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"^Ai.rX \iv, VnAaii"
English Classics,
Classes in English liferatitre, /heading. Grammar, etc.
With Full Explanatory Notes.
HARVARD COLLEGE
LIBRARY
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and Just appreciation of the text, and inddentallr communicate much useful
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1 Byron's Propheey of Dante. (Cantos I. and 11.)
9 Blilton's li'Allesro and II Penseroso.
a I«or(l Baeon's Essays, Cirll and Moral. (Selected.)
4 Byron's Prisoner or Chlllon.
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8 Seotf s I<ay ot tlie Last Minstrel. (Introd action and Canto I.)
9 Bums' Cotter's Saturday Nigtktf and Otber Poents.
10 CrablM's The Vllla«e.
11 Cam^Tbell's Pleasures or Hope* fAbridgmentof Part I.)
19 Macaulfiy's Bssay on Bunyan^s Pilgrim's Prog^ress.
13 Macaulay's Armadat and Other Poems.
14 Shakespeare's Merchant or Venice. (Selectioni from Acta I., III. andlY.)
19 Ooldsmlth's Trareller.
16 Hog^dp's Ctueen's IVake.
17 Colerldg^e's Ancient Mariner.
1 8 Addison's Sir Rog^er De Coverley.
10 Oray's Bleffy In a Country Churchyard.
90 Scott's Liady of the I«akc. (Canto I.)
91 Shakespeare's As You Like It, etc. (Selections.)
99 Shakespeare's Klng^ John and Kluff Richard H. (Se]ectioni>.)
93 Shakespeare's Kins Henry IV., Klnir Henry V.. Kliur Henry
VI. (Selections.) # — • » #
94 Shakespeare's Henry VIII., and Julius Caesar. (Selections.)
95 Wordsworth's Excursion. (Book I.)
96 Pope's Essay on Criticism.
97 Spenser's Faerie ^ueene. (Cantos L and II.)
98 Cowper's Task. (Book I.)
90 Milton's Comus.
30 Tennyson's Enoch Arden.
31 Irvlngr's Sketch Book. (Selections.)
39 Dickens' Christmas Carol. (Condensed.)
33 Carlylc's Hero as a Phrophet.
34 Macaulay's IVarren Hastings. (Condensed.)
35 Goldsmith's Vicar of IVakelleld. (Condensed.)
86 Tennyson's The Two Voices and A Dream ot Fair IVon&en.
87 Memory Quotations.
38 Cavalier Poets.
30 Dryden's Alexander's Feast and McFlecknoe.
40 Keats' The Ere ot St. Afpnes.
From 38 to 48 ^ugem eaeh» 16mo. Others in Prepanktion,
PUBUIHXO BT
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J s
THOMSON'S NEW SERIES OF MATHEMATICS.
NEW
PRACTICAL ALGEBRA;
ADAPTED TO
THE IMPROVED METHODS OF INSTRUCTION
IN
SCHOOLS, ACADEMIES, AND COLLEGES
WITH AN APPENDIX.
BT
JAMES B. THOMSON, LL.D.,
AUTHOR OP A SBRIBS OP MATHEMATICS.
NEW YORK:
Clark & Maynakd, Publishers,
784 Bboadway.
1884.
THOMSON'S Mathematical ISeries.
• ♦•
L A Graded Series of Arithmetics, in three Boohs, viz. :
New nitutrated Table Book, or Juvenile Arithmetic. With oral
and slate exercises. (For beginners.) 128 pp.
New Rudiments of Arithmetic. Combining Mental with Written
Arithmetic. (For Intermediate ClaBses.) 224 pp.
New Practical Arithmetic. Adapted to a complete busineps education.
(For Grammar Departments.) 884 pp.
II. Independent Books.
Key to New Practical Arithmetic. Containing many yaluable sug
gestions. (For teachers only.) 168 pp.
New Mental Arithmetic. Containing the Simple and Compound
Tables. (For Primary Schools.) 144 pp.
Complete Intellectual Arithmetic. Specially adapted to Classes in
Grain mar Schools and Academies. 168 pp.
III. Supplementary Course.
New Practical Algebra. Adapted to High Schools and Academies.
312 pp.
Key to New Practical Algebra. With full solutions. (For teachers
only.) 224 pp.
New Collegiate Algebra. Adapted to Colleges and Universities. By
Thomson & Quimby. 346 pp.
Complete Higher Arithmetic. (In preparation.)
*#* Each book of the Series is complete in itself.
Copyright, 1879, 1880, by James B. Thomson.
Electrotyped by bmith & McOougal, 62 Beekman St., New York.
Uul U 1930
PREFACE.
•«♦•
TT has long been a favorite plan of the author to make a
* Practical Algebra — a Book combining the important
principles of the Science, with their application to metUods
of business.
Several years have elapsed since he began to gather and
arrange materials for this object. Many of the more
important parts have been written and re written and
again revised, till they have found embodiment in the book
now offered to the public.
In the execution of this plan, clearness and brevity in the
definitions and rules have been the constant aim.
A series of practical problems, applying the principles
already explained, has been introduced into the fundamental
rules, thus relieving the monotony of the abstract operations,
and illustrating their use.
The principles are gradually developed, and explained in
a manner calculated to lead the pupil to a full understanding
of the difiSculties of the. science, before he is aware of their
existence.
The rules are deduced from a careful analysis of practical
problems involving the principles in question — a feature so
extensively approved in the author's Series of Arithmetics.
It PREFAOfi.
The arrangement of subjects is consecutive and logical,
their relation and mutual dependence being pointed out by
frequent references.
The examples are numerous, and have been selected with
a view to illustrate and familiarize the principles of the
Science; while puzzles, calculated to waste the time and
energy of the pupil, have been excluded.
Special attention has also been given to Factoring,
Generalization, and the application of Algebra to business
Formulas.
Iti these and other respects, it is believed, some advance
is made beyond other books of the kind. While adapted
to beginners, it covers as much ground as the majority of
students master in their Mathematical course.
In presenting this book to the public, the author ventures
to hope it may receive the approval so generously bestowed
upon his former publications.
J. B. THOMSON.
Brooklyn, N. Y., Septj 1877.
NOTE.
At the suggestion of several teachers, an Appendix has been added
to the present edition of the Practical Algebra, containing a selection
of College Examination Problems used for admission to Yale, Harvard,
and other colleges. These are preceded by a collection of examples of
a similar character, calculated to make experts in Algebra.
CONTENTS.
PAGk
Introduction^ 9
Definitions,  g
Algebraic Notation,     ' 9
Algebraic Operations,  14
Classification of Algebraic Quantities,    19
Force of the Signs, 21
Axioms, 22
Addition^ 23
Subtraction^ 29
Applications of the Parenthesis,   ^ ' ^^
Multiplication^ 35
Demonstration of the Rule for Signs,    37
Multiplying Powers of the Same Letter,    38
Principles and Formulas in Multiplication,   42
Problems,   44
Division^ 46
Cancelling a Factor, 46
Signs of the Quotient, 47
Dividing Powers of the Same Letter,     48
Dividing Polynomials,   50
Problems, 51
Factoring^ S3
Prime Factors of Monomials,  S4
Greatest Common Divisor of Polynomials,   63
Demonstration, 63
Least Common Multiple of Polynomials, •  69
6 CONTENTS.
PAGB
Fractions, 70
Signs of Fractions, 71
Reduction of Fractions, 73
Common Denominators, 77
Least Common Denominator, 79
Addition of Fractions,  .  .  80
Subtraction of Fractions, 82
Multiplication of Fractions, 84
Division of Fractions, 89
Simxde Equations, 95
Transposition, 96
Reduction of Equations,  97
Simultaneous Equations^     112
Elimination by Comparison, 113
Elimination by Substitution, 114
Elimination by Addition or Subtraction,    115
Three or More Unknown Quantities,     120
Generalisation, 124
Formation of Rules, 126
Generalizing Problems in Percentage,    128
Generalizing Problems in Interest,     131
Conjunction of the Hands of a Clock,    i33
Involution,  ^ 134
Reciprocal Powers, 135
Negative Exponents, 135
Zero Power, 136
Formation of Powers, 136
Formation of Binomial Squares,     139
Binomial Theorem,       140
General Rule, 141
Addition and Subtraction of Powers,     144
Multiplication and Division of Powers,    145
Changing Sign of Exponent,      146
Evolution, 147
Decimal Exponents,  •  * • * "^49
CONTENTS. 7
PAGB
Signs of Roots,  150
Square Root of the Square of a Binomial,   151
Square Root of a Polynomial, 152
General Rule,    153
Radical Quantities^ 154
Reduction of Radicals, 155
Addition of Radicals, 159
Subtraction of Radicals,     160
Multiplication of Radicals,      161
Division of Radicals, 163
Involution of Radicals, 164
Evolution of Radicals, 165
Changing Radicals to Rational Quantities,   166
Radical Equations, 169
Quadratic Equations, 171
Pure Quadratics. 172
Affected Quadratics, 175
First Method of Completing a Square,    176
Second Method of Completing a Square,   * '79
Third Method of Completing a Square,    180
Problems, .. 184
Simultaneous Quadratics, 187
RatiOf . 192
Proportion^ 196
Theorems,        198203
Problems, ■» 204
Arithmetical Progression^     205
The Last Term of an Arithmetical Series,   207
The Sum of an Arithmetical Series,     208
Miscellaneous Formulas in Arith. Progression,  211
Inserting Arithmetical Means,     212.
Problems,  212
Geometrical Progression^    215
The Last Term of a Geometrical Series,   216
The Sum of a Geometrical Series,     217
8 CONTENTS.
PAGB
Miscellaneous Formulas in Geometrical Progression,  220
Inserting Geometrical Means,     221
Problems, 221
Harmonical Progression,      223
Infinite Series, .... 226
Logarithms, 232
Finding the Logarithm of a Number,   . 235
To find the Number belonging to a Logarithm,   236
Multiplication by Logarithms,     * 237
Division by Logarithms,   .  .  238
Involution by Logarithms,  238
Evolution by Logarithms,      239
Compound Interest by Logarithms,     240
Table of Logarithms,      241
Mathematical Induction,     243
Business Formulas, 245
Formulas for Profit and Loss, 245
Formulas for Simple Interest,     247
Formulas for Compound Interest,     248
Formulas for Discount,       250
Formulas for Compound Discount,     251
Formulas for Commercial Discount,    252
Formulas for Investments,  253
Formulas for Sinking Funds,     255
Formulas for Annuities, 257
Discussion of Problems^     261
Problem of the Couriers,  262
Imaginary Quantities,  265
Indeterminate and Impossible Problems,    267
Negative Solutions, *268
Horner's Method of Approximation,     269
Test Examples for Review, 274
Appendix, 283
Collegiate Examination Problems,     291
Answers, ••• 295
A L GE BRA,
OHAPTEE I.
INTRODUCTI ON.
Art 1. Algebra* is the art of computing by letters ani
signs. These letters and signs are called Symbols.
2. Quantity is anything which can be measured; as
distance^ weighty time, number, &c.
3. A Measuve of a quantity is a unit of that quantity
established by law or custom, as the Standard UniL
Thus, the measure of distance is the yard; of weight, the Troy
pound; of time, the fnean »ola/r day, etc.
NOTATION.
4. Quantities in Algebra are expressed by letters, or
by a combination of letters and figures; as, a, b, c, ^
4y, szy etc.
The first letters of the alphabet are used to express known
quantities; the last letters, those which are unknown.
Qttbstiohb.— T. What is algebra ? Letters and signs called f a. Quantity f 3. A
measure? 4. How are quantities expressed f
* From the Arabic al and gdbron, redaction of parts to a whole.
10 INTRODUCTION.
5. The Letters employed have no fixed numerical
value of themselves. Any letter may represent any num
ber, and the same letter may represent different numbers,
subject to one limitation; the same letter must always stand
for the same number throughout the same problem.
6. The Relations of quantities, and the operations to
be performed, are expressed by the same signs as in Arith
metic.
7. The &ign of Addition is a perpendicular cross,
called ^Zw5/* as, +.
Thas, a+h denotes the sum of a and &« and is read, " a plus h" or
"a added to 6."
8. The Sign of Subtra^ction is a short, horizontal
line, called minus; f as, — .
Thus, a — 6 shows that the quantity after the sign is to be subtract
ed from the one before it, and is read, " a minus 5," or "a less 6."
9. The Sign of Multiplication is an oblique
cross; as, x.
Thus, axh shows that a and 6 are to be multiplied together, and is
read, "a times b," *' a into 6," or "a multiplied by &/*
10. Multiplication is also denoted by a period be
tween the factors ; as, a • b.
But the multiplication of letters is more commonly ex
pressed by writing them together, the signs being omitted.
Thus, sa&c is equivalent tosxaxftxc.
11. The Sign of Division is a short, horizontal
fine between the points of a colon ; as, v.
Thus, AT j6 shows that the quantity before the sign is to be divided
by the one after it, and is read, " a divided by 6."
5. Valine of the letters ? 6. Relations of quantities expressed? 7. Describe the
sign of addition. 8. Subtraction. 9. Multiplication. 10. How else denoted ?
* Tlie Latin tenn ptus, signifies more.
f The Latin minus, signifies less.
DEFIKITIONS. 11
12. Division is also denoted by writing the divisor
under the dividend^ with a short line between them.
Thus, ^ shows that a is to be divided bj&^and is equivalent to a{&.
13. The Sign of Equality is two short, horizontal
lines, equal and parallel; as, =.
Thus, a = & shows that the quantity before the sign is equal to the
quantity after it, and is read, " a equals h" or " a is equal to 6."
14. The Sign of Inequality is an acute angle, with
the opening turned toward the greater quantity; as, ><.
Thus, a > & shows that a is greater than &, and a < & shows that a
is less than h.
15. The Parenthesis ( ), or Vinculum y
indicates that the included quantities are taken collectively^
or as one quantity.
Thus, 3 (^ + &) and + ^x3, each denote that the sum of a and h
is multiplied by 3.
16. The Double or Ambiguous Sign is a combi
nation of the signs plus and minus; as, ±.
Thus, a±h shows that & is to be added to or subtracted from a, and
is read, " a plus or minus 6."
17. The character .*. , denotes hence, therefore.
18. Every quantity is supposed to be preceded by the
sign plus or minus. When no sign is prefixed, the sign f
is always understood.
19. Like Signs are those which are all plus, or all
minus ; as, + a + b + c, or —x—y — z.
20. Unlike Signs include both plus and minus; as,
a — b + c and —x + y — z.
IX. Describe the sign of division? 12. How else denoted? 13. The sign of
equality? 14. Of inequality ? 15. Use of a parenthesis or vinculum? x6. Double
sign ? 17. Sign for " hence," etc ? x8. By what is every quantity preceded ? When
none is expressed, what is understood? 19. Like signs ? 20. Unlike?
12 IKTBODUCTION.
21. A Coefficient* is a number or letter prefixed to a
quantity^ to show Iww many times the quantity is to be
taken. Hence, a coefficient is a multiplier or factor.
Coefficients may be numeral^ literal^ or mixed,
ThuB, in 5a, 5 is a numeral coefficient of a ; in &6, & is a literal oo<
efficient of c ; in ^dx, 3(2 Ib a mixed coefficient of x.
When no numeral coefficient is expressed, i is always
understood.
Thus, xy means ixy.
EXERCISES IN NOTATION.
22. To exprett a Statement by Algebraic SyniboU*
It is required to express the following statement ii\
algebraic symbols :
1. The product of a, i, and c, divided by the sum of c and
dy is equal to the difference of x and y, increased by the
product of a multiplied by 7.
Ana. axbxC'Tic + d) = (a; — y) + 7a.
Or , = (a: — y) + 7a. Hence, the
EuLE. — For the words, substitute the signs which indicate
the relations of the quantities and the operations to be per
formed.
Express the following by algebraic symbols:
2. The sum of 4c, d, and m, diminished by $x, equals the
product of a and b.
3. The product of 5c and d, increased by the quotient of
a divided by b, equals the product of x and y.
31. A coefficient? Wben no coefficient ie expreeeecL, what is underetood f
83. How tranelate a etatement ftom common lan^ag^ into algebraic Bjmbola ?
* Coeffickrvt, Latin, con, with, and efflcere, to effect ; literally, a
eooperator.
EXEBCISES IN NOTATION. 13
4. The quotient of 3 J divided by 5 c, increased by 4m,
equals the sum of c and 6d, diminished by the product of
7a and x»
5. If to the difference between a and b, we add the
product of X into y, the sum will be equal to m multiplied
by 6n.
6. The difference between x and y, added to the sum of
4a and i minus m, equals the product of c and d, increased
by 15 times m.
fg* These and the following exercises should be supplemented by
dictation, until the learner becomes familiar with them.
23. To translate Algebraic Eicpressions into Common
Language.
Express the following statement in conmion language :
Substituting words for signs, we have the sum of a and b,
divided by d, equals twice the product of a, b, and c, dimin
ished by the sum of x and y, increased by the quotient of
d divided by the product of a and J, Ans. Hence, the
BuLE. — For the signs indicating the given relations and
operations^ substitute words.
Express the following in common language :
2ab , !.« + *, J
2. 1 a — 6 = h axy — 4cd.
X c
3a , , 4a — bcd
4. £^^ax + bc= — h— — 3«'
5 a; 4
abc — a? . cdh 4 x
4axy a — bx+yza^d
o. 4 = •
$a X a 3c
■« .... ij
aj. How translate algebraic ezpressions into common language f
14 INTROD UCTIOK.
ALGEBRAIC OPERATIONS.
24. An Algebraic Operation is combining quanti<i
ties according to the principles of algebra.
25. A Theorem is a statement of a principle to be
proved.
25. a. A Problem is something proposed to be done, as
a question to be solved.
26. The Equality between two quantities id denoted
by the sign = . (Art 13.)
27. The Expression of Equality between two
quantities is called an Equation. Thus, 15 — 3 = 7 + 5
is an equation.
PROBLEMS.
28. The following problems are solved by combining the
preceding principles with those of Arithmetic.
I. A and B found a purse containing 12 dollars, and
divided it in such a manner that B^s share was three times
as much as A^s. How many dollars did each have ?
Bt Abfthmetic. — A had i share and B 3 shares ; now i share +
3 shares are 4 shares, which are equal to 12 dollars. If 4 shares equal
12 dollars, i share is equal to as many dollars as 4 Is contained times
in 12, which is 3. Therefore, A had 3 dollars, and B had 3 times as
much, or 9 dollars.
By Algebra.— We represent opebatioh.
A's share by x, and form an ^^* flJ = A S Share,
equation by treating this letter then 3a: ^ B's share,
as we treat the answer in proving and a: + 3a; = 1 2 dollars,
an operation. If x represent A's ^^^^ ^^ 4a; = 1 2 dollars,
share, %x will represent B's, and tt ^ j i a
J 11 n ^ Hence, a; = 3 dol, A.
a;+3aj = 12 dollars, the sum of ' j i i^
both. Uniting the terms, we 3^ = 9 ^^1., B.
have the equation, 4a? = 12 dollars. To remove the coeflSdent 4, we
34. What is an algebraic operation ? 25. A problem ? A solation ? 26. Equality
denoted ? 27. The expression of equality called ?
ALGEBRAIC OPERATIONS. 15
divide both sides of the equation by it. For, if equals are divided by
equals, the quotients are equal. Therefore, a; = 3 dollars^ A's share,
and 3aJ = 9 dollars, B*s share. (Ax. 5.)
Proof. — By the first condition, 9 dollars, B's share = 3 times 3 dol
lars, A's share. By the second condition, 9 doUars + 3 dollars =
12 dollars, the sum found. Hence,
29. When a quantity on either side of the equation has a
coefficient^ that coefficient may be removed, hy dividing
every term on loth sides of the equation hy it
2. A and B together have 15 pears, and A has twice as
many as B : how many has each ?
By Algebra. — ^If x represents opbratiok.
B's number, 2X will represent A's, Let x = B's number;
and X+2X, or 3a;, will represent ^.^^^ 22: = A's "
the number of both. Dividing _
both sides by the coefficient 3, we ^^ 3^ = I5 P^ars.
have aj = 5 pears, B*s number, and • ' • ^ = 5 P^ars, B s.
2x = 10 pears, A's. 2a; ^ 10 pears, A's.
Note,— It is advisable for the learner to solve each of the follow
ing problems by Arithmetic and by Algebra.
3. A lad bought an apple and an orange for 8 cents, pay
ing 3 times as much for the orange as for the apple. What
was the price of each ?
4. A farmer sold a cow and a ton of hay for 40 dollars,
the cow being worth 4 times as much as the hay. Whafc
was the value of each ?
5. The sum of two numbers is 36, one of which is 3 times
the other. What are the numbers ?
6. A, B, and C have 28 peaches; B has twice as many as
0, and A twice as many as B. How many has each ?
7. A father is 3 times the age of his son, and the sum of
their ages is 48 years. How old is each ?
29. How remove a coeflicient ?
16 INTEODUCTION.
8. A and B trade in company^ and gain loo dollars. If
A puts in 4 times as much as B, what will be the gain of
each?
9. The sum of three numbers is 90. The second is twice
the first, and the third as many as the first and second:
what are the numbers ?
10. A cow and calf were sold for 6^ dollars, the cow being
worth 8 times as much as the calf. What was the value of
each?
11. A man being asked the price of his horse, replied
that his horse, saddle and bridle together were worth
126 dollars; that the saddle was worth twice as much as
the bridle, and the horse 7 times as much as both the otherr
What was each worth ?
12. A man bequeathed $36,000 to his wife, son and
daughter, giving the son twice as much as the d^aghter,
.and the wife 3 times as much as the son and daughter.
What did each receive ?
13. The sum of three numbers is 1872 • the second is
3 times the first, and the third equals the o^ner two. What
are the numbers ?
POWERS AND ROOTS.
30. A Power is the product of two or more eqttdl
factors.
Thus, the product 2 x 2, is the square or second power of 2 ;
2 X a; X a; IB the cube or third power of x,
31. The Index or Exponent of a power is a figure
or letter placed at the right, above the quantity.
Thus, a} denotes a, *or the first power.
a' " ax a, the square^ or second power.
cfi " ax ax a, the ctibe, or third power, etc.
32. A Moot is one of the equal factors of a quantity.
30. What ie a power? 31. How denoted?
ALGEBRAIC EXPRESSIONS. 17
33. Moots are denoted by the Radical Sign y'
prefixed to the quantity, or by a fractional exponent placed
after it.
Thus, y^, a*, or /y/a denote the square root of the quantity a ;
^/a shows that the cube root of a is to be extracted, etc.
34. The Index of the Root is the figure placed
over the radical sign. The index of the square root is
usually omitted.
(For negative indices, see Arts. 256, 258.)
Bead the following examples:
1. a^ + 30. 7. 4 (a — h)\
2. 58 — A 8. a2 + 2al f 5^.
Z> a + V^c. 9. V« + J.
$, 2t^ + ^~Z. II. 2a* + C^.
6. 3 (a2 + J). 12. 4a;* + 2y* '
Write the following in algebraic language :
13. The square of a plus the square of J. . /a
14. The square of the sum of a and h, ■ C ^ ^ 'Cj "^
15. The sum of a and S, minus the square of c.
16. The square root of a, plus the square root Qi x. ^ \^  *^
17. The cube root of x^ minus the fifth power of v, X^
18.. The cube
root of x^ minus the fifth power of y. X^^
root of a, plus the square of S. .  7^ * ^
ALGEBRAIC EXPRESSIONS.
35. An Algebraic Tkcpression is any quantity ex
pressed in algebraic language ; as, 3a, sa — 7S, etc.
36. The Terms of an algebraic expression are those
parts which are connected by the signs + and — .
Thus, in a + &, there are two terms ; in aj+y x g— a there are three.
32. A root? 33. How denoted? 34. What is the figure placed over the radical
Aign called f 35. What is an algebraic ezpree^iiion 7 36. Its terms 7
18 INTRODUCTION.
Note. — Letters combined by the signs x or 4 do not constitute
separate terms. Such a combination, to form a term, must have the
sign + or — prefixed to it, and the operations indicated by the signs
X or f must be performed before the terms can be added to or sub
tracted from the preceding term. (Art. 36.) Thus, a + hxc has two
terms, hxc forming one term and a the other.
37. The Dimensions of a term are its several literal
factors.
38. The Degree of a term depends on the number of
its literal factors^ and is always equal to the sum of theii
exponents.
Thus, ab contains two factors, a and &, and is of the second degree.
a^x contains three factors, a, a, and Xy and is of the thi/rd degree.
1^7? contains five factors, 6, h, x, x, x, and is of the fifth degree.
39. The Numerical Value of an algebraic expres
sion is the number which it represents when its terms are
combined as indicated by the signs. (Art. 36.)
40. To Find the NumericalValiie of an algebraic expression*
1. If a = 5, S = 7, and a? = 9, what is the yalue of
6a + Sb + sx?
Analysis. — Since a = 5, 60 must ovE&Artos,
equal 6 times 5, or 30 ; since & = 7, 8& ^ e v 6 «o
must equal 8 times 7, or 56 ; and since ^y ^ ~
ic = 9, 3X must equal 3x9, or 27. Now ' ^
30 + 56 + 27=113. Therefore, the yalue 3^ ^== 3 X 9 =* ^7
of the given expression is 113. Hence^ Ans» 113
the
EuLE. — For the letters, substitute the figures which the
letters represent, and perform the operations indicated by
the signs,
2. If ft = 3, c = 5, and c? = 8, what is the value of
5b + 7c + 6d?
Suggestion, i 5 + 35 + 48 = 98, Ans.
37. The dimenBions of a term ? 38. Degree ? 39. Numerical value of an alge
braic cxpreBsion ? 40. Hqw foniid ?
ALGEBRAIC QUANTITIES. 19
Find the numerical value of the following expressions,
when a = 2, ^ = 3, c = 4^ c? = 5, and a; = 6.
3. 4^ f 6ab + 5^.= how many ? Ana. 64.
4. (a^+ b^ X c X d — (x ^ c) =: how many ?
5. (.T — fl^) + aif (c ^ a) = how many ?
6. X ^ 2 + {d — c) + be — x = how many?
7. dx— {ax c)+ {a X b) \ x = how many?
8. d + (a; X a) + a — a; + c = how many?
CLASSIFICATION OF ALGEBRAIC QUANTITIES,
41. Quantities in Algehra are primarily divided into
known and unknown,
42. A Shown Quantity is one whose value is given.
An Unknown Quantity is one whose value is not
given.
These quantities are subdivided into like and unlike, posi
tive and negative, simple, compound, monomials, etc.
43. Idke Quantities are those which are expressed
by the same power of the same Utters; as, a and 2a,
20? and a?.
44. Unlike Quantities are those which are expressed
by different letters, or by different powers of the same let
ters ; as 2x and 3^, 2X and a?.
Note. — An exception must be made in cases where letters are re
garded as coefficients. Thus, ax^ and M are like quantities, when a
and b are considered coefficients.
45. A Positive Quantity is one that is to be added,
and has the sign + prefixed to it ; as, 4a + 3 J.
46. A Negative Quantity is one that is to be sub*
traded, and has the sign — prefixed to it; as, 4a — 35.
41. How are qnantltieB in Algebra primarily divided ? 42. A known qnanfcity?
Unknown? 43. Like quantities? 44. Unlike? 45. A positive quantity? 46. A
negative ?
20 INTEODUOTION,
47. The terms Positive and Negative denote oppo
fiiteness of direction in the use of the quantities to which
they are applied. If lines running North from any point
• are positive, those running South are negative. I^ future
time is positive, past time is negative ; if credits are positive,
debts are negative, etc.
48. A Simple Quantity is a single letter^ or seyeral
letters written together without the sign + or — ; as, a,
db, 3xy.
49. A Compound Quantity is two or more simple
quantities connected by the sign + or — ; as 3a f 4^^
2a; — y.
50. A Monomial * has but one term ; as^ a, 3d.
51. A Sinomial f has two terms ; aB, a + b,a — b.
Notes. — i. The expression a — & is often called a retidual, because
it denotes that which remains after a part is subtracted.
2. A Unomial is sometimes called a polynomial*
52. A Trinomial % has three terms; as, « + J — c
53. A Polynomial\ has two or more terms; as,
a + b — c + X.
54. An Homogeneous Polynomial has all its
terms of the same degree.
Thus, 2db +cd+ ^xy is homogeneous ; but i^abe +6^ + 50; is not.
55. The JReeiprocal of a quantity is a unit divided by
that quantity.
Thus, the reciprocal of a is  ; the reciprocal of a+6 is — r •
47. What do the terms positive and negative denote? 48. A simple quantity.'
49. A compound? 50. A monomial? sx. A binomial? Note, The expression
a—b called? 52. A trinomial? 53. A polynomial? 54. When homogeneous^
55. The reciprocal of a quantity ?
* Greek, rnonos, single, and nome, term, having one term,
f Latin, bis, two, and name, name (a hybrid), ttoo terms.
t Greek, treis, three, and nome, name, having three terms.
I Greek, poluSy man^, and r*<?me, name, having many terms.
FOBCE OF THE SIGNS. 21
FORCE OF THE SIGNS.
56. Uach term of an algebraic expression is preceded
by the sign + or — , expressed or understood. (Art i8.)
The Force of each of these signs is limited to the term
which follows it;as, 7 + s — 3 = 12 — 3 = 9; 15 — 6 + 8
= 9 + 8= 17.
57. If a term, preceded by the sign + or — , is combined
with other letters by the sign x or ^, each of these let
ters forms a part of that term, and the operations indicated,
taken in their order, must be performed before any part of
the term can be added to or subtracted from any other term.
«
Thus, the expreeunon 12 + 4 x 2, shows that 4 is to be multiplied by
2 and the product added to 12, and is equal to 20.
In like manner, the expression 16 — 8 ^ 2, shows that 8 is to be
divided bj 2 and the quotient subtracted from 16, and is equal to 12.
58. K two or more terms joined by + or — are to be
subjected to the same operation, they must be connected by
a parenthesis or vinculum.
Thus, if a + 6 or a — 6 is to be multiplied or divided by c, the oper
ations are indicated by {a+b)x c, or c(a+6) ; (o — 5) * c, or "" •
c
EXERCISES.
1. 50 + s X 2 = what number?
2. 50 — s X 2 = what number ?
3. ac + 4^ X 2 = what ?
4. 5 J — 6rf 5 3 = what ?
5 15 + S X 3 + 10 ^ 2 = what?
6. 18 — 2x472 + 10 = what ?
7. 3a; + 8y h 4 + a X S = what?
8. 6J — 7(? X a; + 9a ^ 3 = what ?
9. {b + c) X xy = what ?
56. By what are all algsbraic terms preceded ? The force of each of theee signs t
57. Of the signs x and *i 58. Of the parenthesis and vincnlnm ?
22 INTBOBUOTIOK.
10. 32? X sy 4 2j2r + a = what?
11. (b'a) ^xy + 2Z = what ?
12. sx + xy + 2z X sy = what ?
13. The difference of x and y, multiph'ed by a less ^>, and
the product divided \)Yd = what ?
Find the yalue of the following expressions, in which
a = 3, ft = 4,c = 2, ir = 6, y = 8, and z= 10:
14. a + (axx)^c + yxz = what ?
14 '2b'i'{xi) + a X d X y j 2z = what ?
AXIOMS.
59. An Axiom is a selfevident truth.
1. Things which are equal to the same thing, are equal to
each other.
2. If equals are added to equals, the sums are equal.
3. If equals are subtracted from equals, the remainders
are equal
4. If equals are multiplied by equals, the products are
equal.
5. If equals are divided by equals, the quotients are equal.
6. If a quantity is multiplied and divided by the same
quantity, its value is not altered.
7. If the same quantity is added to and subtracted from
another quantity, the value of the latter is not altered.
8. The whole is greater than its part.
9. The whole is equal to the sum of all its parts,
10. Like powers and like roots of equal quantities, are
tqtial.
Note. — The importance of thoroughly understanding the defini^
turns and principles cannot be too deeply impressed upon the mind of
the learner. The questions at the foot of the page are designed to
direct his attention to the more important points. Teachers, of course
will not be confined to thenL
CHAPTER II.
ADDITION.
60. Addition in Algebra is uniting two or more quan
tities and reducing them to the simplest form.
61. The Result is called the Sum or Amount.
62. Quantities expressed by letters are regarded as
concrete quantities. Hence, their coefficients may be added,
subtracted, multiplied, and divided like concrete numbers.
Thus, 3a and 4a are 7a, df> and 56 are cjb, as truly as 3 apples and
4 apples are 7 apples, or as 4 bushels and 5 bushels are 9 bushels.
PRINCIPLES.*
63. 1°. Like quantities only can he united in one term.
2**. The sum of two or more quantities is the same in
whatever order they are added.
CASE I.
64. To Add like Monomials which have like signs.
I. What is the sum of i^ah + isaS + i^ah ?
Analysis.— These terms are like quantities opbratiow.
and have like signs. (Art. 19.) We therefore , •+ 15^5
add the coefficients, to the sum annex the com ^ i^ai
mon letters, and prefix the common sign. The 1 jg^j
result, + 4706, is the answer required. (Ax. 9.)
f 47aft, Arts.
60. What is additiou t 61. The reealt called ? 63. How are quantities expressed
by letters regarded ? 63. First prisciple ? Second ?
^ The expiesnons i% 2"*, 3% etc., denote >Ir«f, ieeond, third, etc.
24 ABDITIOK.
2. What is the sum of — 142:^, — i6xy, and — iSxy?
Analysis. — Since these terms are like quan — I4^y
tities, and have like signs, we add them as i6xy
before, and prefix the sign — to the result, for iSiPt/
the reason that all the quantities have the
sign — . Hence, the — 4^2:^, Ana.
BuLE. — Add the coefficients; to the sum annex the com
mon letters, and prefix the common sign.
(3.)
(4.)
(5.)
(6.)
(7.)
3«*
s^y
7a*
— "jhcd
^42^2
sab
ixy
3«*
 Sbcd
— 3^y^
6ab
/*y
4^2
^Sbcd
 ^y^
yab
3fey
\^
— Sbcd
— Sa^
8. Add 5aJ2 + i jai^ + 2sat^,
9. Add — Saba^f — sabxh/^— 28afta%8.
10. Add sl^dm^ + Wdm^ + pJ^rfw^ + Wdm\
1 1. If 3a + 5a + a + 7a = 48, to what is a equal ?
SoLTjnoK. 3a + 5a + a + 7a= i6a ; hence, ^=48 h 16, or 3. A^m,
12. If 4SC + gbc + 2bc + sbc = 80, to what is be equal ?
13. If xy + ^y + sxy + /^y = 65, to what is xy equal ?
CASE II.
65. To Add like Monomials wliich have Unlike signs.
14. What is the sum of saJ — 306 — ^db + gab + 6ab
^Sab?
Analysis. — For convenience in operation.
adding, we write the negative terms $0^ — sab
one under another in the right gab — yab
hand column, with the sign — be g^j g^j
fore each, and the posUvoe terms ~ T"^ , .
in the next column on the left. """^^  iSob = 2ah, Am,
We then find the sum of the coefficients of the positive and negative
64. How add monomialB which have like Big&Bt
t
ADDITION. 36
tenns separately ; and taking the leas smn from the greater, the result
206, is the answer. Hence, the
EuLE. — I. Write the positive and negative terms in sepo'
rate columns with their proper signs, and find the sum of the
coefficients of each column separately.
II. From the greater subtract the less ; to the remainder
prefix the sign of the greater, and anjiex the common letters.
Note — If two equal quantities have oppotUe signs, they balance
each other, and may be omitted.
15. Add 4^ + 36? — srf + 6rf — 2d. Ans. 6d.
16. Add — $x + 6x + ^x — 3X + gx — 7a?.
17. Add ^abc + i2abc — 6abc + sabc — loabc •— ^abc.
18. Add 2 J — 5 J + 46 — 6S — 76.
19. Add — 6y + 4y — 8y — 9y + 8y — y.
20. Add 4m + i6m — Sm — 9m + 5m — lom.
21. If Sab + i^ab — "jab + i$ab — i2ab + i6ab = 32, to
what is ab eqnal ?
22. To what is bed equal, if bed — zf>cd + ^bcd + ^bcd
Sbcd=:'j5?
Remabk. — The sum in Arithmetic is ahoays greater than any of its
parts. But, in Algebra, it will be observed, the sum of a posUvoe
and negative quantity is always less than the posUi/oe quantity. It is
thence called Algebraic Sum.
66. Unlike Quantities cannot be united in one term.
Their sum is indicated by writing them one after another,
with their proper signs. (Art. 6$, Prin. i.)
Thus, the sum of jg and sd is neither loi^r nor lod, any more than
7 guineas and 3 dollars are 10 guineas or 10 dollars. Their sum is
7g + 3^. (Art. 63, Prin. i.)
67. Polynomials are added by uniting Uke quantities,
as in adding monomials.
65. How add monomials having nnlike Bigms ? Bern. Wbat ia trae of the Bam in
Arithmetic f In Algebra f 66. How add unlike qoantitlea f 6j, FoIynomialB f
a
26 ADDITION.
23. What is the sum of the polynomial s^b — 3} + 4d
— 3^1 — 5^* + 4X'c— 2d; and bg + d + lab + J ?
Analysis. — ^For opkbatioh.
convenience, we 3a J — 3^ + 4^ — ZX — c
write the quanti — ^ab + b — 2£? f 4^
des BO that like gflj + (Z + bg
terms shall stand ~ ;
one under another.  2& + 3^ + x + hg  c, Ans.
and uniting those which are alike, the result is —2b+3d+x+l>g^e.
68. From the preceding illustrations and principles we
deduce the following
GENERAL RULE.
L Write the given quantities so that like terms shall stand
one under another.
n. Unite the terms which are alike, and to ths result
annex the unlike terms with their proper signs. (Art. 65.)
1. Add 5« — 3a + 6a + 7a + 9a + 2J — 3d.
2. Add 8mw + 3WW — ^7nn + i^mn — xy + be.
3. Add $bc — jbc + xy — mn + iiJc + gbc.
^ Add sab — smn — ab + ^ab + 2Z — ^ab + a^.
5. Add sxy — xy + ab''jxy + b + Sxy — xy + i^xy.
69. Compound Quantities inclosed in a paren
thesis, are taken collectively, or as one quantity. Hence, if
the quantities are alike, their coefficients and exponents are
treated as the coefficients and exponents of like monomials.
(Art 64.)
6. What is the sum of 3 (a+ J) + 5 {a + b) + 7 {a + b) ?
Solution. 3 (<*+ 6) and 5 (a + 6) and 7 (a + 6) are 1 5 (a + 6). Ans.
7. Add 13 {a + b) + 15 {a + b) 7 (a + *).
8. Add 8c {x — y)+ 7c (a;— y) — 5c {x—y) + 9c (a;— y).
9. Add z^Vxy + saVxy — ya^/xy + SaVxy.
10. Add 5 Va + ^Vct — S^/a + 9 Va —  3^^.
1 1. Add sVx — y — ^Vx — y + s Vx — y.
68. The general mle for addition ? 69. How add qnantitiee indaded in a parens
tbetinr
PE0BLEM8. 27
70. The sum of unlike quantities havings common letter
or letters, may be expressed by inclosing the otJier letters,
with their signs and coefficients, in a parenthesis, and an
nexing or prefixing the common letter or letters to the result.
12. What is the sum of <^ax + ilx — 4ca;?
Solution. 5aa;+ 360?— 4ca; = (5a+ 3&— 4<')iP, or a; (5a + 3^—4^). 4n<.
13. Add 7a — dha + 3^^ — 3ma.
14. Add aly + 3^ — 2^y — ^'^ny.
15. Add 9m + ahm — 7C7W + 3^7W.
16. Add i3«a; — 3^0: ■\ ex — '^dx + ??kb.
17. kAA.axy + te^ — cxy.'
PROBLEMS.
71. Problems requiring equal quantities to be added to each
side of the equafion*
1. A has 3 times as many marbles as B, la^jking 6 ; and
both together have 58. How many has each ?
Analysis. — If x represents opbeation.
B's number, then wUl 3a;— 6 I^et X = B's No. ;
represent A's, and a;+3aj— 6 ^^j^ la;  6 = A's "
= 58, the sum of both. To 'a; + 3a:  6 = 58, both,
remove ~6, we add an equaX
po9itive quantity to each side ^ + 3^ ofo 5""r^
of the equation. (Axiom 2.) 4a; = 64
Uniting the terms, we have iC= 16, B's No.
4X = 64> and x = 16, B's, and 3a; — 6 = 42, A's "
3 times 16—6, or 42 = A's No.
72. When a negative quantity occurs on either side of an
equation, that quantity may be removed by adding an equal
positive quantity to both sides.
Note. — In forming the equation, we treat x as we d6 the answer
in proving an operation.
2. A kite and a ball together cost 46 cents, and the kite
cost 2 cents less than twice the cost of the balL What was
the cost of each ? he
70. How may the sum of onlike qaantitieB wbich have a common letter '
presse^i ?
is ADDITION.
3. In a basket there are 75 peaches and pears ; the num
ber of pears being double that of the peaches, wanting 3.
How many are there of each?
4. The sum of two numbers is 85, and the greater is
5 times the less, wanting 5. What are the numbers ?
5. A certain school contains 40 pupils, and there are
twice as many girls, lacking 5, as boys. How many are
there of each ?
6. K 442? + 65a? — 24 = 85, what is the value of x ?
7. K 7a: — 3 + 2a; = 60, what is the value of a; ?
8. If 4y + 2^ + sy — 7 = 70, what is the value oiy?
9. The whole number of votes cast for A and B at a cer
tain election was 450 ; A had 20 votes less than 4 times the
number for B. How many votes had each ?
10. The sum of two numbers is 177 ; the greater is 3 less
than 4 times the smaller. What are the numbers ?
11. What is the value of y, if 4^ + 3y f 27 — 12 = 60 ?
12. A lad bought a top and a ball for 32 cents ; the price
of the ball was 3 times tliat of the top, minus 4 cents.
What was the price of each ?
13. A man being asked the price of his saddle and bridle,
replied that both together cost 40 dollars, the former being
4 times the price of the latter, minus 5 dollars. What was
the price of each ?
14. A lad spent a dollar during a holiday, using three
times as much of it in the afternoon as in the morning,
minus 4 cents ; how much did he spend in each part of the
day?
Find the value of x in the following equations :
15. 3a? + 6a; + 4a; + sar — 8 = 154. Am. 9.
16. 2a; + sa? + 3a: — 10 = 130.
17. 4a; + 3a; + 7a;— 12 =86.
18. loa; — 4a; + 9a; — 25 = 155.
19. isa? — 7a; — 2a; — 60 = 300.
«o. 18a: — 4a: + a? — 75 = 225.
CHAPTER III.
SUBTRACTION.
73. Siibtraction is finding the difference between two
quantities.
The Minuend is the quantity from which the subtrac
tion is made.
The Subtrahend is the quantity to be subtracted.
The Difference is the result found by subtraction.
74. Since quantities expressed by letters are regarded as
concrete^ the coefficient of one letter may be subtracted from
that of another, like concrete numbers. (Art. 62.)
Thus, 7a — 3a = 4a; 85 — 56 = 36.
PRINCIPLES.
75. i^ Like quantities only can he subtracted one from,
another.
2^ The sum of the difference and subtrahend is equal to
the minuend.
3^. Subtracting a positive quantity is equivalent to add*
ing an equal negative one.
Thus, let it be required to subtract + 4 from 6+4.
Taking +4 from 6+4, leave3 6.
Adding —4 to 6 + 4, we have 6 + 4—4.
But (Ax. 7) 6+4—4 5s equal to 6.
4^ Subtracting a negative quantity is the same as adding
an equal positive one,
73. Define subtraction. TbeMinncnd. Sabtrahend. Difibrence. 75. Name the
ilnit principle. Second. Uloetrate Prin. 3 npon the blackboard. Ulastrate Prin. 4,
30
SUBTRACTION.
Thus, let it be required to subtract —4 fiom 10—4.
Taking —4 from 10—4, leaves 10.
Adding +4 to 10—4, we have 10—4+4.
But (Ax. 7) 10—4+4 is equal to 10.
Again, if the assets of an estate be $500, and the liabilities $300,
the former being considered positive and the latter negative, the net
▼alue of the estate will be $500— $300 = $200. Taking $50 from the
assets has the same effect on the balance as adding $50 to the liabilities.
In like manner, taking $50 from the liabilities has the same effect as
adding $50 to the assets,
76. To Find the Difference between two like Quantities.
This proposition includes three classes of examples, as
will be seen in the following illustrations:
I. Prom 25a subtract lya.
Remabk. — I. In this example the signs are
alike, and the subtrahend is less than the min
uend. Subtracting a positive quantity is
equivalent to adding an equal negatim one.
(Prin. 3.) We therefore change the sign of
the subtrahend, and then unite the terms as in addition. Thus,
250— 17a = 8a.
OFBBATIOH.
25^ Mlnnend.
— l']a Subtrahend.
8(7 Difference.
2. Prom 4a subtract ya.
Remabk.— 2. In this example the signs
are alike, but the subtrahend is greater than
the minuend. Changing the sign of the sub
trahend, and uniting the terms as before, the
subtrahend cancels the minuend, and has
—3a left. (Prin. 3.)
