$~dy. + +\Item{3.} $1\frac{1}{20}$~dy. + +\Item{4.} $15$~dy. + +\Item{5.} $12$~hr. + +\Item{6.} $10$~dy. + + +\AnsTo[5]{Exercise}{59.} % Page 113. + +\Item{1.} $7\frac{5}{13}$~hr. + +\Item{2.} $2\frac{2}{5}$~hr. + +\Item{3.} $\frac{10}{11}$~hr. + +\Item{4.} $1\frac{1}{13}$~hr. + +\Item{5.} $30$~hr. + + +\AnsTo[4]{Exercise}{60.} % Page 114. + +\Item{1.} $36$~mi. + +\Item{2.} $26$~hr. + +\Item{3.} $8$~mi. + +\Item{4.} $240$~mi. + + +\AnsTo{Exercise}{61.} % Page 115. + +\Item{1.} $600$. + +\Item{2.} $700$. + +\Item{3.} Dog, $1440$; rabbit, $1800$. +%% -----File: 187.png---Folio 181------- + +\AnsTo[2]{Exercise}{62.} % Page 116. + +\Item{1.} $27\frac{3}{11}$~min.\ past 5~o'clock. + +\Item{2.} $27\frac{3}{11}$~min.\ past 2~o'clock. + +\Item{3.} $43\frac{7}{11}$~min.\ past 2~o'clock. + +\Item{4.} $21\frac{9}{11}$~min.\ past 1~o'clock. + +\Item{5.} $38\frac{2}{11}$~min.\ past 1~o'clock. + +\Item{6.} $38\frac{2}{11}$~min.\ past 7~o'clock. + + + +\AnsTo{Exercise}{63.} % Page 117. + +\Item{1.} $1764$~sq.~ft. + +\Item{2.} $18$~ft.\ by $23$~ft. + +\Item{3.} $14$~ft.\ by $20$~ft. + +\Item{4.} $12$~ft.\ by $15$~ft. + +\Item{5.} $30$~ft.\ by $40$~ft. + + +\AnsTo[4]{Exercise}{64.} % Page 121. + +\Item{1.} $90°\,0'\,30''$; $30°\,30'$. + +\Item{2.} \$$133\frac{1}{3}$. + +\Item{3.} \$$2000$. + +\Item{4.} \$$4000$. + +\Item{5.} \$$3000$. + +\Item{6.} \$$500$. + +\Item{7.} $5$\%. + +\Item{8.} $6$\%. + +\Item{9.} $3$~yr. + +\Item{10.} $9\frac{3}{8}$~yr. + +\Item{11.} \$$25,000$. + +\Item{12.} \$$20,000$. + +\Item{13.} \$$6$\%. + +\Item{14.} $20$~yr. + + +\AnsTo{Exercise}{65.} % Page 124. + +\Item{1.} $x = 2$, $y = 1$. + +\Item{2.} $x = 3$, $y = 2$. + +\Item{3.} $x = 5$, $y = 1$. + +\Item{4.} $x = 2$, $y = 1$. + +\Item{5.} $x = 1$, $y = 2$. + +\Item{6.} $x = 6$, $y = 1$. + +\Item{7.} $x = 3$, $y = 21$. + +\Item{8.} $x = 7$, $y = 7$. + +\Item{9.} $x = 23$, $y = -1$. + +\Item{10.} $x = 2$, $y = 1$. + +\Item{11.} $x = 35$, $y = 20$. + +\Item{12.} $x = 2$, $y = 1$. + +\Item{13.} $x = 1$, $y = 2$. + +\Item{14.} $x = 3$, $y = 2$. + +\Item{15.} $x = 1$, $y = 2$. + +\Item{16.} $x = 4$, $y = 3$. + +\Item{17.} $x = 12$, $y = 4$. + +\Item{18.} $x = 12$, $y = 21$. + +\Item{19.} $x = 5$, $y = 7$. + +\Item{20.} $x = 5$, $y = 2$. + +\Item{21.} $x = 18$, $y = 6$. + +\Item{22.} $x = 3$, $y = 2$. + +\Item{23.} $x = 3$, $y = 2$. + +\Item{24.} $x = 7$, $y = 8$. + +\Item{25.} $x = 8$, $y = -2$. + +\Item{26.} $x = \dfrac{a}{(a - b)}$, $y = \dfrac{b}{(a + b)}$. + + +\AnsTo[2]{Exercise}{66.} % Page 127. + +\Item{1.} A, \$$520$; B, \$$440$. + +\Item{2.} $23$~and~$17$. + +\Item{3.} $20$~and~$16$. + +\Item{4.} Velvet, \$$6$; silk, \$$3$. + +\Item{5.} Wheat, \$$1$; rye, \$$\frac{4}{5}$. + +\Item{6.} Tea, \$$\frac{1}{2}$; coffee, \$$\frac{1}{4}$. + +\Item{7.} Horse, \$$92$; cow, \$$64$. +%% -----File: 188.png---Folio 182------- + + +\AnsTo[5]{Exercise}{67.