3. From 45^? subtract — 2gab,
Remabk. — 3. In this example the signs
are unlike. Subtracting a negative quantity
is the same as adding an equal positive one.
(Prin. 4.) Changing the sign of the sub
trahend and proceeding as before, we have
4506 + 2906 = 740^. Ans,
OPERATION.
4a Minnend.
— 7^ Subtrahend.
— 3a Difference.
OFEBATION.
45 ad Minuend.
+ 2gah Subtrahend.
74^5, Ans.
76. How find the difference between two like quantities f
SUBTRACTION. 31
4. From gbc + yd — 5a?, take 3 fe + 2^ — 42;.
A17ALYSIS. — In snbtraction of ofbbation.
polynomials, for conveniencef we gic + ^d KX
place like tenns under each other. ^j^ 2d ^ dX
Then, changing the signs of all the r
terms in the subtrahend, we unite ^^^ + 5^ — ^y ^^^'
them as before.
77. From the preceding Ulustrations and principles we
deduce the fallowing
GENERAL RULE.
1. Write the subtrahend under the minuend, placing like
terms one under another.
II. Change the signs of all the terms of the subtrahend, or
suppose them to be changed, from + to — , or from — to^
+, and then proceed as in addition, (Art. 75, Prin. 3, 4.)
Notes. — i. Unlike quantUies can be subtracted only by changing
the signs of all the terms of the subtrahend, and then writing them
after the minuend. (Art. 66.)
2. As soon as the student becomes familiar with the principles of
subtraction, instead of actually changing the signs of the subtrahend,
he may simply suppose them to be changed.
EXAMPLES.
1. From 43c + d, take 25c + d. Ans. iSc.
2. From 49X, take 23a; + 3. Ans. 26X — 3.
3. From 2Sxyz, take i4xyz.
4. From — 43ab, take + igab.
5. From 4ab, take — 150^.
6. From 43iry, take + i6a:y.
(7.) (8.) (9.) (10.)
From 2oac 42aa? sya^ — 292:2^2
Take — 230^ saa? — 14«^* + iS^f
77. General role for Bubtraction ? Note. How subtract milike quantities.
(18) .
From
^xy — %a
Take
ixy — 2a
32 BUBTBAGTIOIT.
(ll.) (I2.) (13.) (14.)
From 3ia?J 190*2^ —33*^^ ^^^
Take — 7a'J 190^ + 44^^ — 123^
15. A is worth ioo, and B owes $50 more than he is
worth. What is the difference in their pecuniary standing ?
16. What is the difference in temperature, when the ther
mometer stands 15 degrees above zero^ and when at 10
degrees below ?
17. By speculation, A gained on a certain day $275, and
B lost $145. What was the difference in the results of their
operations ?
(19.) (20.)
8J* + lam 130^ — 7^^
— 552 — 9am — • 5^ if 8y^ ■/ 6fl
21. From iTflb + d^x, subtract sm — 3«. ^ ■ i ^
22. From gcd — ab, take 2m — 3» — 4y.
23. From 13m — 15, take — sm + 8.
24. From 72;^ — 50: + 15, take — 5a? + 8a? + 15.
25. From igab — 2c — yd, take ^ab — 15^ — 8d.
26. From a, take & — c, and prove the work.
27. From II {a + J), take 5 {a + b).
28. From 17 (a ~ J + x), take 8 (a — J + a?).
29. Subtract — 18 (a + b) from — 13 (a + S).
30. Subtract 21 (a;* — y) from 14 (a;* — y).
31. A and B formed an equal partnership and made
$1,000. B's share by right was $1,000 — $500; but wish
ing to withdraw, he agreed to subtract $100 from his share.
What would A*s share be ?
32. What is the difference of longitude between two
places, one of which is 23 degrees due east from the meridian
of Washington, the other 37 degrees due west?
Reicark. — The subtrahend, in Algebra, is often greater than the
minuend, and the dijference hetween a positive and negative quantity
greater than either of them. It is thence called Algebraic Dif"
fereiice.
8UBTBACTI0N. 88
\
78. The Difference of unlike quantities which have a
common letter or letters may be indicated by enclosing all
the other letters, with their coefficients and signs^ in a
parenthesis^ and anneadng, or prefixing the common letter
or letters to the result
33. Prom 3awi, take 2bm.
Analysis, yim = 3a times m, and 2bm = ib times m ; theref oxe^
3am—2bm = (3a— 26) m, or m (3a— 2&). Ana,
34. Prom 2ba?, take ca^ — dofl.
35. Prom abff, take cy + dy ^ xy.
36. Prom 7^2, take ha^ — caK
37. Pipom aJa;, take ^cx + dx + mx.
38. Prom ^xyy take aJar^y — ca;y + dxy.
39. Prom sac + hmc, take 3ac — do.
APPLICATIONS OF THE PARENTHESIS.
79. A parenthesis, we have seen, shows that the quanti
ties inclosed by it are taken collectively, and subjected to
the operation indicated by the sign which precedes it.
(Art. 15.)
80. A parenthesis having the sign + prefixed to it, may
be removed from an expression, if the sig^is of the included
terms remain unchanged.
Thus, a— 6 + (c— d + 6) = a— 6 + c— d + «. Hence,
81. Any number of terms may be inclosed in a parenthe
sis and the sign + placed before it, if the sig^ns of the
inclosed terms remain unchanged.
Thus, a+6— c+d = a+Q>—c\d), or a+6+(— c+cf).
Note. — ^This principle affords a convenient method of indicating
the addUion of polynomials. (Art. 67.)
78. How subtract unlike qnantities having a common letter or letters ? 79. What
is the object of a parenthesis ? 80. How removed when the sign + is prefixed to It.
34 8UBTBACTI0N.
82. A parenthesis having the sign — prefixed to it, may
be removed by changiufj the signs oi all the inclosed terms
from + to — and — to +•
Thus, lemoving it from tlie equal expressions,
\^ , '>■=« — 6 + c — d. Hence,
a — 6 — (d — c) J
83. Any number of terms may be inclosed by a paren
thesis, and the sign — placed before it, if all the signs of
the inclosed terms are changed.
Thus, a—h+c—d = a— (6— c+d!)» or a— 6— (— c+<2), etc
Note. — This principle enables us to express a polynomial in differ
ent forms without changing its value.
1. How express a — x + c, using a parenthesis?
Ans. a ^ X + c = a ^ {x — c)y OT a — {— c + x).
2. How express a — 5 — x — y + z, using a parenthesis ?
Ans. fl — J — (a; + y —;?), or
a — A — (y + a; — 2;), or
a — b — {— z + X + y).
84. When two or more parentheses occur in the same
expression, they are removed by the same rule, beginning
with the interior parenthesis.
Thus, a—[b^c{d+x) + e]=a—(b—c—d—x + e)=a'^b+c + d+x—e.
Note. — Quantities may be included in more than one parenthesis^
by observing the preceding rules.
Eemove the parentheses from the following expressions*.
5. ab — {be —  d). Ans. ab — be + d,
4. b — {c — d + m). •
S Sa? — (— y + «^ — 4cl).
6. 2a — [& + c — {x + y) — d].
7. a — (b — c) — {a — c) \ c — {a — b).
JgP" The principles governing the signs in the use and removal of
parentheses should be made familiar by practice.
82. How when the sign — is prefixed ? 83. How iDcloBe terms in a parenthesis
with — prefixed to it ?
CHAPTER IV.
MULTIPLICATION.
85. Multiplication is finding the amount of a quan
tity taken or added to itself, a given number of times.
TIitLS, 3 times 4 are 12, and 4 taken 3 times (4+4+4) = 12.
The Multiplicand is the quantity to be multiplied.
The Multiplier is the quantity by which we multiply.
The Product is the quantity found by multiplication.
86. The Factors of a quantity are the multiplier and
multiplicand which produce it.
PRINCIPLES.
87. 1°. The multiplier must he considered an abstract
qv^antity.
2°. The product is of the same nature as the multiplicand;
for, repeating a quantity does not alter its nature.
3°. The product of two or more factors is the same in
whatever order they are multiplied.
CASE I.
88. To Multiply a Monomial by a Monomi4Ml.
I. What is the product of a multiplied by c?
Ans. a X Cy or ac.
Note. — ^The product of two or more letters, we have seen, is ex
pressed by writing them one after another, either with or without the
sign of multiplication between them. (Art. 10.)
85. Define mnltlplicatlon. The multiplicand. Moltiplier. Product. 86. Fao
tors. 87. Name Prln. x. Prin. 3. Prin. x.
36 XULTIPLICATIOK.
2. If I ton of iron costs a dollars^ what will x tons cost ?
Analysis. » tons will cost x times as mucli aa i ton ; and x times
a dollars are ax dollars. That is, a dollars are taken x times, and are
equal to a+a+a . . . . , and so on to a; terms.
3. What is the product of 4a by 2 J ?
Analysis. — Since each coefficient and each letter onBA.TxoH:
In the multiplier and multiplicand is a factor, it fol 4a
lows that the answer must be the product of the 2b
coefficients with all the letters of both factors an ^ TT
. rr xi. Ans. oao
nezed. Hence, the
BuLE. — Multiply the coefficients together^ and prefix the
product to the product 0/ the literal factors.
Multiply the following quantities:
4.
4fl J by 5a:.
Ans. 2oabx.
5
6bc by ja.
9. yxy by Sab.
6.
jabc by secy.
10. 6ac by ;&?.
7.
8dm by xy.
II. gbdhyScnu
8.
gbcd by jxyz.
12. jxyz by ga^.
SIGNS OF THE PRODUCT.
89. The investigation of the laws that govern the signs
of the product, requires attention to the following
PRINCIPLES.
i^. Eepeating a qiiantity does not change its sign.
2°. The sign of the multiplier shows whether the repetitions
of the multiplicand are to be added, or subtracted.
90. If the Signs of the factors are alike, the sign of the
product will he positive; if unlike, the sign of the product
will be negative.
88. How multiply a monomial by a monomial ? 89. Name Principle x. Prin. %
90. If signs of factors are alike, what is the sign of the product ? If unlike t
MULTIPLICATION. 87
91. Demonstration.— There are four points to be
proved:
First. That f into + produces +.
Let +a be the multiplicand and +4 the multiplier. It \b plain
that +a taken +4 times is +40. (Prin. i.) The sign of the multi
plier being +, shows that the. product +40, is to be added, whidi is
donp by setting it down without changing its sign. (Art. 66.)
Second. That — into + produces — .
Let —a be multiplied by +4. Now —a taken 4 times is —4a ; for
a negative quantity repeated is still negative, (Prin. i.) But the
sign before the multiplier being + , shows that the negative product
— 4a, is to be added. This also is done by setting it down without
changing its sign. (Art. 66.)
Third. That + into — produces — .
Let +a be multiplied by —4. We have seen above that + a taken
4 times is +40. But here the sign of the multiplier being — , shows
that the product h^a, is to be subtracted. This is done by changing
its sign from + to — , on setting it down. Thus, +a x —4 = —40,
(Art 77.)
Fourth. That — into — produces +.
Let —a be multiplied by —4. It has also been shown that —a
taken 4 times is —4a. But the sign of the multiplier being — , shows
that this negative product —4a, is to be subtracted. This is also done
by changing its sign from — to +» when we set it down. Thus,
—a X —4 =440. (Art. 77.) Hence, universally,
92. Factors having like signs prodtice +, and unlike
signs — .
13. Multiply + 4ab by — *jcd. Ans. — 2Sabcd.
14. Multiply — s^y ^y + 9«fc
15. Multiply + 6ab by + jdc
16. Multiply — Sxy by — igdbc.
17. Multiply + iSabc by — 2^xy.
18. Multiply — ssxy by — Q^jbcd.
gx. Prove the first point from the blackboard. The second. Third. Fonrth
93. Rule for signs.
38 MULTIPLICATIOir.
93. When a letter is multiplied into itself, or taken twice
as a factor, the product is represented by a x a, or aa;
when taken three times, by aaay and so on, forming a series
of powers. But powers, we have seen, are expressed by
writing the letter once only, with the index above it, at the
right hand. (Art 31.)
19. What is the product of aaa into aa?
Analysis, cuia xaa = aaaaa, or a\ Ana. Now aaa = a*, and
041 = a^\ but adding the exponents of a^ and a^ we have a^, the same
as before. Hence,
94. To multiply powers of the same letter together, add
their exponents.
Notes. — i. All powers of i are i.
2. When a letter has no exponent, i is always understood.
Multiply the following quantities:
20. al/^(? by a^c. Ans. c^Vc^.
21. sa^b^hy 206^. Ans. 6cfilAxy.
22. ^xy^ by 52;^. 25. ali^ by a&».
23. 6a^ by ^aV. 26. ^xyz by 2xy.
^> 24. ah^y by ah?y. 27. 6a^l^c by ^aWc.
^ 28. If fl = 3, what is the difference between 3a and a*?
.^ 29. K a; = 4, what is the difference between 4a: and ic* ?
95. The preceding principles illustrating monomials may
be summed up in the following
EuLE. — Multiply the coefficients and letters of both factors
together; to the product prefix the proper sign, and give to
\ each letter its proper index.
^ J Note. — It is immaterial in what order the factors are taken, but it
is more convenient, and therefore customary, to arrange the letters in
f
r L
J
'..^■>
alphabetical order. (Art. 87, Prin. 3.)
30. Multiply — s^y by — 2X.
31. Multiply Sa^i^ by — $a^c.
94. How multiply powers of the same letter together? 95. What U th« rule for
mnltiplying monomieda ?
MULTIPLICATION. 39
(32) (33.) (34.) (35)
Multiply A^y icHU^ Sc^^* ^}^
By ^ %o?l s^ — jab
(36.) (37.) (38.) (39.)
Multiply 3xy jabf^ 4«'c® xys?
By — 2X^ ^db(^ — Toc xyz
CASE II,
96. To Multiply a Polynomial by a Monomial*
1. What is the product of a + J multiplied hy J ?
Analysis. — ^Multiplying each term of the ofbratiot.
multiplicand by the multiplier, we have a x 6 a t
= 06, and 6x6 = 6». The result, 06+6^, is the b
product required. Hence, the Ans. ab + If^
BuLE. — Multiply each term of the multiplicand by the
multiplier; giving each partial prodtict its pi'oper sign, and
each letter its proper index.
Multiply the following quantities :
2. be — ad by ab. Ans. aVc — c?bd.
3. za^ + Acd hy 2c.
4. 5aS» — 2«? + a? by aoaj. ^ '
5. 4«3  lab + 7»2 by  2bdU/M^ ^oU^ "^^^
97. To Multiply a Polynomial by a Polynom^iai.
7. What is the product of a + J into a + b?
Analysis. — Since the multiplicand is to opbkation.
be taken as many times as there are units in a ^ b
the multiplier, the product must be equal to a + b
a times a +6 added to 6 times a +6. Now « 1 ^a
a times a+t = a^+db, and 6 times a +6 , ,•
= +06+6*. Hence, a+h times a+6 must ' "^
be equal to a* +206+62. Ans, a^ + 2ab + V
96. How multiply a polynomial by a monomial ?
5
\ + b.
OnBATIOV.
2a + J — • 3C
•
« + *
2a* + aJ — 3ac
f 2aS + i^ —
Sbc
40 HULTIPLIC ATI027.
8. Multiply b ^ 2a^ ^chj a + b.
Akaltbib. — We multi
plj eacb term in the malti
plicand by each teim Id the
moltipliery giving to each
product the proper sign.
(Art. 89.) Finally, we add
the partial pioductB, and the Am. 20* + 3^4 — 3ac + ^ — 3^c
resalt is the answer required.
98. The yarions principles developed in the precediBg
caaes, may be summed up in one
GENERAL RULE. r
Multiply each term of the multiplicand by each term of
the multiplier, giving each product its proper sign, and each
letter its proper exponent.
The sum of the partial products will be the true product.
Note. — For convenienee in adding the partial products, like terms
should be placed under each other.
Multiply the following quantities:
I. 2a + Jby3a? + y. 5. 3a + 4J — (?by a: — y.
^* 3a? + 4y by a — J. 6. 5a? + 3jr + j? by a + b.
3. 4i — c by 3</ — a. 7. jcdx — 3aJ by 2m — ^n.
4. 6xy — 2a by J + c. 8. Sabc + 4m by ^z — 4y.
9. Multiply 3aJ* by Sefib. Ans. 240^^'*+'.
10. Multiply — jaaf^ by — 8a%f. Ans. 560^"*+*.
11. Multiply $abc^ by xyz"^.
12. Multiply acd*^ by iibccf".
13. Multiply — oa;^ by — ax^.
14. Multiply x{a + by by c (a + by. (Art 15.)
15. Multiply c (a — by by 5 (a  by.
16. Multiply a{x+ y)"" by be {x + yy.
17. Multiply 3a; (a + by by — (a + S).
98. How multiply a polynomial by a polynomial?
MULTIPLICATIOK. 41
99. When the polynomials contain different powers of the
same letter, the terms should be arranged so that the first
term shall contain the highest power of that letter, the
second term the next highest power, and so on to^th&Iast *
term". This letter is called the Zeewim^r fo«cr. ^frCui^^^^^j
(i8.) (19.)
a^+2ab + l^ Sa^ia^b
a + b 4a%— 3aJ
a^+Sa^+saU^+l^, Ans. i2fl»J2_2^j2_3^j2^ ^^j^.
20. Multiply a^ — ab + l^hj ai b.
21. Multiply a^ — ab + ff^hja^^ab^ l^.
22. Multiply a^ix+ihja^ — x+i.
23.^ Multiply ^a^ — 2xy + 5 by ic® + 2xy — 6.
84. Multiply 4ax — 2ay by 6iia; 4 3ay.
2^. Multiply d + bxhj d ^ ex.
100. The Multiplication of polynomials may be indi
cated by inclosing each factor in a parenthesis, and writing
one after the other.
Thus, (a+ 6) {a+b) is equivalent to (a+b) x (a+b).
Note. — Algebraic Expresstotis are said to be defodaped or
eoDpandedf when the operations indicated by their signs and ea;>^;i«72to
are performed.
*
26. Develop the expression (a \ b) {c + d),
Ans. ac + be + ad + bd.
27. Develop {x + y) {x — y).
28. Develop (a' + i) (a + i).
29. Expand (a^ + 2xy + jf^{x\ y).
30. Expand {cT + S») (a™ + 6»).
31. Expand (a; + y 4 ;??) (x f y + 2;).
99. How arrange different powers of the same letter ? xoq. How indicate the mnl*
tiplication of polynomials?
42 MULTIPLICATIOiT.
THEOREMS AND FORMULAS.
101. Theorem i. — The Square of the Sum of two quatir
titles is equal to the square of the firsty plus twice their
product, plus the square of the second.
I. Let it be required to multiply
a+ b into itself.
Analtsis. — Each term of the moltipli ^ a
cand being multiplied by each term of the a f c/
multiplier, we have a times a+b and b times a + a6
a + 6, the sum of which is a* + 2ab + ft*. + ao { tr
Hence, the Ans. a^ + 2ab { i^
Formula. (a + b)^ = a^ + 206 + l^.
102. Theorem 2. — TJie Square of the Difference of two
quantities is equal to the square of the first y minus twice
their product, plus the square of the second.
2. Let it be requii^ed to multiply
fl — Jbya — J. a — h
Analysts. — Reasoning as before, the re a o
suit is the same, except the sign of the mid ^ — ^
die term 2a&, which has the sign — instead ao •\ (^
of +. Hence, the Ans, d^ — idb + ft*
Formula. {a — 6)^ = a^ — . 2ah + 6^.
103. Theorem 3. — The Product of the Sum and Differ
ence of two quantifies is equal to the difference of their
squares.
3. Let it be required to multiply
a + bhj a — b. a+ b
Analysis. — This operation is similar to
the last two ; but the terms H ab and — a&, a \ ad
in the partial products, being equal, balance ^^b cr
each other. Hence, the Ans. a^ — 5*
Formula. (a + 6) (a — 6) = «« — l^.
»— y^^^^,  __ .  ■ . 
loi. Wbat is Theorem i ? xca. Theorem 2 ? 103. Theorem 3 ?
MULTIPLICATION.
43
104. The product of the sum of two quantities into a third,
18 equal to the sum of their products.
4. Let X and y be two quantities, whose sum is to be
multiplied by a. Thus,
The product of the sum (x+y)xa =ax+ay
The sum of the products oixxa+yxa = ax+ay
And ttx+ay = ax+ay. Hence, the
FoBMULA. a{x+ y) = ax + ay.
105. The product of the difference of two quantities into a
third, is equal to the difference of their products.
5. Let X and y be two quantities, whose difference is to be
multiplied by a. Thus,
The^product of the difference (x—y) xa = ax^ay
The difference of the products otxx a^y xa = ax— ay
And ax— ay = ax— ay. Hence, the
FoEMULA. a(x— ij) = (IX — ay.
Remabk. — The application' of the preceding principles is so frequent
in algebraic processes, that it is important for the learner to make
them very familiar.
Develop the following expressions by the preceding for
mulas:
I.
{a+ i){a+ 1).
II.
(4a;— i)(4.r — 1).
2.
{2a + i){2a + i).
12.
(S* + 1) (s* + i).
3. <
[2a — b) {2a — 0).
13
{i^x){i^x).
4.
[x + y) {x + y).
14.
(l 4 2X) (l + 2X).
5
{x  y) (ic  y).
15
(8J — 3a) (8S — 3fl).
6.
{i + x){i —a:).
16.
{ab + cd) {ah + cd).
7.
{if — y)Wyy
17
(3« — 2y) (3a + 2^).
8.
(4m — Zn) (4m + 3w).
18.
(.^ + y) {x^  yy
9
(^y)(^ + y).
19.
{x  f) {x  f).
10.
(i — 72;)(i + ^x).
20.
{2a^ + x) {za^ — x).
X04* What^sfhe product of then^m of two quantitleB into a tbird equal to?
Z05. Of ilied^fferenoef
44 MULTIPIilC JLTION.
PROBLEMS.
106. Problems requiring each side of the equation to be
multiplied by equal quantities.
1. George has 1 third as many pears as apples, and the
number of both is 24. How many has he of each ?
Analysis.— If x represents the num Let x = No. apples ;
ber of apples, then  will represent the
number of pears, and a;+ will equal
24, the number of both. The denomi
nator of a; is removed by multiplying
each term on both sides of the equation
by 3. (Ax. 6.) The result is yt+x,oT
^ = 72. Hence, a; = 18, the apples,
and i8i3 = 6, the pears. Hence,
107. When a term on either side of the equation has a
denominatoTy that denominator is removed by multiplying
every tenn on both sides of the equation by it. (Ax. 4.)
2. What number is that, i seventh of which is 9 ?
Ans, 63.
3. What number is that, 2 thirds of which are 24 ?
4. A man being asked how many chickens he had,
answered, 3 fourths of them equal 18. How many had he ?
5. What number is that, i third, and i fourth of which
are 21 ?
Analysis. — If x represent the number, then j^^ ^ «_ jtq
X X
will  +  = 21, by tlie conditions. Multiplying x X
3 4  + = 21
each term on both sides by the denominators 3 3 4
and 4 separately, we have 40?+ 3a; = 252. (Ax. 4.) 4^ + 3*^ = 25 2
Uniting the terms, ^x = 252, and x =^ 2^6, Ans. . * . X z=. 36
Pboof. i of 36 = 12, and } of 36 = 9. Now, 12 + g = 21.
3""
pears.
24
3a; + a; —
72
4a;
72
.. X
18 apples.
aj_
3""
6 pears.
X07. When a quantity on either Bide .of an equation has a denominator, how re'
moTeit?
MULTIPLIOATIOK. 45
6. What number is that, 2 thirds of which exceed i half
ofUby 8?
7. A general lost 840 men in battle, which equaled
3 sevenths of his army. Of how many men did his army
consist ?
8. If 3 eighths of a yacht are worth I360, what is the
whole worth ?
9. If — equals 20, to what is x equal ?
10. If — is equal to 20, to what is x equal ?
* 1 1. If — is equal to 24, to what is x equal ?
12. If — is equal to 28, to what is x equal ?
13. Henry has 30 peaches, which are 5 sixths the number
of his apples. How many apples has he ?
14. A farmer had 3 sevenths as many cows as sheep, and
his number of cows was 30. How many sheep had he ?
How many of both ?
15. Divide 28 pounds into two parts, such that one may
be 3 fourths of the other.
16. A lad having given 1 third of his plums to one school
mate, and I fourth to another, had 10 left. How many had
he at first ?
17. What number is that, i third and i sixth of which
are 21?
18. What number is that, i fourth of which exceeds
1 sixth by 12 ?
19. Divide 36 into two parts, such that one may be
2 thirds of the other ?
20. One of my apple trees bore 3 sevenths as many apples
as the other, and both yielded 21 bushels. How many
bushels did each yield ?
CHAPTER Y.
DIVISION.
108. Division is finding how many times one quan
tity is contained in another.
The Dividend is the quantity to be divided.
The Divisor is the quantity by which we divide.
The Quotient is the quantity found by division.
The Remainder is a part of the dividend left after
division.
109. Division is the reverse of multiplication, the divi
dend answering to the product, the divisor to one of the
factors, and the quotient to the other,
PRINCIPLES.
110. I**. When the divisor is a quantity of the same hind
as the dividend, the quotient is an abstract number,
2^. When the divisor is a number , the quotient is a quan
tity of the same kind as the dividend.
3°. The product of the divisor and quotient is equal to the
dividend.
4®. Cancelling a factor of a quantity, divides the quantity
by that factor.
CASE I.
111. To Divide a Monomial by a Monomial.
I. What is the quotient of abed divided by cd?
Analysis.— The divisor cd is & factor of the divi opbratior.
dend ; therefore, if we cancel this factor, the other cd ) abcd
factor «&, will be the quotient. (Prin. 4.) Ans. ab.
X08. Define divleion. The dividend. Divisor. Qaotient. Remainder. 109. Oi
what is division the reverse? Explain, zzo. Name ue first principle. The second.
Third. Fourth.
DIVISION. 47
2. What is the quotient of iSab divided by 6a ?
Analysis. — ^Dividing the coefficient of the divi opbbatiow.
dend by that of the divisor, and cancelling the com 6a ) iSab
mon factor a, we have iSabi6a = 3b, the quotient Ans. 3 J.
required, (Prin. i.) Hence, the
EuLE. — Divide one coefficient by the other, and to the re'
suit annex the quotient of the literal parts.
Divide the following quantities :
' (3) (4.) (5) (6.)
2C ) 4abc 4b ) 2obxy Sxy ) 4oxy 16 J ) 3 2a J
(7.) (8.) (9.)
gm ) 4^abm 2omn ) 6obcmn 24xy ) g6mnxy
SIGNS OF THE QUOTIENT.
112. The rule for the signs in division is the same as
that in multiplication. That is, •
If the divisor and dividend have like^ns, the sign of the
quotient will be + ; if unlike, the sign of the quotient
will be — .
Thus, +a X +b = +ab; hence, +db t +h = +0.
—a X +6 = —oft ; hence, —ab * +6 = —a.
+a X —b = —06; hence, —ab i b = +a.
• —a X —6 = +aft; hence, +06 s 6 = —a.
Divide the following quantities :
10.
II.
12.
14.
— 32aJc by — 4ab.
iSabx by — ^.
2iabc by — jai.
■— 2Sbcd by — 4cd.
SScdm by jcm.
Ans. Sc.
J 6abx
Ans. •
z
15. 4Sabc by —  Sac.
16. 6^bdfx by gbx.
17. — 'j2acgm by Scm.
zzz. How divide a monomlAl by a monomial? zz2. What is the mle for the
signs T
48 DIVISION.
113. To Divide IPawers of the same letter.
1 8. Let it be required to divide c^ by cfi,
ANAiiTsis. — The term a^ — 00000, and a' = aaa. Rejecting the
factors aoa from the dividend, the result aa, or o^, is the quotient.
Subtracting 3, the index of the divisor, from 5, the index of the divi
dend, leaves 2, the index of the quotient. That is, a^ « a> = a^.
(Arts. 31, no. Prin. 4.) Hence, the
BuLE. — Subtract the index of the divisor from that of the
dividend.
Divide the following quantities :
19. eP by d*. 22. xyz^ by xyz^,
20. x^ by rr*. 23. i6aJ2 by 4aS.
21. a<^hj ac^. 24. 6xy by 3^2,
114. The preceding principles may be summed up in
the following
EuLE. — Divide the coefficient of the dividend by that of the 
divisor; to the result annex the quotient of the literal fadorsy
prefixing the proper sign and giving each letter its proper
exponent.
Proof. — Multiply the divisor and quotient together, as 171
arithmetic.
Note. — If the letters of the divisor are not in the dividend, the
division is eospressed by writing the divisor U7ider the dividend, in the
form of a fraction.
25. What is the quotient of 5X divided by $y? Ans. ~ —
322:*^ 2 j^a^yz,
gSa^ff^cT i2ab.
S4d^a^ ^ 7dh:y,
loSabx^ r ga^a^.
121 mhM ^ I im^nirt
26.
— 24a2}2^ 2 ^ab.
32.
27.
— 2^^^ h 6xyz.
33
28.
SaW 4 ah
34.
29.
— ^7^^ T — xy.
35
30
a^lfi(^ H a^V^c.
36.
31
i6aW(^ ^ Sa^b*(^.
37.
1x3. How divide powers of the same letter? 114. Bale for diviBion of mono
mialt ? Proof? If ^be letters of the diviBor are not in the dividend, what ig done f
/
DIYISIOIT. 49
CASE II.
115. To Divide a Boiynomial by a IMonomiaL
1. Divide ab + ac + adhja.
ANALYSIS.— Since the factor a enters into opiRAmnr.
eacb term of the dividend, it is plain that « ) ob + OC h od
each term of the dividend must be divisible ^ng, } J * 4. d
by this factor. Henoe» the
EuLE. — Divide each term of the dividend by the divisor,
and connect the results by their proper signs.
Note.— If a polynomial which contains the same factor in ever^
term, be divided by the other quantities connected by their signs, the
quotient will be tTuU factor.
DiTide the following qnau titles:
2. 6a^ + loa' — 14a by 2a. Ans, ^a^ + sa — ?•
3. 40* — 8a* + i2a^ by — 2a^. Ans. — 20^ + 4a — 6,
4. al^ + a(fl + ad^ by a.
6. 6abc — 2fl + Sab by 20.
7 — i6Sy«+4y8by — 8y.
8. 140^ — yxf by — jxy.
9. xt^ + xZ" X by a?.
10. 35a + 28* —42 by —7.
1 1. isa*J — 15a* by 5a.
12. i6a:%i^ + i2accP — 42:aV by — 400.
13. 4a* — 20a' + 8flJ by 40.
14. 3aJ + 15^^* — 27a*W by 306.
151 Sa^c — i6a8^c — 20a J^ by 40^
16. 6a: (a + i)» + 9a? (a + J)2 by 3«.
17. is(^ — y) + 3o(a? — y)bys
18. aofi (J — c) — a^^ (J — c) by aa?.
19. i8a* (a + 6)2 — 12a* {a + bf by 6fl^ (a + ^)*.
20. (r+^ — a^^ + a"+* by o".
■ — ■ — 1 _■ ■■  — r^ I 1^ I _ ■ ij__B__^ rM^ ^^ ^"^^^^
1S5. How divide a polynomial by a moaomialf
60 DIYIBION
CASE III.
UGi To Divide a Polynomial by a Polynornlal.
I. Divide a* + Sa^J + 3aJ» + Vhya^ + 2ab + V.
AKALYBI8. — For com&' ontRATioN.
a*+2aif 8*
a + ^ <tao^
n^i0n<», we arrange the terms cfi + 3a^J + 3aS* + J*
90 that the first or leading a'+2a^d+ dlf^
letter of the divisor shall be I7~7 xTTTw
the first letter of the divi O ^+20^+^
dend. The powers of this fl^ft+2fly+y
letter should be arranged in
order, both in the divisor and dividend, the highest power standing
first, the next highest next, and so on. The divisor may be placed on
the left of the dividend, or on the right, and the quotient under it, at
pleasure.
Proceeding as in arithmetic, we find the first term of the divisor is
contained in the first term of the dividend a times. Placing the a in
the quotient under the divisor, we multiply the whole divisor by it,
subtract the product, and to the remainder bring down as many other
terms as necessary to continue the operation. Dividing as before, a^
is contained in a% +h times. Multiplying the divisor by +& and
anbtracting the product, the dividend is exhausted ; therefore a+Ma
the quotient. Hence, the .t
EuLE. — I. Arrange the divisor and dividend according to
the powers of one of their letters ; and finding how many
times the first term of the divisor is contained in the first
term of the dividend, place the result in the quotient,
II. Multiply the whole divisor by the term placed in the
quotient; subtract the product from tJie dividend, and to the
remainder bring down as many terms of the dividend as the
case in ay require.
Repeat the operation tiU all the terms of the dividend are
divided.
NOTB. — If there is a remainder after all the terms of the dividend
are brought down, place it over the divisor, and annex it to the quotient
zx6. How divide a polynomial by a polynomial ? If tbero i§ a remainder, what is
done wittL iti
PROBLEMS. 61
2. Divide 40^ 4. ^ab + i^ hj 2a +b. Arts. 2a + b.
3. Divide ofi + 2xy + ^ by x + y»
4. Divide o^ — 2aJ + ^ by a —b.
5. Divide cfi — 3^2* + 30*^ — js by « — *.
6. Divide (W f Jc + arf + W by a + J.
7. Divide aa; + Sa? — c^ — id by a + *.
8. Divide 27^ + 72:^ + 6y« by a; + 2y.
9. Divide a* — J^ by a + J.
10. Divide a? — y® by a: — y.
1 1. Divide a^ — V by a — J.
12. Divide 6^2 + 13^6 + 6J2 by 2a 4 3 J.
13. Divide a^ — a — 6 by a — 3.
14. Divide a' — ^c^ + 302:^ — a:^ by a — on.
15. Divide 6a^ — 96 by 3a? — 6.
16. Divide a:* + 7a; + 10 by a; + 2.
17. Divide a:^ — 5a; + 6 by a; — 3.
18. Divide <?— 2cx + a^ by c — a?.
19. Divide a^ + 2ab + y^ hj a + S.
20. Divide 22 (a — S)^ by 11 (a — S).
PROBLEMS.
1. A father being asked the age of his son, replied, My
age is 5 times that of my son, lacking 4 years; and the
sum of our ages is 56 years. How old was each ?
2. John and Frank have 60 marbles, the former having
3 times as many as the latter. How many has each ?
3. The sum of two numbers is 72, one of which is 5 times
the other. What are the numbers ?
4. A man divided 57 pears between two girls, giving one
4 times as many as the other, lacking 3. How many did
each have ?
5. Three boys counting their money, found they had
190 cents; the second had twice as many cents as* the first,
and the third as many as both the others, plus 4 cents.
How many cents had each ?
58 DIVISIOK.
6. A farmer has 9 times as many sheep as cows^ and the
number of both is 200. How many of each ?
7. Diride 57 into two such parts that the greater shall be
3 times the less, plus 3. What are the numbers ?
8. Given 22? + 4a: + a; — 3 = 60, to find x.
9. A and B are 35 miles apart, and travel toward each
othei, A at the rate of 4 miles an hour, and B, 3 miles. In
how many hours will they meet ?
10. Given a + 5a + 6a + 2a + 7 = 119, to find a.
11. Given 8S + 5 J f 7J — 10 = 130, to find h.
12. A lad having 60 cents, bought an equal number of
pears, oranges, and bananas; the pears being 3 cents apiece,
the oranges 4 cents, and the bananas 5 cents. How many
of each did he buy?
13. A cistern filled with water has two faucets, one of
which will empty it in 5 hours, the other in 20 hours. How
long will it take both to empty it ?
14. Given a: +  = 45, to find as,
3
15. What number is that, to the half of which if 3 be
added, the sum will be 8 ?
16. Three boys have 42 marbles ; B has twice as many as
A, and three times as many as A. How many has each ?
r7. If A has 2a? dollars, and B twice as many as A, and
C twice as many as B, how many have all ?
18. Divide 40 into 3 parts, so that the second shall be
3 times the first, and the third shall be 4 times the first.
19. A man divided 60 peaches among 3 boys, in such a
manner that B had twice as many as A, and C as many as
A and B. How many did each receive ?
20. Divide 48 into 3 such parts, that the second shall be
equal to twice the first, and the third to the sum of the first
and second ?
21. What number is that, to threefourths of which if 5
be added, the sum will be 23 ?
OHAPTEE YI.
FACTORING.
117. Factors are quantities which multiplied together
produce another quantity. (Art 2t6,)
118. A Composite Quantity is the product of two
or more integral factors^ each of which is greater than a
unit
Thus, 3a, 55, also o^, are composite quantities.
119. Factoring is resolving a composite quantity
into its factors. It is the converse of multiplication.
120. An Exact I>ivisor of a quantity is one that will
divide it without a remainder. Hence,
Note. — ^The Factors of a quantity are always eocact divisors of it,
and vice versa,
121. A Prime Quantity is one which has no integral
divisor^ except itself and i.
Thus, 5 and 7, also a and h, are prime quantities. Hence,
NOTB. — ^The least diusor of a composite quantity is a prime factor.
122. Quantities are prime to each other when they
have no common integral divisor, except the unit i.
Thus, II and 15, also a and be, are prime to each pther.
123. A Multiple^is a quantity which can he divided
by another quantity without a remainder. Hence,
A multiple is b, product of two or more factors.
1x7. What are tBuctonJ xi8. A coiniK)8ite quantity? 1x9. What ia (hctoringf
xaa An exact divisor? xai. A prime qnantity? xaa. When prime to each other?
Z33. A multiple?
64 FACTORING.
PRINCIPLES.
124. i^ If one quantity is an exact divisor of another,
the former is also an exact divisor of any multiph of the
latter.
Thus, 3 is a diyisor of 6 ; it is also a divisor of 3 x 6, of 5 x 6, etc.
2^ If a quantity is an exact divisor of each of two other
quantities^ it is also an exact divisor of their sum, their dif
ference, or tJieir product.
Thns, 3 is a divisor of 9 and I5» respectively ; it is also a divisor of
9+ 15, or 24 ; of 15—9, or 6 ; and of 15 x 9, or 135.
3**. A composite quantity is divisible by each of its prime
factors, by the product of two or more of them, and by no
otiier quaiitity.
Thus, the prime factors of 30 are 2, 3, and 5. Now 30 is divisible
by 2, by 3, and by2x3;by2x5; by 3x5; by 2x3x5, and by no
other number.
CASE I.
125. To Find the Prime Factors of Monomials.
I. What are the prime factors of i2a^J?
Analysis. — The coefficient 12 = 2x2x3, and o*& = 006. There
fore the prime factors of i2a^h are 2 x 2 x ^ddb. Hence, the
ScjLE. — Find the prims factors of the numeral coefficients,
and annex to them the given letters, taking each as many
times as there are units in its exponent.
Note. — ^In monomials, each letter is a factor. Hence, the prime
factors of literal monomials are apparent at sight.
Besolve the following quantities into their prime factors :
4. lobofi^. 8. 2sai^csfi.
6. 2ixy^s?. 10. 6^wMx. I i\^
X34. Name Principle x. Principle 3. Principle 3. 135. How find the prime &o
tors of monomials ?
VAOTOBING. 56
CASE II.
126. To Factor a Polynomial.
1. Resolve /^b + Sab — 6ac into two factors.
Analysis. — By inspection, we per oebramoh.
ceive the factor 2a is common to 2a ) 4g^& + 8g& — 6ac
each term ; dividing by it, the quo 2flJ + 4? — 3C
tient 2aft+4&3C is the other fw5tor. ^^^ ^a (idb + 4 J — 3c)
For convenience, we enclose this fac
tor in a parenthesis, and prefix to it the factor 2a, as a coefficient
Pkoop. — The factor (206 + 4&— 3c) x 2a=:^^h + 806— 6a(j. Hence, the
BiJLE. — Divide the polynomial iy the greatest common
monomiai factor ; the divisor will be one factor, the quotient
the other. (Art. 115.)
NOTB. — ^Any common factor, or the product of any two or more
common factors, may be taken as a divisor; but the result will very
in form according to the factors employed. (Ex. 2.)
2. Eesolve a^ + aJ^ into two factors, one of which shall
be a monomial Ans. ai (a+b), a {ab+b'^), or b {a^ + ab).
3. Factor a + ab + ac. Ans, a{i + b + c).
4. Factor by + bc+ 38a?.
5. Factor 2ax + 2ay — 4az. y :j ^ /];
6. Factor ^cx — tbcx — Tflbc. ~ /^^ a ^
7. Factor Zdmn — 2^m. sr /^^ — J
8. "
9
. Factor i^am + 14^0:. Jc'^VH. 1 ^ ' '
. Factor 27*^0: — v^/^dmy. ^ Xv ^ a ti !^j ^
10. Factor 6a^ + oaV. ,
11. Factor 2ia/o^ + 35(^> ^ /'''. ^ *,
12. Factor 25 + 152:8 — 202:'^. ;:^ * ; ■^. J'^ 
13. Factor x + 0^^0^. % ^^^'% / '• ^' ^
14. Factor 3a; + 6 — py. 1 »y 4 J — '. ^ \
15. Factor ipa^^a:— 19a*. ^ _ ; ^
196. How factor a polynoznial f
56 FACTOBINa.
CASE III.
127. To Resolve a Trinofniai into two equal Binomial Faotore.
1. Besolye sfl + 2xy + y^ into two equal binomial factors.
Analtbib. — Since the iqwvre of a onbation.
qoantitj is the piodact of two equal V^^= a;, v^ = Jfy
factora (Art. 30), it follows that the ^ . .
square root of a quantitj ia one of the ' ' "" *^'Tjr
two eqnal factora which produce it. (^+y) (^ + 9)9 Ans*
(Art. 32.) Therefore the square root of
tfliaXf that of ^ is y. And since the middle term 2xy is twice the
product of these two terms, a^^2xy^f^ must be the square of the
binomial x\y. (consequently, x+y is one of the two equal binomial
factors.
2. Besolye a? ~ 2xy + t^ into two eqnal binomial factors.
AKALTBis.~Reasoning as before, the opebatioh.
quantity a?— 2a!y+y' is the square of y^ = x, V^ = V
the residual x—y, Tberefore, the two . ^ _i_ #3
equal factora must be x—y and x^y. * * V "V y^
Hence, the (^— y) (^ — y)> •4^«.
BuLE. — Find the square root of each of the square terms,
and connect these roots by the sign of the middle term.
Note. — A trinomial, in order to be resolved into equal hinomiaZ
factora, must have two of its terms squares, and the other term twice
the product of their square roots. (Art. loi.)
Besolve the following into two eqnal binomials:
3. a^ + 2db + J2. 9. y^ + 2y + I.
4. a^ — 2xy + yK 10. i — 2^ f c*.
5. w' + 4m» + ^n\ II. x^ + 27fy^ + y*».
6. i6a^ + 8a + I. 12. 4a^ — 40" + i.
7. 49 + 70 + 25. 13. a* + 2c^V + J*.
8.
^ — i2aS + 98^. 14. ch^ + la^y + y*.
X37. How resolve a trinomial into equal binomial fkctore ?
r
CASE IV.
<
128. To Factor a Binomial consisting of the IHfference of
two Squares.
I. Eesolve 40^ — 94* into two binomial factors;
Akalysib.— Both of these terms omratiok.
are squares ; the root of the first is ^ /\fi? = 2a
2a, that of the second is 36. But the y^ ,
difference of the squares of two quan ^ ^
titles is equal to the product of their •** 4^ 9^ =
sum aud difference. (Art. 103.) Now (2^+3^) (2a — 3d), ^n«.
the sum of these two quantities is
2a + 3&, and the difference is 2a — 36 ; therefore, \a? — <jl^ —
(2a + 36) (20— 3&). Hence, the
EuLE. — Find the square root of e(tch term. The stim of
these roots will he one factor, and their difference the other.
NOTB. — This rule is one of the numerous applications of the for
mula contained in Art. 103.
2. Resolve a^ — a^ into two binomial factors.
3. Eesolve 9a? — 16^ into two binomial factors.
4. Eesolve y^ — 4 into two binomial factors.
5. Eesolve 9 — x^ into two binomial factors.
6. Eesolve a* — i into two binomial factors.
7. Eesolve i — ^ into two binomial factors.
8. Eesolve 25^2 — 166^ into two binomial factors.
9. Eesolve ^ x^ into two binomial factors.
10. Eesolve i — xdc^ into two binomial factors.
II. Eesolve 25—1 into two binomial factors.
1 2. Eesolve a:* — y* into two binomial factors.
13. Eesolve d^ — l^y^ into two binomial factors.
14. Eesolve w* — n^ into two binomial factors.
15. Eesolve a^"* — W^ into two binomial factors.
za8. How flictor a binomial coDBietiiig of the difference of two squares ?
58 FAOTOBINO.
CASE V.
129. Varions classes of examples of higher powers may be
factored by means of the following
PRINCIPLES.
i^ The difference of any two powers of the same degree is
divisible iy the difference of their roots.
Thus, («*— y«) ^ (x—y) = x+y,
(«*— y*) « {x—y) = ir»+aJ^f ajg/'+y".
(«■— y*) « (x—y) =r aj«+a!V+afy +ajy»+y«.
2°. ?%« difference of two evenpotvers of the same degree is
divisible by the sum of their roots.
Thug, (aj^— y*) * (x+y) = x—y.
(aji— y«) ^ (aj+y) = a^— a^y+a^— y».
(aj*— y*) * (aj+y) = aj5— a^+a^^— a;V+a!y*— y*.
3®. 2%« s?/m o/^wo odd powers of the same degree is divi
sible by the sum of their roots.
Thus, (aj»+y») s (a;+y) = a^— a?y+y*.
(aj*+y*) s (aj+y) = a?*— a^f a5*y*— a!y*+y*.
(«''+y') ■*■ («+y) = «•— aj^+a?^*— aJV+a^y*— «!y*+y^ ©tc.
Note.— The indices and w^rn* of the quotient follow regular laws :
ist. The index of ihe first letter regularly decreases by i, while that
of the folUncmg letter increases by i.
2d. When the difference of ttoo powers is divided by the difference
of their roots, the signs of all the terms in the quotient are phi^. When
their sum or differevice is divided by the sum of their roots, the odd
terms of the quotient are plus, and the even terms minus.
B^" If the principles and examples of this Case are deemed too
difficult for beginners, they may be deferred until the Binomial
Theorem is explained. (Arts. 268270.)
X29. Recite Prin. i. Prin. 2. Prin. 3. Note. What ie the index of the first let
ter ? Of the following letter ? What is eaid of the signs ?
PAOTORINa. 69
130. To Factor the Difference of any two Powers of the
same Degree.
1. Eesolve a^^y^ into two factors.
Solution. — The binomial {a?—y^ ^ (a?— y) = a^+xy+j/', .*. x—y
md a^+xy^^ are the factors. (Prin. i.) Heuce, the
Bdxe. — Divide the difference of the powers by the differ
ence of the roots; the divisor will be one factor ^ the quotient
the otJier.
Eesolve the following into two faiCtors :
2. re*— I. 4. a^ — I.
3. a^ — ^. 5. I — 36»».
131. To Factor the Difference of two even Powers of the
same Degree.
6. ' Eesolve a^ — ¥ into two fectors.
Solution.— By Prin. 2, a*— 6* is divisible by a +6. Thus, (a*— M)
* (a+6) = a'— a*6+a6*— &*, the divisor being one factor, the quotient
the other. Hence, the
EuLE. — Divide the difference of the given powers by the
sum of their roots; the divisor toill be one factor, the quo
tient the other. (Art. 129, Prin. 2.)
Eesolve the following quantities into two factors:
7. J2_aj2.yy.t.'5A 10. a;*— 1.
8. d* — «*. ♦ II. I — a«.
9. cfi — V. 12. c? — I.
132. To Factor the Sum of two odd Powers of the same
Degree.
13. Eesolve c? + Ifi into two factors.
Solution. — Dividing a^+H^ by a+&, the factors are a+& and
fl2— a5+6», (Prin. 3.) Hence, the
EuLE. — Divide the sum of the powers by the sum of the
roots; the divisor and quotient are the factors.
130. How fiactor the difference of any two powers of the eame degree ? 131. How
ftictor the difference of two eyen powers of the same degree ? 13a. The sum of two
odd powers of the eame degree ?
60 FACTORING.
Besolye the following quantities into two factors:
14 ^ + !^' 17 i+y*
iS» cfl+ 1. i8. I + a\
i6. a*+ I. 19. I + Jl
133. It will be observed that in the preceding examples
of this Gase^ binomials have been resolved into ttoo factors.
These factors may or may not be prime factors.
Thus, in Ex. 6, a*— &* = (a+6)(a»— fl»6+a6«— &»). But the factor
(a^—a*b+at^—V) is a composite quantity = (a— &)(»*+ 6*).
134. When a binomial is to be resolved into prime xiac
tors, it should first be resolved into two factors, oup of
which is prime; then the composite factor should be
treated in like manner.
20. Let it be required to find the prime factors of a* ^ ¥.
Solution.— The ya* = a*, and y^ = 6*. (Art. 128.)
Nowff»&* = (a«62)(a«+52). Buta8&« = (a+6)(a6). (Art. 104.)
Therefore the prime factors of o*— &* are (a* + J*) (a +6) (a— 6).
Eesolve the following quantities into their prime facton '
21. a*— I. Ans. {a^ + i){a+ i){a— 1).
22. I— y*. Am. {i+y^){i+y){i'!/).
23. a^ — /. *
Ana. {a?xy + f){ofi + xy + y^){x + y){xy).
24. ai^ — 2X^ + y*.
Ans. {x^fy={x + y){x + y){X'y){x'y).
25. a:*— I.
Ans. (x+ i)(a; i){a^ + x+ i)(a«ar+ i).
26. «• + 2fl»J» + J«.
27. «» + 9a + 18. ^7^5. (a + 6) (a + 3).
28. 4^2 — i2aS + 9J2. ^;w. (2a — 3^») {2a — 3*).
(See Appendix, p. 284.)
CHAPTER YII.
DIVISORS AND MULTIPLES.
135. A Common Divisor is one that will divide two
or more quantities without a remainder.
136. Commensurable Quantities are those which
have a common divisor.
Thus, 05" and dbc are oommensurable by ab.
137. Incom^mensurahle Quantities are those
which have no common divisor. (Art. 122.)
Thus, otb and asyz are incommensurable.
138. To Find a Common Divisor of two or more Quantities.
1. Find a common divisor of abxy acyy and adz.
Analysis. — Resolving the given quanti oebratioh.
ties into factors, we perceive the factor a, is dbx =z axbxx
common to each quantity, and is therefore a acy :=: axcXV
common divisor of tliem. (Art. 119.) Hence, ^^ __ ^ ^ ^ ^ ^
® Ans. a.
Rule. — Resolve each of the given quantities into factors^
one of which is common to all.
Find a common divisor of the following quantities :
2. $abcd and gabm, Ans. $ab.
3. a^yz and 2abx. 6. 2aXy 6bXy i^cx,
4. d^by body al^xy, 7. 3S»iw, yw^, 42/w^.
5. 2abCy acx\ aHy. 8. 2/^c^b, \2aWy 6aW,
X35. What is a common divipor ? 136. Comniensnrable quantities? 137.100001
mensumble quantities ? 138. How find a common divisor of two or more qaantities f
62 DIVISORS.
139. The Greategt Common Divisor of two or
more quantities is the greatest quantity that will divide
each of them without a remainder.
Notes. — i. A common divisor of two or more quantities is always a
common faetor of those quantities, and the g, c. d.* is their greatut
common factor,
2. A commo7i divisor is often called a common measwre, and the
greatest common divisor, the greatest common meatfure.
PRINCIPLES.
140. I**. The greatest common divisor of two or more
quantities is the product of all their common prime factors.
2°. A common divisor of ttoo quantities is not altered by
multiplying or dividing either of them by any factor not
found in the other.
Thus, 3 is a common divisor of i8 and 6 ; it is also a common divisor
of i8, and of (6 x 5) or 30.
3°. Hie signs of a polynomial may he changed by divid
ing it by — I.
Thus, (— 3a+4&— 5<5) * — I = 3a— 4&+5C. (Art. 112.) Hence,
4°. I7ie signs of the divisor, or of the dividend, or of both,
may be changed without changing the common divisor,
141. To Find the Greatest Common Divisor of Monomials by
Prime Factors.
I. What is the gr. c; d!. of 35aca:, 2Sabc, and 2iay?
Analysis. — Resolving the ofbbatiov.
given quantities into their prime 3 c acx := 5x7 Xaxcxx
factory 7 and a only are com. 2Sabc = 2 X2 XJ Xaxbxc
mon to each; therefore their
product 7 X a, is the ^. c. d. re ^^^^ 3X7Xaxy
quired. (Prin. i.) •'• 7 X a = 7« Ans.
139. What is the greatest common divisor of two or more quantities ? Ifote x.
What is true of a common divisor of two or more quantities ? Of the g, c. d, f
140. Name Principle x. Principle'2. Principle 3.
* The initials gr. c. d, are used for the greatest common divisor.
DITISOES. 63
2. Find the g. c. d. of ^Vc^ locRfiy and i^abdx.
Analysis. — ^Resolving these quan opkratioh.
titles into their prime factors, the ^C^l^C = 2 x 2 X ooahbc
factor 2 is common to the coefficients; loa^ifi == 2 X 5 X aabib
also, a and b are common to the lit lAubdx = 2 X 7 V ahdx
eral parts. Now multiplying these a , ,
^ ■ . ,■. •■ JxTlS* 2 X tf X u — 2(10
common factors together, we have
2 X a X & = 2aby which is the g, c. d. required. (Prin. i.) Hence, the
BuLE. — Resolve the given quantities into their prime f ac
tors ; and the product of the factors common to ally will he
the greatest common divisor. (Prin. i.)
Note. — In finding the common prime factors of the literal part,
give each letter the least exponent it has in either of the quantities
3. Find the g. c. d. of 6a^(^ and gabc.
4. Of i6ah:y and iSacs^.
5. Of i2(^lrhfis^ and i6ah^.
6. Of 6affh!^sfiy i2a^3^s?, ana iSah^.
142. To Find the Greatest Common Divisor of Quantities by
Cfantintied jDivimott*
I. Eequired the greatest common divisor of ^ox and 420?.
Analysis. — If we divide the greater opebation.
quantity by the less, the quotient is i, 30X ) 42a: ( i «
and 122? remainder. Next, dividing the 30a;
first divisor 30a?, by the first remainder ~"x ^
12a;, the quotient is 2 and the remainder ^ ^ ^
6x. Again, dividing the second divisor ^
by the second remainder, the quotient 6x) I2x{ 2
is 2 and no remainder. The last divisor, j 22;
6a;, is the g. c. d.
Demonstration. — Two points are required to be proved :
ist. That tx is a common divisor of the given quantities.
2d. That 6a; is their greatest common divisor.
First. We are to prove that 6a; is a common divisor of 30a? and 422?.
By the last division, 6a; is contained in 12a;, 2 times. Now as 6a; is a
, ,  1  —  ■  ■
X41. How find the g.c.d, of monomials by prime factors ? Ifote. In Ending th«
prime ^tors of the literal part, what exponents are given ?
64 DIYISOBS.
divisor of las, it is also a divisor of the product of lax into 2, or 241s.
(Art 124, Prin. i.) Next, since 6a; is a divisor of itself and 24a;, it
most be a divisor of the sum of &c+ 24^, or 30X, which is the smaller
quantity. For the same reason, since to is a divisor of 12a; and 3ar, it
mast also be a divisor of the sum of i2a;+30ir, or 42a;, which is the
larger quantity. Hence, 6jc is a eommitn divisor of 301; and 42a;.
Second, We are to prove that tx is the greatest common divisor of
3ar and 42a;.
If the greatest common divisor is not 6a;, it must be either greater
or less than 6a;. But we have shown that 6a; is a common divisor of
the g^ven quantities ; therefore, no quantity less than 6a; can be the
greatest common divisor of them. The assumed quantity must there
fore be greater than 6a;. By supposition, this assumed quantity is a
divisor of 30a; and 42a; ; hence, it must bo a divisor of their difference,
42a;— 30a;, or 12a;. And as it is a divisor of i2a;j it must also divide the
product oii2X into 2, or 24^.
Again, since the assumed quantity is a divisor of 30a; and 24^, it
must also be a divisor of their difference^ which is 6a; ; that is, a greater
quantity will divide a less without a remainder, which is impossible.
Therefore, 6a; must be the greatest comm^on divisor of 30a; and 42a;,
the second point to be proved. Hence, the
Bjjlk— Divide the greater qtcantity by the less, then divide
the first divisor by the first remainder ^ the second divisor by
the second remainder, and so on, till there is no remainder.
The last divisor will be the greatest common divisor.
Note. — If there are more than two quantities, find the g. c. <f«
of the smaller two, then of this common divisor and a third quantity,
and so on with all the quantities.
2. What is the g. c. d. of 48a, 72a, and loScr?
Suggestion.— The g. c. d. of 48a and 72a is 24a : and that of 24a
and iQ&a is 12a. Therefore, 12a is the g, c. d* required.
142,0. The Greatest Common J>ivi8or of Poly"
nomials is found by the above rule, as illustrated in the
following examples:
X42. Show upon the blackboard the trnth of this nile ? 142, a How find the
• g. k3. (/. of polynomials ?
DIVISORS.
3. What is the g. c. d. of 40* — 21a' + 150 + ao
aa — 6a + 8 ?
OFBKATION.
fl^ — 6g + 8 zstdiyisor.
65
and
40^ — 2ia2 + 15a f 20
4a' — 24a^ + 32a
4« + 3
z8t qaotleiit
+ 3«^ — 17a + 20
f 3^2 — i8a + 24
a — 4 zBt remainder and ad dltlior.
a — 2 ad quotient
a* — 6a + 8
a^ — 4a
— 2a 4 8
— 2a + 8
^915. a — 4.
Alf ALYBis. — Dividing the greater quantity by the less, the remain
der is a~4. Again, dividing the first divisor by the first remainder,
the quotient is a— 2, and no remainder. The last divisor, a— 4, is the
greatest common divisor.
143. It is sometimes necessary, in order to avoid frac
tions, to introduce a factor into one or both the given
quantities, or to cancel one before finding the greatest com
mon divisor.
It is also sometimes necessary to change the signs of
the divisor or dividend, or of both. (Art. 140, Prin. 4.)
4. What is the g. c.d, of ar^ — 2xy + t^ and x^ — y^?
AlfALTSis. — Dividing the
greater by the less, the first
remainder is — 2xy + 2^.
Cancelling from it the com
mon factor 2y, we have for
the second divisor — a? + y.
Changing the signs, it be
comes x^y. (Art. 140, Prin. 4.)
Dividing as before, the quo
OPSSATION.
rr'— 2a?yf y^
aj8 — ^
2y)—2xy+2y^
Divisor, — X + y
or, X'y
Quotient, X + y
X^—y^ Divisor.
I Quotient.
ofi'y^
3^ — y^
tient is a; +y, and no remainder. The last divisor, aj— y, is the greatest
common divisor.
X43. How does it affect the g. c. d. if a factor is introduced into either or both the
given qnantitie? ? How if one is cancelled ? What is tnic of the t^i^^ie ?
96
DIVISORS.
5. What is ihe g.cd. ot 40;^ — 6a^ — 4X + 3 and
22^ + a^ + X''l?
OFEBATUnr.
XBt
dlTidend, 4^ — 62? — 4a; + 3
4Sfi + 2S^ + 2X — 2
•d diTiflor, — 8:g^ — 6a; + 5
•d quiOtleiit,
— a;
addivldeiid, — 8a? — 6aj + 5
^Saf^ + 36a; — 16
— 21 ) — 42a; + 21
4tli divisor,
4th quotient,
22?— I
— ir + 4
20^ + 0^ + Z — I xfltdlTisor.
2 xBt qnotient
Sa;^ + 40^ \ 4^C — 4 2d dividend
8a? + 6a;2 — 5X
— 23? + 9a? — 4 3d dlTisor.
3d quotient
22? + pa; — 4 4thdivideiid.
22? + a;
2a; — I is the gr. c. d.
8a; — 4
8a; — 4
Akalybib. — Dividing the greater bj the less, the first term of the
first remaiiider, —Safi, is not contained in 20^, the first term of the
second dividend. We therefore multiply this dividend by 4, and it
becomes 8:1;' +4^ + 4^— 4, and dividing this by the second divisor, the
second remainder is — 2aJ*+9a;— 4. Dividing the preceding divisor by
this remainder, we see that the third remainder, — 420^+21, is not
contained in the next dividend. Cancelling the factor —21, the fourth
divisor becomes 23; — i, the greatest common divisor required.
Find the g. c* d. of the following quantities:
6. a? — y8 and a? — 2xy + ^.
7. cfl + 1^ and a^ + 2ah + V.
8. i» — 4 and ja + 4} + 4,
9. a? — 9 and a? + 6a; + 9.
to. a^ — 3« + 2 and a* —  a — 2.
11. fl* + 3a* f 4a + 12 and a^ 4 4^2 (.4^ + 3,
12. a? + I and a? + Twa? + wa; } 1.
13. c^ — ¥ and a' —  J^.
14. a^ _ ^^i _^ 4J2 apd (j^ ^ ci^j) + ^dt^ — 3&*.
^5 3a? — loa? + 15a; + 8 and 3^ — 2a;*— 6a? +42?
4 13a; + 6.
(See Appendix, p. 384.)
MULTIPLES. ^7
MULTIPLES.
144. A Multiple is a quantity which can be divided
by another quantity without a remainder. (Art 123.)
145. A Coitiiinon Multiple is a quantity which can
be divided by two or more quantities without a remainder.
Thus, i8a is a common multiple of 2, 3, 6, and 9.
146. The Least Common Multiple of two or
more quantities is the least quantity that can be divided by
each of them without a remainder.
Thus, 21 is the least common multiple of 3 and 7 ; 30 is the least
common multiple of 2, 3, and 5.
PRINCIPLES.
147< i^. ^ multiple of a quantity must contain all the
prime factors of thai quantity.
Thus, 18 is a multiple of 6, and contains the prime factors of 6,
which are 2 and 3.
2^. A common multiple of two or more quantities must
contain all the prime factors of each of ths given quantities.
Thus, 42, a common multiple of 14 and 21 /contains all the prim«
factors of those quantities ; viz., 2, 3, and 7.
3°. The least common multiple of two or mme quantities
is the least quantity which contains all their prime factors^,
each factor being taken the greatest number of times it occurs
in either of the given quantities.
Thus, 30 is the least common multiple of 6 and 10, and contains all
the prime factors of these quantities ; viz. , 2, 3, and 5.
144. What is a multiple? 145, A common maltipje? 146. The least common
multiple ? 147. Name Principle x. Principle 3. Principle 3.
68 MULTIPLES.
148. To Find the Least Common Multiple of IMonomlals by
I^itne Factors.
I. Find the L c. m.* of isahfi, sl^cx, and gbc^
Analysis. — The prime foctora onauTioK.
of the coefficients are 5, 3, and 3. 150*2? = 3X5Xa*Xa?
The prime factors of the letter a ^Vcx =z ^ xVxcxx
are a, a, a, a, which are denoted by gid^z = 3X3X6xc'x«
a*. In like manner, the prime fac ^^^ Aca^e^d^z.
tors of X are denoted by a^, those of *^
b by ^, and those of r by <^ ; 2 is prime. Taking each of these factors
the greatest number of times it occurs in either of the given quanti
ties, the product, 450^0^2, is the I, c* in. required. (Art 147, Prin. 2.)
Hence, the
Rule. — Resolve the quantities into their prime factors ;
multiply these factors together, taking each the greatest num
ler of times it occurs in either of the given quantities. The
product is the I. c. m. required.
Or, Find the least common multiple of the coefficients, and
annex to it all the letters, giving each letter the exponent of
its highest power in either of the quantities.
Note. — Tn finding the I, c, tn. of algebraic quantities, it is often
more expeditious to arrange tUem in a horizontal line, then divide,
etc., as in arithmetic.
Eequired the I. c. rti. of the following quantities :
2. 9^8, i2a^a?^ and 2^ax^y. Ans. "jzahh/.
3
4
5
6
7
8
^ah^c, 2Sbc^y and $6a^l?d.
i6x^^z, 2oy^z, and Zxy^.
i^a^V^c, gal^i^, and i^a^bd^.
2Sab'^, i4a^*, zS^'^^^y ^^^ 420*3.
ym^n^y, i2mhiy\ and :^mn^y^.
X48. How find the I. e. n». of moDomialB by prime Actors f What other method 7
* The Initials U c. //!• are used for the least common multiple.
MULTIPLB8. 69
149. To Find the Least Common Multiple of Polynomials.
9. Eequired the I. c. m, of a* + J^ and c? — V,
AlffALYBiB. — Resolving ofbbation.
ihe quantities into their fl'—^' = (flfft) X («— J)
prime factors, as in the 0*+ J^ = (a + J) X («^— oA + J'')
margin, (a+6) is common (^ + j) ^ /^_jx ^ (a^^ah+V) z=
toboth,andistheirrycrf. «4_«8j + ^«j4. ^^:
(Art. 139.) >iow multiply
ing these factors together, taking each the greatest number of times
it occurs in either of the given quantities, the product al^—a^ + db^^l^
18 the I. c. tn* required. (Art 148.)
Second Method.
Stnce the ff, c. d, contains all the factors common to both quan
tities (Art. 147, Prin. 2), it follows if one of them is divided by the
(/• c. d, and the quotient multiplied by the other, the product will
be the I. e. tn. Hence, the
BuLE. — Resolve ihe quantities into their prime factors
and multiply these factors together^ faking each the greatest
number of times it occurs in either of the given quantities.
Their product is the I. c. tn. required.
Or, Find the greatest common divisor of the given quanti
ties^ and divide one of them by it. The quotient, multiplied
by the other, will be their I. c. m.
10. Find the ?. c. m. of 2a — i and ^a^ — i.
Solution. —The g. c. d. is 2a— i. Now (4a'— i)*(2a— i)=(2a + 1) :
and (20 + 1) X (2a— i) = 4a*— I, Ans.
Find the I. c, tw.. of the following quantities :
11. a^ — y^ and a^ — 2xy + yK
Ans. Q^ — ^y — iry® + y*.
12. «' — 53 and flS _ j8^
13. ofi^ I and a^ + 2X + 1.
14. 2a' + 3a — 2 and 6a^ — a — i.
^15. m* + m — 2 and m* — 1.
(See App<»adli, p. 284.)
CHAPTER YIII.
FRACTIONS.
150. A Inaction is one or more of the equal parts into
which a unit is divided.
151. FractioDs are expressed by two quantities called the
numerator and denominator, one of which is written below
the other, with a short line between them.
152. The Denominator is the quantity below the
line, and shows into how many equal parts the unit is
divided.
153. The Numerator is the quantity above the line,
and shows how many parts are taken.
Thos, the expression  shows that the quantily is divided into b
equal parts, and that a of those parts are taken.
154. The Unit or Sase of a fraction is the quantity
divided into equal parts.
155. The Terms of a fraction are the numerator and
denominator.
156. An Integer is a quantity which consists of one
or more entire units only ; as a, 3a, 5, 7.
157. A Mixed Quantity is one which contains an
integer and a fraction.
Thus, a •{•  is a mixed quantity.
c
X50. What is a faction? 151. How expressed? 153. What does the denomina
tor show? X53. The nnmerator? 154. V^hat is the base of a fraction f 155. Th«
leim? of a fraction t 156. An integer? 157. A mixed quantity ?
FBAOTIOKS. 71
158. Fractions arise from division, the namerator being
the dividend and the denominator the divisor. Hence^
159. The Value of a fraction is the quotient of the
numerator divided by the denominator.
Thus, the value of 6 thirds is 613, which is 2 ; of —  is ^m.
SIGNS OF FRACTIONS.
160. JEvery Fraction has the sign + or — , expressed
or understood, before the dividing line.
161. The I>ividing Line has the force of a vincu^
lum or parenthesis^ and the sigii before it shows that the
value of the tohole fraction is to be added or subtracted.
162. Every Numerator and Denominator is
preceded by the sign + or — , expressed or understood.
In this case, the sign MiTects only the single term to which
it is prefixed.
163. If the Sign before the Dividing Line is
changed from + to — , or from — to +, the valvs of the
fraction is changed from + to — , or from — to +.
_, hx — cx — dx  jM X. . hx — ex — dx
Thus, a + = a + o — e — a; but a =
X X
a'b + c + d.
164. If all the Signs of the Numerator are
changed, the value of the fraction is changed in a corre
sponding manner.
Thus, = + a + & ; but = — a — 6. w
XX ^
158. From what do firactiouB arise? 159. What is the value of a firaction*
z6o. What is prefixed to the dividing line of a fraction ? x6z. What is the force of
the dividing line? 162. By what is the namerator and denominator preceded?
How fiir does the force of this sign extend ? 163. If the sign before the dividing
line id changed, what Is the efi'ect ? 164. If all the signs of the namerator are
changed ?
72 FBACTIOKS.
165. If all the Signs of the Denominator are
changed^ the value is also chaoged in a corresponding
manner.
Thus, — =+a; but — = — a. Henoe,
166. If any two of these changes are made at the same
time, they will balance each other, and the Talae of the
fraction will not be altered.
ab "Ob —db db "
■"»'». j = ^ = — = — »=+«•
db — db db —db
PRINCIPLES.
167. The principles for the treatment of fractions in
Algebra are the same as those in Arithmetic.
1°. Multiplying the numerator y or ) Multiplies the
Dividing the denominatovy ) fraction.
_ 2X2 4 2 . , 2 2
Thus,  = 5 = . And  =  •
6 63 6*2 3
2°. Dividiiiq the numerator y or ) r^ 'j al ^ \i
^r 12' 1 ' 2T. J '2 > Divides the fraction.
Multiplying the denommatory )
,« 2t2 I . , 2 2 1
Tbufl,  = . And  = — =  .
6 6 6x2 12 6
3°. Multiplying, or dividing both ) Does not change its
terms by the same quantity ) value.
^ 2x2 4 2 I ., 2*2 I
Thus, = I. =  = . And = 
6x2 12 6 3 6i2 3
4°. Multiplying and dividing a ) Does not change its
fraction by the same quantity j value.,
2X2S2 2
Thus,
6x2^2 6
165. If an the BignB of the denomiuator are changed ? xG6. If both are changed t
167. Name Principle z. Principle a. Principle 3. Principle 4.
REDUCTION OF FRACTIONS. 73
REDUCTION OF FRACTIONS.
168. JReduction of Fractions is changing their
terms without altering the value of the fractions.
CASE I.
169. Jo Reduce a Fraction to its Lowest Terms.
Def. — The Lowest Terms of a firaction are the
smallest terms in which its numerator and denominator
can be expressed. (Art. 122.)
1. Reduce  — r — to its lowest terms.
iSaocx
Analysis.— By inspection, we perceive the opbeatiok.
factors 5, a, &, and x are common to both terms. 5^ ^^^ aua ^
Cancelling these common factors, the fraction i^abcx 3c
becomes — . Now since both terms have been divided by the
same quantity, the valne of the fraction is not changed. (Art. 167,
Prin. 3.) And since these terms have no common factor, it follows
that — are the lowest terms required. (Art. 122.)
3^
Note. — It will be observed that the factors 5, a, b, and x are prime ;
therefore, the product sdbx is the g, c. d, of the numerator and
denominator. (Art. 121.) Hence, the
EuLE. — Cancel all the factors common to the numerator
Sind denominator.
Or, Divide loth terms of the fraction by their greatest
common divisor, (Art 167.)
2. Eeduce  — ^ to its lowest terms. Ans. «^«
i2aoc 3
3. Seduce to its lowest terms. Ans. —
sac c •
z68. What is rednctlon of fractions ? X69 What are the lowest terms of a frac
tion ? How reduce firactiooF to the lowest terms ?
.
74 REDUCTION OF FRACTIOKS.
Bedace the following fractions to the lowest terms:
4. r=« 10. 5 • •
^ ^xh^ 2x^y — 2xyz
i2a%<? 7 u ^ a + hc
^ 4a^>cd ^ (a + Jc) X a?
^ i^jlf^cxy " ^^ ^ — f
Sil^cxy' ' a^ — y*'
S4aW(^' ax — ofl
o? — V a — I
a2 + 2a^> + J3 ^ aa — 2a + I
a^ — 2a;y + y«' ^' a? + 2a?y + y*'
CASE II.
170. To Reduce a Fraction to a Whole or Mixed Quantity.
I. Beduce — ^^Ji to a whole or mixed quantity.
Analtbib.— Since the value opibatiok.
of a fraction is the quotient of 2a + 4b + C , c
the numerator divided by the 2 ,2
denominator, it follows that per
forming the division indicated will give the answer required. Now
2 is contained in 2a, a times ; in 4Z), 25 times. Placing the remainder
e over the denominator, we have a + 2& + , the mixed quantity
required. Hence, the
RvLB,— Divide the numerator by the denominator, and
placing the remainder over the divisor, annex it to the
quotient.
None. — This rule is based upon the principle that both terms are
(Umded by the sani'e quantity. (Art. 167, Prin. 3.)
170. Howredace a fhictlon to a whole or mixed quantity? JVWtf. Upon what
principle Is this rule baeed?
BEDUOTIOK OF FRACTI0K8. 75
Bednce the following to whole or mixed quantities :
ax "a? ^ a^—2ah + l^ 0^
2. • 6. —
X
aS3»
a
ja + (j»
8.
a^ + a^—aa?
CASE III.
171. To Reduce a Mixed Quantity to an Improper Fraction.
I. Reduce a +  to the form of a fraction.
3
Analysis. — Since in i unit there are operation.
three thirds, in a units there must be a , ^ 3^ . ^
6 b a +  = ^^ + 
times 3 thirds, or — ; and — +  = ^ — , 3 3 3
3 33 3 3(1 b ^a + b
the fraction required. Hence, the "I" "i" 1 — 1
^ 3 3 3
BuLE. — Multiply the integer by the denominator ; to the
product add the numerator, and place the sum over the
denominator. (Art. 65.)
Notes. — i. An integer may be reduced to the form of a fraction by
making i its denominator. Thus, a =  .
* 2. If the sign before the dividing line is — and the denominator is
removed, all the signs of the numerator must be changed. (Arts. 163, 82.)
Reduce the following to improper fractions :
y cd . abd — cd ,
3. ao — i Ans, j = od — c.
a a J
xy — b ^ i'X rr ^
4*
y I +x
, 2X VI — Cfi
•s
76 BEDUCTION OF FKACTIONS.
CASE IV.
172. To Reduce an Integer to a Fraction having any required
Denominator.
1. Bednce 3a to fifths.
Analtbib. — Since in la there are 5 fifths, in opsiuTioir.
3a there must be 3 times 5 fifths, or ^. 3^5 = —
Or, reducing the integer 3a to the form of a ^^ ^ ^ ^ ^
fraction, it becomes ^; multiplying both ^ X 5 5 *
terms by the required denominator, we have ^. Hence, the
EuLE. — Multiply the integer by the required denominator^
and place the product over it.
2. Bedace 22; to a fraction having 6m for its denominator.
3. Keduce 602; to a fraction having 4a J for its denominator.
4. Reduce 3a + 4J to a fraction having 6(? tor its
denominator.
5. Bedace a? — y to a fraction having a: f y for its
denominator.
6. Eeduce 27^y to a fraction having 3^^ — 2b for its
denominator.
173. To Reduce a Fraction to any Required Denominator.
I. Change  to a fraction whose denominator is 12.
Analysis. — Dividing 12, the required de opbbation.
nominator, by the given denominator 3, the 12 — 3 =4
quotient is 4. Multiplying both terms of a X 4 ^a
the given fraction by the quotient 4, the 3X4 12
Ml
result, ^, is the fraction required. Hence, the
EuLE. — Divide the required denominator by the denomi
nator of the given fraction, and multiply both terms by the
quotient
172. How rednce an integer to a fVaction having anj required denomiitttor?
REDUCTION OF FRACTIONS. 77
Bediice the following to the required denominators:
2. Eeduce — to thirtyfifths.
_ __ 3^x5 15a ,
Solution. 35 « 7 = 5. Now  — = = =^. Ans.
7x5 35
3. Eeduce  to the denominator ac.
4. Eeduce — to the denominator 49^.
5. Eeduce ^ to the denominator a^ — 2xy + y*.
X — y
6. Eeduce • to the denominator 2>a^ Ix + y)\
x + y \ :f/
COMMON DENOMINATORS.
174. A Common Denominator is one that belongs
equally to two or more fractions.
PRINCIPLES.
1°. A common denominator is a multiple of each of the
denominators ; for every quantity is a divisor of itself and
of every multiple of itself (Art. 124, Prin. i.) Hence,
2°. The least common denominator is the least common
multiple of all the denominators.
CASE V.
175. To Reduce Fractions to Equivalent Fractions having a
Conitnon Denominator.
a c X
1. Eeduce rr, ^, and , to equivalent fractions having
a common denominator.
Z73. How reduce a fraction to any required denominator ? 174. What is a com*
mon denominator ? Principle x ? Principle 2 ?
78 BBDtTCTION OF FRACTIONS.
SoLimoN.^MnltiplTing the denominatora b, d, and y together, we
have hdy, which is a oommon denominator.
b X d X y =^ hdy The common denominator.
a X d X y = ady )
ex b X y =• bey /• The now nomeraton.
X X b X d=: bdx )
M a _ ady c _ bcy^ x _ bdx
b"^ bdy* d^bdy' y"^ bdy
To lednoe the given fractions to this denominator, we multiply
each numerator by all the denominators except its own, and place the
resolts over the common denominator. Hence, the
Rule. — Multiply all the denominators together for a com
man denominatory and each mtmerator into all the denom
inators except its owuyfor the new numerators.
Notes. — i. It is advisable to reduce the fractions to their lotoest
terms, before the rule is applied. (Art. 169.)
2. This rule is based on the prindple» that the mdtis of a fraction is
not changed by multiplying both its terms by the same quantity.
(Art. 167, Prin. 3.)
Bedace the following to equivaleiit fractions having a
common denominator:
, « a; 3 A„. ^y 4«» 3cy
c y 4 4cy 4cy Acy
c b 2d ^ 2a c + I
^ 5' x' 7* ». ^> J >
a b X 20 c + d
^ Vx' "?' y ^ 3' V' ^"^d"
la X ^y ^ 2a
5» — t"* — : — 5"* io» — 9 ~9 "i"'
J ^i.> ^ I A z 2' b
6. ~^. ^L^. „. '" • ^ + ^
12.
Zb' a
+ b
x — y
^ + y^
x + y'
x^y
a ^ h
sa— I
4' ^' ^ + y
a — x a + X
a '^'" a + x^ a — x '
Z75. How reduce fractions to equivalent fractions having a common denomina
tor ? Note. Upon what principle is thia rule based ?
BEDUCTION OF FfiAOTIONS.
79
CASE VI.
176. To Reduce Fractions to the Least Common Denominator.
I. Eeduce • — , and — to the I, c.d.
X xy yz
Solution.— The ImCtn, of the denominators is xyz. (Art, I4n
xyz = Lc d.
xyzTx =yz
xyz Txy = z
xyz iyz =2X
The
moltipllen.
a X yz _^ayz
X X yz"^ xyz
b X z bz
xy X z
d XX
xyz
dx
The
factions
required.
yz X X xyz
To change the given fractions to others whose denominator is xyz, we
mnltiplj each numerator by the quotient arising from dividing this
mnltiple bj its corresponding denominator. Hence, the
Rule. — I. Find the least common multiple of all tJie
denominators for the least common denominator,
II. Divide this multiple by the denommator ofeachfra4>
tiony and multiply its numerator by the quotient.
Note. — All the fractions must be reduced to their lowest terms
before the rule is applied.
Reduce the following fractions to the I. c. d. :
2.
a
be
x'
1.
4C
3.
cd
ar
2X
3«'
xy 
ac
4.
a
2'
b c
~3' 4
X
5
a^c
ab'
2Cd
Ve
' ^bc
6.
idb
3
' 4'
X
a\'
I
8
7.
2a
4b'
cd
be'
x^y
bcx
J76.
How reduce fh
wt
.jt
8.
y 9
lO.
II.
12.
13'
a + b a — b a^ + V
2(x+y) a ab
3{x+y)' xy' 6{x+y)'
'c^' Wb
X m y
ac' ¥c' ^'
X a + b
m+n m—n
d
xz
m
3
3^2 ^ 2a.7^ * 4cx
'/
7<
80 ADDITION OFFRACTIONS,
ADDITION OF FRACTIONS.
177. When fractions have a common denominator^ their
nufnerators express like parts of the same unit or base, and
are like quantities. (Art. 43.)
178. To Add Fractions which have a Cknnfnon Denominator.
1. What is the snm of , , and  ?
80LOTIOH. I and } are ^, and f are Y» ^^^^ Hence, the
BuLE. — Add the numerators, and place the sum over the
common denominator.
2. Add — , — , and — • Ans. ^•
m m m m
3. Add  — , , — , and  —
2xy 2xy 2xy 2xy
A iji idxz iidxz iidxz , Adxz
4. Add '—J, , y — J , and =t
5aoc ^abc saoc saoc
5. Add ^tt^ to 3*=f . 6. Add 3^±^ to 4^^II^.
179. To Add Fractions which have Different Denominatorsc
_. a c fn
7. What is the sum of t, „ and — ?
' b d X
bxdxx = bdx, c. d.
a adx c box
Analysis. — Since these fractions
have different denominators, their
numerators cannot be added in their b bdx d bdx
present form. (Art. 66.) We there m bdm
fore reduce them to a common denom x bdx
inator, then add the numerators. ^^ _(. j^^ ^ j^^
(Art. 175.) Hence, the y^ , Ans.
Rule. — Reduce the fractions to a common denominator,
and place the sum of the numerators over it.
Note.— AH answers should he reduced. to the lowest terms.
177. When iVactione have a common denominator, what is tme of the nnmeratorB*
X78. How add each fracUona ? 179. How, when they have different denominators ?
ADDITION OF FRACTIONS. 81
Find the sum of the following fractions :
y X m ' mxy
3« . a? . y cd ±y , bx
453 3«2rf5
20? I 3y a gy^ + d
3 2a^ 4 a 3A
a . a: a . d
11. ^— \"^ . 17. H
X + y ^ X'y o — a;,— A
12. — ^^ H ^. i8.
2xy xy y m — n
T, g +^ . 3 + ga; — 4 . — 16
13. • 19. •
y ay 27 — 3
a ^ ab m 6c vm
^^ :;rrz + z — 7/ ^o x + T "" ^'
aj + ya; — y d a 3a;
180. To Add Mixed Quantities.
h vn
1. What is the sum of a 4  and d ?
c n
SoLUTlOK. — Adding the integral and fractional parts separately,
the result is a + <2 H , the sum required. Hence, the
en
EuLB. — Add the integral and fractional parts separately y
and unite the results, (Art. 179.)
Note. — Mixed quantities may be reduced to improper fractions, and
then be added by the rule. (Art. 171.)
h a.
2. "What is the sum oi a •\ and c +  ?
2 X
3. What is the sum of a; + ^ and ?
^ b m — y
4. What is the sum of xd and a H ?
^21
a "^ ti
5. What is the sum of 5a; + t and — 
t8o. How add mixed qnantltiee ? Note. How else may they be added ?
82 SUBTRACTION OF FBACTIOKS.
181. To Incorporate an Integer with a Fraction.
6. Incorporate the integer ab with the fraction — "^ — •
Sx + y
.SOLUTTON.<— Reducing ab to the denominator of the fraction, we
3«+y 3«+y d^+y 3^+y
Hence, the
Rule, — Reduce the integer to the denominator of the f roc
tion, and place the stim of the numerators over the given
denominator. (Art. 172.)
Oft
7. Incorporate the integer 3d with the fraction r'
8. Incorporate — 4y with «
9. Incorporate — a with — ^^•
10. Incorporate 3^ + y with
11. Incorporate — a + 5J with ^•
12. Incorporate 2X + 2y with »
i*/ "~~ I
SUBTRACTION OF FRACTIONS.
182. The numerators of fractions which have a common
denominatorj we have seen, are like quantities. (Art 177.)
Hence, they may be subtracted one from another as integers.
1. Subtract  from f .
7 *? 7—1 2 1 1
Solution. ^ — ^ = /_i =  or . 4n«. •
8 8 8 8 4 4
2. Prom Y subtract t* 4^^« t — x = — i — •
b b b b b
z8z. How incorporate an integer with a fraction ? xSa. What is trae of the
numeralora of fhtctiona haying a common denominator ? How suhtract such fhu>
tione ?
SUBTRACTION OF FBACTIOKS. 83
3. From —7— subtract ^r*
a a
4. Prom ^^ subtract  —
a a
183. To Subtract Fracticns which have IHfferent
Denominators.
5. Prom  subtract — •
3 2
Analysis — Since these fractions
have different denominators, they can
not be subtracted one from the other
3 X 2 = 6, C. A
7a_ 14a
in their present form. We therefore 3^ 9^
reduce them to a eomman denominator, 2 6
which is 6, and place the difference of jAa ga Ka
the numerators over it. Hence, the "^ ^ =^ ~^ > A1I8.
ScjLE. — Reduce the fractions to a common denominator^
and subtract the numerator of the subtrahend from that
of the minuend, placing the difference over the common
denominator.
Notes. — i. The integral and fractional parts of mixed quantities
should be subtracted separately, and the results be united.
Or, mixed quantities may be reduced to improper fractions, and then
be subtracted by the rule. (Art. 171.)
2. A fraction may be subtracted from an integer, or an integer
from a fraction, by reducing the integer to the given denominator,
and then applying the rule.
6. From 5^, take 3£^. Am. ^^^ " 3^.
X y xy
7. Prom — , take "" ■
m' y
» 8, Prom , take •
m y
9. Prom ?+^, take ^^:=^.
4 3
183. How when they have different denominators ? Note z. How eabtract mixed
qnantities f Note 2. How a Ihiction from an integer, or an integer from a Ihiction ?
84 MULTIPLICATION OF FKACTIOKS.
10. From A,take*+^.
m y
11. From , take fw.
y
12. Prom 4a + , take xa — 3*
c a
ij. From a + , take "" ■
a 3
14. From = , take ^ •
ft — a; rf + y
15. From a , take — •
y 3
16. From "^y take "^ «
10 ip + y
17. From a? — ^"" , take "~^ g.
2 3
MULTIPLICATION OF FRACTIONS.
CASE I.
184. To Multiply a Fraction by an Inteffer*
a
I. Multiply ^ by m.
Analysis. — ^Multiplying the numerator of the
OFBRATION.
fraction by the integer, the product is am.  v Wl = ^^
(Art. 167, Prin. i.) * ft
2. What is the product of jxx?
a a
X Xz=z
Analtsis. — A fraction is multiplied by
dividing its denominator ; therefore, if we bx bx ^ X
divide hx by x, the result will be the product a a
required. (Art. 167, Prin. i.) Hence, the g^. . ^ ^^ j*
Rule. — Multiply the numerator by the integer.
Or, Divide the denominator by it.
Notes. — i. A fraction is multiplied by a quantity equal to its
denominator, by cancelling the denominator. (Art. no, Prin. 4,)
X84. How multiply a fraction by an integer?
MULTIPLIOATIOi^ OF FRACTIONS. 85
2. A fraction is also multiplied by bxlj factor in its denominator, by
eanceUing tliat factor, (Art. no, Prin. 4.)
Find the product of the following quantities:
3. —5^ X (a — J). Ans. 3a:.
ah , \ aJ)
cd c
5. — r^ X (3 + ♦w). 12. — ^ X (12a; 4 18).
3 + wi 6
6. x6. 13. ^^x(it.).
7. 7^^ X (3c + 2d), 14. X 4i«^
8. ax — X 6a;. 15. — x i^z — 2).
3a; 40;? — 10 ''
a + h ^ 30 d vf
T X 5a;. 16. X ^5.
20a; + 250;^
/
10. ^— ! X 2ac, 17. — ^ — X (y^— i).
5c 4 c y — I
II. — i^ X 20a;. 18. ^ s X (a; + z).
CASE II.
185. To Multiply a Fraction by a Fraction.
d d
1, What is the product of  by •
Analysis.— Multiplying the numerator of opbbatton.
the fraction  by d, the numerator of the _ z= —
c '' c C
multiplier, we have — . But the multiplier is ^^ ^^
 ; hence the product  is w times too large. C X m'' cm
m c a d ad
To correct this, multiply the denominator  x — = —
by m. (Art. 167, Prin. 2.) ^ ^ ^^
NoUx. How is a fraction multiplied by a quantity equal to ita denominator J
Note 2. How by any factor in its denominator ?
86 IfULTIPIiICATIOK OF FRACTIONS.
2. Beqnired the product of r~^ multiplied bj — •
Analybjs. — ^The factors 2, a, and e are onvunmr.
ocnnmon to each tenn of the given fiao 2db cm 2abcfn
tiona. Cancelling theae common f actors^ 6cd OX tocdx
the reault is ^, the product required. 2ahcm _ ftm ^^
(Art 167, Prin. 3.) Hence, the **
BuLE. — Cancel the common factors; then muUipltf the
numerators together for the neto numerator, and the denom
inators for the new defwminator.
KoTBS. — I. Mixed quantities should be reduced to improper fractions,
and then be multiplied as abova
Or, ihe fractional and integral parts may be multiplied separately,
and the results be united.
2. Cancelling the common factors shortens the operation, and gives
the answer in the loioest terms.
3. The word of in compound fractions has the force of the sign x .
Therefore, reducing compound fractions to simple ones is the same as
multiplying the fractions together. Thus, foff = xf=:A.