} % Page 128. + +\Item{1.} $\frac{3}{7}$. + +\Item{2.} $\frac{13}{25}$. + +\Item{3.} $\frac{3}{20}$. + +\Item{4.} $\frac{5}{21}$. + +\Item{5.} $\frac{7}{22}$. + + +\AnsTo[4]{Exercise}{68.} % Page 129. + +\Item{1.} $45$. + +\Item{2.} $72$. + +\Item{3.} $75$~and~$57$. + +\Item{4.} $54$. + + +\AnsTo{Exercise}{69.} % Page 130. + +\Item{1.} \$$2500$ at $4$\%. + +\Item{2.} \$$1600$ at $6$\%. + +\Item{3.} \$$6000$ at $4$\%; \$$4000$ at $5$\%. + + +\AnsTo[4]{Exercise}{70.} % Page 131. + +\Item{1.} $22$~and~$18$. + +\Item{2.} $60$~and~$8$. + +\Item{3.} $\frac{14}{17}$. + +\Item{4.} Wheat, \$$1$; barley, \$$\frac{4}{5}$. + +\Item{5.} A, \$$235$; B, \$$65$. + +\Item{6.} A, \$$70$; B, \$$30$. + +\Item{7.} Lemon, $2$~cts.; orange, $3$~cts. + +\Item{8.} A, $30$~apples; B, $10$~apples. + + +\AnsTo[4]{Exercise}{71.} % Page 134. + +\Item{1.} $±2$. + +\Item{2.} $±3$. + +\Item{3.} $±5$. + +\Item{4.} $±8$. + +\Item{5.} $±7$. + +\Item{6.} $±5$. + +\Item{7.} $±5$. + +\Item{8.} $±3$. + +\Item{9.} $±3$. + +\Item{10.} $±3$. + +\Item{11.} $12$~and~$16$. + +\Item{12.} $12$~oranges at $3$~cts. + +\Item{13.} $3$~rods. + +\Item{14.} Width, $12$~rd.; length, $48$~rd. + + +\AnsTo{Exercise}{72.} % Page 137. + +\Item{1.} $9$~or~$3$. + +\Item{2.} $4$~or~$2$. + +\Item{3.} $3$~or~$1$. + +\Item{4.} $1$~or~$-\frac{1}{5}$. + +\Item{5.} $1$~or~$-3$. + +\Item{6.} $\frac{4}{3}$~or~$\frac{4}{3}$. + +\Item{7.} $1$~or~$-\frac{1}{6}$. + +\Item{8.} $3$~or~$-1$. + +\Item{9.} $\frac{3}{4}$~or~$\frac{1}{4}$. + +\Item{10.} $3$~or~$\frac{1}{3}$. + +\Item{11.} $17$~or~$-3$. + +\Item{12.} $25$~or~$9$. + +\Item{13.} $4$~or~$-5$. + +\Item{14.} $4$~or~$-3$. + +\Item{15.} $5$~or~$1$. + +\Item{16.} $2$~or~$-6$. + +\Item{17.} $2$~or~$2$. + +\Item{18.} $5$~or~$-11$. + +\Item{19.} $2$~or~$-5\frac{1}{3}$. + +\Item{20.} $4\frac{1}{3}$~or~$-3\frac{2}{3}$. + +\Item{21.} $2$~or~$\frac{1}{3}$. + +\Item{22.} $4$~or~$-\frac{2}{5}$. + +\Item{23.} $2$~or~$-3$. + +\Item{24.} $10$~or~$2$. + +\Item{25.} $3$~or~$-2\frac{1}{3}$. + +\Item{26.} $2$~or~$2$. + +\Item{27.} $\frac{1}{2}$~or~$-3$. + +\Item{28.} $5$~or~$\frac{1}{2}$. + +\Item{29.} $7$~or~$2$. + +\Item{30.} $4$~or~$-\frac{2}{3}$. + +\Item{31.} $8$~or~$2$. + +\Item{32.} $4$~or~$-7$. + +\Item{33.} $0$~or~$3$. + +\Item{34.} $0$~or~$7$. + +\Item{35.} $5$~or~$2$. + +\Item{36.} $4$~or~$-1$. +%% -----File: 189.png---Folio 183------- + + +\AnsTo{Exercise}{73.} % Page 140. + +\Item{1.} $6$~and~$5$. + +\Item{2.} $5$~and~$15$. + +\Item{3.} Son, $8$~yr.; father, $40$~yr. + +\Item{4.} $9$. + +\Item{6.} $5$~rd.\ by $7$~rd. + +\Item{7.} $12$~ft. + +\Item{8.} $20$~ft.\ by $18$~ft. + +\Item{9.} $10$~rd.\ by $12$~rd. + +\Item{10.} Son, $10$~yr.; father, $54$~yr. + + +\AnsTo{Exercise}{74.} % Page 141. + +\Item{1.} $6$~miles an hour. + +\Item{2.} $7$. + +\Item{3.} $5$. + +\Item{4.} $8$. + +\Item{5.} $36$. + + +\AnsTo[4]{Exercise}{75.