Find the products of the following fractions:
2X ^xy 2dy ,1 3
3. — X ^ X ^ 6. — ; X ^•
^ y 2d X + 30:8
ho X d (a+97i)xA 4V
4. — X r X — 7 ^ — X 7—r\ —
a oy Tfin (a+97i)xc
aj — y a? + Jf o a + J cd
^ yz y + » (^ X
9. What is the product of ~^^ by ^^ZJ^ ?
Solution. — ^Factor and cancel. Ans, ^(x^y),
185. How multiply a fraction by a fhiction? yoU t. How mixed qnaotitieat
Vote 3. How shorten the operation? yiote 3. What is the force q( the wof^^ in
compound fractions ?
MULTIPLICATION OF FBAOTIOK«. 87
ID. Multiply by ■^ Ans. — ^^^*
II. Multiply ?^ by jE^'
, 12. Multiply ^^^^ by ^.
13. Mnltiply a + 5^^±— by ^.
22^ X \ ft
14. Multiply a; + — by — ^
IS
Maltiply x
X
by  + •
16.
Maltiply a
2C?
^ ah
CASE III.
186. To Multiply an Integer by' a FradA^in.
dx
1. Multiply the integer a by —
Ai7AiiTBi8. — Changing the integer to the ^
a a = 
form of a fraction, we have  to be multiplied i
dx adx ^ ^ ^^
by — , which equals — . Hence, the  X — — ~T~r*
•^ <y cy 1 cy cy
Rule. — Reduce the integer to a fraction ; then multiply
the numerators together for the new numerator, and the
denominators for the new denominator.
Notes. — i. Multiplying an integer by 9kfr<iction is the same as find
ing K fractional part of a quantity. Thus, xx\\s the same as finding
} of X, each being equal to — . That is»
4
2. Three times i fourth of a quantity is the same as i fourth of
3 times that quantity.
z86. How multiply an integer by a Auction ? Note, To what I9 this operation
similar}
88 MULTIPLICATION OF FKACTI0N8.
Find the product of the following quantities :
. 9. {^^1/^)X '^
2.
, dx
aoc X —
cy
3
. b\c
ad X
^y
4.
m + n
ax X •
4a
S
{a + h)xf
6.
(3«y) x^y
10, (a* + ab) X.
II. (a;* + i) X
3 (« + y)
2{a+ b)
2ax
3(a?^'
12. zxyia — b) x ^3 _ y '
13 3«(2?— i) X
a^8— I
7. (a:* + i) X ^ 14. (2fl* + S*) X "^
2;— I "^v '4a + ft
8. (i — a2)x^ 15. (in2)x— ^ —
187. The principles developed in the preceding cases may
be summed up in the following
GENERAL RULE.
Reduce whole and mixed quantities to improper fractions ,
then cancel the common factors, and place the product of the
numerators over the product of the denominators,
1. Multiply ^—— by 152?.
%x
2. Multiply — ^— by y* — i.
3. Multiply * + ^7 by ^^•
4. Mulhply  X ^ by f^.
5. Multiply ^ by g.
^2 }^ a X
6. Multiply X — :T by t
187. What is the general rule for mnltiplying firactions ?
• /
DIVISION OF FRACTIONS. 85
7. Multiply ^— ^ by a; + «.
8. Multiply ^ by 2f.
if
ic I ty
9. Multiply ^ by a^ — 2a:y + y».
•
10. Multiply ^^ ^ by 85; — 2.
11. Multiply a: — — by  + ^.
12. Multiply a + — by ^.
13. Multiply ^^+^>%y ^
2a {c •\ d)
14. Mniltiply ^ by ^^.
15. Multiply * + ^^ by 5  ^^.
2fl (Xj^ — i/
16. Multiply by —*
^ "^ a; — y '^ ax
DIVISION OF FRACTIONS.
CASE I.
188. To Divide a Fraction by an Integer.
I. If 3 oranges cost — dollars, what will i cost ?
Analysis.— One is i tbird: of 3 ; therefore, °^
I orange will cost i third of — dollars, and ~ "^ 3 — "T"
 of — dollars is — dollars. (Art. 167, Prin. 2.)
3 71 n
2. Divide  by m.
Analysis.— Since we cannot divide opebatton.
the numerator of the fraction by m, ^ _i_ ^ __ ^ «_ ^
we multiply the denominator by it. C ' C X m cm
90 DIVISION OF 7BACTI0K8.
The result 10 — . For, in eecli of the fractions  and — , the same
C1U G em
nnmber of parts is taken ; but. since the unit is divided into m times
as many parts in the kUt&r as in the former, it follows tliat each part
in the latter is onlj — th of each part in the former. Hence, the
Bulk — Divide the numercUor by the integer.
Or, Multiply the denomhuitor by it.
Note. — If the dividend is a miaed qnantity, it should be reduced
to an improper fraction before the rule is applied. (Ex. 3.)
Divide the following qnantitiea:
. Jc ,  . ax+bc
3 « +  by A Ana. —^ —
4 ^ + ^byay. ^n^. __._.
601^ . a^ + ax . ,
s —  by s^y 9. —^ ^7 « + ^•
6. by 0. 10. r— ;; DV d — C.
Zac ^ b+ c ^
7. a 4 y by a. 11. ^a + c ^^ ^'^^'
8. ax { ^ by a?. 12. ^ by a + ft.
a a —
CASE II.
189. To Divide a Fraction by a FnwHon*
This ease embraces two classes of examples :
First, Those in which the fractions have a common
denominator.
Second. Those in which they have different denominator.
x88. How divide a fraction by an integer? Note* If tbe diyidend iB a mixed
qnantity, how proceed ?
DIVISION OF FBAOTIONS. 91
1. At — dollars apiece^ how many kites can a lad bny for
— dollars ?
Analysis. — Since these fractions have a eonu oteratioh.
mon denominator, their numerators are like 12a %a
quantities, and one may be divided by the m * m
other, as integers. (Art. 177.) Now 3a is con Ans, 4 kites,
tained in 12a, 4 times. (Art. iio^ Prin. i.)
2. How many times is  — contained in ^ ?
^ X X
3. What is the quotient of ^^ divided by ^
/» /J
4. It is required to divide  by •
X y
Analysis. — Since these frac
tions have different denomina
tors, their numerators are unUke
quantities ; consequently, one
cannot be divided by the other
in this form. (Art. ii4,7W««.) We
therefore reduce them to a covi^
mon denominator ; then dividing
one numerator by the other, the
result is the quotient.
Or, more briefly, if we iwcert the divisor, and multiply the dividend
by it, we have the eavn^ combinoEtuma and the eame result as before.
(Art. 185.) Hence, the
BuLE. — Multiply the dividend by the divisor inverted.
Or, Reduee the fractions to a eommon denominator, and
divide ths numerator of the dividend by that of the divisor.
Notes. — ^i. A fraction is inverted, when its terms are made tc
change placea Thus, ^ inverted, becomes  •
2. The object of inverting the divisor is convenience in multiplying.
3. After the divisor is inverted, the common faetore should be can
celled before the multiplication is performed.
189. How divide a fraction by a fhiction when they have a common denominator ?
When the denominators are different, how?
VmST OPEBATION.
a _
x"
_ay
xy'
c ex
y'^^y'
xy •
ex
xy"
^ay
ex
SBOOHD OFBBATIOV.
a
•
•
X
c _a
y^x
y
X 
c
ex
92 DIVISION OF FBACTIONS.
Divide the following fractions :
ah ^ X ^ h y ^ o^V
6. 4^by^. 14. — /by^^.
6aoy ''ax 4ca ^ 2d
7. 3^by^. IS. ^^by^.
4''2 ^a; + y'4y
8. ^by^. 16. ^by3^*.
^' ic ^ oa;' '' aofi ^ hx
10. ^— ^ by ^^ 18. ^Y by — r
\oab ^ 2oaox ^ 5J "^ lojy
^^ i2(g+y) . A(^+y) _sft_ . loSy
«J ^ 2ab ^' 36flrf ^ i8flS
12 rT^^y:r 20. _^by ^
.♦
a + h ^ a ' iSab ^ s^ad
CASE III.
190. To Divide an Integer by a Fraction.
I. Divide the integer ydc hj ^^•
* Analysis.— Having reduced the opbbatiok.
integer to a fraction, and inverted ^ j^ _i_ 3^^ _,
the divisor, we cancel the common b
factor d, and proceed as in the last jdc h 78c
case. Uence, the "T ^ 3ad ~ 3a ' '
Rl'LE. — Reduce the integer to a fraction^ and multiply it
hy the divisor inverted.
Divide the following quantities:
ex 1/
^ m + n SCI
4. (a + x)^^. 7. (ifl2)^Lt?.
y ^ 3iC
zga How divide an integer by a fraction ?
DIVISIOK OF FRACTIOKS.
93
191. Complex Fractions are reduced to simple ones,
by performing the division indicated.
a
8. Eeduce — to a simple fraction.
3
Analysis. — The given fraction is equiv
(t *?
alent to r 5 • Performing the division
indicated, we have —, the simple fraction
required. (Art. 189.)
V
=, the dividend.
, the divisor.
4
033^
Reduce the following fractions to simple ones :
10.
_
b
cd
g + I ,
g — I
a— I
Ans.
a^
hcd
12.
II.
a + I
a — h
I±JL.
a + b
13'
^ — y
a — h
x + y
x + y
a — b
a + b '
xy
3
192. Th« various principles developed 'in the preceding
cases may be summed up in the following
GENERAL RULE.
Reduce integers and mixed quantities to improper frac
tions, and complex fractions to simple ones : then multiply
the dividend by the divisor inverted.
191. How redacc complex fractions ? 192. What is the general role for dividing
fractions ?
94 DIVISION OF 7BAGTI0K8.
1. Divide ^ — by sbc
A/xyz
2. Divide ^^ by 9^.
27y ^ ^^
^. . , i6xy . 2cd
3. Divide — ^ by — •
4. Divide ^ — 5 by
a« — y» ^ a: — y
5. Divide y^^^r« ^V ^
^ a* + 2aft + J^ * a + J
6. Divide tzzjm±t by ^.
7. Divide « ; — • by •
SOLunOK. — ^Fuctoring and cancelling, we have,
a^—m* (a* + f» *)(a+f»)(g— m) a^+am __ a(anyi) 
a*— 2a»i+i»' "" (a— m)(a— w) ' a—w ~" a—m * '
o. Divide  — "^ — by — ^^«
10. Divide ; by ^7^ — (•
ac I aa: "^ 4(a + cc)
11. Divide , . , bv — ; —
„. Divide *i^^ by li^^).
T^. .J a — b ^ a^^W
13. Divide ^ 7 i„ by — — i—
14. Divide = by
•n. . J Jc' + Serf V h{c '\ d) ^ ,
15. Divide \ by — ^ '• (See Appendix, p. 285.;
a; + aaJi+a
CHAPTER IX.
SIMPLE EQUATIONS.
193. An Equation is an expression of equality between
two quantities. (Art. 27.)
194. Every equation consists of two parts, called the
first and second members.
195. The Mrst Member is the part on the left of the
sign =.
The Second Member is tlie part on the right of the
sign =.
196. Equations are divided into degrees, according to the
exponent of the unknown quantity; as the first, second,
third, fourth, etc.
Equations are also divided into Simple, Quadratic,
Cubic, etc.
197. A Simple Equation is one which contains
only thej^rs^ power of the unknown quantity, and is of the
first degree ; as, aa; == d.
198. A Quadratic Equation is one in which the
highest power of the unknown quantity is a square, and is
oi the second degree ; as, aa^ + ex = d.
199. A Cubic Equation is one in which the highest
power of the unknown quantity is a cube, and is of the
third degree ; ba, aafi + ba^ — ex = d.
Z93. What iB ao equation? 194. Uow many parts? 195. Which is the flrpt
memher ? The second ? 196. How are equations divided ? What other diTlslons ?
197. What Is a simple equation ? 198. A quadratic ? 199. Cubic?
96 SIMPLE EQUATIONS.
200. An Identical Mquation is one in which both
members have the same form^ or may be reduced to the
same form.
ThuB, oft— c = ab—c, and Sa?— 30? = 52?, are identical.
Note. — Such an equation is often called an identity.
201. The Transformation of an equation is chang
ing its /orm without destroying the equality of its members.
Note.— The members of an equation will retain their equality, 00
long as they are equaUy increased or diminished, (Ax. 25.)
TRANSPOSITION.
202. Transposition of Terms is changing them
from one side of an equation to the other without destroy
ing the equality of its members.
203. Unknown Quantities may be combined with
known quantities by addition, subtraction, multiplication,
or division.
Note. — ^The object of transposition 3s to obtain an equation in which
the terms containing the unknown quantity shall stand on one side,
and the known terms on the other.
.204. To Transpose a Term f^om one Member of an
Equation to the other.
1. Given a; + J = a, to find the value of x.
Solution.— By the problem, a? + & = a
Adding —6 to each side (Ax. 2), a; +6— 6 = a—5
Cancelling (+6— &) (Ax. 7), .*. x = a—h
This result is the same as changing the sign of h from + to — in
the first equation, and transposing it to the other side.
2. Given x — d = c, to find the value of x.
Solution.— ;By the problem, x—d = e
Adding \dio each side (Ax. 2), x—d+d = c+d
Cancelling {—d+d), (Ax. 7.) .*. x = c+d
200. Identical ? 201. What is the transformation of an equation ? Note. Bqnality.
209. What is transposition of terms ? 203. How combine unknown quantities ?
Jfote. Object cftranspositton?
ONE UNKNOWN (JUANTITY. 97
ThiB result is also the same as changing the sign of d from — lu 4,
and transposing it to the other side. Hence, the
Bjjlk— Transpose the term frmn one member of the equa*
tion to the other, a7id change its sign.
Note.— In the first of the preceding examples, the unknown
quantity is combined with one that is known by addition; in the
second, with one by 9M,r<uSt%on,
3. Given i — c + ar= a — rf, to find x.
4. Given a; + aJ — (?=a + &, to find x.
205. The Signs of all the terms of an equation may be
changed without destroying the equality. For, all the
terms on each side may be transposed to the other, by
changing their signs.
206. If all the terms on one side are transposed to the
other, each member will be equal to o.
Thus, if x+c = d,iX follows that x^e—d = o.
REDUCTION OF EQUATIONS.
207. The Reduction of an equation consists in finding
the value of the unknown quantity which it contains.
208. The Value of an unknown quantity is the number
which, substituted for it, will satisfy the equation. Hence,
it is sometimes called the root of the equation.
209. The reduction of equations depends on the following
PRINCIPLE.
Both members of an equation may be increased or dimirir
ished by the same quantity vrithout destroying the equality.
204. How traoBpose a term from one member of an equation to the other?
305. What is the effect of changing all the signs ? 906. Of transposing all the terms t
007. In what does the redaction of an equation consist ? ao8. What is the value of
annnknown quantity? What sometimes called? 809. TTpci) wl.at principle does
the reduction of equations depend ?
s
I
98 SIMPLE EQUATIONS.
210. This principle may be illustrated by a pair of
Bcales: If 4 balls, each weighing i lb., are placed in each
scalei they balance each other.
Adding 2 lbs. to eacli scale,
4+2 = 4+2
Subtiacting 2 lbs. from each,
4—2 = 4—2
Multiplying each by 2,
4x2 = 4x2
Dividing each by 2,
4H2 = 4+2
211. To Reduce an Equation containing One Unlcnown
Quantity by Transposition.
5. Given 2a; •— 3a + 7 = a; + 35, to find x.
Solution. —By the problem, 2a— 3a + 7 = a? + 35
Transposing the terms (Art. 204), 2X^x = 35—7 + 3a
Uniting the terms, (Ax. 9), Ans, a? = 28 + 3a
Therefore, 28 + 3a is the value of x required. Hence, the
Rule. — Transpose the unknown quantities to one sidt,
and the hwwn quantities to the other, and unite the terms.
Notes. — i. Transposing the terms is the same, in effect, as adding
equal quantities to, or subtracting them from each member; hence,
it is often called redaction of equations by addition or siibtractiofk
2. If the unknown quantity has the sign — before it, change the
signs of all the terms. (Art. 205.)
212. When the same term, having the same sign^ is on
,. opposite sides of the equation, ifc may be cancelled. (Ax. 3.)
" 6. Eeduce 3a; + « •— 6 = J — 4 + 22;.
^ '. . 7. Reduce a; — 3 + c = 2ic + fl — J.
8. Reduce 2^ + Jc — aeZ = y + 2m •— 8.
9. Reduce ^ah — y + rf = — 2^ + 17.
10. Reduce ^cd + 27 — 4a; + rf = 28 — 3a; + 3JA.
11. Reduce h + c — j^=i^2 + i — $x + d.
12. Reduce a; + 4 — 2a; — 3 = 3a? + 4 + 8 — 5a;.
azo. ninstrate this principle? axz. What is the rale for redocing equftlioDfr?
Note. To what is traii8po9it1oii eqaivalent? 212. When the scms term, having tht
fOiTM siffn^ is on opposite sides, what may he done ?
ONE tJNKNOWN QtJAKTlTY. 99
CLEARING OF FRACTIONS.
213. To Reduce an Equation containing Fractions.
1. Given  f 2 = 27, to find the value of x,
2 o
Solution.— By the problem, _ + _z = 27
2 6
Multiplying each tenn by 6, the L c. w, ) __
of the denominators (Art. 148), ) 3^+20?  162
Uniting the terms, ^x = 162
Dividing each side by the coefficient, Ans. x = 32{
X X 10
2. Given  +  = ^ , to find the value of x.
234
Solution.— By the problem, ? + ? = ??
234
Mult, by 12, the L c. in. of denominators, 607+40; = 90
Uniting terms, and dividing (Art. 211), Ans, 0; = 9
Therefore, the value of 0; is 9. Hence, the
BuLE. — Multiply each term of the equation by the least
common multiple of the denominators; then, transposing
and uniting the terms, divide each member by the coefficient
of the unkjiown quantity.
Notes. — i. An equation may also be cleared of frad^ionSt by multi
plying both sides by each denominator separately.
2. The reason that clearing an equation of fractions does not destroy
the equation, is because both members are miUtiplied by the same
quantity. (Ax. 4.)
3. A fraction is multiplied by its denominator by canceling the
denominator. (Art. 184, Note i.)
4. Removing the coefficient of a quantity divides the quantity by it.
5. If any given numerator is a multiple of its denominator, divide
the former by the latter before applying the rule.
6. The unknown quantity in the last two problems is combined
with those that are known by multipUcation and division. Hence, the
operation is often called, reduction of equations by multiplication and
division,
9x3. Rule for clearing an equation of fractions ? Notes, z. In what other way may
ftactionB be removed ? a. Why does not thie process destroy the equation ? 3. What
is the effect of cancelling a denominator? 4. Effect of removing a coef9cient'
5. If n numerator is a multiple of its denominator, how proceed t
100 SIMPLE EQUATIOKS.
_ _ ^rc AX
3. Eeduce ^^ + 12 = ^ + i.
5 3
4. Beduce = 6a? — 66.
36
5. Reduce ^ +  = 35 — «.
214. When the sign — is prefixed to a fraction and the
denominator is removed, the signs of all the ierms in the
numerator must be changed from + to — , or — to +.
6. Eeduce ^x = 20.
^ 5
Solution. — By the problem, 30? = 20
Removing the denominator 5, isaj— a;+2 = 100
Uniting and transposing the terms, 1/^ = 98
Dividing by the coefficient, Ans. x= 7
7. Given " ~ = — ^ , to find x.
X a
8. Given 3a; — — = a , to find x,
5 ^o
9. Given —x\ 1— = — , to find x.
3 4 24
X X
10. Given a + b { 6=^ \ [ a + b[ cr Sy '^ fii^d x.
2 4
215. The principles developed by the preceding illustra
tions may be summed up in the following
GENERAL RULE.
I, Clear the equation of fractions. (Ari 213.)
11. Transpose and unite the terms. (Art. 204.)
III. Divide both sides by the coefficient of the unknown
quantity. (Art. 213, Note 4.)
Proof. — For the unknoion quantity substitute its value,
and if it satisfies the equation, the work is right.
2x4. If sign — is prefixed to a fraction ? 215. What is the general rule for simple
equations ? How proved ?
ONE UNKNOWN QUANTITY. 101
EXAMPLES.
1. Given x \ 1  = 14, to find x.
2 4
X *JX
2. Given  + x=: — f 40, to find x.
2 10
z. Given — + 10 = 1 13, to find x.
5 10 ^
4. Given —  —  6 = 8, to find Xi
X — 2
5. Given x H = 10, to find x.
6. Given 2x H = 18 + ?, to find x
5 4 '
7. Given  H f  = 78, to find x.
234
8. Given ^— ^ 8 = j 5, to find a?.
o 4
9. Given x + ^ ^ + 7^ = 12, to find x.
2 o
10. Given 2a? — 16 = ^ , to find x.
3
11. Given —^ +  = 30, to find x.
4 2 '3
12. Given  + 7 = 16 + , to find x.
20 o
n 3^+ ^ 2a: , a?— I . ^ ,
13. Given  — ■ 10 = — , to find x,
2 36
14. Given f ^ = — — i A, to find x.
^ 10 6 15 ^^'
15. Given 8a: + 6J —  = 8i  — + — , to find x.
Note. — Sometimes there is an advantage in uniting similar terms,
before clearing of fractions. Thus, uniting 6^ with S^ ; also with
2
^=— , we have, 8aj = 2 + 8a? ; /. a? = 7, ^n«.
2 7
102 SIMPLE EQUATIONS.
i6. Given ^ — 6f7 = Tf^j to find x.
o o o
17. Given — = — + 15 — 12, to find x.^ ^
18. Given 2a; — 4 =  + 2, to find x.
2
19. Giyen ^ + ^ 1^ = ^ + ij^, to find x.
4 5 5 ^ C,
20. Given  = J + c, to find x.
(I
CLX
21. Given — = rf, to find x.
n
22. Given 1 = c, to find ic
2 3
23. Uiven = • to nnd rr.
^ a c
24. Given 1 =  + , to find x.
ax 2 a
^ 25. Given h — + ^=%H , to find x.
^ 3526*2'
26. Given ^ + ?? = 4S ^ 15? ^ g^j ^
2 3 S3
27. Given ^ 5 =, to find x.
' ■ X ^ x^
X ' I I
28. Given — ; h i = > to find x,
X + 1 a
X X o,
20. Given  H =  + a, to find x.
^ a c ^ a c
a?
30. Given a? + J = 7 , to find x.
\ 31. Given a? — a = , to find x.
^ '^ X — a
3.. Given 3 (^*) + (^) = 4 (^*). to find
^. ^fl? 2a; — 1;6 « ,
3;> Given ^ = a; ^ , to find a?.
9 3
a;
«
OKE UNKK0W2Sr QUANTITY. 103
34. Given — + x = 25, to find x.
4 2
2/ X
35. Given 801=42; 2# ^^ ^^^ *•
^6. Given ^—  = 22? , to find a.
3 4
37. Given 10 — 2a; = ^ — — — ^, to find x.
3 3
38. Given a; — 3 = 15 — , to find a?.
39. Given a? + 2 = 3a? H ■ — , to find o^
4 3
n $x x'4 a? — 10 /rx^j
40. Given ^ H = a; — 6, to find a?
42 2
^. iia; — I 5a;— II a;— I . «,
41. Given = , to find x.
12 4 10
42. Given ^ — 1 = 120, to find x.
5 10
' 2X "4" I
43. Given a? — 20 = , to find x.
44. Given ^^ = ^— f 12 — a*, to find x.
2 3
/N. I — X 2x 1 — ^a? ^,
45. Given — 2 h 10 = ^* to find x,
6 32
46. Given — ; = J, to find a^
a+ I a— I • '
rt* ^ 2 + a? C xisj
47. Given 7 rT = 5 — s> to A^d asr
.48. Given = 5 f , to find x.
X X
49. Given — = rf , to find x.
2 3
I ^~ X
50. Given 8a = , to find x.
I + a?
5,. Given t±^L±A = ^^ to find*.
104 SIMPLE EQUATIONS.
PROBLEMS.
216. The Solution of a problem is finding a quantity
which will satisfy its conditions. It consists of two parts :
First, — The Formation of an equation which will
express the conditions of the problem in algebraic language*
Second. — The Sediiction of this equation.
217. To Solve Problems in Simple Equations containing
one unicnown Quantity.
I. A fanner divided 52 apples among 3 boys in such a
manner that B had i half as many as A, and C 3 fourths as
many as A minus 2. How many had each ?
1. PoBMATiON— Let X = A's number.
By the conditions,  = B's "
2
4
Therefore, by Ax. 9, aj +  + — — 2 = 52, the Tvhole.
2. Reduction— 4^+2fl;+3a;— 8 = 208
Transposing, etc., 90; = 216
Removing the coefficient, a; = 24, A's number.
Ans. A had 24, 6 had 12, and C had 18—2 = 16.
From this illustration we derive the following
GENERAL RULE.
I. Represent tJis unhnown quantity iy a letter, then state
in algebraic langtiage the operations necessary to satisfy the
conditions of the problem.
II. Clear the equation effractions ; then, transposing and
uniting the terms, divide each member by the coefficient of the
unknown quantity. (Art. 213.)
Note. — A careful study of the conditions of the problem will soon
enable the learner to discover the quantity to be represented by the
better, and the method of forming the equation.
2x6. What is the solution of a problem ? Of what does it consist f 3x7. What is
the f^eneral rnje ?
ONE UNKNOWN QUANTITY. 105
2. The bill for a coat and vest is $40 ; the value of the
coat is 4 times that of the vest. What is the value of each ?
3. A bankrupt had $9000 to pay A^ B^ and C ; he paid B
twice as much as A, and as much as A and B. What
did each receive ?
4. The whole number of hands employed in a factory was
1000 ; there were twice as many boys as men, and 11 times
as many women as boys. How many of each ?
5. Two trains start at the same time, at opposite ends of
a railroad 120 miles long, one running twice as fast as the
other. How far will each have run at the time of meeting ?
6. A man bought equal quantities of two kinds of flour,
at $10 and $8 a barrel. How many barrels did he buy, the
whole cost being $1200 ?
7. If 96 pears are divided among 3 boys, so that the
second shall have 2, and the third 5, as often as the first
has I, how many will each receive ?
8. A post is onefourth of its length in the mud, one
third in the water, and 12 feet above water; what is its
whole length ?
9. After paying away J of my money, and then  of the
remainder, I have $72. What sum had I at first ?
10. Divide $300 between A, B, and C, so that A may
have twice as much as B, and C as much as both the others.
11. At the time of marriage, a man was twice as old as
his wife ; but after they had lived together 18 years, his age
was to hers as 3 to 2. Eequired their ages on the wedding day.
12. A and B invest equal amounts in trade. A gains
$1260 and B loses $870; A's money is now double B's.
What sum did each invest ?
13. Eequired two numbers whose difference is 25, and
twice their sum is 114.
14. A merchant buying goods in New York, spends the
first day  of his money ; the second day,  ; the third day,
J; the fourth day, i; and he then has $300 left. How
much had he at first ?
106 SIMPLE EQUATIONS.
15. What number is that, from the triple of which if 17
be subtracted the remainder is 22 ?
16. In fencing the side of a field whose length was 450
rods, two workmen were employed, one of whom built 9 rods
and the other 6 rods per day. How many days did they
work ?
17. Two persons, 420 miles apart, take the cars at the
same time to meet each other; one travels at the rate of 40
miles an hour, and the other at the rate of 30 miles. What
distance does each go ?
18. Divide a line of 28 inches in length into two such
parts that one may be  of the other.
19. Charles and Henry have $200, and Charles has seven
times as much money as Henry. How much has each ?
20. What is the time of day, provided ^ of the time past
midnight equals the time to noon ?
21. A can plow a field in 20 days, B in 30 days, and C in
40 days. In what time can they together plow it ?
22. A man sold the same number of horses, cows, and
sheep; the horses at $100, the cows at $45, and the sheep
at $5, receiving $4800. How many of each did he sell ?
23. Divide 150 oranges among 3 boys, so that as often as
the first has 2, the second shall have 5, and the third 3.
flow many should each receive ?
24. Four geese, three turkeys, and ten chickens cost $10 ;
a turkey cost twice as much as a goose, and a goose 3 times
as much as a chicken. What was the price of each ?
25. The head of a fish is 4 inches long; its tail is 12 times
as long as its head, and the body is onehalf the whole
length. How long is the fish ?
26. Divide 100 into two parts, such that one shall be 20
more than the other.
27. Divide a into two such parts, that the greater divided
by c shall be equal to the less divided by d.
28. How much money has A, if i, f, and i of it amount
to $1222 ?
ONE UNKNOWN QUANTITY. 107
29. What number is that, i, J, J, and i of which are
equal to 60 ? *
30. A man bought beef at 25 cents a pound, and twice as
much mutton at 20 cents, to the amount of $39. How
many pounds of each ?
31. A says to B, " I am twice as old as you, and if I were
15 years older, I should be 3 times as old as you/' What
were their ages ?
32. The sum of the ages of A, B, and C is 1 10 years ; B is
3 years younger than A, and 5 years older than C. What
are their ages ?
33. At an election, the successful candidate had a
majority of 150 votes out of 2500. What was his number
of Totes ?
34. In a regiment containing 1200 men, there were
3 times as many cavalry as artillery less 20, and 92 more
infantry than cavalry. How many of each ?
35. Divide $2000 among A, B, and C, giving A $100
more than B, and I200 less than G. What is the share of
each?
36. A prize of $150 is to be divided between two pupils,
and one is to have f as much as the other. What are the
shares?
* When the conditions of the problem contain fractional expressions*
as ^, I, i, etc., we can avoid these fractions, and greatly abridge the
operation, by representing the quantity sought by such a number of
<K^B as can be divided by each of the denominators without a remainder.
This number is easily found by taking the least common multiple of
all the denominators. Thus, in problem 29,
Let 12a; = the number.
Then will 6a! = i half.
" « 4a? = I third.
** " 3aj = I fourth.
" " 2aj = I sixth.
Hence, 6aj+4aj+3aJ+2ar=6o
.*. a; = 4
Finally, aj x 12 or i2aj = 48, the number required
108 SIMPLE EQUATIONS.
37. Two horses cost 6616, and 5 times the cost of one was
6 times the cost of the other. What was the price of each ?
38. What were the ages of three brothers, whose united
ages were 48 years, and their birthdays 2 years apart ?
39. A messenger travelling 50 miles a day had been gone
5 days, when another was sent to overtake him, travelling
65 miles a day. How many days were required ?
^~ 40. What number is that to which if 75 be added, f of
the sum will be 250 ?
41. It is required to divide 48 into two parts, which shall
be to each other as 5 to 3.*
42. *What quantity is that, the half, third, and fourth of
which is equal to a ?
43. a! and B together bought 540 acres of land, and
divided it so that A's share was to B'sas 5 to 7. How many
acres had each ?
44. A cistern has 3 faucets; the first will empty it in
6 hours, the second in 10, and the third in 12 hours. How
long will it take to empty it, if all run together ?
45. Divide the number 39 into 4 parts, such that if the
first be increased by i, the second diminished by 2, the third
multiplied by 3, and the fourth divided by 4, the results
will be equal to each other.
46. Find a number which, if multiplied by 6, and 12 be
added to the product, the sum will be 66.
47. A man bought sheep for $94 ; having lost 7 of them,
he sold i of the remainder at cost, receiving $20. How
many did he buy ?
"48. A and B have the same income ; A saves J of his, but
B spending $50 a year more than A, at the end of 5 years is
lioo in debt What is their income ?
* When the quantities sought have a given ratio to each other, the
solution may be abridpfed by taking such a number of a;'s for the
unknown quantity, as will express the ratio of the quantities to each
other without fractions. Thus, taking 5a; for the first part, 3a: will
represent the second part ; then 5X+ 30? = 48, etc.
ONE UNKKOWN QUANTITY. 109
49. A cistern is supplied with water by one pipe and
emptied by another; the former fills it in 20 minutes, the
latter empties it in 15 minutes. When full, and both pipes
run at the same time, how long will it take to empty it ?
50. What number is that, if multiplied by ??i and n
separately, the difference of their products shall he d?
51. A hare is 50 leaps before a greyhound, and takes
4 leaps to the hound^s 3 leaps; but 2 of the greyhound's
equal 3 of the hare's leaps. How many leaps must the
hound take to catch the hare ?
52. What two numbers, whose difference is S, are to each
other as a to c ?
53. A fish was caught whose tail weighed 9 lbs.; his head
weighed as much as his tail and half his body, and his body
weighed as much as bis head and tail together. What was
the weight of the fish ? •
 54. An express messenger trayels at the rate of 13 miles
in 2 hours ; 12 hours later, another starts to overtake him,
travelling at the rate of 26 miles in 3 hours. How long and
how far must the second travel before he overtakes the first ?
55. A father's age is twice that of his son ; but 10 years
ago it was 3 times as great. What is the age of each ?
56. What number is that of which the fourth exceeds the
seventh part by 30 ?
57. Divide $576 among 3 persons, so that the first may
have three times as much as the second, and the third one
third as much as the first and second together.
58. In the composition of a quantity of gunpowder, the
nitre was 10 lbs. more than f of the whole, the sulphur
4J lbs. less than J of the whole, the charcoal 2 lbs. less than
f of the nitre. What was the amount of gunpowder ? *
* The operation will be shortened by the following artifice :
Let 4205+48 = the number of pounds of powder.
Then 28a? +42 = nitre ; 7a;+3j = sulphur ; 405 + 4 = charcoal.
Hence, 39a; +49 J = 42a? + 48.
/. a? = J, and 420? +48 = 69 pounds, Ana.
110 SIMPLE EQUATIONS.
59. Divide ^6 into 3 parts, such that J of the first, J of
the second, and J of the third are all equal to each other.
60. Diyide a line 2 1 inches long into two parts, such that
one may be J of the other.
61. A milliner paid $5 a month for rent, and at the end
of each month added to that part of her money which was
not thus spent a sum equal to i half of this part; at the
end of the second month her original money was doubled.
How much had she at first ?
62. A man was hired for 60 days, on condition that for
every day he worked he should receive 75 cents, and for
every day he was absent he should forfeit 25 cents; at the
end of the time he received $12. How many days did he
work ?
63. Divide $4200 between two persons, so that for every
$3 one received, the other shall receive $5.
64. A father told his son that for every day he was perfect
in school he would give him 15 cents; but for every day he
failed he should charge him 10 cents. At the end of the
term of 1 2 weeks, 60 school days, the boy received $6. How
many days did he fail ?
65. A young man spends J of his annual income for
board, ^ for clotblng^i^ in charity, and saves $318. What
is his income ?
66. A certain sum is divided so that A has $30 less than J,
B $10 less than ^, and 18 more than J of it. What does
each receive, and what is the sum divided ?
67. The ages of two brothers are as 2 to 3; four years
hence they will be as 5 to 7. What are their ages ?*
»
Note. — To change a proportion into an equation, it is necessary to
assume the truth of the following well established prindple :
If four quantities are proportional, the product of the eostremes is
* A strict conformity to system would require that this and similar
problems should be placed after the subject of proportion ; but it is
convenient for the learner to be able to convert a proportion iAto an
equation at this stage of his progress.
ONE UNKNOWN QUANTITY. Ill
eq^ial to the product of the means. Hence, in such cases, we have only
to make the product of the extremes one side of the equation, and the
product of the means the other.
Thus, let 2X and yc be equal to their respective ages.
Then 2a;+4 : 325+4 :: 5 : 7.
Making the product of the extremes equal to the product of the
meana we have,
i4aj+28 = i5a?+2o.
Transposing, uniting terms, etc., a; = 8.
/. 2X = 16, the younger ; and 3« = 24, the older.
«
6S, What two numbers are as 3 to 4, to each of which if
4 be added, the sums will be as 5 to 6 ?
69. The sum of two numbers is 5760, and their difference
is equal to J of the greater. What are the numbers ?
70. It takes a college crew which in still water can pull at
the rate of 9 miles an hour, twice as long to come up the
river as to go down. At what rate doe^the river flow ?
71. Onetenth of a rod is colored red, ^ orange, ^ yellow,
^ green, ^ blue, ^^ indigo, and the remainder, 302 inches,
violet. What is its length ?
72. Of a certain dynasty, J of the kings wete of the same
name, J of another,  of another, f^ of another, and there
were 5 kings besides. How many were there of each name ?
73. The difference of the squares of two consecutive
numbers is 15. What are the numbers?
• 74. A deer is 80 of her own leaps before a greyhound;
she takes 3 leaps for every 2 that he takes, but he covers as
much ground in one leap as she does in two. How many
leaps will the deer have taken before she is caught ?
75. Two steamers sailing from New York to Liverpool, a
distance of 3000 miles, start from the former at the same
time, one making a round trip in 20 days, the other in
25 days. How long before they will meet in New York,
and how far will each have sailed ?
(See Appendix, p. 286.)
OHAPTEE X.
SIMULTANEOUS EQUATIONS.
TWO UNKNOWN QUANTITIES.
218. Simultaneous * Equations consist of two oi
more equations, each containing two or more unknovm
quantities. They are so called because they are satisfied by
the same values.
Thus, x+y = 7 and 5a?— 4^ = 8 are simultaneous equations, for in
each JB = 4 and ^ = 3.
219. Independent JEquations are those which
express different conditions, so that one cannot be reduced
to the same form as the other.
Thus, 6a;— 4y=i4 and 2a;+3y = 22 are independent equations.
But the equations x+y = s and 3ic + 3y = 15 are not independeiU, for
one is directly obtained from the other. Such equations are termed
dependent.
Note. — Simultaneous equations are ususDj independent ; but inde^
pendent equations may not be simultaneous ; for the letters employed
may have the same or different values in the respective equations.
Thus, the equations x\y = 'j and 2a;— 2^=14 are independent,
but not simultaneous ; for in one x = 7—y, in the other x= 7+y, etc.
220. Problems containing more than one unknown quan
tity must have as many simultaneous equations as there are
unknown quantities.
K there are more equations than unknown quantities,
some of them will be superfluous or contradictory,
3x8. What are BimultaneouB equations ? 2x9. Independent equatlonB f sao. How
many equations must each problem have ?
* From the Latin simul, at the same time.
^■'
TWO UNKXOWN QUANTITIES. 113
If the number of equations be less than the number of
unknown quantities, the problem will not admit of a definite
answer, and is said to be indeterminate,
221. Elimination* is combining two equations
which contain two unknown quantities into a single equa
tion, having but one unknown quantity. There are three
methods of elimination, viz.: by Comparison, by Substitution,
and by Addition or Subtraction.
aj+y= i6
(I)
aJy= 4
(2)
X = i6— y
(3)
x= 4+y
(4)
4+y=i6— y
(5)
2y= 12
(6)
.*. y= 6
a?= lo
CASE I.
222. To Eliminate an Unknown Quantity by CkMnpaHsan,
I. Given x \ y = 16, and x — y = 4, to find x and y.
SolWion. — By the problem,
it t*
Transposing the y in (i),
** the y in (2),
By Axiom i,
Tzansposing and uniting terms^
Substituting the value of ^ in (4),
t^* In (5) it will be seen we have a new equation which contains
only one unknown quantity. This equation is reduced in the usual
way. Hence, the
EuLE. — I. From each equation find the value of the quan
tity to be eliminated in terms of the other quantities.
II. Form a new equation from these equal values, and
reduce it by the preceding rules.
Note. — This rule depends upon the axiom, that things which are
equal to the same thing are equal to each other. (Ax. i.)
g^ For convenience of reference, the equations are numbered (i),
(2), (3), (4), etc.
■ ■ ■ ■ .M. 1 I ■ ■■■ ■■ ■ I I ■ ^ — ■.— ■ fc ■■■■ —■ — — ^^.i, ■■■■ ^ —
221. What is elimination? Name the method?. 222. How eliminate jm unknown
quantity by comparison? j^ote. Upon what principle does this rule depend?
* From the Latin eliminare, to ccust out
L14 SIMPLE EQUATIONS.
2
3
. Given a; + y = 12, and x—y+4 = 8, to find x and y,
3. Given ^x+2y =z 48, and zx — $y = 6, to find x and y.
4. Given x+y = 20, and 2a;+3y = 42, to find x and y.
5. Given 4X+^y = 13, and 32;+ 2y = 9, to find x and y.
6". Given 30;+ 2^=1 18, and a?+ 5^=191, to find a: and y.
7. Given 4x+^y = 22, and 7a;+3y=27, to find x and 2^.
CASE II.
223. To Eliminate an Uni(nown Quantity by StibHtituHon.
8. Given x{2y = 10, and 32:+ 2y = 18, to find x and y.
Solution. — By the problem, a;+2y = to (i)
" " 3a;+2y=i8 (2)
Transposing 2p in (i), x = 10— 2y (3)
Sabstituting the value of a; in (2), 30— 4y = 18 (4)
Transposing and uniting terms (Art 2ii), 4^ = 12 (5)
 y= 3
Substituting the value of ^ in (i), a; = 4
g^ For convenience, we first find the value of the letter which ia
least involved. Hence, the
EuLE. — I. From one of the equations find the value of the
unknown quantity to be eliminatedy in terms of the other
quantities,
II. Substitute this value for the same quantity in the
other equation^ and reduce it as before.
Notes. — i. This method of elimination depends on Ax. i.
2. The given equations should be cleared of fractions before com
mencing the elimination.
9. Given a;+3y = 19, and 5x—2y = 10, to find x and y,
X U X u
0. Given  + ^ = 7, and  +  = 8, to find x and y.
23 32
1. Given 2x+^y = 28, and 3.T+ 2y = 27, to find x and y,
2. Given 42;+^ = 43, and 50;+ 2y = 56, to find x and y.
3. Given 5a;+8 = 7^, and 5^ + 32 = 7X, to find x and y.
4. Given 4X+^y = 22, and 7x+;^y = 27, to find x and y,
333. How eliminate an nnknown quantity by substitution? Nate. Upon what
principle does this method depend ?
TWO UNKNOWN QUANTITIES. 115
CASE III.
224. To Eliminate an Unlcnown Quantity by Addition or
Svibtraction.
15. Given 4^+3^=18, and sar— 2y=ii, to find x and y.
Solution. — By the problem, 4aj+3y = i8 (i]
" " Sx—2y = 11 (2,
Multiplying (i) by 2, the coef. of ^ in (2), 8aj+6y = 36 (3)
Multiplying (2) by 3, the coef. of y in (i), 15a?— 6y = 33 (4)
Adding (3) and (4) cancels 6^, 23^; =69 (5)
.\ X =3
Substituting the value of 2; in (i), 12 + 3^ = 18
y= 2
l^iT In the preceding solution, y is eliminated by addition.
x6. Given 6a; +5^=2 8, and 8a; + 3^=30, to find x and y.
Solution. — Bj the problem, 62;+ sy = 28 (i)
" 8a?+ 3y= 30 (2)
Multiplying (i) by 8. the coef. of a? in (2), 48aj+4oy = 224 (3)
Multiplying (2) by 6, the coef. of 2; in (i), 482?+ i8y = 180 (4)
Subtracting (4) from (3), 22y = 44
.*. y= 2
Substituting the value of y in (2) 80;+ 6 = 30
.'. x= 3
@* in this solution, x is eliminated by svbtracHon, Hence, the
EuLE. — I. Select the letter to he eliminated; then multiply
or divide one or both equations by such a number as will
make the coefficients of this letter the same in both. (Ax. 4, 5.)
II. If the sig7is of these coefficients are alike, subtract one
equation from the other ; if unlike, add the two equations
together. (Ax. 2, 3.)
Notes. — i. The df^ect of nvuUiplying or dividing the equations is to
equalize the coefficients of the letter to be eliminated.
2. If the coefficients of the letter to be eliminated are prime num
bers, or prime to each other, multiply each equation by the coefficient
of this letter in the other equation, as in Ex. 15. *
224. What is the mle for elimination by addition or subtraction ? Notes.— 1. The
object of multiplying or dividing the equation ? 3. Xf coe£cientB are prime ?
116 SIMPLE EQUATIONS.
3. If not prime, divide the L c. fn. of the coefficients of the lett«i
to be eliminated by each of these coefficients, and the respective
quotients will be the multipliers of the corresponding equations.
Thus, the I. c. in, of 6 iind 8, the coefficients of x in Ex. 16, is 24;
hence, the multipliers wcild be 3 and 4.
4. If the coefficients of the letter to be eliminated have common
factors, the operation is shortened by cancelling these factors before
the multiplication is performed. Thus, by cancelling the common
factor 2 from 6 and 8, the coefficients of x in the last example, they
become 3 and 4, and the labor of finding the /• c. ni. is avoided.
17. Given 3a?4 4^=29, and 7a;+iiy=76, to findarandy.
i8f Given 9a;— 4^=8, and 13^;+ 7^=101, to find a? and y.
19. Given 3a:— 7^=7, and i2a;f 5^=94, to find x and y.
20. Given 30;+ 2^=1x8, and a:+5y=i9i, tofind xandy.
21. Given 43^+5^=22, and 7a;+ 3^=27, to find x and y.
Hem. — The pn^ceding methods of elimination are applicable to all
simultaneous simple equations containing two unknown quantities,
and either may be employed at the pleasure of the learner.
The first method has the merit of clearness^ but often gives rise to
fractions.
The second is convenient when the coeflSdent of one of the unknown
quantities is i ; if more than i, it is liable to produce fractions.
The third never gives rise to fractions, and, in general, is the most
simple and expeditious.
•
EXAMPLES.
Find the values of x and y in the following equations:
1. 2X+ sy= 23, 5. 5x\r jy = 43>
^x — 2y = 10. iia: + 9^ = 69.
2. 4^4 «/= 34, 6. 8a; — 2iy= 33,
4y+ ^= 16. 6a: + 35^ = 177.
3. 30a; + 4oy = 270, 7. 2iy + 20a; = 165,
50a: + 3oy = 340. 77y — 30a: = 295.
4. 2X+ 7y= 34, 8. iia;— ioy= 14,
SX+ 9y=: 51. 5X+ jy= 41.
Notes.— 2 If not prime, how proceed? 4. If the coefflcicnte have commou Uxo
tore, how eborten tbe operation ?
TWO UNKNOWN QUANTITIES. 117
15. Sz+ y= 42,
2X + 4y = 18.
10. AX + 3y = 22, 16. 2x + 4y = 20,
4X+ sy= 28.
". 3^— sy= 135 17. 4^+ 3y= 50,
3«— 3y= 6
12. 5a;— 7^= 33, 18. 32;+ 5y= 57,
5^+ sy— 47.
13.  + ^ = i8> 19 f + f = 7,
34 ^
14. 16a; + i7y = 500, 20. 2a;+ y = 50,
6' ^ 7
6y —
2X=:
208,
loa; —
4y =
156.
4a? +
3y =
22,
52^
ly —
6.
3^ —
5y =
13.
2X +
72/ =
81.
5^
7y =
33'
iia; +
X2y —
100.
I
y ^
6 ""
18,
X
2
y
4
21.
17a?— 3y= "o ^ . y
PROBLEMS.
1. Eequired two numbers whose sum is 70, and whose
difference is x6.
2. A boy buys 8 lemons and 4 oranges for 56 cents; and
afterwards 3 lemons and 8 oranges for 60 cents. What did
he pay for each ?
3. At a certain election, 375 persons voted for two candi
dates, and the candidate chosen had a majority of 91. How
many voted for each ?
4. Divide the number 75 into two such parts that three
times the greater may exceed seven times the less by 15.
5. A farmer sells nine horses and seven cows for I1200 *,
and six horses and thirteen cows for an equal amount.
What was the price of each ?
. 6. From a company of ladies and gentlemen, 15 ladies
retire; there are then left two gentlemen to each lady.
After which 45 gentlemen depart, when there are left five
ladies to each gentleman. How many were there of each at
first?
118 SIMPLE EQUATIONS.
/
. 7. Find two numbers, such that the sum of five times the
first and twice the second is 19; and the difference between
seven times the first and six times the second is 9.
8. Two opposing armies number together 21,110 men;
and twice the number of the greater army added to three
times that of the less is 52,219. How many men in each
army?
9. A certain number is expressed by two digits. The
sum of these digits is 11 5 and if 13 be added to the first
digit, the sum will be three times the second. What is the
number ?
10. A and B possess together $570. If A's share were
three times and B's five times as great as each really is, then
both would have $2350. How much has each ?
11. If I be added to the numerator of a fraction, its value
is i; and if i be added to the denominator, its value is .
What is the fraction.
12. A owes $1200; B, $2550. But neither has enough to
pay his debts. Said A to B, Lend me  of your money,
and I shall be enabled to pay my debts. B answered, I
can discharge my debts, if you lend me  of yours. What
sum has each ?
13. Find two numbers whose difference is 14, and whose
sum is 48.
14. A house and garden cost $8500, and the price of the
garden is ^ the price of the house. Find the price of each.
' 15. Divide 50 into two such parts that J of one part,
added to f of the other, shall be 40.
16. Divide $1280 between A and B, so that seven times
A's share shall equal nine times B's share.
17. The ages of two men differ by 10 years ; 15 years ago,
the elder was twice as old as the younger. Find the age of
each.
18. A man owns two horses and a saddle. If the saddle,
worth $50, be put on the first horse, the value of the two
is double that of the second horse ; but if the saddle be put
TWO UNKNOWN QUANTITIES. 119
on the second horse, the value of the two is $15 less than
that of the first horse. Eequired the value of each horse
19. A warsteamer in chase of a ship 20 miles distant,
goes 8 miles while the ship sails 7. How &r will each go
before the steamer overtakes the ship ?
20. There are two numbers, such that ^ the greater added
to J the less is 13 ; and if J the less be taken from i the
greater, the remainder is nothing. Find the numbers.
21. The mast of a ship is broken in a gale. Onethird of
the part left, added to i of the part carried away, equals
28 feet; and five times the former part diminished by
6 times the latter equals 12 feet. What was the height of
the mast? ^
22. A lady writes a poem of half as many verses less two
as she is years old ; and if to the number of her years that
of her verses be added, the sum is 43. How old is she ?
How many verses in the poem ?
23. What numbers are those whose difference is 20, and
the quotient of the greater divided by the less is 3 ?
24. A man buys oxen at $65 and colts at $25 per head,
and spends $720 ; if he had bought as many oxen as colts,
and vice versa, he would have spent $1440. How many of
each did he purchase ?
25. There is a certain number, to the sum of whose
digits if you add 7, the result will be 3 times the lefthand
digit ; and if from the number itself you subtract 18, the
digits will be inverted. Find the number.
26. A and B have jointly $9800. A invests the sixth part
of his property in business, and B the fifth part of his, and
each has then the same sum remaining. What is the entire
capital of each ?
27. A purse holds six guineas and nineteen silver dollars.
Now five guineas and four dollars fill J of it. How many
will it hold of each ?
28. The sum of two numbers is a, and the greater is n
times the less. What are the numbers ?
L'Z2
SIMPLE EQUATIONS.
227. The solution of equations containing many unknowQ
quantities may often be shortened by substituting a single
letter for several
w+x+y=is (i)
w+x+z=iiy (2)
w+y+z =:iS j^3)
x+y + z = 21 (4)
10. Required the value of w, x, y,
and z in the adjoining equations.
tt
4€
t*
s—z = 13
8y = 17
s—x — 18
«— «? = 21
Note. — Substituting « for the sum of the four quantities, we have.
Equation (i) contains all the letters but z,
(2) " . " '* y,
(3) " " " «,
(4) " " '* WJ,
Adding the last four equa \ , x v ,
tions together. \ "' 4*(e+y +*+«) \ = 69
•* V or 4«— «
That is, 3« = 69
.*. « = 23
Substituting 23 for s in each of the four equations, we have,
tr = 2, a? = 5, y = 6, e = 10.
II. Required the value of v,
Wy X, y, and z, in the adjoining «
equations.
V +w+x+y=z 10
V +W\X^Z = II
V +w^y\z = 12
V \xiy + z = 13
(^«^4^+«/+^ = 14
Note.— Adding these equations, 4tJ + 4«j + 4a; +4^+42 = 60
Dividing (6) by 4, u+ mj+ a;+ y+ » = 15
Subtracting each equation from (7), we have,
2 = 5, y = 4, a? = 3, w = 2, and © = i.
11. w
+ X + Z
—
10,
X
+ y + ^
—
12,
w
+ x\y
^=
9.
w
+ y + z
1^
II.
13
I I
 + 
X y
y ^
X Z
5
6'
12
3
4
(5)
(6)
(7)
(9)
(10)
(2)
(4)
(s)
(6)
(7)
(liM Appendix, p. M.)
THBEE OB MORE UKKNOWN QUANTITIES. 123
PROBLEMS.
1. A man has 3 sons ; thoNUim of the ages of the first
and second is 27, that of the first and third is 29, and of
the second and third is 92. What is the age of each ?
2. A butcher bought of one man 7 calves and 13 sheep
for $205 ; of a second, 14 calves and 5 lambs for $300; and
of a third, 12 sheep and 20 lambs for I140, at the same
rates. What was the price of each ?
3. The sum of the first and second of three numbers is
13, that of the first and third 16, ana that of the second
and third 19, What are the numbers ?
4. In three Battalions there are 1905 men: i the first
with I in the second, is 60 less than in the third ; i of the
third with I the firsts is 165 less than the second. How
many are in each ?
5. A grocer has three kinds of tea: 12 lbs. of the first>
13 lbs. of the second, and 14 lbs. of the third are together
worth $25 ; 10 lbs. of the first, 17 lbs. of the second, and
1 1 lbs. of the third are together worth $24 ; 6 lbs. of the
first, 12 lbs. of the second, and 6 lbs. of the third are together
worth 1 1 5. What is the value of a pound of each ?
6. Two pipes, A and B, will fill a cistern in 70 minutes,
A and C will fill it in 84 minutes, and B and C in 140 min.
How long will it take each to fill the cistern ?
7. Divide I90 into 4 such parts, that the first increased
by 2, the second diminished by 2, the third multiplied by 2,
and the fourth divided by 2, shall all be equal.
8. The sum of the distances which A, B, and C have
traveled is 62 miles; A's distance is equal to 4 times C^s,
added to twice B's ; and twice A's added to 3 times B^s, is
equal to 17 times O's. What are the respective distances ?
9. A, B, and C purchase a horse for 5ioc5. The payment
would require the whole of A's money, with half of B's ; or
the whole of B's with  of C's ; or the whole of C's with J
of A's. How much money has each ?
CHAPTEE XI.
GENERALIZATION.
228. Generalization is the process of finding a
formuUiy or general rule, by which all the problems of a
class may be solved.
229. A Problem is generalized^ when stated in
general terms which embrace all examples of its class.
230. In all General JProhlems the quantities are
expressed by letters.
I. A marketman has 75 turkeys; if his turkeys are mul
tiplied by the number of his chickens, the result is 225.
How many chickens has he ?
Note. — This problem maybe stated in the following general terms:
231. The Product of two Factors and one of the Factors
being given, to Find the other Factor.*
SuG^BSTlON. — If 2i^ product of two factors is divided by one of them,
it is evident the quotient must be the other factor. Hence, substituting
a for the product, h for the given factor, we have the following
Genebal Solution. — ^Let x = the required factor.
By the conditions, a* x ft, or fer = a, the product. Hence, the
FOBMULA. X = i
h
Translatii^ this Formula into common language, we have the
following
Rule. — Divide the product ly the given factor ; the quo
tient is the factor required.
aaS. What is ^neralization ? 939. When is a problem generalized? 930. Ho^
are qaantitiee expressed in general problems ?
* New Practical Arithmetic, Article 93.
GENERALIZATION. 125
Generalize the next two problems :
2. A rectangular field contains 480 square rods, and the
length of one side is 16 rods. What is the length of the
other side ?
3. Divide 576 into two such factors that one shall be 48.
4. The product of A, B, and C's ages is 61,320 years; A
is 30 years, B 40. What is the age of ?
Note. — ^The items here given may be generalized as follows :
232. The Product of three Factors and two of them being
given, to Find the other Factor.
SuGGBSTiON.— Substituting a for the product, h for one fiictor, and
c for the other, we have the
Oeneral Solution.— Let x = the required factor.
By the conditions, aj x 6 x c, or hex = a, the product.
Removing the coeflBcient, we have the
Formula. x = i—
be
EuLE. — Divide the given product by the product of the
given factors ; the quotient is the required factor,
5. The contents of a rectangular block of marble are 504
cubic feet ; its length is 9 feet, and its breadth 8 feet. What
is its height ?
6. The product of 3 numbers is 62,730, and two of its
factors are 41 and 45. Required the other factor.
7. The amount paid for two horses was I392, and the
difference in their prices was $18. What was the price of
each ?
Note. — ^From the items given^ this problem may be generalized
Bfl follows :
831. When the product of two factors and one of the factors are given, how find
the other foctor? 332. When the product of three factors and two of them are
given, how find the other &ctor ?
126 GENERALIZATION.
233. The Sum and DifTerence of two Quantities being giVMi,
to Find the Quantities.
Suggestion. — Since the mm of two quantities equals the greater
piv8 the less ; and the less plus the difference, equals the greater ;
it follows that the »um plus the difference equals twice the greater.
Substituting h for the sum, d for the difference^ y for the greater,
%nd I for the less, we have the following
General Solution. — Let g = the greater number,
and 1= ** less "
Adding, g^l^s, the sum.
Subtracting, g—l = d, the difference.
Adding sum and difference^ 2g = 8+d
shd
Removing coefficient, g = , greater.
Subtracting difference from sum, 2l = s—d
8—d
Removing coefficient, I = , leas. Hence, the
Formulas.
ff =
1 =
2
8 + d
2
S — d
2
This problem may be solved by one unknown quantity.
FORMATION OF RULES.
234. Many of the more important rules of Arithmetic
are formed by translating Algebraic Formulas into common
language. Thus, from the translation of the two preceding
formulas into common language, we have, for all problems
of this class, the following general
Rule. — I. To find the greater, add half the sum to half
the difference,
11. To find the less, subtract half the difference from half
the sum.
8. Divide $1575 between A and B in such a manner that
A may have I347 more than B. What will each receive ?
?33. When the eum and diflference of two quantities are given, how find the
qoaniiiies ? 334. Give the rule derived from the laet two fonnula»«,
GENERALIZATION. 1^7
9. At an election there were 2150 votes cast for two
persons ; the majority of the successful candidate was 346.
How many votes did each receive ?
10. If B can do a piece of work in 8 days, and m 12
days, how long will it take both to do it ?
Note.— Regarding the work to be done as a unit or i, the problem
may be thus generalized :
235. The Time being given in which each of two Forces can
produce a given Result, to Find the Time required by the united
Forces to produce it.
Suggestion.— Since B can do the work in 8 days, he can do i eighth
of it in I day, and C can do i twelfth of it in i day. Substitulang a
for 8 days, and & for 12 days, we have the
General Solution.— Let « = the time required.
Dividing x by a, we have  = part done l>y B.
« X by &, we have, & ^ ** "
X
b
X X
By Axiom 9,  +  = i, the work dono.
Clearing of fractions, bx{ax = ab
Uniting the terms, (a + ft) ar = 06, B and C*s time.
Removing the coefficient, we have the
ab
Formula. x = ZTTTb'
Bulb.— ZWtnde the product of the numbers denoting the
time required by each force, iy the sum of these numbers;
the quotient is the time required by the united forces.
II. A cistern has two pipes; the first will fiU it in
9 hours, the second in 15 hours. In what time will both
fill it, running together ? ^_^__
«S The time being given in which two or more forces can produce a result, how
find the time required for the united forces to produce it ?
128 GENEBALIZATION.
12. A can plant a field in 40 hours, and B can plant it in
50 hounu How long will it take both to plant it, if they
work together ?
236. In Generalizing JProblems relating to Per
centage^ there is an advantage in representing the quantities,
whether known or unknown, by the initials of the elements
or factors which enter into the calculations.
Non;, — ^The eUmenU or factors in percentage are,
iBt. The JB€Ui€, or nnmber on which percentage is calcalated.
2cL The Rate per cent, which shows how many hundredths of
the base are taken.
3d. The Percentage, or portion of the base indicated by the rate.
4th. The Amount, or the basep^t/^ or minus the percentage. Thns,
Let 6 = the base. p =■ the percentage.
r = the rate per cent. a = the amount.
13. A man bought a lot of goods for $748, and sold them
at 9 per cent above cost. How much did he make ?
NoTB. — ^The items in this problem m&j be generalized as follows :
237. The Base and Rate being given, to Find the
Percent€i>ge.*
Suggestion. — Per cerU signifies hundredtTis; hence, any given
per cent of a quantity denotes so many hundredths of that quantity.
But finding a fractional part is the same as mvUipHying the quantity
by the given fraction. Substituting b for the cost or hose, and r for
the number denoting the rate per cent, we have the
General Solution. — ^Let p = the percentage.
Multiplying the base by the rate, hr = p. Hence, the
FoBMULA. p = br.
BuLE. — Multiply the base by the rate per cent ; the product
is the percentage. Hence,
Note. — Percentage is a product, the factors of which are the base
and rate.
236. Note, What are the elements or fetors in percentage? 937. When base
aud rate are given, how find the percentage ?
* New Practical Arithmetic, Arts. 336—340.
GBKERALIZATION. 129
14. The population of a certain city in 1870 was 45,385 ;
in 1875 it was found to have increased 20 per cent. What
was the percentage of increase ?
15. Find 37^ per cent on $2763.
16. A western farmer raised 1587 bushels of wheat, and
sold 37 per cent of it. How many bushels did he sell ?
17. A teacher's salary of $2700 a year was increased $336,
What per cent was the increase ?
Note. — The data of this problem may be generalized as foUowB :
238. The Base and Percentage being given, to Find the Rate*
Suggestion. — Percentage , we have seen, is a product, and the base
is one of its factors (Art. 237, note); therefore, we have the product
and (??i« factor given, to find the other factor. (Art. 231.) Substituting
p for the product or percentage, and b for the salary or haae, we
have the
General Solution.— Let r = the required rate.
Then (Art. 231), pf6 = r. Hence, the
Formula. r = ^.
o
BuLE. — Divide the percentage hy the base; the quotient
is the rate.
18. From a hogshead of molasses^ 25.2 gallons leaked out.
What per cent was the leakage ?
19. A steamship having 485 passengers was wrecked, and
291 of them lost. What per cent were lost?
20. A man gained I750 by a speculation, which wai^
25 per cent, of the money invested. What sum did hh
invest ?
Note. — The particular statement of this problem may be iranb
formed into the following general proposition :
■ — ■ — — x ^^ ■ — —
338. When the base and percentage are given, how find the rate f
130 GBNERALIZATIOX.
239. The Percentage and Rate being given, to Find the Base*
Sttgobstion. — We have the product and one of its factors given,
to find the ot?ier factor. (Art. 237, note,) Substituting p for the
pereefUage, and r for the rate per cent gained, we have the
General Solution. — Let b = the base.
Then (Art. 237), p+r = b. Hence, the
Formula. b = —'
r
BuLE. — Divide the percentage by the rate, and the quotient
is the hose.
21. A paid a tax of $750, which was 2 per cent of his
property. How much was he worth ?
* 22. A merchant saves 8 per cent of his net income, and
lays up I2500 a year. What is his income ?
23. At the commencement of business, B and C were
each worth $2500. The first year B added 8 per cent to
his capital, and C lost 8 per cent of his. What amount
was each then worth ?
Note. — The items of this problem may be generalized thus :
240. The Base and Rate being given, to Find the Anamnt.
SuoOBSTiON. — Since B laid up 8 per cent., he was worth his origi
nal stock plv^ 8 per cent of it. But his stock was 100 per cent, or once
itself ; and 100 per cent. + .08 = 108 per cent or 1.08 times his stock.
Again, since C lost 8 per cent, he was worth his original stock
minus 8 per cent of it. Now 100 per cent minus 8 per cent equals
100 per cent — .08 = 92 per cent, or .92 times his capital. Substituting
b for the base or capital of each, and r for the number denoting the
rate per cent of the gain or loss, we have the
General Solution.— Let a = the amount.
Then wiU b{i{r) = a, B's amount.
And b (I— r) = a, C*s amount.
Combining these two results, we have the
Formula. a = b{i ±r).
BuLE. — Multiply the base by 1 ± the rate, as the case may
require, and the result will be the amount,
939. When percental and rate are given, how find the base ? 240. How find th«
amonnt when the base and rate are given ?
M
(«
GBNBBALIZATIOIS^. 131
Note. — ^When, from the nature of the problem, the amount is to be
greater than the base, the multiplier is i pltis the rate ; when less^ the
multiplier is i ,ninus the rate.
24. A man bought a flock of sheep for $4500^ and sold it
25 per cent above the cost What amount did he get for it ?
25. A man owned 2750 acres of land, and sold 33 per
cent of it. What amount did he have left ?
241. The elements or factors which enter into computa
tions of interest are the principal, rate, time, interest, and
amount. Thus,
Let p = the principal, or money lent.
** r 2= the interest of $1 for i year, at the given rate.
t = the time in years.
i = the interest, or the percentage.
" a = the amount, or the mm of principal and interest
26. What is the interest of I465 for 2 years, at 6 per cent?
Note. — The data of this problem may be stated in the following
general proposition :
242. The Principal, the Rate, and Time being given, to Find
the Interest.
General Solution. — Since r is the interest of $1 for i year,
pxr must be the interest of p dollars for i year ; therefore, pi* x t
must be the interest of p dollars for t years. Hence, the
Formula. i = prt*
BuLE. — Multiply the principal by the interest of ti for
t%e given time, and the result is the interest
27. What is the interest of 1 1586 for i yr. and 6 ra., at
8 per cent ?
28. What is the int of I3580 for 5 years, at 7 per cent ?
29. What is the amt. of $364 for 3 years, at 5 per cent ?
Note.— This problem may be stated in the following general terms *.
Note, When the amonnt is greater or leHB than the base, what ii* the multiplier ?
34X. What are the elements or factors which enter into computations of interest?
342. When the principal, rate, and time are given, how find the interest?
132 GElfEBALIZATIOia^.
243. The Principal, Rate, and Time being given, to Find the
AniounU
General Solutiok. — Reasoning as in the preceding aitide,
the interest = prt.
But the anumnt is the sum of the principal and interest.
/. p¥prt = a. Hence, the
FoBMULA. a = 2> + prt.
BuLE. — Add the interest to the principal, and the result is
the amount.
30. Find the amouDt of I4375 for 2 years and 6 months,
at 8 per cent.
31. Find the ami of $2863.60 for 5 years, at 7 per cent.
244. The jRelation between the four elements in the
FoEMULA, a = 2> + prt,
is such, that if any three of them are given, the fourth may
be readily found. (Art 243.)
245. Tiie Amount, tiie Rate, and Time being given, to Find
the Principal.
Transposing the members and factoring, we have the
FOEMULA. p = •
32. What principal will amount to $1500 in 2 years, at
6 per cent ?
33. What sum must be invested at 7 per cent to amount
to $300 in 5 years ?
246. The Amount, the Principai, and the Rate being given,
to Find the Time.
Transposing p and dividing by pt\ (Art. 244), we have the
Formula. t = — ^^^^.
pr
34. In what time will $3500, at 6 per cent, yield I525
interest ? *
343. The amoant? 344. What in the relation between the four elements in the
preceding formnia. 345. When the amoant, rate, and time are ^iven, state the
fhrmnla. 246. When the amount, principal, and rate are griven, state the formala.
GENERALIZATION".
133
247. The Hour and Minute Hands of a Clock being together
at 12 M., to Find the Time of their Conjunction between any
two Subsequent Hours.
35. The hour and minute hands of a clock are exactly
together at 12 o'clock. It' is required to find how long
before they will be together again.
AifALYSis. — The distance around the dial of a clock is 12 hour
spaces. When the hourhand arrives at I, the minutehand has passed
12 hour spaces, and made an entire circuit. But since the hourhand
has moved over one space, the minutehand has gained only 1 1 spaces.
Now, if it takes the minutehand i hour to gain 11 spaces, to gain
I space will take ^ of an hour, and to gain 12 spaces it will take 12
times as long, and 12 times yV hr. =r ^ hr. = i^^ hour. Or,
Let X = the time of their conjunction.
Then 11 spaces : 12 spaces :: i hour : a; hours.
Multiplying extremes, etc., iia; = 12
Removing coefficient, x = i^ hr., or i hr. 5^ min.
36. When will the hour and
minute hand be in conjunction
next after 3 o'clock ?
Suggestion. — Substituting a for i^
hr., the time it takes the minutehand to
gain 12 spaces, /t for the given number
of hours past 12 o'clock, t for the time of
conjunction, we have the following
General Solution, a x h = t, the
time required. Hence, the
PoEMULA. t = ah.
EuLE. — Multiply the time required to gain 12 spaces by
the given hour past 1 2 o^cloch ; the product will he the time
of conjunction.
37. At what time after 6 o'clock will the hour and minute
hand be in conjunction ?
38. At what time between 9 and 10 o'clock ?
(See Appendix, p. 287.)
247. What is the formula for finding when the hands of a clock will be in
conjunction ? Translate this into a rule.
CHAPTER XIL
INVOLUTION *
248. Involution is finding a power of tf^nanfcity.
249. A Power is the product of two or more equal
&ctors.
Thus, 3x3 = 9; axaxa = a*; ^ and a' are ix>wei8.
250. Powers are divided into different degrees ; as firsty
second, third, fourth, etc., the name corresponding with the
number of times the quantity is taken as Si factor to produce
the power.
251. The First Power is the quantity itself.
The Second Power is the product of two equal factors,
and is called a square.
The Third Power is the product of three equal factors,
and is called a cube, etc.
Note.— The quantity caUed the Jvrst power i«f, strictly speaking, not
a power, but a root Thus, a} or a, is not the product of any ttoo equal
factors, but is a quantity or root from which its powers arise.
252. The Index or Exponent f of a power is b, figure
or Utter placed at the right, above the quantity. Its object
is to show how many times the quantity is taken as a factor
to produce the power.
Thus, a* = a, and is caUed the prst jwwer.
a' = a X a, the second power, or sgua/re,
a^ = axaxa, the third power, or ct^.
a^ = axaxaxa, the fourth power, etc.
248. What is involution? 249. A power? 250. How divided? 251. The first
power ? Second power ? Third ? 252. What is the index or exponent ? Its object f
* Involution, from the Latin intdvere, to roll up.
f Index (plural, indices), Latin indica/re, to indicate.
Exponent, from the Latin exponere, to set forth.
INVOLUTION. 136
N0TB8. — I. The index of tbe f/ni ix>wer l)eing i, (a oommoDly
omitted.
2. The expression «* is read " a fourth," ** the fourth ix>wer of a,"
or " raised to the fourth power ;" oS^ is read, *' x nth," or " the
»th ix)wer of x**
Bead the following: «», c*, a?^, y^^ z^, J"*, d*.
253. Powers are also divided into direct and reciprocal.
254. Direct Powers are those which arise from the
continued multiplication of a quantity into itself.
Thus, the continued multiplication of a into itself gives the series,
a, a', a", a*, a*, «•, etc
255. Reciprocal Powers are those which arise from
the continued division of a unit by the direct powers of that
quantity. (Art. 55.)
Thus, the continued division of a urtU bj the direct powers of a
gives the series,
I I I I I I
a' ^' "^»* ^' ¥»' Ifi' ^^'
256. Reciprocal Powers are commonly denoted by
prefixing the sign — to the exponents of direct powers of
the same degree.
Thus,  = (T^ T = «', —5 = AT*, 7 = a*, etc
a a* a* a*
257. The difference in the notation of direct and recip^
rocal powers may be seen from the following series :
(I.) a«, fl*, fl^, «», a^ I, i, ^, i, ^, ^, etc.
(2.) cfij a*, a'*, a^^ a\ cfi^ ar\ ar^, ar% a~*, a"^, etc.
Note. — The first half of each of the above expressions is a series
of direct powers ; the last half, a series of redproeal powers.
258. Negative Exponents are the same as the
exponents of direct powers, with the sign — prefixed to
them.
NoU, The index i? 253. How else are powers divided? 254. Direct poweret
255. Reciprocal? 356. How is a reciprocal power denoted?
136 IKVOLUTIOI?^.
Notes. — i. This notation of reciprocal powers is derive3 from the
continued dvoision of a series of direct powers by their root; that is, by
subtracting i from the successive exponents. (Art. 113.)
2, The use of negative exponents in expressing reciprocal powers
avoids fractions, and therefore is convenient in calculations.
3. Direct powers are often called podtive, and recivrocal powers,
negative. But the student must not confound the quantities whose
exponents have the sign + or — , with those whose coefficientshSiYe the
sign + or — . This ambiguity will be avoided, by applying the term
direct, to powers ^^Aiposiivoe exponents, and reciprocal^ to those with
negative exponents.
259. The Zero Power of a quantity is one whose
exponent is o ; as, «° ; read, " the zero power of a."
Every quantity with the index o, is equal to a unit or i.
a'* a^
^or, — = a"" = a° (Art. 113) ; but — = i ; hence, o® = i.
a" a"
SIGNS OF POWERS.
260. When a quantity is positive, all its powers are
positive.
Thus, axa = a^; axaxa = a^, etc
When a quantity is negative, its even powers are positive,
and its odd powers negative.
Thus, —a X —a = a^ ; —a x —a x —a = —a', etc.
FORMATION OF POWERS.
261. AU Powers of a quantity may be formed by
multiplying the quantity into itself. (Art. 249.)
262. To Raise a Monofnial to any Required Power.
The process of involving a quantity which consists of
several factors depends upon the following
259. What is the zero power? To what is a quantity of the zero power equal?
, a6o. Rule for the signs ?
INVOLUTION. 137
PRINCIPLES.
1°: The power of the product of two or more factors is
equal to the product of their powers.
2°. The product is the same, in whatever order the factors
are taken. (Art. 87, Prin. 3.)
1. Given ^ai^ to be raised to the third power.
BOLUTIOK.
(3a52)3 — 3^j2 X 3flJ2 X 3a*2 (Art 261),
or, sxsxsxaxaxaxV^xl^xV^ (Prin. 2),
.. (3aS2)3 _ 27a8j6^ j^^^.
Involving eacli of these factors separately, we have, (3)* = 27 ;
{of = a' ; and {¥Y = 6* ; and 27xa^x¥ = 270^, An8. Hence, the
Rule. — Raise the coefficient to the power required, and
multiply the index of each letter by the index of the power,
prefixing the proper sign to the result. (Art. 90.)
Notes. — i. A single letter is involved by giving it the index of the
required power.
2. A quantity which is already a potoer is involved by multiplying
its index by the index of the required xwwer.
3. The learner must observe the distinction between an index
and a coefficient. The latter is simply a multiplier, the farmer shows
how many times the quantity is taken as Si factor.
4. This rule is applicable both to positive and negative exponents.
2. What is the square of abc ?
3. What is the square of — abc ?
' 4. What is the cube of xyz ?
5. What is the fifth power of abc?
6. What is the fourth power of 2x^y ?
7. What is the third power of 6a^i^ ?
8. What is the fourth power of $aWc?
9. What is the sixth power of 2C^b(^?
10. What is the eighth power of abed?
11. What is the wth power of ocyz ?
a62. How raise a monomial to any power? Note, A Bingle letter ? A quantity
already a power ? Distinctiou between index and coefficient ?
138 INVOLUTION.
12. Find the fifth power of {a f hy.
13. Find the second power of (o f J)*.
14. Find the T^th power of {x — y)™.
15. Find the nth power of {x + y)\
16. Find the second power of (a^ + S*).
17. Find the third power of {aW¥).
263. To Involve a Fraction to any required Power.
18. What is the square of ^^?
QoLvno^. (^V^?^' x^^ = ^. Hence, the
BnLE. — £ai8d io^A tJis numerator and denominator to the
required power.
10. Find the cube of  — •
^ 2a
20. Find the fourth power of ^ ,
x*y^
21. Find the square of ^grn*
22. Find the mth power of —
23. Find the wth power of — ^ •
xy
264. A compound quantity consisting of two or more
terms, connected by + or — , is involved by actual multi
plication of its several parts.
24. Find the square of 3a + V, Ans. 90^+ 6aJ*+ft*.
25. What is the square of a + ft + c ?
Ana. d? f 2al + 2ac + J* + ihc f c3.
26. What is the cube otx + 2y + 2?
265. It is sometimes sufficient to express the power of a
compound quantity by exponents.
Thus, the sqaare of a + 6 = (a +6)* ; the wth power of ab+c+ ^cP =
363. How IdtqIvo » ftfictlon? 364. How involve a compound qnanti^ 1
INTOLUllOH.
FORMATION OF BINOMIAL SQUARES.
266. To Find th« Square of a Btnomial in the T«m» of
Hi Parts.
I. Giveo two numbers, 3 and 2, to find the square of
their Bom in the terms of its parts.
IujUBTBATIOit.— Let the shaded part of the diagnun repnsent the
eqnare of 3 ;— each side being divided Into
3 Incbea, its contents are eqoal to 3 k 3, or
giq. In.
To preaerve the form of the eqnare, it is
plain equal additions must be made to liBO
atljaeent udes ; for, if made on one tide, or on
oppose HideB, the Ggnre will no longer be a
ffinee S la 2 more than 3, it foUova that
two rmn of'3 aqaorea each mast be added at
the top, and 3 rows on one of the adjacent sides, to make ils length
and breadlA each equal to ;. Now 2 into 3 ploa 2 Into 3 are iz sqnareB,
or tviiee the product of the ttvo parts 3 and 3.
But the diagram wants two tjmes a small squares, to fill the comer
on the right, and 3 tames s, or4, is the square of the second part. We
have then 9 (the sqtiare of the first part), is (t^ce the product of the
two parts 3 aiid 2), and 4 (the square of the second part). Therefore,
(3+2)' = 3'+3y(3x2)+2'.
a. Required the sqoare of x+j/. Ans. ^+2xxy+i^.
Hence, nDirersally,
The square of the sum of two quantities is equal to the
square of the first, plus twice their product, plus the square
of the second.
NoTK. — The tguare of a Hnotaidl always has three terma, and con
Koqaentl; is a tnnomiiU. Heuce,
No binomitd can be a perfect tqaare. (Art. loi.)
140 BINOMIAL THEOREM.
267. All Binomials may be raised to any required
power by continued multiplication. But when the expo
nent of the power is large, the operation is greatly abridged
by means of the Biiiomial Theorem.^
268. The Binomial Theorem is a gemral formula
by which any power of a binomial may be found without
recourse to continued multiplication.
To illustrate this theorem, let us raise the binomials a+5 and a—h
to the second, third, fourth, and fifth powers, by continued multipli
cations :
{a + hf = a« 4 2fl56 4 6*.
{a + 5)8 = cfi + zaV) + 3a6» + &».
{a + hf = o* + 4a«& + 6a«6« + 4a&« + &*.
(a 4 hf = a** + 5a*& 4 loa^ft* 4 loaV^ 4 5a&* 4 6*.
{a  hf = a«  206 4 6*.
{a — hf = a»  za^h + 306' — 6«.
{a  6)* = a*  4a*6 4 6a«6«  4a5» 4 6*.
(a — 6)5 = a* — 50*6 + ioa^» — ioa«6» + 506* — 6*.
269. Analyzing these operations, the learner will discover
the following laws which govern the expansion of binomials :
1. The number of terms in any power is one more than the
index of the power.
2. The index of the first term or leading letter is the
index of the required power, which decreases regularly by x
through the other terms.
The index of the following letter begins with i in the
second term, and increases by i through the other terms.
3. The sum o/the indices is the same in each term, and is
equal to the index of the power.
r~
368. What is the Binoraial Theorem? 369. What is the law respecting the Dum
her of terms in a power ? The indices of each quantity ? The sum of the indices
iu each term ?
* This method was discovered by Sir Isaac Newton, in 1666.
BIKOMIAL THEOREM. 141
4. The coefficient of th^ first and last term of every power
is I ; of the second and next to the last, it is the index of
the power; and, universally, the coefficieDts of any two
terms equidistant from the extremes, are equcd to each other.
Again, the coefficients regularly increcLse in the first half
of the terms, and decrease at the same rate in the last half.
5. Hie signs follow the same rule as in multiplicationo
270. The preceding principles may be summed up in the
following
• GENERAL RULE.
I. Indices. — Oive the first term or leading letter the index
of the required power, and diminish it regularly by i through
the other terms.
The index of the following letter in the second term is i,
and increases regularly by 1 through the other terms.
II. Coefficients. — The coefficient of the first term is i.
To the second term give the index of the power; and,
universally, multiplying the coefficient of any term by the
index of the leading letter in that term, and dividing the
product by the index of the following letter increased by i,
the result will be the coefficient of the succeeding term.
III. Signs. — If both terms are positive, make all the terms
positive ; if the second term is negative, make all the odd
terms, counting from the left, positive, and all the even terms
negative.
THE binomial FORMULA.
n — " I
{a + by zzza"" + n x a""^ b + n x a""^ J^^ etc.
Note. — The preceding rule is based upon the supposition that the
index is a positive whole number ; but it is eqiially txue when the index
is eiihfft podtive or negative^ integral or fractumal.
The coefficients of the flrst and last terms ? The law of the signs ? 370. What
is the genenl rale t
142 BIKOHIAL TfiEOHEM.
Expand the following binomials :
I.
{a + by.
6.
(y + «)"».
2.
(a  by.
7
(a  by.
3
{c + dy.
8.
{m + w)"
4
(^ + vY
9
{X  y)«
5
(^  yy
lO.
(a + by.
271. When the terms of a binomial have coefficients or
exponents, the operation may be shortened by substituting
for them single letters of the first power. After Ijie opera
tion is completed, the value of the terms must be restored.
11. Eequired the fifth power of a? + 3^.
Solution. — Substitute a for aj*, and 6 for 3^ ; then
(flkby = a» + 5a^+ioa»&»4ioa26a45a&*+&R.
Restoring the valnes of a and b,
(a^ + 3y*)' = «'•+! 5«V + go^V + 27oa^ + 405^ V + 243y*»
12. Expand (2:^ — 3^)*.
^W5. ic8 — i23fii + s^W — io8a;2^ + 81&*.
272. ^t'^ry potoer of i is i, and when a factor it has no
effect upon the quantity with which it is connected. (Art. 94,
note,) Hence, when one of the terms of a binomial is i, it
is commonly omitted in the required power, except in the
first and last terms.
Note. — In finding the exponents of such binomials, it is only
necessary to observe that the sum of the two exponents in each term
is equal to the index of the power.
13. Expand (x + i)\ 15. Expand (i — a)\
14. Expand {t — i)*. 16. Expand (i + d)\
(See Appendix, p. 387.).
371. When the terms havo coefflcieuts or exponents, how prooeedV
POWERS OF POLYKOMIAliS. 143
273. A FolynomicU may be raised to any power by actual
multiplication, taking the giyen quantity as a factor as many
times as indicated by the exponent of the required power.
But the operation may often be shortened by reducing the
several terms to twoi by substitution^ and then applying the
Binomial Formula.
1*7. Required the cube ot x + y + z.
Solution. — Substituting a for (y+«), we have aJH(y+2) = aj+a.
By fonnula, (a? + a)* = aj" + ^a + gaw* + a\
Restoring the value of a,
(a:+y+2)« = 35^+30^ (y+2)+3iB(y+2)' + (y+«)'.
274. To Square a Polynomial without Reoourae to
Multiplication.
18. Sequired the square of a + b + c.
SOLXTTIOK. — By actual multiplication, we have,
{akh+ef = a* + 2aft+2flKJ+6* + 26c+fl^.
Or, changing the order of terms,
a^+¥{c* + 2ab+2ac+2be.
Or^ factoring, we have, a* + 2a (ft + c) + 6* + 26c + c*.
19. Required the square of a + b + c + d.
Solution. — By actual multiplication, we have,
a^+l^ + c'^ + (P + 2db+2ac+2ad+2bc + 2bd+2rd.
Or, changing the order of the terms, and factoring, we have.
a^ + 2a(p+c\d)\l^ + 2bCc+d) + e^ + 2cd\cP, Hence, the
BuLE. — To the sum of the squares of the terms add twice
the product of each pair of terms.
Or, To the square of each term add ttoice its product into
the sum of all the terms which follow it.
20. Eequired the square oix + y + z.
21. Eequired the square oi a — b + c.
23. Required the square oi a + x + y + z.
873. now tuny 1 polynomial Im nieed to any required power T 374. What it the
rale for sqnai iu^ u polynomial ?
144 ADDITION OF POWERS.
275. When one of the terms of a binomial is dLfraction^
it may be involved by actual multiplication, or by reducing
the mixed quantity to an improper fraction, and then
involving the fraction. (Art. 17 1)
23. Eequired the square of a: + ^ ; and a; — J.
x + \ aji
g + i g — i
7?+ ^x a?— ix
+ ia? + i fg + i
a^+ X + { a^— aj + J
Or, reduce the mixed quantities to improper fractions. Thus,
I 2X41 , I 2a;— I ,. . .
X +  = ; and X = . (Art. 171.)
22 22
(2a;+iy_ 4a^ + 4X+i / 2g— i y_ 4^^— 40?+ 1
"^J" 4 ' "" \.2J" 4 * •
Expand the following mixed quantities :
24. (a + J)2. 26. (— f + 2abc)\
276. Powers are added and subtracted like other
quantities. (Arts. 67, 77.) For, the same powers of the
same letters are like quantities ; while powers of different
letters and different powers of the same letter are unlike
quantities, and are treated accordingly. (Arts. 43, 44,)
28. To ^a^ + S{a + lf — 6x + z^ + a^
Add — sa^ + 4 {a + b)^ + 4a; + 42?^ — a*
Arts. 4^2 + 9 (fl + i)* — 2a: + 30:8 + 42)2 + a*^ — ^
29. From 3fl8 f 5^ — 4/^ + 4a;2 — fl^
Take  4g» + 3^^^ + 3c» — 5a:" + «^
^W5. 7a^ f 2J2 — 7^8 ^ j/i;8 ^ ^a;? _ ^5 _ ^4
^^M^— ■■ ■ ■■ ■ ■■■.M ■ ■^a,l.M ■ ■■ M ■  ■■! Ill I ■ ■ I I. ^— ■■.■■■I. ■■■^■M—IM ■ ^» ^^^M^^^^^^^^M^^i^fcJi
375. How involve a biiiomial, when one term is a fraction ? 376. How are
added and subtracted ? Why ?
DIVISION OP POWEES. 145
MUL7 [PLICATION OF POWERS.
277. To Multiply Powers of the Same Moot.
1. What is the product of 30* J3 multiplied by a^J^?
SOLunoK. — ^Adding the exponents of each letter, we hare 3a*
and 6*. Now $afix¥ = yiVj^, Ans. (Art. 94.)
2. Multiply 3a«J* by w^b'^.
Solution. — Adding the exponents of each letter, as before, we
have 3a*6', Ans. Hence, the
EuLE. — Add the exponents of the given quaniitieSy and the
result mil be the product, (Art. 94.)
Notes. — i. This role is applicable to positive and negative exponents.
2. Powers of different roots are multiplied bj writing them one
after another.
Multiply the following powers:
3. cfi by a^. 7. ar^b by a~*J*.
4. ar^ by ar^. 8. ar^cd by aVeP.
5. J^ by S*. 9. ¥c^y^ by J^cy.
6. cC^ by a\ 10. cfiy^ffi by ar^y^A
DIVISION OF POWERS
278. To IH'Vide Powers of the Same Root.
II. Divide c^ by cfi.
SOLXTTiOir. — Subtracting one exponent from the other, we have
a^iofi = a». the quotient sought. (Art. 113.) Hence, the
EuLE. — Subtract the exponent of the divisor from that of
the dividend; the result is the quotient (Art. 113.)
Note. — This rule is applicable to positive and negative exx>onents.
■ I   ■ T I ^   . ■■ r fc I . ■ ■ ■■■>^^_ , ■ ■■,._,■■
^T. How multiply powers oT th6 Baxne root r Mte. Of different roots f 978. What
\9 tne role for diiidtng powers of the same root ?
7
146 TBAKSFEBBING fACIOKS.
Diyide the foUowing powers :
12. d^ by AT*. i6. 7^^ by ar^y^''^.
13. ar* by a^. 17. i2cfib~^c by ^(fb''^c^,
14. J* by 5«. 18. 6a;<yV by 2ar^yz\
15. (T* by ar^. 19. 6od^¥(^ by 5a~*J*<?~A
279. The Method of denoting Reciprocal Powers shows
that any factor may be transferred from the numerator of a
fraction to the denominator, and vfc^ versa, by changing the
Wjj^w of its exponent from + to — , or — to +. (Art. 256.)
20. Transfer the denominator of —5 to the numerator.
Solution. ^ = «' ^ "^ = <** ^ ^"^ = a'ar*, 4fw.
21. Transfer the denominator of —5 to the numerator.
Solution. 5 = — =a^J = axaJ• = oaJ*. iltw.
22. Transfer the numerator of — to the denominator.
y
a ^ «• I . I I I . I ^
Solution. ~= x a» = ^J = ^flr' = =j ^.i*
y y y a^ y ar^
23. Transfer the numerator of — to the denominator.
y
Solution. — = ^ x  = 5 , ^tw.
y a^ y a^y
aocT^
24. Transfer or' to the denominator of •
V
25. Transfer y^ to the numerator of ^i
^^
26. Transfer rf~« to the denominator of :3—
•i
27. Transfer af* to the numerator of — •
oar
079* What inference nuiy be drawn from Che method of denotinf; roc^pk^al
powen? HowtnmiferalJMtor?
CHAPTER XIIL
EVOLUTION *
280. Evolution is finding a root of a quantity. It is
often called the Bxtraction of roots.
281. A Soot is one of the equal factors of a quantity.
Notes. — i. Powers and roots are correlative terms. If one quantity
is a power of another, the latter is a root of the former.
Thus, a^ is the cube of a, and a is the cube root of a*.
2. The learner should observe the following distinctions :
ist. By involution a product of equal factors is found.
2d. By ewLution a quantity is resolved into equal fa/dors. It is the
reverse of involution.
3d. By division a quantity is resolved into two factors.
4th By subtraction a quantity is separated into two parts.
282. Roots, like powers, are divided into degrees ; as, the
square, or second root; the cube, or third root; the fourth
root, etc.