} % Page 144. + +\Item{1.} $75$. + +\Item{2.} $38$. + +\Item{3.} $4\frac{1}{7}$. + +\Item{4.} $-8\frac{3}{4}$. + +\Item{5.} $23$. + +\Item{6.} $0$. + +\Item{7.} $156$. + +\Item{8.} $20$th. + +\Item{9.} $101$st. + +\Item{10.} $26$. + +\Item{11.} $a$. + +\Item{12.} $21$, $22$,~etc. + + +\AnsTo[4]{Exercise}{76.} % Page 146. + +\Item{1.} $440$. + +\Item{2.} $201$. + +\Item{3.} $4frac{1}{6}$. + +\Item{4.} $128$. + +\Item{5.} $-378$. + +\Item{6.} $187\frac{1}{2}$. + +\Item{7.} $1$, $3$, $5$. + +\Item{8.} $156$. + +\Item{9.} $300$. + +\Item{10.} $2550$~yd. + +\Item{11.} $5812.1$~ft. + +\Item{12.} $144.9$~ft. + +\Item{13.} $11$, $15$. + +\Item{14.} $7$, $9$, $11$. + +\Item{15.} $12$~miles. + +\Item{16.} $5$, $12$, $19$. + + +\AnsTo[4]{Exercise}{77.} % Page 151. + +\Item{1.} $243$. + +\Item{2.} $192$. + +\Item{3.} $\frac{3}{64}$. + +\Item{4.} $256$. + +\Item{5.} $±4$. + +\Item{6.} $±4$. + +\Item{7.} $1092$. + +\Item{8.} $765$. + +\Item{9.} $11\frac{29}{32}$. + +\Item{10.} $15\frac{15}{16}$. + +\Item{11.} $127\frac{3}{4}$. + +\Item{12.} $44$. + +\Item{13.} $1\frac{11}{54}$. + +\Item{14.} \$$1.27$. + +\Item{15.} \$$81.90$. + +\Item{16.} $14,641$. + + +\AnsTo{Exercise}{78.} % Page 153. + +\Item{1.} $a + b + c$. + +\Item{2.} $x^{2} + x + 1$. + +\Item{3.} $x^{2} - 2xy + y^{2}$. + +\Item{4.} $2a^{2} - 3ab + 5b^{2}$. + +\Item{5.} $4x^{3} + 3x^{2}y - 2y^{3}$. + +\Item{6.} $2x^{3} - xy^{2} + 3y^{3}$. + + +\AnsTo[4]{Exercise}{79.} % Page 156. + +\Item{1.} $18$. + +\Item{2.} $21$. + +\Item{3.} $23$. + +\Item{4.} $31$. + +\Item{5.} $3.2$. + +\Item{6.} $7.3$. + +\Item{7.} $232$. + +\Item{8.} $785$. + +\Item{9.} $1225$. + +\Item{10.} $589$. + +\Item{11.} $5601$. + +\Item{12.} $1234$. + +\Item{13.} $1.4142\dots$ + +\Item{14.} $1.7320\dots$ + +\Item{15.} $2.2360\dots$ + +\Item{16.} $2.4494\dots$ + +\Item{17.} $0.7071\dots$ + +\Item{18.} $0.9486\dots$ + +\Item{19.} $0.8164\dots$ + +\Item{20.} $0.8660\dots$ + +\Item{21.} $0.8944\dots$ + +\Item{22.} $0.7905\dots$ +%% -----File: 190.png---Folio 184------- + + +\AnsTo{Exercise}{80.} % Page 159. + +\Item{1.} $x + y$. + +\Item{2.} $2x - 1$. + +\Item{3.} $2x - 3y$. + +\Item{4.} $4a - 3x$. + +\Item{5.} $1 + x + x^{2}$. + +\Item{6.} $x^{2} - x + 1$. + + +\AnsTo[4]{Exercise}{81.} % Page 163. + +\Item{1.} $36$. + +\Item{2.} $35$. + +\Item{3.} $45$. + +\Item{4.} $65$. + +\Item{5.} $48$. + +\Item{6.} $637$. + +\Item{7.} $478$. + +\Item{8.} $638$. + +\Item{9.} $503$. + +\Item{10.} $728$. + +\Item{11.} $12.34$. + +\Item{12.} $12.25$. + +\Item{13.} $0.2154\dots$ + +\Item{14.} $0.3684\dots$ + +\Item{15.} $0.5848\dots$ + +\Item{16.} $1.5874\dots$ + +\Item{17.} $2.1544\dots$ + +\Item{18.} $4.4310\dots$ + +\Item{19.} $1.3572\dots$ + +\Item{20.} $1.2703\dots$ + +\Item{21.} $1.4454\dots$ + +\Item{22.} $0.8735\dots$ + +\Item{23.} $0.9085\dots$ + +\Item{24.