283. The Square Root is one of the two equal factors
of a quantity.
Thus, 5 X 5 = 25, and axa = a*; therefore 5 is the cquare root of
25, and a the square root of a\
284. The Cube Soot is one of the three equal factors
of a quantity.
Thus, 3 X 3 X 3 = 27, and axaxa = (^; therefore, 3 is the cube
root of 27, and a is the cube root of a*,
a8o. What is evolation ? aSi. A root ? Note. Of what is evolution the reverse?
283. What la the square root ? 2S4. Cnhe root?
* From the Latin evolvere, to unfold.
148
EVOLUTIOK.
k Soots are denoted in two ways :
ist. By prefixing the radical sign ^ to the quantity.*
2d. By placing a fractional exponent on the right of the
quantity.
Thus, Y^ and a* denote tlie square root of a.
^/a and a* denote the cube root of a, etc.
Notes.— I. The figure placed over tlie radical sign, is called the
Index of the Boot, because it denotes the name of the root.
Thus, ^/a, and y a, denote the square and cube root of a,
2. In expressing the square root, it is customary to use simply the
radical sign /y/~*, the 2 being understood.
Thus, the expression y^25 = 5, is read, "the square root of 25 = 5."
3, The method of expressing roots hy fractional exponents is derived
from the manner of denoting powers by integral indices.
Thus, a/^=axaxaxa; hence, if a^ is divided into four equal
factors, one of these equal factors may properly be expressed by a,
286. The numerator of a fractional exponent denotes the
poweVy and the denominator the root.
Thus, a^ denotes the cvbe root of the f/rst power of a ; and o* denotes
the fourth root of the thi/rd power of a, or the third power of the
fourth root, etc.
Bead the following expressions:
1. a^.
2. al.
4. S^.
S c*.
6. x^.
7
dl
10.
8.
m^.
II.
9.
n^.
12.
m
13. Write the third root of the fourth power of a.
14. Write the fifth power of the fourth root of x,
15. Write the eighth root of the twelfth power of y.
287. A Perfect Power is one whose exact root can
be found. This root is called a rational quantity.
385. How are roots denoted? 286. What docs the nnmerator of a fractioiutf.
exponent denote ? The denominator ? 287. What is a perfect power ?
* From the Latin radix, a root.
The sign ^^ is a corruption of the letter r, the initial of radix.
EVOLUTION. 149
288. An Imperfect Power is a quantity whose exact
root cannot be found.
289. A Surd is the root of an imperfect power. It is
often called an irrational quantity.
Thus, 5 is an imperfect power, and its square root, 2.23 + , is a surd.
Note. — All roots as well as powers of i, are i. For, a root is a
factor, which multiplied into itself produces a power ; but no number
except I multiplied into itself can produce i. (Art. 272.)
Thus, I, i', I", and /y/T, ^1, ^^i, etc., are aU equal
290. Negative Exponents are used in expressing roots as
well as powers. (Arts. 255, 257.)
Thus, r = a^; — r = a~i; j = ai.
a* a* flt
291. The value of a quantity is not altered if the index
of the power or root is exchanged for any other index of
the same value.
Thus, instead of x\ we may employ a^, etc. Hence,
292. A fractional exponent maybe expressed in decimals.
Thus, a* = a" = a®*^ . That is, the square root of a is equal to the
fifth power of the tenth root of a.
Express the following exponents in decimals :
16. Write ai in decimals. 19. Write b^ in decimals.
17. Write fli in decimals. 20. Write x^ in decimals.
18. Write a^ in decimals. 21. Write yi in decimals.
22. Express ai in decimals. Ans. ai = ^'^•ssaassi ,
23. Express x^ in decimals. Ans. x^ = rc0.6e6fl6+ ^
24. Express y^ in decimals. Ans. y^ = y^**
25. Write a^ in decimals. Ans. a^ = a^®. •
Note. — In many cases, afractianal expment can only be expressed
approximately by decimcUs.
288. An imperfect power ? 289. A surd ? 290. Are negative exponents used in
ezpresBtag roots ? 29a. How are fractional exponents sometimes expressed f
150 EVOLUTION.
293. The Signs of Roots are gc remed by the f oUowing
PRINCIPLES.
i^. An odd root of a quantity has the same sign as the
quantity.
2°. An even root of a positive quantity is either positive
or negative, and has tlie double sign, ±.
Thus, tlie square of + a is a^, and the square of —a m a^ ; thereibie
the square root of a^ may be either 4a or —a ; that is, y^= ± a.
3°. The root of the product of several factors is equal to
the product of their roots.
Notes. — i. The ambiguity of an even root is removed, when it is
known whether the power arises from a positive or a negative quantity.
2. It should also be observed that the two square roots of a pogiti^
quantity are nu7neii>cally equal, but have contrary signs.
294. An JEven Root of a negative quantity cannot be
found. It is therefore said to be impossible.
Thus, the square root of —a^ is neither +a nor —a. For, +a x +a
= 4a' ; and —a x —a = 4a'. Hence,
295. An even root of a negative quantity is called an
Imaginary Quantity.
Thus, \/^, 'v/— a', \^—a^, are imaginary quantities.
296. To Find the Moot of a Monomial.
I. What is the square root of a*?
AKAiiTBiB. — Since a^ = axa, it follows that one opbbation.
of the equal factors of o' is a ; therefore, a is its >y^^ ^^^ ^
square root. (Art. 283. )
Again, since multiplying the index of a quantity by a number
raises the quantity to a corresponding power, it follows that dividing
the index by the same number resolves the quantity into a correspond
ing root. Thus, dividing the index of «* by 2, we have a' or a, which
is the square root of a*.
293. What principles poyern the eignB of roots ? When Is the doable siorll used f
niustrate this. iVio^. Wlien is the ambiguity removed ? 394. What is an even root
of a negative quantity ? Illustrate. 395. What is it called ?
BVOLUTIOir. 151
2. What is the square root of ga^I^?
Ai7ALY8i8.~Slnce 9 = 3 X 3, the index of opbbatiok.
a* = 2 X 2, and the index of ft* = i x 2, it 's/qaSl^ = 2fi^}>
follows that the square root of 9 is 3, that of
«* is a*, and that of ¥ is h^ or &. Therefore, yf^/a^ = yj^h. Hence, the
Rule. — Divide the index of each letter by the index of
the required root; to the result prefix the root of the
coefficient with the proper sign. (Art. 293.)
NOTB. — This rule is based upon Principle 3. If a quantit7 ^ <^
imperfect power, its root can only be indicated.
296, a. The root of a Fraction is found by extracting the
root of each of its terms.
Find the required roots of the following quantities!
3. wcfi. 10. y/z^a^V.
4. Va* or cu
5. V ^4^.
6. V^^.
7. V^yaic.
8. VTec^.
9. ^3flftr^.
1. ^23^y^.
2. V64aW
4. V49icV.
5. y/rjc^.
49^ ,
64y»*
6.
1/
297. Ta Extract the Square Boot of the Square of a
Binomial.
I. Bequired the square root of a^ + 2db + 8*.
AiiALYBiB. — Arrange the terms opbbatiot.
according to the powers of the letter a^ + 2aS + S^ ( a + J
a ; the square root of the first term d^
is a, which is the first term of the .~T\ T~~t m
root. Next, eubtraeting its «iuare »« + * ) ^«J + ^
from the given quantity, bring down 2CTg f y
the remainder, 2ab + &'.
896. How find the root of a monomial ? iVb^.— Upon what principle U this ml^
b&se^ ? 396 a. How find the root of a thiction ?
152 EVOLUTION.
Divide the ist term of this rem. by 2a, doable the root tha« found, the
quotient b is the other term of the root. Place b both in the root and on
the right of the divisor. Finally, multiply the divisor, thus increased,
by the second term of the root, and subtracting the product from
2ab'^b^, there is no remainder. Therefore, a+& is the root required.
The square root of a^—2ab+b^ is found in the same manner, the
terms of the root being connected by the sign — . Hence, the
Rule. — Mnd the square roots of the first and third ter
and conned them by the sign of the middle term.
2. What is the square root of a^ + 4X + 4?
3. What is the square root of a^— 2a + 1?
4. What is the square root o{ i i 2X + x^?
5. What is the square root ot a^ + ^x + ^?
6. What is the square root of a^ — a + J ?
■*i"»^^
7. What is the square root of a^ + bx ] — ?
4
298. To Extract the Square Root of a Polynomial.
8. Required the square root of 40*— i2a*+sa^+6a+ i.
OnSBATIOll.
40* — i2a« + sa^ + 6a + I ( 2a^ — 3a — i
40*
4a?
3«) ~
12C^ +
12Cfi +
90*
+ 6a + t
40^
»
^6a
0
4a*
4a^
+ 6a + i
+ 6a + i
Analysis. — ^The square root of the first term is 2a\ which is the
first term of the root. Subtract its square from the term used
and bring down the remaining terms. Divide the remainder by
double the root thus found ; the quotient —3(1 is the next Xerm of the
root, and is placed both in the root and on the right of the partial
divisor. Multiply the divisor thus increased by the term last placed
in the root, and subtract the product as before.
Next, divide the remainder by twice the part of the root already
found, and the quotient is — i, which is placed both in the root and
on the right of the divisor.
397. How extract the square root of the square of a blnoinial?
EVOLUTION. 153
Finally, multiply the diyisor, thus increased, by the term last
placed in the root, and subtracting the product, as before, there is no
remainder. Therefore, the required root is 2a'— 3a— i. Hence, the
*
Bulb. — I. Arrange the terms according to the powers of
some letter, beginning with the highest, find the square root
of the first term for the first term of the root, and subtract
its square from the given quantity,
11. Divide the first term of the remainder by double the
root already found, and place the quotient both in the root
and on the right of the divisor.
ILL Multiply the divisor thus increased by the term last
placed in the root, and subtract the product from the last
dividend. If there is a remainder, proceed with it as before,
till the root of all the quantities is found.
Peoof. — Multiply the root by itself, as in arithmetic.
Note. — This rule is essentially the same as that used for extracting
the square root of numbers.
Extract the square root of the following quantities:
9. 2^ + 2xy + y^ + 2XZ + 2yz + A
10. a^ —  /^h + 2a + 4S2 __ 4J + I.
11. a* 4 A<^^b f 4^ — 4^2 — 85 + 4.
12. I — 4J2 f 4J4 + 2a; — 4J2a; + 7^.
13. 40* — i6a^ + 240^ — i6a + 4.
14. a^ab^W.:^ Ck.[fj^l
ic — — 2 4 — • ^ ^r^ ^ ■ 'jf
^^ f ^ ^ "^ ^^
299. The fourth root of a quantity may be found by
extracting the square root twice ; that is, by extracting the
square root of the square root
Thus, ^/i6a* = 4a*, and ^^40* = 2a, Therefore, 2a is the fourth
root of 16a'*.
Proof. 2a x 2a = 4a' ; 4^^ x 4a* = i6a*.
The eighth, the sixteenth, etc., roots may be found in like
manner.
298. Of a polynomial ? 299. How find the fourth root of a qaantity ? The eighth ?
li.<^.^i p% i^'>:'?/h,^^/Mii
m *
?^9.
OHAPTEE XIV.
RADICAL QUANTITIES.
300. A Madical is the root of a quantity indicated by
the radical sign or fractional exponent.
Notes.— i. The figures or UUers placed before radicals are eaeffleienta.
2. In the following investigations, all quantities placed nnder the
radical sign, or having a fractional exponent, whether perfect or
imperfeet powers, are treated as radicals^ unless otherwise mentioned.
301. The Degree of a radical is denoted by its index,
or by the denominator of its fractional exponent. (Arts.
285, 286.)
Thus, ^^ax, a^, and (a+&)^ are radicals of the same degree.
302. Idke Rildicals are those which express the
same root of the same quantity. Hence, like radicals are
like quantities. (Art. 43.)
Thus, s^a^—b and 3^a^—b, etc., are like radicals.
REDUCTION OF RADICALS.
303. JBeductian of Radicals is changing their
form without altering their yalue.
304. The Simplest Form of radicals is that which
contains no factor whose indicated root can be extracted.
Hence, in reducing them to their simplest form, all exact
powers of the same name a^ the root must be removed from
under the radical sign^
■ '■' '■ ■ . . ..   — .■■_. — — ■ — — .
yoo. Wliat is a radical ? 301. How is the decree of a radical denoted ? 30a. What
are like radicals ? 303. Define redaction of radicals. 304. What is the Bimplest form
}t radicals r
BBDUOTIOlir OF BADIOALS. 165
CASE I.
30& To Reduce a Radical to its Simplest Form.
1. Eeduce Vi8a^ to its simplest form.
Analysis. — By inspection, we orasATioif.
perceive that the given radical is "^iSa/kc = ^ga^ x 2X
composed of two factors, 9a' and ^^ r—^ t —
2ic, the first being a perfect square ^ X V 2a;
and the second a surd. (Art. 289.) /. a/i8^ =r 3a VaS
Removing 9a' from under the rad
ical sign and extracting its square root, we havie 3a, which prefixed
to the other factor gives 3a y^, the simplest form required.
2. Eeduce 4 V«* — aH to its simplest form.
Analysis. — Factoring opbratioh.
the radical part, wehaye 4^/^4=^ = ^^a^^(ax)
the two &ctors, i/d^ and ay — ^.
^a—x, the first bems: a ^____ ___
perfect cube^and the sec /. 4W(l^ — a^ = ^Clya — X
ond a surd. Remove a'
from under the radical sign, and its root is a, which multiplied by
the coefficient 4, and prefixed to the radical part, gives i^^a—x, the
simplest form. Hence, the
BuLE. — I. Resolve ilie radical into two factors^ one of
which is the greatest power of the same name as the root,
11. Extract the root of this power, and multiplying it by
the coefficient, prefix the result to the other factor, with the
radical sign between them.
Notes.— I. This rule is based upon the principle that the root of
the product of two or more factors is equal to the product of their
roots.
2. When the radicals are smaU, the greatest «xact power they
contain may be readily found by inspection.
3. Eeduce sVsoa^ to its simplest form. Ans. isaVixi
4. Beduce 6\/s4:^ to its simplest form. Ans. iSxV^y.
305. Beclte the role. MU. UpoD what based?
V 156 BEDUOTION OF RADICALS.
306. To Find in large Radicals the Greatest Power
corresponding to the indicated Root.
>w
5. Bednce V1872 to its simplest form.
OFEBATION.
4 ) 1872 \/i872 = V4 X 4 X 9 X V13
4 ) 468 " = a/i44 X V13
9)117(13 V1872 = i2Vi3> 4ws.
Analysis. — ^Divide the radical by the smallest power of the same
degree that is a factor of it ; the quotient is 468. Divide this quan
^ ^ tity by 4 ; the second quotient is 117. The smallest power of the same
N degree that will divide 117, is 9. The quotient is 13, which is not
divisible by any power of the same degree. The product of the
divisors, 4x4x9 = 144, is the greatest square of the given radical.
Extracting the square root of I44« we have 12^^13, the simplest form
required. Hence, the
BiTLE. — Divide the radical by the smallest power of the
same degree which is a factor of the given radical
Divide this quotient as before; and thus proceed till a
quotient is obtained which is not divisible by any power of
the same degree. The continued product of the divisors vdtl
be the greatest power required.
Note. — This rule is founded on the principle that the product of
^ any two or more square numbers is a agttare, the product of any two
or more cubic numbers is a cttbe^ etc.
Thus, 2* X 3' = 36 = 6* ; and 2' x 3* = 216 = 6^.
Reduce the following radicals to their simplest form :
V^. 12. v^54a^c.
7. VSfl^. 13. 7^9^^ — 27^2^.
8. 2V9^. 14. ^64a:^y.
9
10
II
3V24. 15. '^Sio^i.
6\/252^. 17. Vi584a2.
306. How find the greatest power corresponding to the indicated root, In larse
radical y Note. On wliat principle is this rule founded?
BEDUCTIOir OF BADICALB. 157
>
CASE II.
. 307. to Reduce a Rational Quantity to the Form of a
^ Radical,
I. Eeduce 3^2 to the form of the cube root.
AiTALTSis. — The cube root of a quantity, we opsiLkTioK.
iiave seen, is one of its three equal factors. (3^^)^ = 27a*
(Art. 284,) Now 3a^ raised to the third power . ^2 — . ^27fl*
is 27a*. Therefore 3a = ^2*ja^, Hence, the
EuLE. — Raise the quantity to the power denoted hy the
given rooty and to the result prefix the corresponding radical
sign.
Note. — The coefficient of a radical, or any factor of it, may be
placed under the sign, by raising it to the corresponding poioer, and
placing it as a factor under the radical sign.
2. Reduce 20*6 to the form of the cube root.
3. Eeduce {2a + J) to the form of the square root.
4. Eeduce (a — 2b) to the form of the square root.
5. Place the coefficient of saVb under the radical sign.
6. Place the coefficient of lo^^fab under the radical sign.
7. Eeduce 20^1^^ to the form of the fourth root.
8. Eeduce \a})c to the form of the cube root.
9. Eeduce 3 (a — S) to the form of the cube root
10. Eeduce d^ to the form of the cube root.
Note. — When a power is to be raised to the form of a required
root, it is not the given letter that is to be raised, but the power of the
letter.
II, Eeduce o^& to the form of the fourth root.
12. Reduce a — 5 to the form of the square root.
13. Eeduce a/^ to the form of the wth root.
307. How reduce a rational quantity to the form of a radical ? Note. How place
a coefficient under the radical ei^fn ? Note. How raise a power to the form of a
required root? •
158 BEBUCTIOlf OF BABICALS.
CASE III.
308. To Reduce Radicals of diffSerent Degrees to othera of
equal Value, having a Cknnmon Index.
1. Seduce a^ and b^ to equiyalent radicals of a common
index.
Analysis.— The fractional opbbatioh.
indices J and J, reduced to a i = A ^^^ i = A
common denominator become • J^ _. r,^ ^nd b^ = &A
*r.l** f*«'' = («^)'*' aA=(a*)iVandSA=(j8)A
and 6^ ^^^ (js)^ (Art, 174.) "^ ^^ ' **"'"'' ^'^^
Therefore (a*)^ and (6*)^^ are the radicals required. Hence, the
BuLE. — I. Reduce the indices to a common denominator.
II. Raise each quantity to the power expressed by the
numerator of the new index, and indicate the root expressed
by the common denominator. (Art 174.)
Bednce the following radicals to a common index :
2. a^ and {bc)^. 7, V4a* and ^/Ic^.
3. 3^ and 5*. 8. o^ and ji.
4. a* and 6^. 9. ji and c».
S* V7> 'V^, and ^. 10. {a + S)* and (a — J)t.
6. 'v^2a;* and Vsa^. 11. (a? — y)t and (a?+ y)i.
CASE IV.
309. To Reduce a Quantity to any ^Required Index.,
I. Keduce a^ to the index J.
Analysis. — ^Divide the index J by^; operation.
we have } or J. Place this index over a; i"5"i = iXJ = ^
it becomes a*, and setting the required i "^ "J = i = i
index over this, the result, {a})^, is the .«, (ai)t, Ans.
answer. Hencei the
308. How reduce radicals to a common index ?
ADDITIOl^^ OF BADICALS. 159
SuLE. — Divide the index of the given qimntity by the
required index, and placing the quotient over the quantity,
set the required index over the whole.
NoTB. — This operation is the same as resolving the original index
into tiMfcLctorSt one of which is the required index. (Art. 126, TU)te.)
2. Reduce a^ and b^ to the index }.
SoLTjnoN. J* J = J X f = I, the first index.
f *t = 1 X f = I, the second index.
Therefore, (o*)* and (ft*)* are the quantities required.
3. Eeduce 3^ and 4^ to the common index \.
4. Reduce a^ and S* to the common index \.
5. Reduce a' and b^ to the common index J.
6. Reduce a^ and &» to the common index \,
7. Reduce a« and J"» to the common index J.
ADDITION OF RADICALS.
310. To Find the Sutn of two or mo^e Radicals.
1. What is the sum of ^Va and 5 V« ?
ANAiiYSis.^— Since these radicals are opbrattoh.
of the same degree and have the same xy/a f 5 Vfl^ = SVa
radical part, they are like quantities.
(Art. 43.) Therefore their coefficients may be added in the same
manner as rational quantities. (Art. 67.)
2. What is the sum of 3^8 and 4a/i8 ?
ANAiiYSis. — These radicals opbbatioh.
are of the same degree, but 3\/8 = 3V4 X Vi = 6\/2
the radical parts are unUke ; /—z /~ n /
therefore, they cannot be 4V^ = 4V^ X V^ = 12^/2
united in their present form. /. 6v2 + I2V2 = 18A/2
Reducing them to their sim
plest form, we have 3^^ = 6\/2, and 4^78 = 12^, which are
like radicals. (Arts. 302, 305.) Now t^2 and I2y2 = 18^^/2, Ane.
309. How rednce qnaiitities to any required index? Ncie. To what ie this oper*
ation similar?
160 SUBTAACTIO^i^ OF BADIGAL6.
3. What is the sum of 31/18 and 4V^«
Analysis. — Reducing the ofkbatiok.
radicals to their simplest form, 3V^ = 3V9 X \/2 = gV^
we have 3 V^ = 9^2, and ^^^^ 4^ x \/3 = S^/]
4/^24 = 8>v^3, which are un a /~ o ^/~
like quantities, and can only ^^^' 9V2 + 8V3
be added by writing them one after the other, with their proper sigiu.
(Arts. 43, 67.) Hence, the
BuLE. — ^I. Reduce the radicals to their simplest form.
11. If the radical parts are alike, add the coefficientSy and
to the sum annex the common radical
If the radicals are unlike, write 'them one after another^
with their proper signs.
Note. — To determine whether radicals are alike, it is generallj
necessaiy to reduce them to their simplest form. (Art. 305.)
Find the sura of the following radicals :
4. VY2 and 1/27. 9. 3"^^ and 4V^i28.
5. V20 and V48. 10. 7 A/243 and 5^363.
6. 2V^ and ^Vcfib' u. aVSiS and saV49b.
7. aVsa^ and cV2'jab. 12. bV^S^ and x^^Gx^
8. 3Vi8«i»^ and 2^/3^(1^ 13. 4\^^ and Sa/S^.
SUBTRACTION OF RADICALS.
311. to Find the Difference between two Radicals.
I. From 3 V45 subtract 2a/2o.
Analysis. — Reducing to the oebratio*.
simplest form, we have g^/s 31/45 = 3^/9 x Vs = 9V5
and 4a/5, which are like / — f r~ r
J.. XT r r 2V20 = 2V4 X V5 = 4V5
quantities. Now gy 5 — 4y 5
= 5^/5, the diflFerence re .'. 3V45 — 2V20 = 5 Vs
quired. HencOi the
3x0. How add radicals ? Note. How determine whether they are like quantitiiefl!
MULTIPLICATION OF RADICALS. 161
BuiiE. — Reduce the radicals to their simplest form ; change
the sign of the subtrahend, and proceed as in addition of
radicals. (Art. 310.)
(4.)
4V320
— sVs^
«
(^•)
(3.)
.From
4V112
V480
Take
V448
4^/63
5. From 3V49a^ take 2^/2^ax.
6. From $^a 4 b take ^^a + b,
7. From 3'V^ ta^e — 4V^^.
8. From ^\^2$obhi take 2V^W».
9. From  AT* take  za^. p— ^ i^r~r. ^ 4
10. From 5 VI take 2^*. A/'o''^ ^ ' ^ ' ^ ' ;?
MULTIPLICATION OF RAmCALS.
/
312. To Multiply Radical Quantities.
1. What is the product of 3\/a by 2 V*.
Analysis. — Since these radicals are opbratioh.
of the same degree, we maltiply the 3^^ X 2 VJ = 6\/aJ
radical parts together, like rational
quantities, and to the result prefix the product of the coefficient&
2. Multiply 3\/« by 2^c.
AiTALYSTS. — As these radicals are of differ opbbatioh.
ent degrees, they cannot be multiplied together ^ V^ = 3 {o^
in their present form. We therefore reduce s/— / i\
ihem to a common index, and then, multiply — ^^ — i—
ing as before, we have e^a*^. ^^ ^ {a^C^P
3. Multiply ai by a^.
Analysis. — These radicals are of different degrees, but of the same
radical part or root ; we therefore multiply them by adding their
fractional exponents. ^+^ = f. Therefore, a^xa* = a^. Hence, the
3XX. How subtract radicals V
162 MULTIPLICATION OF RADICALS.
BuLE. — L Reduce the radicals to a common index.
II. — Multiply the radical parts together as rational quan
tities, and placing the result under the common index, prefix
to it the product of the coefficients.
Notes. — i. Roots of like quantities are multiplied together by
adding their fractional exponents, (Art. 94.)
2. This rule is based upon the principle that the product of the
roots of two or more quantities is the same as the root of the product,
(Art. 293, Prin. 3.)
3. The product of radicals becomes rational^ whenever the nmner
ator of the index can be divided by its denominator without a
remainder.
4. If rational quantities are connected with radicals bj the sigiis +
or — , each term in the multiplicand must be multiplied by each term
in the multiplier. (Art. 98.)
Multiply the foUewiDg radicals:
4. s\/i8 by 3V20. 10. a» by a?».
5. aVx by bVx. 11. 7V^ by 3V^4.
6. VcT+b by Va — & 12. Vga by ViSa.
7. Vox by Vcy. 13. V18 by Va
8. a* by c*. 14. Vs^ by V^cu^
(9) ~ (15.)
Multiply a+ Vb Mult a + Va?
By c+ Vd By i + iVg
ac + cVb a + Vx
+ aVd + Vbd f abVx + hx
Ans. ac+cVb+aVd+Vbd Ans, a+Vx+abVx + dz
16. 2VI by 2\/. 18. (m + w)* by (m + »)^
17. 4\/ by 3VJ. _ ^ /paJ ^^ ^ /t^
^9 x/i^^yy^
3za. How multiply radicals? JI^Mm. How are rootB of like qoantities mvlti
plied? Upon what principle is tbls mle based? When does the prodact of
radicals become rational? If radicals are connected with rational quantities} tow
multiply them 7
DIVISION OP BADICAL8. 163
DIVISION OF RADICALS.
313. To Divide Radical Quantities.
1. Divide 4^24^ ^J 2 VSa.
Analysis. — Since the given radicals are operitiok.
of the same degree, one may be divided by ^^^T^ac / —
the other, like rational quantities, the quo r^ = 2V 3^
tient being \/3C. (Art. 11 1.) To this result
prefixing the quotient of one coefficient divided by the other, we have
2 \/3C, the quotient required.
2. Divide 4Vac by 2\^a,
Analysis.— Since these radicals are aa/oc a (aci^
of different degrees, they cannot be — ^z= = —^ — jr
divided in their present form. We ^Va 2 (a)*
therefore reduce them to a common A^^ac A (cfid')^
index, then divide one by the other, and •*• s/"" ^^ . .1
to the result prefix the quotient of the 2 V a ^ \^ )
ooeffidents. The answer is 2^a^. or 2 {cu^^, Ans.
3. Divide a^ by «i
Analysis. — These radicals are of different ofbbation.
degrees, but have the same radical part or root ^yi :::: ^1
We therefore divide them by subtracting the 1 ■
fractional exponent of the divisor from that of
the dividend. (Art. 113.) Reducing the expo fl* y a* rr a*
nents to a common denominator, a* = a*, and .^ ^J .2. /»i ^: njk
al=a*. Now a^^a^^c^y Am. Hence, the
Rule.— I. Reduce the radical parts to a common index.
IL Divide one radical part by the othery and placing the
quotient under the common index, prefix to the result the
quotient of their coefficients.
Note. — Boots of like quantities are divided by svbtracting thefrac*
tional exponent of the divisor from tliat oftji^ dividend. (Art. T13.)
3x3. How divide radicale»? Ifote, How divide roots of like quantitiee T
164 INVOLUTION OF RADICALS.
Divide the following radicals:
4. \/72a^ by A/4C. 10. i^a^/xy by^7Vy.
5
6
7
8
6V^d^ by 2Vtfo?. II. (a 4 })» by (a 4 J)«.
(a* + aa;)i by ai 12. 3\/5oS* by V^x.
I2(fl2y2)i by (ay)i. 13. ^/WHT^ by \^x+y.
2/^'sfax by sVfl. 14. 16^/32 by 2\/4.
\^acy/hx by 2c\/^. 15. 8^5x2 by 4^2.
INVOLUTION OF RADICALS.
314. To Involve a Radical to any required Power.
I. Find the square of a^.
OPERATION.
Analysis. — As a square is the product of two 1 1 1
equal factors, we multiply the given index by
the index of the required power. Hence, the .*. fl*, A.7hS,
EuLE. — Multiply the index of the root by the index of the
required power, and to the result prefix the required power
of the coefficient.
Note. — A root is raised to a power of the same name by removing
the radical sign or fractional exponent. (Ex. 2.)
2. Find the cube of v« + b. Ana. a + b.
3. Find the cube of ai
4. What is the square of ^V2X.
5. What is the cube of 2V^
6. Eequired the cube of v^.
7. Eequired the cube of 4A / — •
8. Find the fourth power of Z\ ~'
9. Whafc is the square of a + Vy ?
314. How involve radicals to any required power ? NoU, How raise a lOot to a
Vower of the same name ?
BVOLUTIOK OF EADIOALS. 165
EVOLUTION OF RADICALS.
315. To Extract the Boot of a Radical*
I. Find the cube root of a^if^.
Analysis. — Finding the root of a radi om^nxm,
cal is the same in principle as finding the WcP^Wzzz ^ cfibi
root of a rational quantity. (Art. 296.) ,
Reducing the index of the radical to an V fljSji = ab\^ Am.
equivalent fractional exponent, we extract
the cube root by dividing it by 3. The result is a&^. Henoe, the
Bulk — Divide the fractional exponent of the radical by
the number denoting the required root, and to the result
prefix the root of the coefficient.
Notes. — i. Multiplying the index of a radical by any number is the
same as dividing ihQ fractional e^eponent by that number.
Thus, /^a = a*. Multiplying the former by 2, and dividing the
latter by 2, we have /^a = a*.
2. If the coefficient is not a i)erfect power, it should be placed under
the radical sign and be reduced to its simplest form. (Art. 305.)
2. Eequired the square root of pV^
3. Required the square root of 4^^
, 4. Find the cube root of sVxy.
5. Find the cube root of 2j\/2j.
6. What is the cube rpot of a (bc)^ ? ^
7. What is the* fourth root of fv^?
8. What IS the fourth root of V^ \/?t
9. Find the seventh root of i28a/«.
no. Find the fourth root of Va{b^).
II. Find the fifth root of ^a^V^
12. Find the nth root of a\^.
315. How extract the root of a radical ? Notes. To what is mnltipiyinj^ the index
of a radical equivalent f If the coefficient is not a perfect power, what is done i
166 CHAKGIKG BABICALS TO BATI0KAL8
CHANGING A RADICAL TO A RATIONAL
QUANTITY.
CASE I.
316. To Change a Radical Monomial to a Rational Quantity.
1. Change Va to a rational quantity.
Akaltbis.— Since multiplyiDg a root of a opehation.
quantity into itself produces the quantity, it ^^^ ^ <y/^ ^. q
follows that \^a x \/a = a, which is a rational
quantity. (Art. 287.)
2. It is required to rationalize a^.
AnaIlysis. — A root is multiplied by another opsRATioir.
root of the same quantity by adding the expo ffi X ff^ — a
nents ; therefore we add to the index ^ such a
fraction as will make it equal to i. (Art. 94.)
Thus, a^ X a' = a^'*' { = a' = a, the rational quantity required.
3. It is required to rationalize xi.
Solution.— Multiplying «* by «*, the result is operation.
w, which is a rational quantity. Hence, the x^ X <^^ — T
BuLE. — Multiply the radical by the same qtuintity Jiaving
8vch a fractional exponent as, when added to the given
exponent, the sum shall be equal to a unit, or i.
4. Required a factor which will rationalize a^.
5. What factor will rationalize \^a^c?
6. What factor will rationalize '^(a + b)K
7. What factor will rationalize ^^aWct
8. What factor will rationalize ^Jx^Tyf^
9. What factor will rationalize y/{a + d)^?
10. What factor will rationalize \/(a + J + c) ?
3x6. How reduce a redioal monomial to a rational quantity T
0HAK6INQ RADICALS TO BATI0KAL8. 16?
CASE II.
317. To Change a Radical Binomial to a Rational Quantity.
1. It is required to rationalize ^/a + VJ.
Analtsib. — The product of the sum and opsBAnoir.
difference of two quantities is equal to the #y^ j. /y/J
difference of their squares (Art. 103) ; there r r=
fore, (^a^^/o) multiplied bj (/y/a— ^ft)
s= a— &, which is a rational quantity. a + ^/ab
Therefore, the factor to employ as a multi ^\/ab — > h
plieris V5V^. a^h,Am.
2. What factor wiD rationalize Vx — Vy ?
Analysis.— If the binomial v^  V^ ^ <htoatkw.
multiplied by the same terms with the sign of yX — \y
the latter changed to +, we have ,^/J , ^z:
(Art. 103.) Therefore, /y/5 + ^y is the fao — ^
tor required. Hence, the ^^^ V a? + V y
BiJLE. — Multiply the binomial radical by the correspond'
ing binomial with its connecting sign changed.
3. What fector will rationalize x + 4^/9?
4. Bationalize V9 — V6
5. What factor will rationalize VT + Va?
6. Bationalize 6 — Vs
7. What factor will rationalize Vsa — Vp.
8. Bationalize Va — \/S'
9. What factor will rationalize ^Va + Vs.
10. What factor will rationalize 4^/2^ — 5 V?
3x7. flow redooe a radical binomial to a rational qaantily t
168 RADICAL FBACTIONS.
CASE III.
318. To Change a Radical Fraction to one whose Numerator
or Denominator is a Rational Quantity.
1. Change^ to a rational denominator.
Analysis. — Multiply both terms of the opebatioh.
fraction by the denominator ^^h, and the q x ^b a^h
result is ¥» whose denominator is V^ X wb
rational. (Art. 167, Prin. 3, note,) Hence, the
Rule* — Multiply both terms of the fraction by such a
factor as will make the required term rational.
Note. — Since the product of the sum and difference of two quan
tities is equal to the difference of their squares, when the radical
fraction is of the form — :: , if we multiply the terms by
(\/5 + \/^)» ^® h&ve a — & for the denominator. (Art. 103.)
2. Rationalize the denominator of ~^« Ans. ~ — »
3. Rationalize the numerator of —=• Ans.
e
4. Rationalize the denominator of ^ 
vx
5. Rationalize the denominator of jt
wc
^ X 4 A^t/
6. Rationalize the denominator of =r p«
7. Rationalize the denominator of 7= p«
Va — wo
8. Rationalize the denominator of
9. Rationalize the denominator of
I + Vs
s — Vs
3x8. How reduce a radical fraction to one whose numerator or denominator Is t
rational quantity ? When the fractions contain compound quantities, what piia
clple entere into their reduction f
BAPIOAL EQUAIIONB. 169
RADICAL EQUATIONS.
319. A Sadical Hquation is one in which the
anknown quantity is under the radical sign.
320. To Solve a Radical Equation.
1. Given Vx + 2 = 7, to find x.
Analysis. — Transposing 2, we have, \/5 + 2 = 7
y^aj = 5. Since 5 is equal to the ^x, it /—
follows that the square of 5, or 25, must be ^ ' 5
the square of y^. Therefore, a? = 25. * — S •" ^5
2. Given 2a + Vx = ga, to find x.
Solution. — ^By the problem, 2a + ^x = 9a
By transposing, ^i = ja
By involution, x = 49a'
3. Given 5 va? + i = 35> to find !».
Solution.— By the problem, 5/y^« + i = 35
Removing coefficient, ^x + 1 = 7
Involving, « + i = 343
Transposing, « = 342. Hence, the •
EtTi^B. — Involve both sides to a power of the same name as
the^ root denoted by the radical sign.
Note. — ^Before involving the quantities, it is generally best to dear
of fractions, and transpose the terms, so that the quantities under t]*d
radical sign shall stand alone on one side of the equation.
Beduce the following radical equations :
4. a + ^fx + c=^d. 8. ^2X + 3—6 = 13.
5. \^x + 2=:$. 9. ^x — 4 = 3.
6. 3y«?^4 + S = 7i 10. 2'^a? — 5 = 4.
3x9. What is a radleal equation ? 39a flow solved ? Noie. Wliat ihoold be dom
before inyolving the quantities f
8
170 &ADICAL EQUATI0K8.
12. Oiyen — ■ , ■ — = 8*, to find x.
13. Eeduce a/o^ + Vx = —y^±±—'
Analtbir— By removing the opbratkw.
denominator the first member is / « '7= 3 + C
squared. But x is still under y "TV •/ « • ^~\
the radical sign. This is re ^
moved by involving both mem ^ + V^? = 3 + ^
bers again. 4fW. a? = (3 + c — a^"
14. Given ^""_^ = — ^, to find y.
vy y
15. Given x + V«^ + ^ = — r *o tod a;.
Note. — ^If the equation has two radical expressions, connected with
other terms by the signs + or — , it is advisable to transpose the terms
so that one of the radicals shall stand alone on one side of the equation
By involving both members, one of the radicals becomes rational ; and
by repeating the operatiouj the other will also disappear.
16. Given V4 + 5'^ — Vyo = 2, to find x.
Transposing, ^4 + 52? = \/^ + 2
Involving, 4 + 5^5 = 4 + 4V'3« »• 39
Tiansposing and dividing by 4, ^\/Jx = 
Involving, 38^ = r
4
Transposing and multiplying by 4» s^ = 129
Hence. « = la. Ana.
17. Given Vx + 12 = 2 + Vx, to find x.
18. Given V5 x Vx + 2 = 2 + VJx, to find x.
^. Vx X  ax ^ ,
19. Given — = — 7=— , to nnd x.
X ya*
(9o« App«ndlz, p. a99.>
^/]/M (Z^^l^i/v^'^^WA) <Uy^tM/^^U4i 3ydL. /^
CHAPTER XV.
QUADRATIC EQUATIONS.
321. Equations arejdivided into different degrees^ as the
first, second^ third, etc., according to the powers of the
unknown quantity contained in them.
An equation of the First Degree is called a Simple
JEguation, and contains only the^r^^ power of the unknown
quantity.
' An equation of the Second Degree is called a Quad
' ratio Equation^ and the highest power of the unknown
^ quantity it contains is a square.
An equation of the T/iird Degree is called a Ctibic
JSquatioUy and the highest power of the unknown quantity
it contains is a cube.
An equation of the Fourth Degree is called a
BiquadratiCy etc.
322. Quadratic Fqtuitians are divided into pure
and affected.
323. A Pure Quadratic contains the square only of
the unknown quantity ; as, a? = b.
324. An Affected Quadratic contains both ^q first
aiidsee?ow(f powers of the unknown quantity; as, a^\ax^=cd.
Notes. — i. Pure qnadratics are sometimes called incomplete equa
^ tlons ; and affected quadratics, complete equations.
331. How are eqnationB dirided ? What Ib an equation of the flret degree ? The
aecsondf Thitd? Fourth? 323. How arc qnadratic equations divided ? 335. What
iB u pnre quadratic r 334. An affected quadratic ? Note. What are they Bometimea
called?
172 PTTBE QUADEATIC8.
2. A Chmpkte Equation contains every integral power of the tin
known quantity from that which denotes its degree down to the zero
power. •
An IheompleU EqtuxUon is one which lacks one or more of these
power&
PURE QUADRATICS.
325. Every pure qtcadraiic may be reduced to the form
.For, by transposition, etc., all the terms containing a^ can be
reduced to one term, as ba^ ; and all the known quantities to one
term, as c. Then will
Dividing both members by h, and substituting a for the quotient of
etb, the result is the form,
a^ = a.
326. Pure quadratic equations have two roots, which are
the same numerically, but have opposite signs. (Art. 293, n,)
Thus, the square of + a and of —a is equally a^, Hencet
^/^= ±a,
327. To Solve a Pure Quadratic Equation.
I. Find the value of a? in ^^ 6 = \ 2.
9 3
Solution.— Given 6 = — + 2
9 3
Clearing of fractions, saj^ — 54 = 3a^ + 18
Transposing, etc., 2^ = 72
Removing coefficient, ic^ = 36
Extracting sq. root, (r = ± 6, Ana,
. Substituting h for 2, and c for 72, in the third equation^ we have
the form, bofi = c.
Removing coefficient, etc., a? = a. Hence, the
KuLE. — Reduce the given equation to the form a? =z a, and
extract the square root of both members. (Art. 296.)
^6. How many roots has a pare quadratic?
PUBB QUADRATICS. 173
Find the value of 2; in the following equations :
2. 3^ — 5 = 70 lo 20?+ 12 = ^a^ — ^j.
3. 92^ + 8 = 30;? + 62. II, 7iB8 — 7 =32;? + 9.
4. 52:^ + 9 = 23^ + 57 12. ah!? = a^.
5. 63? + 5i=z4a? + S5' i^*^ (a? + 2)^ = 4a: + 5.
6. " + 35 = 325^ + 7 14. aj^— 1= :
4 4
23^ + 8 _ 3!^ — 6 flg(2rg + 9) _ $x + 6
10 10 •* •* 30 10
8. ? = ?! 16. S_ + ^ = l
42 a? 4 — a? 4 + x 3
9.  +  =  + ^« 17. — > ^ = I — a;.
^22:32; ' i+a
328. Badical equations, when cleared of radicals, often
become pure quadratics.
_ •
18. Given Vo? + 11 = V^a? — 5, to find a:.
Solution. — Clearing of radicals, a^ + 11 = 2a^ — 5
Transposing and extracting ioot« rs ± 4
4a;
19. Given 2Va? — 5 = — , to find a?.
20. Given 2Va^ — 4 = 4Va' — i> to find a?.
— — — /7
21. Given Vx + c = , to find a?,
V a; — {?
22. Given \/ — '^^— = Vi, to find x.
23. Given  v'i+^, to find a;.
. V (x — a)
24. Given = Vx — 10, to find x.
Vx + 10
337. What is the rule for tbe solation of pnre quadratics? 338. What may
radical equations become ?
174 PUBS QUA0BATIOB.
PROBLEMS
1. The product of onethird of a namber multiplied b;
onefourth of it is io8. What is the number ?
2. What number is that, the fourth part of whose square
being subtracted from 25, leaves 9 ?
3. How many rods on one side of a square field whose
area is 10 acres ?
4. A gentleman exchanges a rectangular piece of land
50 rods long and 18 wide, for one of equal area in a square
form. Eequired the length of one side of the square.
5. Find two numbers that are to each other as 2 to 5, and
whose product is 360.
6. If the number of dollars which a man has be squared
and 7 be subtracted, the remainder is 29. How much
money has he ?
7. Find a number whose eighth part multiplied by its
fifth part and the product divided by 16, will give a quotient
of 10.
8. The product of two numbers is 900, and the quotient
of the greater divided by the less is 4. What are the
numbers ?
9. A merchant buys a piece of silk for I40.50, and the
price per yard is to the number of yards as 3 to 54.
Eequired the number of yards and the price of each.
10. Find a number such that if 3 times the square be
divided by 4 and the quotient be diminished by 12, the
remainder will be 180.
11. A reservoir whose sides are vertical holds 266,112
gallons of water, is 6 feet deep, and square on the bottom.
Eequired the length of one side, allowing 231 cubic inches
to the gallon.
12. What number is that, to which if 10 be added, and
from which if 10 be subtracted, the product of the sum and
difference will be 156 ?
▲ FFSOIED QUAOBAIIOS 176
AFFECTED QUADRATICS*
329. An Affected Quadratic Mquation is one
which contains i\kQ first and second powers of the unknown
quantity; as, aa? + te = {?.
330. Every affected quadratic may be reduced to the form,
ic* i oa? = J,
in which a, by and x may denote any quantity, either
positive or negative^ integral or fractional.
For, bj transposition, etc., all the terms containing afi can be
reduced to one term, as ea^ ; also, those containing x can be reduced
to one term, as dx ; and all containing the known quantitiefl can be
reduced to one term, as g. Then, ea^ + dx = g.
Dividing both members by c, and substituting a for the quotient of
■d^e, and b for the quotient otg tCy we have,
0? + oa? = &.
Take any numerical quadratic, as^ 8 = a? + ^ — 4.
Clearing of fractions, &i^ — 40; — 48 = 60^ + 4:1; — 24
Transposing, etc., 20? — Sa? = 24
Removing the coefficient, a? — 4X = i2
Substituting a for 4, and b for 12 in the last equation, we have,
oj^ — oo? = 6. Hence,
AU affected quadratics may be reduced to the gejieral form,
a? ±ax=:b.
331. The First Member of the general form of an
affected quadratic equation, it will be seen, is a Bhiomial,
but not a Complete Square, One term is wanting to make
the square complete. (Art. 266, note.) The equation,
therefore, cannot be solved in its present state.
339. What is an affected quadratic equation? 33a To what general form may
erery affected quadratic he reduced ? 331. What is true of the first memher of the
SienenX form of an affected quadratic?
* Qnadratiefrom the Latin quadrare, to make square.
Affected, made up of different poioers ; from the Latin ad and/acK>,
tp make or join Vh
17^ AFFECTED QUADEATICS.
332. There are three methods of completing the square
and solving the equation.
FIRST METHOD.
1. Giyen a? + zax = J, to find the value of x.
Analysis.— The first ofbratioh.
and tlwrd tenns of the ^ + 2ax = i
square of a binomial are ^ J^ 2ax + a^ =z a^ i i
complete powers, and the x + a=z ± ^dJ^ 4 h
second term is twice the ^ « i ^ y o , j.
X zzz — a + V fit +
product of their roots;
or the prodact of one of the roots into twice the other. (Art loi.)
In the expression, x^ + 200;, the first term is a perfect square, and
the second term 200; consists of the factors 2a and x. But x is the
root of the first term a^ ; therefore, the other factor 2a must be ttciee
the root of the third term which is required to complete the square.
Hence, half of 2a, or a, must be the root of the third term, and a' the
term itself. Therefore, a? + 2ax+a^ is the square of the first member
completed.
But since we have added a^ to the first member of the equation,
we must also add it to the second, to preserve the equalitj.
Extracting the sq uare ro ot of both members, and transposing a, we
have aj = — a ± ^y/a* +6, the value sought. (Art. 297.)
2. What is the value of a? in 20^ + a? = 64 — 7a??
AlTALTSis. — Transposing —7a! opbratiok.
and removing the coefficient of a^, 2a;? + a? = 64 — 73?
we have the form a^ + 43? = 32. But 2a;^ + 8aj = 64
the first member, a!* + 43?, is an incom /jjj ^ ^^ —  ^2
plete square of a binomial. ^j^,^^^^^^
In order to complete the square, ^
we add to it the square of half the "*" ^
coefficient of x. (Art, 266.) Now, a; = — 2 ± 6
having added 4 to one member of t. 6., a? = 4 or — 8
the equation, we must also add 4 to
the other, to preserve their equality. Extracting the root of both,
and transposing, we have a? = 4, or — 8. (Art. 297.)
333. How many methods of completing the sqnare ?
AFFECTED QUADBATIOS. 177
Notes. — i. Adding the sqtuire of half the eoeffldent of the second
tenn to both members of the equation is called computing the square,
2, The first member of the fourth equation is the sgua/re of a bino
mial ; therefore, its root is found by taking the roots of the firft and
third terms, which are perfect powers. (Art. 297.) From the process
of squaring a binomial, it is obvious that the middle term (4^) forms
no part of the root. (Art. 266.)
333. From these Ulnstrations we derive the following
Eule. — I. Reduce the equation to the fornix afl ±c& = b.
II. Add to each member the square of half the coefficient ofx
III. Extract the square root of each, and reduce the resuU
ing equation.
3. Find the value of a; in — 22^ + Sax = — 6 J.
Solution.— B7 the problem, — 2a^ + 800; = —66
Removing coefficient of aj*, ^cfi+4ax = —36
Making a? positive (Art, 140, Prin. 3), cfi—^ax =36
Completing square, A^— 4007+40^ = 4a' +36
Extracting the root, x^2a = ± \/4a^ + 3&
.% w = 2a± 'v/4«'+3*
4. Given a^ + ax + bx = d, to find x,
OFKBATiaH,
sfi + ax + bx = d
a? + (a + b)xz:id
a + b ^
X=z ±
Analysis. — Factoring the terms which contain the first power of x,
we have ax+bx = (a\b)x ; hence, (a+b) may be considered a com*
pound coefficient of x. By adding the square of half this coefficient to
both members, and extracting the root, the value of a; is found.
333. What is the rule for the first method of Bolying affected quadratics f
L78 AFFECTED QUABBATIGS.
5. Given 3a? — 2a;* = — 9, to find x.
SOLTTTiOK. — By the problem, jaj — 2iC* = — 9
Making a? podtiye, etc., af — =^ = ^
Completing square, *'? + ^=! + ^ = rz
2 16 2 16 16
30
Extracting root, oj —  = ± 
/. OJ = f ± f, ». «., a? = 3 or — ij.
6. Given yifi — 24a? = — 36, to find x.
Ans. + 6 or +2.
Note.— The two roots of an affected quadratic may have the same
or different signs. Thus, in the 6th and 12th examples they are the
pame ; in the ist, 2d, 3d, 4th, and 5th, they are different.
7. Given s^ — 4^^ = 45> to find x.
8. Given a? — 6ax = rf, to find x.
9. Given 22:^ + 2ax = 2 (J + c), to find x.
Solution. — Completing the square, a? + aa?+— = — +( + &
4 4
Extracting root, aj+=±y — + 6+0
Transposhigi «= f =*= /~ + ft + «
10. Given 2^* — 22a? = 120, to find o^
11. Given o^ — 140 = 13a;, to find x.
12. Find the value of re in a:^ — 3a; + i = 52: — 15.
Solution. — By the problem, a?— 33?+ 1 = 5^^15
Transposing, a^— 8aj = — 16
Completing square, a*— 8aj+ 16 = o
Extracting root, a?— 4 = o
.'. a? = 4
t70TB. — In this equation, both the figm and the numerical wUttes of
two roots are al^. Such equations are said to have equal rooU»
NoU.^WbaX Bigns have the rooU of an affected qnadfatlcf
AFFECTED QUADBATICS. 179
SECOND METHOD.
334. When an afiected quadratic equation has l}een
reduced to the general form^
its root may be obtained without recourse to completing the
square.
I. Given a? + Sx=: 65, to find x.
Analysis. — After the square ovMosuLTKot.
of an affected quadratic is com /r2 4. ga? 6 c
pleted and the root extracted, H _i_ /a ^
the root of the third term is ^ —— 4 ± VoS + 16
transposed to the second mem •'• ^ = "" 4 it 9
ber, by changing its sign. (Art. 1. e., a? = S Or — 13
204.)
Now, if we prefix half the coefSdent of x, with its sign changed, to
plus or minus the square root of the second member increased by the
square of half the coefficient of x, the second member of the equation
will contain the same combinations of the same terms, as when the
square is completed in the ordinary way. Hence, the
EuLE. — Prefix half the coefficient of x, with the opposite
sign^ to plus or minus the square root of the second member,
increased hy the square of half the coefficient ofx.
SoIyo the following equations:
2* 3a? — 9a? — 3 = 207. 8. a? + 4aaj = J.
3. 4a:» + 15a; + S = 45 9 3^ — 74 = 6a: + 31.
4. 3a? — 14a; + 15 = o. 10. «* + 13 = 6x,
5. /^ — 9a? =28. II, (a;— 2) (a; — i)= 20.
6. —^ ; =2. 12. ; = ^
2a;a? + 2 x x + 1 6
7. «* + 7 — db=:d. 13. Q^ y ch = ld.
b "^ c
,  
334. What is the second method of Bolying affected qiudratics f
180 AFFECTEI) QUABBATIOS.
THIRD METHOD.
335. A third method of reducing an affected quadratic
equation may be illustrated in the following manner:
1. Given aa? + bx = Cf to find x.
Analtbib. — Multiply ofkbatkox.
Ing the given eqoation by oofi + bxz=zc
a, the coefficient of a^, and 4^%!^ + 4(ibx = ^ac
by 4, the smallest square ^tfy^ + ^abx + V = 400 + V
number, wehave 2ax + b = ± Vi^TT^
4/va? + 4aox = 4ae,
the first term of which la _ J±V_^+^
an exact square, whose 2 a
root is 2ax, Factoring
the second term, we have ^abx = 2 (200; x &). (Art 119.)
As the factor 2ax is the square root of 40,^0^, it is evident that 40*0^
may be regarded as the first term, and 4abx the middle term of the
square of a binomial. Since 4abx is twice the product of this root 2ax
into b, it follows that b is the second term of the binomial ; conse*
quently, 5^ added to both members will make the first a complete
square, and preserve the equality. (Axiom 2.) Extracting the square
root, transposing, etc., we have,
X = V45£ ^ ^g value of x required.
2. Given 20^ + ^x =z 27, to find x.
Solution.— By the problem, 2a? + 3* = 27
Multiplying by 4 times coef. of afi, ita^ + 24X = 216
Adding square of 3, coef. of x, i(afi + 24X + 9 = 225
Extracting root, 4«+ 3 = ± 15
Transposing, 43; = — 3 ± 15
.; x = 3 or —44.
336. From the preceding illustrations, we derive the
EuLE. — ^I. Reduce the equation to the form, as? ±bx = c.
II. Multiply both members by 4 times the coefficient ofsfi.
III. Add the square of the coefficient of x to each member ^
extract the root, and reduce the resulting equation.
396. Wlutt is the rule for the third method of redacing aflflected qxiadmtics ?
AFFECTED QUADEATICS. 181
Notes. — i. When the coefficient of 2 is an even numbei It is
sufficient to multiply both members by the coefficient of ofi, and add
to each the square of half the coefficient of x,
2. The object of multiplyiDg the equation by the coefficient ofa^ia
to make the first term a perfect square without removing the coefficient.
(Art. 251.)
3. The reason for multiplying by 4, is that it avoids fractums in
completing the square, when the coefficient of a; is an odd number.
For, multiplying both members by 4, and adding the square of the
entire coefficient of a; to each, is the same in effect as adding the square
of half the coefficient of a; to each, and then clearing the equation of
fractions by multiplying it by the denominator 4.
4. This method of completing the square is ascribed to the Hindoos.
3. Given 3a:? + 4a; = 39, to find x* Ans. 3 or — 4^.
Beduce the following equations:
4. a;^ — 30 = — ic. 8. 2a?— 6a? = 8.
S 5a; + 3ar* = 2. 9. 3212 + 5a; = 42.
6, 4a;^ — 7a; — 2 = o. 10. a;^ — 15a; z= — 54.
7. saj^+2a; = 88. 11. 93^ — 7a: =116.
337. The preceding methods are equally applicable to all
classes of affected quadratics^ but each has its advantages in
particular problems.
The first is perhaps the most naturaly being derived from
the square of a binomial ; but it necessarily involves frao
tionSf when the coefficient of x is an odd number.
The second is the shoriiest, and is therefore 2l favorite with
experts in algebra.
The advantage of the third is, that it always avoids
fractions in completing the square.
t^" The student should exercise his judgment as to the method
best adapted to his purpose.
NoUs. When the coefficient of a; is an even nnmber, how proceed ? Object of
multiplying by coefficient of a;' ? By 4 ?
182 AFFEOTED QUADBATIOS.
EXAMPLES.
Find the ralae of re in the following equations :
1. a:* — 42;=— 3. 17. 3a;* — 70?— 20 = a
2. a^^ ^x = — 4. 18. ja^ — 160 = 3a;.
3. 22^ — 7a; = — 3. 19. 22^ — 235 = I J.
4* a?+ioa?=24, 20. (a? — 2) (a;— i) = 6.
5. 6a;'— 133? + 6 = 0. 3Z. 4(3:*— i) = 4a; — i.
6. 14a: — a^ = 33. 32. (2a? — 3)* = &R
7. a:?— 3 = — _£. 23. 3a; — 2 =
6 o a ^_ J
7^5 140 ^^aj+i ^
16 100 — oa; • . 3 43?
9. 9^ = 3. 25. a* + ^ = ^.
^ar 42^ '' ^ 25 s
a a? 2 . a? I
©•  +  = — 26. a? +  = —
X a a '22
I. a? + 2ma? = J^. 27. a^— 2/w: = m' — »•.
a^ — lox^ +1
3 a^.6a? + 9 ^3
42? a?— I _ 93? + 7
14 — 35"^ 3a; "" a?
5. 2V«^ — 4a: — I = — 43?,
6. Va?+ S + 6 = a: + 5.
338. An Equation which contains but two powers of the
unknown quantity, the index of one power being ttcice that
of the other, is said to have the Quadratic Form.
The indices of these powers may be either integral or
fractional,
Thns, «•— a^ = 12 ; o^+aJ* = A ; and ^x — ^« = c, are eqnatioiui
of the quadratic form.
Note. — Equations of this character are sometimeB called trinomial
equations.
338. When has ao equation the quadratic formf Not$, What ar^ snch eqnatiOQa
eaUedf
AFFEOTED QUADBATIOS. 183
339. Equations of the quadratic form may be solyed by
the roles for affected quadratics.
I. Given jr* — 2a;2 = 8, to find «•
SoLTjnoN — B7 the problem, a?* — 2a? = 8
Completing square, a^ » 2a^ + x = 9
Extracting square root, a^ — i ss ± 3
Transposing, (B* = 4 or — 2
Extracting square root again, a; = ± 2, or ± ^^^
2. Given a:* — 40:8 = 32, \^ g^^ ^.^
Solution. — ^By the problem, a^ — 4aj* = 32
Completing square, a^ — 4aj* + 4 = 36
Extracting square root» a:* — 2 = ± 6
Transposing, etc., a^ = 8 or ^4
Extracting cube root, a; = 2 or ^—4
3. Given a^ — 4&»* = a, to find x.
Solution. — By the problem, a^» — 463^ = a
Completing square, o?* — 4ftaj» + 45^ = a + 4^^
Extracting square root, a?» — 26 = ± >y/o+4y
•
Transposing, aj» = 26 ± y'a+4^
Extracting the nth root, x = V 2& ± \/a+
4. Given a;* + 8 = 6a^, to find a::
.5. Given a;* — 2a:? = 3, to find a?.
^. Given a^ — 7a:* = o, to find as.
/. Given 1  = ^, to find a?.
2 4 32
8. Given v^ + f V^a? = i, to find as.
9. Given 4a; + /^^/x + 2 = 7, to find x.
10. Given \_ = ^ — pr— , to find as.
4 + V a? V «
184 AFFECTED QUABBATICS.
PROBLEMS.
1. Find two numbers such that their sum is 12 andtheii
product is 32.
2. A gentleman sold a picture for I24, and the per cent
lost was expressed by the cost of the picture. Find the cost
Note. — Let aj = the cost.
Then — = the per cent.
100 *^
X
We now have x — xx — = 24, to find the value of x,
« 100
3. The sum of two numbers is 10 and their product is 24.
What are the numbers ?
4. A person bought a flock of sheep for I80 ; if he had
purchased 4 more for the same sum, each sheep would have
cost 1 1 less. Find the number of sheep and the price of
each.
5. Twice the square of a certain number is equal to 65
diminished by triple the number itself. Eequired the
number.
6. A teacher divides 144 oranges equally among her
scholars ; if there had been 2 more pupils, each would haye
received one orange less. Eequired the number in the
school.
7. A father divides $50 between his two daughters, in
such a proportion that the product of their shares is $600,
What did each receive ?
8. Find two numbers whose sum is 100 and their product
2400.
9. The fence enclosing a rectangular field is 128 rods
long, and the area of the field is 1008 square rods. What
are its length and breadth ?
10. A colonel arranges his regiment of 1600 men in a
solid body, so that each rank exceeds the file by 60 soldiers.
How many does he place in rank and file ?
AFFECTED QUADBATIOS. 186
11. A droyer bnys a number of lambs for I50 and sells
them at $5.50 each^ and thus gains the cost of one lamb.
Required the number of lambs.
12. The sum of two numbers is 4 and the sum of their
reciprocals is i. What are the numbers ?
13. The sum of two numbers is 5 and the sum of their
cubes 65. What are the numbers ?
14. The length of a lot is i yard longer than the width
and the area is 3 acres. Find the length of the sides.
15. A and B start together for a place 300 miles distant;
A goes I mile an hour faster than B^ and arriyes at his
journey's end 10 hours before him. Find the rate per hour
at which each.trayels.
16. A and B distribute I1200 each among a certain
number of persons. A relieyes 40 persons more than B, and
B giyes to each person $5 more than A. Eequired the
number relieyed by each.
17. Diyide 48 into two such parts that their product may
be 252.
18. Two girls, A and B, bought 10 lemons for 24 cents,
each spending 12 cents ; A paid i cent more apiece than B:
how many lemons did each buy ?
19. Find the length and breadth of a room the perimeter
of which is 48 feet, the area of the floor being as many
square feet as 35 times the difference between the length
and breadth.
20. In a peach orchard of 180 trees there are three more
in a row than there are rows. How many rows are there,
and how many trees in each ?
21. Find the number consisting of two digits whose sum
is 7, and the sum of their squares is 29.
22. The expenses of a picnic amount to $10, and this sum
could be raised if each person in the party should give 30 cts.
more than the number in the party. How many compose
the party ?
186 AFFECTED QUABBATIOS.
23. Find two numbers the product of which is 120, and
if 2 be added to the less and 3 subtraxjted from the greater^
the product of the sum and remainder will also be 120.
24. Divide 36 iuto two such parts that their product shall
be 80 times their difference.
25. The sum of two numbers is 75 and their product is
to the sum of their squares as 2 to 5. Find the numbers.
26. Divide 146 into two such parts that the difference of
their square roots may be 6.
27. The forewheel of a carriage makes sixty revolutions
more than the hindwheel in going 3600 feet ; but if the
circumference of each wheel were increased by three feet, it
would make only forty revolutions more than the hind*
wheel in passing over the same distance. What is the
circumference of each wheel ?
28. Find two numbers whose difference is 16 and their
product 36.
29. What two numbers are those whose sum is i^ and the
sum of their reciprocals 3 J ?
30. Find two numbers whose difference is 15, and half
their product is equal to the cube of the less number.
31. A lady being asked her age, said, If you add the
square root of my age to half of it, and subtract 12, the
remainder is nothing. What is her age ?
32. The perimeter of a field is 96 rods, and its area is
equal to 70 times the difference of its length and breadth.
What are its dimensions ?
33. The product of the ages of A and B is 120 years. If
A were 3 years younger and B 2 years older, the product of
their ages would still be 120. How old is each ?
34. A man bought 80 pounds of pepper, and $6 pounds
of saffron, so that for 8 crowns he had 14 pounds of pepper
more than of saffron for 26 crowns; and the amount he
laid out was 188 crowns. How many pounds of pepper did
he buy for 8 crowns ?
SIMULTAlSrEOnS QUADBATIOS. 187
SIMULTANEOUS QUADRATIC EQUATIONS.
TWO UNKNOWN QUANTITIES.
340. A Homogeneous JEquation is one in which
the sum of the exponents of the unknown quantities is the
same in every term which contains them.
Thus, a^— ^ = 7, and «*— ay+y* = 13, are each homogeneous.
341. A Symmetrical JEquation is one in which
the unknown quantities are involyed to the same degree.
ThiiB, aj' +y* = 34, and T^y—x^f = 34, are each sTmmetrical.
342. Simultaneous Quadratic Equatio^is con
taining two unknown quantities, in general involve the
principles of Biquadratic equations, which belong to the
higher departments of Algebra.
There are three classes of examples, however, which may
be solved by the rules of quadratics.
ist. When one equation is quadratic, and the other simple.
2d. When both equations are quadratic and homogeneous.
3d. When each equation is symmetrical.
343. To Solve Simultaneous Equationt oontitfing of a
Quadratic and a Simple Equation.
I. Given aj* + y^ = 13, and x + y = $, to find x and y.
Solution.— By the problem, a^+y* = 13 (i)
x+y= 5 (2)
By transposition, x = s—y (3)
Squaring each side of (3) (Art. 102), aj» = 25iqy+^ (4)
Substituting (4) in (i), 25 — iqy + y« + y« = 13 (5)
Uniting and transposing, 2y*— ipy = — 12 (6)
Comp. sq. (Art. 336, note), 4^— 2qy + 25 = — 24 + 25 (7)
Extracting root, 2^—5 = ± i
.♦. y = 3 or 2.
Substituting value of y in (3), a; = 2 or 3. Hence, the
^■1^^— ^■^B ■ ■ ■■■■■■■■■I ■■■ ■ ■■■.■ . ■ — I ■ ■■■l^■ ■ ■■ ■»■■»■ ■— ^^M— ^^W^i^M^^^—
340. Wbat is a homogeneoaH equation ? 341. A Bymmetrical eqnatioii ?
188 8IH17LTAKEOT78 QUADRATICS.
BuLE. — Find the value of one of the unknown quantities
in the simple equation by transposition, and substitute this
value in the quadratic equation, (Arts. 221^ 223.)
Solve the following equations:
2. a« + ^=2S, 5 ^ + y* = 244,
a?+y=7 y— a:=2.
3. a? + fz=l^4^ 6. 3^— y*=2Si,
a; + y=i2. « + 4y = 38
4. a«y3=28, 7. 82«+5y»=728,
a: — y = 2. 6y — a; = 15,
344. To Solve Simultaneous Equations whioii are botii
Quadratic and Homogeneous.
8. Given o^+xy = 40, and j^+xy = 24, to find x and y.
SoLTTTiON. — ^By the problem, a? +0^ = 40 (i)
^+ay = 24 (2)
Let « = py (3)
SubBtituting py in (i), >V +p^ = 40 (4)
" (2X ^•+1^ = 24 (5)
Factoring, etc, (4), ^ = ^^ (6)
" (5), ^ = 5^ ^^^
Equating (6) and (7). ^^ = ^ (8)
Clearing of fractions, 5 + 5P = 2P' + 31> (9)
Transposing, etc., SP''^p = 5 (10)
Comp. sq., 3d metli. (Art. 336), gp*— 6p + 1 = 15 + 1=16 * (11)
Extracting root, 2P— i = ± 4 (12)
Transposing, 3p r= i ± 4
Dropping the negative value, !> = t
Substituting value of ;> in (7), y* = 24»(i +)=9
Extracting root, y = ± 3
Substituting value ot p and y in (3), jc = x ±3=±5.
Hence, the
343. Bale for solution of eqaatione consisting of a quadratic and simple equa^
tion?
SIMULTAKBOUS QUADEATICS. 189
Bulb. — I. For one of the unknown quantities substitute
the product of the other into an auxiltary quantity, and then
find the value of this auxiliary quantity.
II. Find the values of the unknown quantities by substu
tuting the value of the auxiliary quantity in one of the
equations least involved.
KoTB. — ^An auxiliary quantity is one introduced to aid in the Bola«
tiob of d problem, as p in the above operation.
9. Given x +y = 9>)x/jj j
. :, ^ > to find a; and v«
And AP^y'=' 24, ) ^
12. Given ^ —xy + t^=z
And
+ »*i9. Kofindrcandy.
345. To Solve Simultaneous Quadratic Equations when each
Equation is Symmetrical.
13. Given a: + y = 9, and xy = 20, to find x and y.
Solution. — By the problem, aj+y= 9 (i) C
•* " jry = 20 (2)
Squaring (i), a^+2ay+y* = 81 (3) ^
Multiplying (2) by 4, /^ = 86 <4>
Subtracting (4) from (3), a?— 2a!y+y* =1 fe)
Extracting sq. root of (5), a?— y = ± i (6)
Bringing down (i), x\y= 9
Adding (i) and (6), 2X = 10 or 8 (7)
Removing coefficient, a; = 5 or 4
Substituting value of a; in (i), y = 4 or 5
Notes. — i. The values of x and y in these equations are not equal,
but interchangeable ; thus, when a? = 5, y=4 ; and whena? = 4, y = 5.
344. How Bolye equations which are both quadratic and homogeneoiu f NoU,
What is an auxiliary quantitjt
190 BIXCLTaKSOUS ^cadbatics.
2. Hw ■J u tfcm Off tfajta dMB Off problems cones ace mdlu g to tlie
given tqmtAmm. Co n aeqaently, no m>edfir> roles csn be glren that
wOl meet ereiy cue. Bat jndgmMrt snd pnctioe will rendflj supply
enedientiL ^uis^
L When the sum and prodnct are giren. (Ex. 13, 15.)
Find the difference and combine it with the sum. (Art. 2 24.)
n. When the diflEeienoe and prodnct are given. (Ebc 16.)
Find the sum and combine it with the difference.
nL When the sum and diffeience of the same powers are
giyen^ (£x. 14, 17.)
Combine the two equations by addition and subtraction.
lY. When the memhers of one equation are multiples of
the other. (Ex. 18.)
Divide one by the other, and then reduce the resulting
equation.
i4.Qhenxi + f^ = S. 0) ) ^ find tr and y.
And rc*.y*=i, (2)) ^
SoLimoH. — ^Adding (i) and {2), and diyiding, o^ = 3
Involying, « = 27
Babtiacting (2) from (i)» eta« ^ = 9
InyolTing, jf = 8
15. Given a? + y=a7, )x^j jj
16. Giren ary= '4, 1 ^o, gnd <nmd y.
And ay = i47> )
17. Given a.i + j^ = 7,Kg^^,3„^y
And aji — yt = 3, J
18. Given ^ + aY=ijU,fi„a^andy.
And sfiy +xf = 6y) ^
NoT<.WliatiBtrneoftheBolation of BimaltaneoGB quadratics f When the
and product are ^ven,how proceed? When the difference and prodnct? When
the ram and dlflbrance of the rame powen are ^reh 7 When the members of ons
fqnatUm are mnltiplet of the other ?
8IHULTAKE0US QUAOBAXICB. 191
PROBLEMS.
1. The difference of two numbers is 4, and the difTerenoe
of their cubes 448. What are the numbers ?
2. A man is one year older than his wife, and the product
of their respective ages is 930. What is the age of each ?
3. Bequired two numbers whose sum multiplied by the
greater is 180^ and whose difference multiplied by the less
18 16.
4. In an orchard of 1000 trees, the number of rows
exceeds the number of trees in each row by 15. Bequired
the number of rows and the number of trees in each row.
5. The area of a rectangular garden is 960 square yards,
and the length exceeds the breadth by 16 yards. Bequired
the dimensions.
6. Subtract the sum of two numbers from the sum of
their squares, and the remainder is 78 ; the product of the
numbers increased by their sum is 39. What are the
numbers ?
7. Find two numbers whose sum added to the sum of
their squares is 188, and whose product is 77.
8. A surveyor lays out a piece of land in a rectangular
form, so that its perimeter is 100 rods, and its area 589
square rods. Find the length and breadth.
9. Bequired two numbers whose product is 28, and the
sum of their squares 65.
10. A regiment of soldiers consisting of 1154 men is
formed into two squares, one of which has 2 more men on a
side than the other. How many men ai*e on a side of each
of the squares ?
11. Bequired two numbers whose product is 3 times their
sum, and the sum of their squares 160.
12. What two numbers are those whose product is 6 times
their difference, and the sum of their squares 13 ?
(See Appendix, p. eeob)
CHAPTER XVII.
RATIO AND PROPORTION.
346. Ratio is the relation which one quantity bears to
another with respect to magnitude.
347. The Terms of a Ratio are the quantities
compared. The first is called the Antecedent, the second
the Consequent,"^ and the two together, a Couplet,
348. The Sign of ratio is a colon : f placed between the
two quantities compared.
Ratio is also denoted by placing the consequent under the
antecedent^ in the form of diffraction.
Thus, the ratio of a to & is written, a : 6, or = •
349. The Measure or Value of a ratio is the quotiem
of the antecedent divided by the consequent, and is equal
to the value of the fraction by which it is expressed.
Thus, the measure or value of 8 : 4 is 844 = 2.
Note. — That quantities may have a ratio to each other, they must
be so far of the same nature, that one can properly be said to be equal
to, or greater, or leas than the other.
Thus, a foot haa a ratio to a yard^ but not to an hour, or a pound
350. A Simple Matio is one which has but two terms ;
ifts, a : d, 8 : 4.
346. What is ratio ? 347. What are the termH of a ratio ? 348. The sign ? How
also is ratio denoted ? 349. The measure or value ? Note. What quantitieB have a
ratio to each other ? 350. What is a simple ratio ?
* Antecedent, Latin ante, before, and cedere, to go, to preceae.
Consequent, Latin con^ and sequi, \jofoUow.
f The sign of ratio : is derived from the sign of division +, the
horizontal line being dropped.
BATIO. 193
351. A Compound Ratio is the product of two or
more simple ratios.
ThuSj 4:2) are each edmple But 4x9: 2x3
9:3) ratioB. is a oompoand ratio.
Note. — The nature of compound latiofii is the same as that of sim
ple ratios. They are bo called to denote their origin, and are usaally
•expressed by writing the corresponding terms of the simple ratios one
nnder another, as above.
352. A Direct Matio arises from dividing the ante
cedent by the consequent
353. An Inverse* or Reciprocal Itatio arises
from dividing the consequent by the antecedent, and is the
same as the ratio of the reciprocals of the two numbers
compared.
Thus, the direct ratio of a to a& = = , or t , and that of 4 to
ah h ^
4 I
12 = ~ , or  •
12 3
€ib 12
The inverse ratio of a to a& = — , or 6 ; of 4 to 12 = — , or 3.
a 4 "^
It is the same as the ratio of the reciprocals.  to — r , and  to — •
a ab 4 12
Note. — A reciprocal ratio is expressed by inverting the fraction
which expresses the direct ratio. When the edton is used, it is
expressed by inverting the order of the terms.
354. The ratio between two fractions which have a
common denominator, is the same as the ratio of their
numerators.
Thus, the ratio of ( : f is the same as 6 : 3.
NoTB. — When the fractions have diff^ent denominators, reduce
them to a common denominator; then compare their numerator.
(Art. 175.)
355. A Ratio of Equality is one in which the quan
tities compared are equal, and its value is a unit or i.
35X. What is a compound ratio? Note, Whyeo called? 359. Wliat is a direct
ratio ? 353. A reciprocal ? 355. What ia a ratio of equally ?
* Inverse, from the Latin in and terto, to turn upside down, to inserts
9
194 RATIO.
356. A Ratio of Greater Inequality is one whose
antecedent is greater than its consequent^ and its value k
greater than i.
357. A Ratio of Less Tneqtuility is one whose
antecedent is less than its consequent^ and its value is less
than I.
358. A Duplicate Ratio is the square of a simph
ratio. It arises from multiplying a simple ratio into itself,
or into another equal ratio.
359. A Triplicate Ratio is the cube of a simple ratiO;
and is the product of three equal ratios.
Thus, the duplicate latio of a to 6 is a< : &*.
The triplicate ratio of a to & is a' : &^.
360. A Subduplicate Ratio is the square root of a
simple ratio.
361. A Subtriplicate Ratio is the cube root of a
simple ratio.
Thus, the subdupUeate ratio of 2; to ^ is ^x : ^y^.
The subtriplicate ratio of a; to y is ^x : ^y^, etc.
362. Since ratio may be expressed in the form of a
fractiouy it follows that changes made in its terms have the
same effect on its value, as like changes in the terms of a
fraction. (Art. 167.) Hence, the following
PRINCIPLES.
i^
. Multiplying the antecedenty or \ ^^ ,.. ,. .,
r^, .f. J . r Multiplies the rat%o.
Dividing the consequent^ )
2^. Dividing the antecedent, or \ n j u. a*
Br lA t ' AL ± \ Divides the ratio.
Multiplying the consequent^ )
3**. Multiplying or dividing loth) Does not alter the vahu
terms by the same quantity, ) of the ratio.
356. Of greater inequality? 357. Of leB? ineqnality? 358. A dnpUeate imtiol
359. Triplicate? 360. SiiMuplicate ? 3$!. Sal)tripUcAte? 3^. Kamo PHndpto u
Pi^ncipla 9 Principl* >
BAXXO 196
EXAMPLES.
1. What is the ratio of 4 yards to 4 feet ?
SoLTTTiON. 4 yards = 12 feet ; and tlie ratio of 12 ft. to 4 ft. is 5
2. What is the ratio of 6aflto2x? Ans. $x.
3. What is the ratio of 40 square rods to an acre?
4. What is the ratio of i pint to a gallon ?
5. What is the ratio of 64 rods to a mile ?
6. What is the ratio of Ba^ to 4a ?
7. What is the ratio of isabc to 5a J?
8. What is the ratio of $5 to 50 cents ?
9. What IS the ratio of 75 cents to 16 ?
10. What is the ratio of 35 quarts to 35 gallons ?
11. What is the ratio of 20^ to 4a ?
12. What is the ratio ota^^y^tox + y?
13. What is the compound ratio of 9 : 12 and 8 : 15 ?
Solution. 9 x 8 = 72, and 12x15= i8a Now 721 180 = t^, Ana
Or, 9: 12=^, and 8: 15 =^y. Now^x^ = ^o = A*^'*«
14. What is the compound ratio of 8 : 15 and 25 : 30 ?
15. What is the compound ratio of o : J and 2 J : 300??
16. Eeduce the ratio of 9 to 45 to the lowest terms.
Solution. 9 : 45 = ^, and ^j = J, Ana.
17. Eeduce the ratio of 24 to 96 to the lowest terms.
18. Eeduce the ratio of 144 to 1728 to the lowest terms.'
19. What kind of ratio is 25 to 25 ?
20. What kind of a ratio is ab: ab?
21. What kind of ratio is 35 to 7 ?
22. What kind of ratio is 6 to 48 ?
23. Which is the greater, the ratio of 15 : 9, or 38 : 19?
24. Which is the greater, the ratio of 8 : 25, or V4 ' V^S.
25. K the antecedent of a couplet is 56, and the ratio 8
what is the consequent ?
26. If the consequent of a couplet is 7, and the ratio 14,
what is the antecedent ?
196 PBOPOBIIOlf.
PROPORTION.
363. Proportion is an equality of ratios.
Thus, the ratio 8 : 4 = 6 : 3, is a proportion. That is,
Foar quantities are in proportion, when the jvnt is the same mvlH
p^ or part of the second that the third is of ihe fourth,
364. The Sign of Propoirtion is a double colon : :,*
or the sign =. Thus,
The eqnality hetween the ratio of a to 5 and e to (? is expressed hy
a : 6 : : c : ei, or hy ■= = 3
a
The former is read, "aisto&ascistod;" the latter, "h\a contained
in a as many times as (2 is contained in c,**
Note. —Each ratio is caUed a couplet , and each term a praportkmal,
365. The Terms of a proportion are the quantities
compared. The^r^^ and fourth are called the extremes, the
second and third the means.
366. In every proportion there must be at least four
terms ; for the equality is between two or more ratios^ and
each ratio has two terms.,
367. A proportion may, however, be formed from three
quantities, for one of the quantities may be repeated, so as
to form two terms ; as, a : ft : : & : c.
Note. — Care should be taken not to confound propoiiion with ratio.
In common discourse, these terms are often used indiscriminately.
Thus, it is said, '* The income of one man bears a greater proportioD
to his capital than that of another/' etc. But these are loose expressions
In a simple ratio there are but tiro terms, an antecedent and a
consequent; whereas, in a proportion there must at least be Jbur
terms. (Arts. 350, 366.)
363. What is proportion? 364. The sign of proportion? Ifote, What is each
ratio called ? 365. What are the terms of a proportion ? 366. How many terms in
every proportion ? 367. How form a proportion l^om three qnantities ?
* The sign : : is derived from the sign of equality =, the four
points being the terminations of the lines.
PROPOETION. 197
Again, one ratio may be greater or less than another, but one
proportion is neither greater nor less than another. For equality does
not admit of degrees. In scientific investigations, this distinction
should be carefully observed.
368. A Mean Proportional between two quantities
is the middle term or quantity repeated, in a proportion
formed from three quantities*
369. A Third Proportional is the last term of a
proportion having three quantities.
Thus, in the prox>ortion aih : : & : c, Ms a mean proportional, and
e a third proportionaL
370. A Direct Proportion is an equality between
two direct ratios; as^ a : i : : cid, ^i6 : : 4:8.
371. An Inverse or Reciprocal Proportion is an
equality between a direct and reciprocal ratio ; as,
372. Analogous Terms are the antecedent and con
sequent of the same couplet.
373. Homologous Terms are either two anteceden ts
or two consequents.
PROPOSITIONS.
374. A Proposition is the statement of a truth to be
proYed, or of an operation to be performed.
Propositions are of two kinds, theorems and problems.
375. A Theorem is something to be proved.
376. A Problem is something to be done.
377. A Corollary is a principle inferred from ^
preceding proposition.
368. What Ib a mean proportional ? 369. What Is a third proportional ? 370. A
direct proportion? 371. An inverse or reciprocal proportion? 372. What are
analo^ons terms ? 373. Homologous ? 374. What is a proposition ? How divided f
375. What is a theorem ? 376. A problem ? 377. A ccroUary t
198 PBOPOBTION.
378. The more important theorems in proportion are the
following:
Theobek L
If four quantities are proportional^ the product of th9
extremes is equal to the product of the means.
Let a : 5 :: ei d
Clearing of fracdoiui, ads^he,
Yebification by Nttmbebs.
QlTeiif 2 : 4 : : 8 : i6 ; and 2 x z6 = 4 x &
OoB. — ^The relation of the fonr terms of a proportion to
each other is such^ that if any three of them are giyen, the
fourth may be found.
Thus, since ad = be, it follows that
a = bcid, b = tul+c, c = adib, and d = be^a*, (As. 5.)
KoTES. — ^i. The role of Simple Proportion In Arithmetic is founded
npon this principle, and its operations are easily proved hj it.
2. This theorem furnishes a very simple test for determining
whether any four quantities are proportional. We have only to
multiply the extremes together, and the meana
Theobek IL
Jf three quantities are proportional, the product of tin
extremes is equal to the square of the mean.
Let a: b :: b • e
By Art. 363, ^ = 
Clearing of fractions, 00 = &*.
Again, 9:6 : : 6:4, and 4x9 = 6*.
Cob. — A mean proportional between two quantities is
equal to the square root of their product.
PBOPOBTIOH* 199
Theobem nL
If the product of two quantities is equal to the product of
two others, the four quantities are proportional ; the factors
of either product being taken for the extremes, and the factors
of the other for the means.
Let odsio
Dividing b7 M; 1 = 1
Or, by Art 363, a : 5 : : : A
Again* 4x6 = 3x8» and 4:3:: 8 ;6^
Theobeh IV.
If four quantities are proportional, they are proportional
when the means are inverted.
. Let a: h :: ei d^ then ai e :: h : d
For, by Art, 363. j = g
Multiplying by ,  = j
Or, ai e ti bi d.
Again, 3 : 6 : : 4 : 8, and 3 : 4 : : 6 : 8. (Th. x.)
Note. — This change in the order of the meana ia called
** Altematianm''
Theobem V.
If four quantities are proportional, they are proportional
when the terms of each couplet are inverted.
Le» a I b :: c i d^ then b i a .: die
By Theorem i, ad = be
By Theorem 3, b '. a 11 d i e.
Again, 6 : a :: 15 : 5« then 2 : 6 :: 5 : 15. (Th. x.)
Co EL — If the extremes are inverted, or the order of the
terms^ the quantities will be proportional.
200 PKOPOBTioir.
N0TB8.— z. If the terms of only one of the ooaplets are inyerted,
the proportion becomes reeiproeal.
3. The change in the order of the terms of each oonplet is called
3. This propositioii supposes the quantities compared to he of the
same kind. Thus, a line has no relation to weight. (ArLy^noU.)
Theorem YL
If four quantities are proportional, two analogous or tioo
homologous terms may be multiplied or divided by the sam
quantity without destroying the proportion.
Let aih 11 e i d
MnltiplTlng analogous terms, am thm 11 c id
and a ih 11 cm i dm
b™™. (!«, 36«, Pito. J), ^=j. «a =£
Multiplying homologous terms, am i b : : cm : d
And a : bm : : c : dm.
— . . . ^ am cm « a c
Hence. (Ax. 4. 5), s=j. «d ^=^
Dividing analogous termsy — : — :• c : d^
J ^ e d
and a : : : — : —
' m m
($ c
Dividing homologous temu, — : 5 : : — \ d
, h d
and a : — : • c : —
m m
Clearing of fractions (Th, i), ad = be
OoB. — All the terms of a proportion maybe multiplied
or divided by the same quantity without destroying the
proportion.
Notes. — i. When the Tiomciogaus terms are multiplied or divided,
both ratios are equaUy iruyreased or dminished.
2. When the analogous terms are multiplied or divided, the ratios
are not aUered.
PBOPOKTION. 201
Theorem VIL
If four quantities are proportional, the sum of the first
and second is to the second, as the sum of the third and fourth
is to the fourth.
Let a b II e : d, then a+h : b i:'c+d : d
_ a e
Adding i to each member, v + i = ^ + i. (Ax. a.)
Incorporatmg i, — r— = —r
Therefore (Art. 363), akb :b :: c+d i d
Again, 4:2 :: 6:3, then 4+2 : 2 :: 6+3 : 3
Note. — This combination is sometimes called " Gompodtion,*'
Theobem VIII.
If four quantities are proportional, the difference of the
first and the second is to the second, as the difference of tJie
third and fourth is to the fourth.
Let a lb i\ c : d, then a—b : b :: e^d : d
a tj
Subtracting i from each member, r — i = ^ — i
a
, ^. a— 6 c—d
Incorporatmg — i, — =— = — r—
a
Therefore, 0—6 : 6 : : c—d\d
Again, 4:2 : : 6:3, then 4—2 : 2 : : 6—3 : 3
Note. — This comparison is sometimes called " Division" *
* The technical terms. Composition and Dimsion, are calculated
rather to perplex than to aid the learner, and are properly falling into
disuse. The objection to the former is, that it is liable to be mistaken
for the .composition or compounding of ratios, whereas the two
202 PBOPOBTION.
Theokbm IX.
If two ratios are respectively equal to a third, they an
equal to each other.
Let a lb :: m i n, and c \ d :: m \ n
Then ^ = — » «^d ; = —
on d n
€t C
By Ax. I, T = :?• Tliat is, a : b = e : d
d
Again, 12 : 4 = 6 : 2, and 9:3 = 6:2
.*. 12 : 4 = 9 : 3
Theobem X.
WJien any number of quantities are proportional^ any
anteA)edent is to its consequent, as the sum of all the ante^
cedents is to the sum of all the consequents.
Let a : h
: : c : d :: e : f, etc
Then a : b
:: a+e+e : 6+d+/, etc.
For (Th. i).
ad = be
And, "
itf=be
Also,
ab = ha
Adding (Ax, 2),
ab\ad+qf^ ha\bc\be
Factoring,
aQ>+d+f) = b{a+e+e)
Hence, (Th. 3), a : b :: {a^e+e, etc.) : (&+(?+/, etc.)
operations are entirely different In one the terms are added, in the
other they are multiplied together. (Art. 351.)
The objection to the latter is, that the change to which the term
division is here applied, is effected by subtraction, and has no
reference to division, in the sense the word is used in Arithmetic and
Algebra. Moreover, the alteration in the terms of Theorem 6 is
produced by actual division. Usage, however ancient, can no longer
justify the employment of the same word in two different senses, in
explaining the same subject.
PBOPOBTIOSr. SOS
Theobem XL
If the corresponding terms of two or more proportions are
multiplied together, the products will be proportioned.
Let a : b :: c : d, and e :f :: g : h
Then ae : hf :: eg : dh
rn a e ji e g
For. j = g and ^ =
Mult, ratios together (Ax. 4), h?^^
Hence, (Th. 3), ae : bf :: eg : dh.
theobeh xn.
If four quantities are proportional, like powers or roots
of these quantities are proportional
Let a : b : : e : d^ then T = ^
Hence (Th. 3), a?* : 6» : : <^ : <^
Extracting eq. root, a* : 6* : : (J* : d*
Again, 2:3 : : 4:6, then 2" : 3" : : 4* : 6*
Li like manner, ^^4 : y^ : : ^^ : \/36.
Note.— The index n may be either integral or fraetumaL
Theorem XIII.
Equimultiples of two quantities are proportional to the
Quantities themselves.
Smce ^ = 5' ^y ^^' 362, Pnn. 3, 5^ = 5
Hence, am : bm :: 6 : (2.
204 PBOFOBTION.
PROBLEMS.
1. The first three terms of a proportion are 6, 8^ and 3.
What is the fourth ?
Let X = the fourth tenn»
Then 6:8 :: 3:9;
.*. tx = 24, and a; = 4.
2. The last three terms of a proportion are 8, 6, and 12.
What is the first ?
3. Beqnired a third proportional to 25 and 400.
4. Beqnired a mean proportional between 9 and 16.
5. Find two numbers, the greater of which shall be to
the less, as their sum to 42 ; and as their difference to 6.
6. Divide the number 18 into two such parts, that the
squares of those parts may be in the ratio of 25 to 16.
7. Divide the number 28 into two such parts, that the
quotient of the greater divided by the less shall be to the
quotient of the less divided by the greater as 32 to 18.
8. What two numbers are those whose product is 24, and
the difference of their rubes is to the cube of their difference
as 19 to I ?
9. Find two numbers whose sum is to their difference as
9 is to 6, and whose difference is to their product as i to 12.
10. A rectangular farm contains 860 acres, and its length
is to its breadth as 43 to 32. What are the length and
breadth ?
11. There are two square fields; a side of one is 10 rods
longer than a side of the other, and the areas are as 9
to 4. What is the length of their sides ?
12. What two numbers are those whose product is 135,
and the difference of their squares is to the square of their
difference as 4 to i ?
13. Find two numbers whose product is 320 ; and the
difference of their cubes is to the cube of their difference
as 61 to I.
CHAPTER XVIII.
PROGRESSION.
379. A JProgression is a series of quantities which
increase or decrease according to a fixed law.