} $0.9352\dots$ +%[** TN: Environment opened by \AnsTo] +\end{multicols} + + +%%%%%%%%%%%%%%%%%%%%%%%%% GUTENBERG LICENSE %%%%%%%%%%%%%%%%%%%%%%%%%% + +\BackMatter +\begin{PGtext} +End of the Project Gutenberg EBook of The First Steps in Algebra, by +G. A. (George Albert) Wentworth + +*** END OF THIS PROJECT GUTENBERG EBOOK THE FIRST STEPS IN ALGEBRA *** + +***** This file should be named 36670-t.tex or 36670-t.zip ***** +This and all associated files of various formats will be found in: + http://www.gutenberg.org/3/6/6/7/36670/ + +Produced by Andrew D. Hwang, Peter Vachuska, Chuck Greif +and the Online Distributed Proofreading Team at +http://www.pgdp.net. + + +Updated editions will replace the previous one--the old editions +will be renamed. + +Creating the works from public domain print editions means that no +one owns a United States copyright in these works, so the Foundation +(and you!) can copy and distribute it in the United States without +permission and without paying copyright royalties. Special rules, +set forth in the General Terms of Use part of this license, apply to +copying and distributing Project Gutenberg-tm electronic works to +protect the PROJECT GUTENBERG-tm concept and trademark. 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When a bi-no-mial is the dif-fer-ence of two squares. + [] + +[95] [96] [97] [98] +Underfull \hbox (badness 10000) in paragraph at lines 4959--4963 + + [] + +[99] [100] [101] [102] [103] [104] +Overfull \hbox (1.29138pt too wide) in paragraph at lines 5268--5270 +[]\OT1/cmr/m/n/10.95 Two num-bers whose prod-uct is $30$ are $1$ and $30$, $2$ +and $15$, $3$ and $10$, + [] + +[105] [106] [107] [108] [109] [110 + +] [111] [112] [113] [114] [115] [116] [117 + +] [118] [119] [120] [121] [122] [123] +Overfull \hbox (0.88312pt too wide) in paragraph at lines 6065--6066 +[][] $[]$\OT1/cmr/m/n/12 , $[]$. + [] + +[124] +Overfull \hbox (3.79523pt too wide) in paragraph at lines 6118--6118 +[] + [] + +[125] [126] [127] [128] [129] [130] [131] [132] [133] [134] [135] +Overfull \hbox (12.6967pt too wide) in paragraph at lines 6449--6449 +[] + [] + +[136 + +] +Overfull \hbox (13.73837pt too wide) in paragraph at lines 6508--6508 +[] + [] + +[137] [138] [139] [140] +Overfull \hbox (127.43008pt too wide) in paragraph at lines 6644--6644 +[] + [] + +[141] +Overfull \hbox (82.71672pt too wide) in paragraph at lines 6685--6685 +[] + [] + +[142] [143] +Overfull \hbox (50.27081pt too wide) in paragraph at lines 6733--6733 +[] + [] + + +Overfull \hbox (2.05074pt too wide) in paragraph at lines 6749--6749 +[] + [] + +[144] [145] +Overfull \hbox (60.99086pt too wide) in paragraph at lines 6827--6827 +[] + [] + +[146] [147] +Overfull \hbox (41.2357pt too wide) in paragraph at lines 6885--6885 +[] + [] + +[148] +Overfull \hbox (33.08694pt too wide) in paragraph at lines 6936--6936 +[] + [] + +[149] [150] +Overfull \hbox (30.04593pt too wide) in paragraph at lines 6991--6991 +[] + [] + +[151] +Overfull \hbox (57.59785pt too wide) in paragraph at lines 7039--7039 +[] + [] + +[152] +Overfull \hbox (24.40863pt too wide) in paragraph at lines 7052--7052 +[] + [] + +[153] +Overfull \hbox (54.40073pt too wide) in paragraph at lines 7088--7088 +[] + [] + +[154] +Overfull \hbox (55.29538pt too 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paragraph at lines 8136--8136 +[] + [] + +[179] +Overfull \hbox (75.6592pt too wide) in paragraph at lines 8158--8158 +[] + [] + + +Overfull \hbox (81.72191pt too wide) in paragraph at lines 8180--8180 +[] + [] + +[180] [181] [182] [183] +Overfull \hbox (14.36713pt too wide) in paragraph at lines 8349--8349 +[] + [] + + +Overfull \hbox (39.82089pt too wide) in paragraph at lines 8367--8367 +[] + [] + +[184] +Overfull \hbox (48.35448pt too wide) in paragraph at lines 8374--8374 +[] + [] + + +Overfull \hbox (14.62827pt too wide) in paragraph at lines 8395--8395 +[] + [] + +[185] [186] +Overfull \hbox (36.34747pt too wide) in paragraph at lines 8465--8465 +[] + [] + +[187] [188] [189 + +] [190] +Overfull \hbox (41.95776pt too wide) in paragraph at lines 8610--8610 +[] + [] + +[191] [192] [193] [194] [195] [196] [197 + +] +Overfull \hbox (5.98813pt too wide) in paragraph at lines 8846--8859 +[]\OT1/cmr/m/n/10.95 The 5th and 7th terms are $\OML/cmm/m/it/10.95 ar[]$ \OT1/ +cmr/m/n/10.95 and $\OML/cmm/m/it/10.95 ar[]$\OT1/cmr/m/n/10.95 , re-spec-tively +. + [] + +[198] +Overfull \hbox (1.00482pt too wide) in paragraph at lines 8878--8880 +[]\OT1/cmr/m/n/12 Hence\OT1/cmr/m/it/12 , the ge-o-met-ri-cal mean of any two n +um-bers is the square + [] + + +Overfull \hbox (84.25482pt too wide) in paragraph at lines 8894--8894 +[] + [] + +[199] +Overfull \hbox (35.57372pt too wide) in paragraph at lines 8921--8921 +[] + [] + + +Overfull \hbox (36.34883pt too wide) in paragraph at lines 8935--8935 +[] + [] + +[200] +Overfull \hbox (2.11494pt too wide) in paragraph at lines 8950--8952 +[][] \OT1/cmr/m/n/12 Find the com-mon ra-tio if the 1st and 3d terms are $2$ an +d $32$. + [] + +[201] [202] [203 + +] [204] [205] [206] [207] [208] [209] [210] [211] [212] [213] [214] [215] [216] +[217] [218] [219] [220] +Overfull \hbox (2.5148pt too wide) in paragraph at lines 9785--9786 +[][]$\OT1/cmr/m/n/10 14$. + [] + + +Overfull \hbox (2.5148pt too wide) in paragraph at lines 9787--9788 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Anyone seeking to utilize +this eBook outside of the United States should confirm copyright +status under the laws that apply to them. diff --git a/README.md b/README.md new file mode 100644 index 0000000..8c29382 --- /dev/null +++ b/README.md @@ -0,0 +1,2 @@ +Project Gutenberg (https://www.gutenberg.org) public repository for +eBook #36670 (https://www.gutenberg.org/ebooks/36670) |