380. The Terms of a JProgression are the quan
tities which form the series. The first and last terms are
the extremes ; the others, the means,
381. Progressions are of three kinds: arithmeticaif
geometrical, and harmonicah
ARITHMETICAL PROGRESSION.
382. An Arithmetical JProgression is a series
which increases or decreases by a constant quantity called
the common difference,
383. In an ascending series, each term is found hy adding
the common difference to the preceding term.
If the first term is a, and the common difference d, the series is
a, «+(?, a+2(?, a+3rf, eta
If a = 2, and <2 = 3, the series is 2, s> 8> 11* I4> etc.
384. In a descending series, each term is found by
subtracting the common difference from the preceding term.
If a is the first term, and d the common difference, the series is
a, a — dy a — 2d, a — seZ, etc.
In this case, the common difference may be considered —d. Hence,
the common difference may be either positive or negative. And, since
adding a negative quantity is equivalent to subtracting an equal
379. What is a progression? 380. The terms? 381. How many kinds of
progression ? 382. An arithmetical progresE^ion ? What is this constant quantity
called? 383. An ascending series ? 384. A descending series?
206 ARITHMETICAL PEOGRESSION.
poiUive one, it may therefore properly be said that each successive
term of the series is derived from the preceding by the addUianoi the
common difference. (Art. 75, Prin. 3.)
Notes.— I. The €ommo?i difference was formerly called arithmetical
ratio; but this term is passing out of use.
2. An Arithmetical Progression is sometimes called an Bguidifferent
Series, or a Progression by Difference, In every progression there may
be an infinite number of terms.
If four quantities are in arithmetical progression^
the sum of the extremes is equal to the sum of the means.
Let a, a+d, a+2(2, a^^d, be the series.
Adding extremes, etc., 2a+3(2 = 2a+3d,
Or, let 2, 2+3, 2 + 6, 2+9, be the series.
Then 2 + 2 + 9 = 2 + 3 + 2+6.
386. If three quatitities are in arithmetical progression,
the sum of the extremes is equal to double the mean.
Let a, a+d, a+2d, be the series.
Then 2a+2d = 2(a\d),
Again, let 2, 2 + 4, 2 + 8, be the series.
Then 2 + 2 + 8 = 2(2+4).
Cob. — ^An Arithmetical Mean between two qnantitiefi
may be found by taking half their sum.
387. In Arithmetical JProgression there are fiv^
elements to be considered: the first term, the common
difference, the last term, the number of terms, and the sum
of the terms.
Let a = the first term.
d = the common difference,
I = the last term.
n = the number of terms.
8 = the sum of the terms.
The relation of these five quantities to each other is such
that if any three of them are given, the other two can be
found.
385. VThat 18 trae of fonr qnantitioB in arithmetical progression ? 386. Of three
quantities ? 387. Name the elements in aritbmetical progression ? What relation
have they to each other ?
▲ BITHHETICAL PBOGBESSION. 207
qaii
ca:
05
CASE I.
388. The Fip$t Term, the Common Difference, and Number
of Terms being given, to Find the Last Tenn.
Each Buoceeding term of a prQgresBion is found by adding the
common difference to the preceding term. (Art. 384.) Therefore the
terms of an ascending series are
a, a¥dt a'\2d, a+3(f, etc
The terms of a descending series are
a, CL — d, a^^d, a — 3^1 ®tc*
It will be seen that the coeffident of d in each term of both aeries
is one less than the number of that term in the series. Therefore,
putting I for the last or nth term, we have
FoEMULA I. Z = a ± (n — i) A
BuLE. — ^L Multiply the common difference by the numher
of terms less one.
I!L When the series is ascending^ add this prodtu>t to the
first term ; when descending, subtract it from the first tertn*
1. Given a = 3, rf = 2, and n = 7, to find I
l=:a± («— I) d = 3 + (7—1) 2 = 15, Ans,
2. Given a = 25, rf = — 2, and n = g, to find I
3. Given « = 12, d = 4, and w = 15, to find L
4. Given « = i, rf= — ^, and n = 13, to find t
5. Given a = J, d = i, and w = 9, to find L
6. Given a = i, d = — .01, and w = 10, to find L
7. Find the 12th term of the series 3, 5, 7, 9, 11, etc.
Note; — In this problem, a = 3, d = 2, n = 12. Ans. 25.
8. Find the 15th term of i, 4, 7, 10, etc.
9. Find the 9th term of 31, 29, 27, 25, etc.
10. What is the 30th term of the series i, 2^, 4, 5 J, etc.
11. Find the 25th term of the series x+^x+^x + jx, etc.
12. Find the 72th term of the series 2a, ^a, Sa, nay etc.
388. What is the rale for finding the last termf
208 ABIXHHETICAL PBOOBESSION.
CASE II.
389. The Extremes and Number of Terms being given, to
Find the Sum of the Series.
Let a, a¥d, a+2d, a+3<2 . . . i, be an arithmetical progiesBion.
the earn of which is reqnired.
Since the sum of two or more quantities is the same in whatever
order they are added (Art. 63, Prin. 2), we have
8 = a+ (a+ d) + (a+ 2d) + (a+ 3d) + , . . +1
Inverting, 8 = 1 + (I — d) h (l — 2d) + (I— 3d) + . . . +«
Adding, 2« = a + ; + (a + + («+0 + (^+0 + • • . +a+^
.*. 2« = (a+0 taken n times, or as many times as there are
terms in the series.
That is, 28 = (a+l)n, Hence, the
Formula II. s = ^ — ^^^ x n.
2
Rule. — Multiply half the sum of the extremes b'g the
number of terms.
Cor. — From the preceding illustration it follows that the
sum of the extremes is equal to the sum of any two terms
equally distant from the extremes.
Thns, in the series, 3, 5, 7, 9, 11, 13, the smn of the first and last
terms, of the second and fifth, etc., is the same, viz., 16.
1. Given a = 4, Z = 148, and w = 15, to find 5.
Solution. 4+148 = 152, and (15252) x 15 = 1140, Ans.
2. Given a = ^, ? = 30, and w = 50, to find s.
3. Given a = 6, Z = 42, and w = 9, to find s,
4. Given a = 5, Z = 75, and n = 35, to find s.
5. Given a = 2, Z = i, and w = 17, to find s.
6. Find the sum of the series 2, 5, 8, 1 1, etc., to 20 terms.
7. Find the sum of the series i, i}, 2, 2^, etc., to 25 terms.
8. Find the sum of the series 75, 72, 69, 66, 6^^ etc.,
to 15 terms.
ARITHMETICAL PROGRESSION. 209
390. The two preceding formulas are fundamental, and
furnish the means for solving all the problems in Arith
metical Progression. From them may be derived eighteen
other formulas.
By Formula L
391. This formula contains /bwrrff/ferew^ quantities; the
first term, the common difference, the last term, and the
numier of terms. If any three of these quantities are given,
the other may be found. (Art. 388.)
I. l = a ±{n — ^)d; a,d, and n being given.
3. Given d, I, and n, to find a, the first term.*
Transposing (w— i) d in (i),
a = l±(n—i)d.
4. Given a, I, and n, to find d, the common difference.
Transposing in (i), and dividing by (h— i),
i — a
d =
n — I
5. Given a, d, and Z, to find w, the number of terms.
Clearing of fractions and reducing (4),
I — a .
n = , + I.
a
1. Given a = 25, e? = 3, and w = 12, to find I.
2. Given a = 58, d = 5, and ?i = 45, to find Z.
3. Given d= 3, ^ = 35, and nz= 9, to find a.
4. Given 1 = ^j, d= 5, and w = 21, to find a.
5. Given a = 15, 1= 85, and n = 31, to find d.
6. Given a = 28, Z = 7, and w = 26, to find d.
7. Given a = 2^, d = 5, and Z = 5138, to find Wo
8. Given a= 6, e?= 6, and ?=ii52, to find w.
* For Formula 2, see Art. 389.
210 ABITHMETIGAL PBOGBESSIOH.
Bt Formula IL
392. In tbifl formula there are four different quantities:
the first terniy the last term^ the number of terms, and the
sum of the terms. If any three of these quantities are
given, the other may be found. (Art. 389.)
2. 8 = — ^^ X n, a, I, and n being given.
2
Note. — ^For Formulas 35, see Article 391.
6. Given I, n, and s, to find a, the first term.
Clearing (2) of fractions, dividing and transposing,
28 .
a = 1.
n
7. Given a, n, and s, to find I, tbe last term.
Transposing in (6), we have
; = a.
n
8. Given a, ly and 8, to find n, the number of terms.
Clearing (7) of fractions, transposing, factoring, and dividing,
28
n = 1.
a + l
X. Given a = g, I = 4i> and n = 7, to find 8.
2. Given a = }, ? = 45, and 7* = 50, to find «.
3. Given Z = 50, rf = 4, and w = 12, to find a.
4. Given a = 9, Z = 41, and « = 150, to find m
5. Given rf = 7, ? = 21, and w = 35, to find a.
6. Given a = 46, Z = 24, and s = 455, to find w.
7. Given a = 27, n = 9, and 5 = 72, to find I
8. Given a = 72, w = 8, and s = 288, to find L
9. Find the sum of the series 3, 5, 7, 9, etc., to 15 terms.
10. Find the twentieth term of 5, 8, 11, 14, 17, etc.
11. If the first term of an ascending series is 5, and the
common difference 4, what is the 15th term ?
ARITHMETICAL PBOGRESSIOK.
211
393. The remaining twelve formulas are derived by
combining the preceding ones in such a manner as to
eliminate the quantity whose value is not sought. They
are contained in the following
TABLE.
No
lO
II
12
14
15
i6
X7
i8
20
Given.
d n, 8
df I, 8
a, I, 8
I, n, 8
tty n, 8
d, fly 8
a, d, 8
a, d, 8
dy ly 8
tty dy n
tty dy I
dy ly n
Rbqitibed.
a
a
d
d
d
I
I
n
8
8
8
FOBMTTLAl.
a =
28 — dv? 4 dn
271
rf = =
25 — / — a
w (» — i)
, 25 — 2an
d = — 5
n^ — n
j_s (ni)rf
n 2
^""' 2d
_ 2l+d ± V{2l +(Qa~8 dg
"" 2d
« =  [2a 4 (w — i) efl
2
1+ a . P — a»
2
2d
n
5 = [2Z(ni)d]
2
Of the twenty formulas in Arithmetical Progression, the first
two are indiapenmble, and should be thoroughly commuted to memory ;
the next six are important in the solution of particular problems. The
remaining twelve are of less consequence, but will be foun^
interesting to the inquisitive student.
212 ARITHMETICAL PEOGEESSION.
394. By the fourth formula in Art. 391, any number of
arithmetical means may be inserted between two glTen
terms of an arithmetical progression. For, the number of
terms consists of the two extremes and all the intermediate
terms.
Let m = the number of means to be inserted.
Then «i+2 = », the whole number of terms.
Substituting m + 2 for n in the fourth formula, we liave
d = . Hence,
The required number of means is found by the continued
addition of the common difference to the successive terms.
1. Find 4 arithmetical means between i and 31.
2. Find 9 arithmetical means between 3 and 48.
PROBLEMS.
1. If the first term of an ascending series is 5, the common
difference 3, and the number of terms 15, what is the last
term?
2. If the first term of a descending series is 27, the
common difference 3, and the number of terms 12, what is
the last term ?
3. If the first term of an ascending series be 7, and the
common difference 5, what will the 20th term be ?
4. Find s arithmetical means between 2 and 60.
5. What is the sum of 100 terms of the series J, f, i, f,
6. If the sum of an arithmetical series is 18750, the least
term 5, and the number of terms 20, what is the common
difference ?
7. Required the sum of the odd numbers i, 3, 5, 7, 9, n,
etc., continued to 76 terms?
8. Required the sum of 100 terms of the series of even
numbers 2, 4, 6, 8, 10, etc.
■1' »
ARITHMETICAL PROGRESSION. 213
9. The extremes of » series are 2 and 47, and the number
of terms is 10. What is the common difference ?
10. Insert 8 means between 6 and 72.
11. Insert 9 means between 12 and 108.
12. The first term of a descending series is 100, the
common difference 5, and the number of terms 15. What
is the sum of the terms ?
Note. — i. In Arithmetical Progression, problems often occur in
which the terms are not directly given, but are impUed in the
conditions. Such problems may be solved by stating the conditions
algebraicaUy, and reducing the equations.
13. Find four numbers in arithmetical progression, whose
4um shall be 48, and the sum of their squares 656.
Let X = the second of the four numbers.
And y = their common difference.
By the conditions, (aJ— y) + x+(x+y) + (x+ 2y) =48 (i)
And (x^yf\ix^\{x+yf+{x+2y',*=z6s6 (2)
Uniting terms in (i), 4a:+2y= 48 (3)
'* "(2), 4^\4xy+6y^ = 6s6 (4)
Transposing and dividing in (3), y = 24 — 2a; (5)
Dividing (4) by 2, 2a^ + 2ajy + 3^ = 328 (6)
Substituting value of y, 2a? + 20^24— 2a?) + 3(24 — 2aj)* = 328
Reducing, aj*— 240;= —140
Completing square, etc., x= 14 or 10
Substituting in (5) y = —4 or 4
Hence the required numbers are 6, 10, 14, and 18.
NeTB. — 2. The first two values of a; and y produce a descending series ;
the other two an ascending series. In both the numbers are the same.
14. Find three numbers in arithmetical progression whose
sum is 15^ and the sum of their cubes is 495.
15. K IOC marbles are placed in a straight line a yard
apart, how far must a person travel to bring them one by
one to. a box a yard from the first marble ?
214 ABITHliEIICAL PBOaBESSIOK.
i6. How many strokes does a common clock strike in
24 hoars?
17. A student bought 25 books^ and gave 10 cents for the
first; 30 cents for the second, 50 cents for the third, etc.
What did he pay for the whole ?
18. A boy puts into his bank a cent the first day of the
year^ 2 cents the second day, 3 cents the third day, and so
on to the end of the year. What sum does he thus lay np
in 365 days ?
19. The clocks of Venice go on to 24 o'clock. Howmany
strokes does one of them strike in a day ?
20. What will be the amount of $1, at 6 per cent simple
interest, in 20 years ?
21. What three numbers are those whose sum is 120, and
the sum of whose squares is 5600 ?
22. A trayeller goes 10 miles a day ; three days after^
another follows him, who goes 4 miles the first day, 5 the
second, 6 the third, and so on. When will he oyertake the
first?
23. Find four numbers, such that the sum of the squares
of the extremes is 4500, and the sum of the squares of the
means is 4100.
24. A sets out from a certain place and goes i mile the
first day, 3 miles the second day, 5 the third, etc. After he
has been gone 3 days, he is followed by B, who goes 1 1 miles
the first day, 12 the second, etc. When will B overtake A?
25. The first term of a decreasing arithmetical progression
is 10, the common difference J, and the number of terms 21.
Required the sum of the series.
26. A debt can be discharged in 60 days by paying $1 the
first day, $4 the second, $7 the third, etc. Eequired the
amount of the debt and of the last payment
eEOMETBIGAL PBOGBE88IOK. 216
GEOMETRICAL PROGRESSION.
395. A Geometrical Progression is a series of
quantities which increase or decrease by a constant multiplier
called the ratio. Hence,
The ratio may be an integer or Vi, fraction.
KOTB, — ^When the ratio \& fractiondlt the series will decrease. For
xnxQtiplying by a fraction is taking a certain part of the multiplicand
as many tmes as there are Uke parts of a unit in the multiplier.
396. In a geometrical series, each succeeding term is
found by multiplying the preceding one by the ratia
Thus, if a is the first term, and r the ratio, the series la
a, a/Ty a/jfiy ar^, ar^^ ar^, ar*, etc.
If the ratio is 3, the series is
a, ax 3, «X3', ax 3', etc.
If the ratio is ^, the series is
a, axj, ax^x^, ax^x^x^, etc
397. An Ascending Series is one which increases
by an integral ratio ; as, 2, 4, 8, 16, 32, etc.
398. A Descending Series is one which decreases
by a fractional ratio; as, 64, 32, 16, 8, etc.
399. When the ratio is a positive quantity, all the terms
of the progression are positive; when it is negative, the
terms are alternately positive and negative.
Thus» if the first term is a, and the ratio —3, the series is
a, —sa, +9^, —27a, +8ia, etc.
395. What is a geometrical progreBsion ? 397. What Is an ascending series f
998. Desctndiiigf 399. What law goyems the signs ?
216 OEOMETBIGAL PBOGBESSIOJS^.
400. In geometrical progression there are five elements,
the first term, the last term, the fiumber of terms, the
common ratio, and the sum of the terms.
Let a = the first term,
I = the last term,
n = the number of terms,
r = the ratio,
8 = the sum of the terms.
The relation of these five quantities to each other is such
that if any three of them are given, the other two can be
found.
CASE I.
401. The First Term, the Number of Terms, and the Ratio
being given, to Find the Last Term.
In this problem, a, n, and r are given, to find I, the last term.
The successive terms of the series are
a, or, aj^, ar^, ar^, etc., to ar^K (Art. 397.)
By inspection, it will be seen that the ratio r consists of a regular
series of powers, and in each term the index of the power is one Iw
than the number of the terms. Therefore, the last or nth term of the
series is ar'^K Hence, we have
PoEMULA I. Z = ar*^^.
Rule. — Multiply the first term by that power of the ratio
whose index is one less than the number of terms.
CoE. — Any term in a series may be found by the preceding
rule ; for the series may be supposed to stop at that term.
1. Given a= 5, w = 6, and r = 2, to find I
2. Given a= 2, ti = 8, and r = 3, to find L
3. Given a = 72, n = 5, and r = ^, to find I
4. Given a= 5, /i = 4, and r = 4, to find I.
5. Given a= 7, w = 5, and r = 2, to find I.
6. Given a = 10, n = 6, and r = — 5, to find 2.
400. Name the eiemeuU In geometrical progression. 401. How And the hwt
term?
GEOMETRICAL PROGBBSSIOIT 211'
CASE II.
402. The First Term, the Last Term, and the Ratio being
given, to Find the Sum of the Terms.
In this problem, a, l^ and r are given, to find s. ^
Since » = the sum of the terms, we liave
<=a+ar+ar*+ar'+ . . .. +ar*« + ar"*, (ij
Multiplying (i) by r,
r« = ar+a/i^+ar^^ar^^ .... +<w*'+ai*. (2)
Subtracting (i) from (2), ra—s = ar^—a, (3)
Factoring and dividing, a = — "^—^ (4)
T — I
In equation (4), or" is the last term of (2), and is therefore th^
product of the ratio by the last term in the given series.
Substituting Ir for or", we have
Ir — a
Formula II. « =
r— I
BuLE. — Multiply the last term by the ratio, from the
product subtract the first term, and divide the remainder by
the ratio less one*
B^ For the method of finding the sum of an mflnite descending
series, see Art. 435.
1. Given a = 2, / == 500, and r = 3, to find the sum.
_, Ir—a 500 X 3 — 2 .
Solution, s = = = 749, Ans,
r— I 2 '^^*
2. Given a = 3, Z = 9375, and r = 5, to find s.
3. Given a = 9, Z = 9000, and r = 10, to find s.
4. Given a = 5, Z = 20480, and r = 4, to find s,
5. Given = 15, Z = 3240, and r = 6, to find s.
6. Given a = 25, Z = 6400, and r = 4, to find s,
40a. How find the Bum of the terms f
ID
218 GBOKETBICAL PBOGBESSIOK.
403. The two preceding formulas famish the means for
solving all problems in geometrical progression. They may
be varied so as to form eighteen other formulas.
By Fobmula L
404. The first formula contains /owr different quantities:
the first term, the last term, the ratio, and the number of
terms. If any three of these quantities are given, the other
may be found. By the first formula,
I. J = ar*^^ ; a, n, and r being given. (Art. 401.)
For formula 2, see Article 402.
3. Given Z, n, and r, to find a, the first term.
Factoring (i), and dividing by r»~',
I
a =
•1
4« Given a, Z, and n, to find r, the ratio.
Dividing (i) by a, and extracting the root denoted \xy tlie index,
5. Given a, Z, and r, to find n, the number of terms.
Dividing (i) by a, r"> =  •
By logarithms, log r (»— i) = log Z — log a
Dividing, etc., n = ^^g^""^^^ + x.
logi*
Note. — Since this formula contains logarithms, it may b« defemd
till that subject is explained.
1. Given a = 3, » = 5, and r = 10, to find Z.
2. Given a = 5, w = 6, and r = 5, to find Z.
3. Given I == 256, w = 8, and r = 2, to find a.
4. Given I = 243, w = 5, and r = 3, to find a.
5. Given a = 2, Z = 2592, and ^ = 5, to find n
6. Given a = 4^ Z =: 2500^ and n = 5, to find r.
GBOKETEIGAL PB06BBS)3I0K. 319
By Foemula II.
405. This formula contains four different quantities: the
first term, the last term, the ratio, and *the sum of the
terms. If any three of them are given, the other may be
found. By the second formula,
I'M* /» *
2. « = , tty I, and r being given. (Art. 402.)
For formulas 35, see Article 404.
6. Given ?, r, and s, to find a, the first term.
Clearing (2) of fractions, etc.,
a = ir — » (r — i)
7. Oiven a, r, and s, to find Z, the last term.
Transposing in (6),
Ir = a + «(r — i).
Dividing by r,
"" r
8. Given a, Z, and ^, to find r, the ratia
Clearing (2) of fractions,
8r — .9 = 2r — a.
Transposing in tlie last equation*
«r — Ir = « — a.
Factoring, etc,
1. Given a = 2, I = 162, and r = 3, to find 5.
2. Given Z = 54, r = 3, and s = 80, to find a. .
3. Given a = 4, r = 5, and s ^ 624, to find L
4. Given a = 4, Z = 12500, and s = 15624, to find 9\
5. Given a = 5, Z = 180, and r = 6, to find s.
6. Given a = 7, r = 3, and s = 847, to find Z.
220
GEOMETRICAL PROaRESSIOlT.
406. The remaining twelve formulaa are derived by
combining the preceding ones in such a manner as to
eliminate the quantity whose value 'is not sought
TABLE.
No.
lO.
II.
12.
13
14.
15
16.
17.
x8.
19.
20.
Gimr.
n, r, 8
I, Uy 8
Oy fly 8
n, r, 8
a, I, 8
a, r, 8
h r, 8
a, n, 8
I, n, 8
a, fly r
I, fly r
tty ly fl
Rbqttibbd.
a
a
I
I
fl
fl
n
T
8
8
8
FOBKULAB.
(r— i)s
« = ^^i '
r" — I
?(^ — ?)«i=:a(^ — a)»^
• = ^ •
r"— I
log I — logg
log(5— a)— log(50
^ _ log[g + (ri)g]logfl
log r
_ logZlog[/r(ri)^] ^^
logr
» =
+1
fl
. 8 8
r» r = I «
a ' a
r» +
8
^rft— 1 — .
1S' "1^8
__ a(r" — 1)
~ r — I '
Ir^ — l
8 =
5 =
8 =
^l — ''^/a
Of the twenty formulas in Geometrical Progression, the fri
two are fwndamental, and should be thoroughly commuted to memory ;
the next six are important in the solution of particular problems. Tbe
remainder are less practical.
6E0METEIGAL PK06BESSI0K. 221
407. By the fourth formnla (Art 404), any number of
geometrical means may be found between two giren
Quantities. '
Let m = the number of means required.
Then m+2=zn,
Substitutmg m+ 2 for n in the formula, we have
{^
The ratio being found, the means required are obtained by continued
multiplication.
1. Find two geometrical means between 3 and 192,
Solution, r = Q^ = /^ = ^6i = 4.
The ratio being 4, the first mean is 3 x 4 = 12 ; the second is
12x4 = 48.
2. Find three geometrical means between ^ and 128.
PROBLEMS.
1. In a geometrical progression, the first term is 6, the
last term 2916, and the ratio 3. What is the sum of all the
terms?
2. In a decreasing geometrical series, the first term is ^,
the ratio i, and the number of terms 8. What is the sum
of the series ?
3. What is the sum of the series i, 3, 9, 27, etc., to 15
terms ?
4. Find the sum of 12 tenns of the series, i, , f, ^V» ®^
5. If the first term of a series is 2, the ratio 3 and the
number of terms 15, what is the last term ?
6. What is the i6th term of a series, the first term of
which is 3, and the ratio 3 ?
Note. — When the terms of the series are not stated directly, they
may be represented algebraically.
222 OXOMETBICAL PROGRESSIOIir.
7. Find three nnmbere in geometrical progreBsion^ sacb
that .their sum shall be 2 1, and the sum of their squares 189.
Let the three noinben be x, ^/xy, and y.
By the oonditioiis, iF+ y^+y = 21 (i)
And a!*+ay+y* = i89 (2)
Transposing and eq. (i), a? + 2ay+y* = 44i~42y^+a:y (3)
Subtracting (2) from (aX ay = 252— 42\^+ay (4)
Tranapoeing, etc., V^ = 6 (5)
Involving, ay = 36 (6)
And say = 108 (7)
Subtracting (7) from (2), a^— 2ay+y* = 81
Extracting root, x—y = 9 (g)
Substituting (5) in (i), a?+y= 15 (9)
Combining (8) and (9), a; = 12
y= 3
Hence the numbers, 12, 6, and 3, Am.
B. A father gives his daughter $1 on New Tear's day
tG^ards her portion, and doubles it on the first day of every
month through the year. What is her portion ?
9. A dairyman bought 10 cows, on the condition that he
should pay i cent for the first, 3 for the second, 9 for the
third, and so on to the last. Wbat did he pay for the last
cow and for the ten cows?
10. A man buys an umbrella, giving i cent for the first
brace, 2 cents for the second brace, 4 for the third, and so
on, there being 10 braces. What is the cost of the
umbrella?
11. The sum of three numbers in geometrical progression
is 26, and the sum of their squares 364. Find the numbers.
12. What would be the price of a horse, if he were to be
sold for the 32 nails in his shoes, paying i mill for the first,
2 mills for the second, 4 for the third, and so on ?
13. Fiud four numbers in geometrical progression, snch
that the sum of the first three is 130, and that of the last
three is 390.
GEOMETRICAL PROGRESSION. 223
14. A man divides $210 in geometrical progression among
three persons; the first had $90 more than the last How
much did each receive ?
15. There are five numbers in geometrical progression.
The sum of the first four is 468, and that of the last four is
2340. What are the numbers?
16. The sum of $700 is divided among 4 persons, whose
shares are in geometrical progression; and the difference
between the extremes is to the difference between the means
as 37 to 12. What are the respective shares?
17. The population of a town increases annually in
geometrical progression, rising in four years from loooo to
14641. What is the ratio of annual increase ?
18. The sum of four numbers in geometrical progression
Is 15, and the sum of their squares 85. What are the
numbers ?
HARMONICAL PROGRESSION.*
408. An Harmonical JProgression is such, that
of any three consecutive terms, the first is to the third as the
difference of the first and second is to the difference of the
second and third.
Thus, 10, 12, 15, 20, 30, 60,
are in harmonic progression ; for
10 : 15 : : 12—10 : 15—12
12 : 20 :: 15—12 : 20—15
15 : 30 : : 20—15 : 30—20
20 : 60 : : 30—20 : 60—30
Let a, b, c, d, 6,f, ^, be an harmonical progression, then
a : c : a—h : h—c, etc.
Note.— When three quantities are such, that the first is to the
third as the difference of the first and second is to the difference of the
second and third, they are said to be in Harmoniccd Proportion.
Thus, 2, 3, and 6, are in harmonical proportion.
408. What is an harmonica] pro^eBsion ?
* If a musical string be divided in harmonical proportion, the
different parts will vibrate in harmony. Hence, the name.
224 HARMOKIOAL PBOGBESSION.
409. To Find the Third Term of an Harmonical
Progression, the First Two being given.
Let a and b be the first two terms, and x the third term.
Then a : x :: a—b : 6— »
Multiplying extremes, etc., ab—ax = ax—bx
Transposing, etc., 2ax—bx = ab
Factoring, and dividing by 2a— b, we have the
ah
FOBMdLA. fiC =
2a — ft
BuLE. — Divide the product of {he first two terms by tvrics
the first mintis the second term ; the quotient will be tht
third term.
Note. — ^This role famishes the means for extending an harmomc
progression, by adding one term at a time to the two preceding terms.
1. Find the third term in the harmonic series of which
12 and 8 are the first two terms. Ans. 6.
2. Find the third term in the harmonic series of which
12 and i8 are the first two terms. Ans. 36.
3. If the first two terms of an harmonic progression are
15 and 20, what is the third term ? Ans. 30.
4. Continue the series 12, 15, 20, for two terms.
Ans. 30 and 60.
5. Continue the series 7^^ g, 12, for two te^rms.
Ans. 18 and ^6.
410. To Find a Mean or Middle Term between Two Terms
of an Harmonic Progression.
Let a and e be the first and third of three consecutive terms of an
harmonic progression, and m the mean.
Then a : c :: a— m . m— c
Mult, extremes and means, am—ac = ae^cm
Transposing and uniting, am+cm = 2ae
Factoring and dividing by a+c, we have the
200
Formula. m
af e
409. How find the third term of an harmonical progreeeion, the first two being
given ^
HARMONICAL PBOGRESSXOK. 226
Rule. — Divide twice the product of the first and third
terms by their sum; the quotient will be the mean or middle
term.
6. The first and third of three consecutive terms of an
harmonic progression are 9 and 18. Bequired the mean or
middle term.
Solution. 2x9x18 = 324, and 9 + 18 = 27.
Now 324127 =12, Ans.
7. Find an harmonic mean between 1 2 and 20. Ans, 15.
8. Find an harmonic mean between 15 and 30. Ans. 20.
411. The Reciprocals of the terms of an harmonic
progression form an arithmetical progression.
Thus, the redprocalfl of 10, i^, 15, 20, etc., viz.,
i^> A* A> Vir» A» ®^"
are an arithmetical progression, whose common difference is ^^j.
Again, let a, &, <; be in harmonic progression.
Then a : c : : a—h : b^e
Mnlt. extremes and means, db—ac = ae—bc
Dividing by oftc, c~"6~6""a ' ^^^' ^^'^
Conversely, the reciprocals of an arithmetical progression
form an harmonic progression. Thus,
The reciprocals of the arithmetical progression i, 2, 3, 4, 5, etc.,
^'^y h h h i* h ®^*> ^i^ ^ harmonic progression.
412. If the lengths of six musical strings of equal weight
and tension, are in the ratio of the numbers
i» h h h h h etc.,
tlie second will sound an octave above the first; the third
will sound the twelfth j the fourth the double octave, etc.
4xo> How And a mean between two terms of an harmonic progression?
41 X. What do the reciprocals of an harmonica! progression formf
226 INFIKITE 8£BI£8.
INFINITE SERIES.
413. An Infinite Series is one in which the successive
terms are fonned by some regular law^ and the number of
terms is unlimited.
414. A Converging Series is one the sum of whose
terms, however great the number, cannot numerically
exceed a finite quantity.
415. A Diverging Series is one the sum of whose
terms is numerically greater than bjij finite quantity.
416. To Expand a Fraction into an Infinite Series.
Remabk. — Any common fraction whose exact value cannot be
expressed by decimals, may be expanded into an infinite series.
1. Expand the fraction  into au infinite series.
Solution, if3 = .333333, and so on» to infinity.
Or, n3 = A+dhr + Ti/W + TTr*Tn7. etc. Hence, the
Rule. — Divide the numerator by the denominator.
2. Eeduce to an infinite series.
1 —X
I— SB )i (i+aj+oJ^+ic'+aj*, etc, the quotient. (Art. 17a)
I— a?
+a^
+a!'— a?*
+aj*, etc.
I
Therefore, = i+aj+a?+a^+a5*+aj", etc., to infinity.
413. What is an infinite Beriesf 4x4. A conveiging Beriesf 4x5. DivergfogT
416. How expand a fraction into an infinite series r
INFINITE SERIES. 227
Let x = i; then will = 7 = 2 ; and the series will be
" I — aj I — i
' + i + J + i + TV + A» ®*c., the sum of which = 2.
U x = i, then will = r = ft, and the series will beoome
■ I — a; I — J *
i+i+t+^+«^+iiT> etc. = f.
Notes. — i. If x is less than i, the series will be con'oergent,
¥gt, when x is less than i, the remainder must continually decrease ;
therefore, the further the division is carried, the less will be the
qoantitjto be added to the last term of the quotient in order to
express the exact value of the fraction.
2. If X is greaJter than i, the series will be dwergent.
For, when x is greater than i, the remainder must constantly
increase ; therefore, the/ar^Aer the division is carried, the greater will
be the quantity either positive or negative to be added to the quotient.
3. Bednce the fraction to an infinite series.
1 + X
Solution, t i(i + aj) = i — a? + a^— «» + a?*— aj" + , etc.
This series is the same as that in Ex. 2, except the odd powers
of X are negative.
Let a; = ^ ; then will =  ; which is equal to the series
I "4 Qj
4. Beduce the fraction to an infinite series.
I —a?
Ans. 1 + 2X + 20? + 20^+ 2rc*, etc.
417. A fraction whose denominator has more than two
tenns, may also be expanded into an infinite series.
5. Expand ^ into an infinite series.
I— aj+a?) I (i+aj— a^— a:*+a^, etc., J.n&
I— ay+a:*
aj— aJ*
a;— a:* fa!*
— a^, etc.
228 INFINITE SERIES.
418. To Expand a Compound Surd into an Infinite Series,
6. Bedaoe Vi ^x to an infinite series.
OPKBATIOH
X
I
2 + 
2
+ «
a*
+ « + —
4
a + *g
0? a*
2 + OJ + T
4 i6
4
a^ aj* flj*
""T""'8"'*"64
ic' a?*
+ ^ — 2~ * 6*<5. Hence, the
8 64 *
BULE. — Extract the square root of the given mird.
(Art 298.)
7. Expand V^i:?. ^.«..a:^g^,etc
8. Expand \/2, or Vi + i. 4/w. i + i — i h jV^ etc
419. The Binomial Theorem applied to the Formation of
Infinite Series.
The Binomial Theorem may often be employed with
advantage, in finding the roots of binomials. For a root is
expressed like a power, except the exponent of one is an
integer, and that of the other is effraction.
9. Expand {x + y)^ into an infinite series.
Solution. — ^The terms without coe£3cients are
^> «~*y» <x^^^> <xr^t^, (C^P*, etc.
The coefficient of the second term is + J ; of the 3d, = — } *,
of the 4th term, — 2 ■ = + ^, etc
The series is a?* + ix~^y — JaJ^'y' + iVa;~*y', etc.
fci ■ ■■■ — — ■ ■■■ I » — _ 
418. How ezpaod a surd into an infinite seriet f
INFINITE SERIES. 229
420 When the index of the required power of a binomial
is a positive integer, the series will terminate. For, the
index of the leading quantity continually decreases by i ;
and soon becomes o ; then the series must stop. (Art. 269.)
421. When the index of the required power is negative^
the series will never terminate. For, by the successiye
subtractions of a unit from the index, it will never become
o ; and the series may be continued indefinitely.
10. Expand (s^+y)^ into an infinite series, keeping the
fiswjtors of the coefficients distinct
11. Expand V^* or (ifi)^, keeping the factors of the
coefficients distinct
Ans. i+i ^+—^7 ^^ + ^ 'i 'J , etc.
2 2.4 2.4.0 2.4.0.0 2.4.0.0.IO
422. An Infinite Series must not be confounded with an
Infinite Quantity.
423. An Infinite Quantity is a quantity so great
that nothing can be added to it
424. An Infinite Series is a series in which the
number of terms is unlimited,
425. The magnitude of the former admits of no increase;
while in the latter the number of terms admits of no
increase, and yet the sum of all the terms may be a small
quantity.
Thus, if the series i + }+}+iV+ih> ^^m ^ which each succeeding
term is half the preceding, is continaed to infinity y the sum of all the
terms cannot exceed a unit,
426. When one quantity continual^ approximates
another without reaching it, the latter is called the Idniit
of the former.
830 INFINITE SEBIES.
427. An Infinitesimal is a quantity whose Talae is
less than any assignable quantity.
428. The Sign of Infinity^ or of an infinite
quantity, is a character resembling an horizontal figure
eight ( 00 ).
The Sign of an Infinitesimal is zero ( o ).
429. One infinite series may be greater or Jess than
another.
Thus, the series i+i+^+i+i^, etc, whose limit is 2, is greatei
than the series i+i+^ + ^ + i^, etc., whose limit is i.
430. Since an infinitesimal is less than any assignable
quantity, and in its limit approaches zero, when connected
with finite quantities by the sign + or — , it is of so little
value that it may be rejected without any appreciable error.
431. An infinite series may be multiplied by a finite
quantity.
Thus, if the series 222222, etc., is multiplied by 3,
the product 666666, etc., is three times the multiplicand.
432. An infinite series may also be divided by a finite
quantity.
Thus, if the series 888888, etc., is divided by 2,
the quotient 444444, etc., is half the divideiuL
433. If a finite quantity is multiplied by an infiniiesimaly
the product will be an infinitesimal. For, with a given
multiplicand, the less the multiplier, the less will be the
product. Thus, a; x o = o.
434. If a fi7iite quantity is divided by an infinitesimal,
the quotient will be infinite. Thus, a; ^ o = 00 .
If a finite quantity is divided by an infinite quantity, the
quotient will be an infinitesimal. Thus, a; f 00 = o.
If an infinitesimal is divided by a finite quantity, the
quotient is an infinitesimal. Thus, o = a; = o.
Note.— In higher mathematics, the expression q^q a4iuits of
various interpretations.
IKFIJI^ITE SEBIES. 231
435. To Find the Sum of a Converging infinite Series, the
First Term and Ratio being given.
By tlie second formula in geometrical progression, we have for an
increasing series (Art. 402),
(r — o or" — a
s = , or •
r — I r — I
In a decreasing series, the ratio r is less tlian i ; therefore, / or ar^ 1
is less than a. (Art. 398.)
Iv—^a aT'*^a
That both terms of the fraction , or , may be positive,
we change the signs of both (Art. 166), and
a — Ir
s = •
I — r
Bnt, in a decreasing infinite series, I becomes an infinitesimal, or o;
therefore, Ir = o. (Art. 427.) Hence, rejecting the infinitesimal from
• = , we have the
I — r '
Formula. s = •
I — r
Bulb. — Divide the first term ly i minus the ratio.
1. Find the sum of the infinite series
X + i + J + T?T + ifr, eta
2. Eequired the sum of the infinite series
I — J + i — i+, etc.
3. Find the sum of the series i + J + 1 +, etc
4. Find the sum of the infinite series i + i + t+> eta
5. Find the sum of the series 1 + ^ + A +> etc.
6. Find the sum of the series 3 + 2 +  +> eta
7. Find the sum of the series 4 + V" + M +> eta
8. Find the sum of the series .3333, etc.
9. Find the sum of the series .66666, etc.
 +  + 
II. Suppose a ball to be put in motion by a force which
impels it 10 rods the first second, 8 rods the next, and so on,
decreasing by a ratio of ^ each second to infinity. Through
wliat space would it move ?
10. Find the sum of the series  H — 5 H — = + eta
OHAPTEE XIX.
LOGARITHMS*
436. The Logarithm of a number is the exponent of
the power to wfiich a given fixed number must be raised to
produce that number.
437. This Moced Number is called the Sase of the
system.
Thua, if 3 ifi the base, then 2 is the logarithm of 9, because 3' = 9;
and 3 is the logarithm of 27, because 3' = 27, and so on.
Again, if 4 is the base, then 2 is the logarithm of 16, because
4' = 16 ; and 3 is the logarithm of 64, because 4' = 64, and so on.
438. In forming a system of logarithms^ any number,
except I, may be taken as the base, and when the base is
selected, all other numbers are considered as some power or
root of this base. Hence, there may be an unlimited
number of systems.
Note. — Since all powers and roais of i are i, it is obvious that other
numbers cannot be represented \xj its powers or roots. (Art. 289.)
439. There are two systems of logarithms in use, the
Napierian system,! the base of which is 2.718281828, and
the Common System, whose base is 10. J
The abbreviation log. stands for the term logarithm.
436. What are Ibgarithms ? 437. What ie this fixed nnmber called? 439. Name
the eyBtems in use. The base of each.
* The term logarithm is derived from two Greek words, meaning
the relation of numb^a,
f So called from Baron Napier, of Scotland, who invented log
arithms in 1614,
X The common sjstem was invented by Heniy Briggs, an English
mathematician, in 1624.
LOGABITHMS. 233
440. The Sase of common logarithms being lo, all
other numbers are considered as powers or roots of lo.
Thus, the log. of i is o ; for lo^ equals z (Art. 259) ;
** " 10 is I ; for 10* " 10 ;
'* " 100 is 2 ; for 10* " 100 ;
" *' 1000 is 3 ; fop lo* " 1000, etc. Henoe»
The logarithm of any number between i and 10 is a
fraction; for any number between lo and loo, the logarithm
is I plus a fraction ; and for any number between 100 and
1000, the logarithm is 2 plus a fraction, and so on.
441. By means of negative exponents, this principle may
be applied to fractions.
Thas (Art. 256), the log. of .1 is —i ; for 10^' equals .1 ;
'* ** .01 is —2 ; for 10' " .01 ;
** " .001 is —3 ; for lo"* *' .001.
Therefore, the logarithms for all numbers between i and
0.1 lie between o and — i, and are respectively equal to — i
plus a fraction; for any number between o.i and 0.0 1, the
logarithm is —2 plus a fraction ; and for any number
between 9.01 and 0.00 1, the logarithm is — 3 plus a fraction,
and so on.
Hence, the logarithms of all numbers greater than lo or
less than i, and not exact powers of 10, are composed of
two parts, an integer and 3,fractio7i.
Thus, the logarithm of 28 is 1.44716 ;
and of .28 is 1 44716.
442. The integral part of a logarithm is called the
Characteristic; the decimal part, the Mantissa.
443. The Characteristic of th^ logarithm of a whole
number is one less than the number of integral figures in
the given number.
Thus, the characteristic of the logarithm of 49 is i ; that of 495 is
2 ; that of 4956 is 3 ; that of 6256.414 is also 3, etc.
440. What i» the logarithm of any number from i to lo? From lo to loo? Prom
TOO to 1000? 442. What Ib the intepn^l part of a logarithm caDed? The decimal
part f 443. What is the characteristic of the logarithm of a whole namber ?
234 LOOARITHHS.
444. The Characteristic of the logarithm of a
decimal is negative^ and is one greater than the number of
ciphers before the first significant figure of the fraction*
Thus, the characteristic of the logarithm of if)^ or .i is — i ; that of
jJtj, or .01, is —2 ; that of ^^^, or .001, is —3, etc. (Art. 256.)
The logarithm of .2 is ~i with a decimal added to it ; that of .05
8 —2 with a decimal added to it, etc.
Note. — ^It should be chsenred that the eharacteristie only is negative,
while the mantissa, or decimal part, is always positive. To indicate
this, the sign — is placed over the characteristic, instead of before it.
Thus, the logarithm of .2 is 1.30103,
'• '* •* .05 is 2.69897, eta
445. The Decimal Part of the logarithm of any
nnmber is the same as the logarithm of the number
muUiplied or divided by 10, 100, 1000, etc.
Thus, the logarithm of 1876 is 3.27325 ; of 18760 is 4.27325, etc.
TABLES OF LOGARITHMS.
446. A Table of Logarithms is one which contains
the logarithms of all numbers between given limits.
447. The Table found on the following pages gives the
mantissas of common logarithms to five decimal places for
all numbers from i to 1000, inclusive.
The characteristics are omitted^ and must be supplied by
inspection. (Arts. 443, 444.)
Notes. — i. The first decimal figure in column is often the same
for several successive numbers, but is printed only once, and is
understood to helong to each of the blank: places below it.
2. The character ( ♦ ) shows that the figure belonging to the place
it occupies has changed from 9 to o, and through the rest of this line
the first figure of the mantissa stands in the next line below.
444. What is the characteristic of the logarithm of a decimal ? 445. What is the
effect upon the decimal part of tha \o^. of a number, if the number is multiplied or
divided by xo, xoOf xooo, etc. 446. What 1b a table of logarithm? ¥
LOGARITHMS. 235
448. To Find the Logarithm of any Number from I to 10.
BuLE. — Look for the given number in the first line of the
table; its logarithm mil be found directly below it.
1. Find the logarithm of 7. Ana, 0.8451a
2. Find the logarithm of 9. Ans. 0.95424.
449. To Find tlie Logarithm of any Number fk*om 10
to 1000, inclusive.
Rule. — Look in the column marked K for the first two
figures of the given number, and for the third at the head
of one of the other columns.
Under this third figure, and opposite the first two, will
be found the last decimal figures of the logarithm. The first
one is found in the column marked 0.
To this decimal prefix the proper characteristic (Art. 443.)
Note. — If the number contains 4 or more figures, multiply the
tabular difference by the remaining figures, and rejecting from the
right of the product as many figures as you multiply by, add the rest
to the log. of the first 3 figures.
3. Find the logarithm of 108. Ans. 2.03342.
4. Find the logarithm of 176. Ans. 2.24551.
5.^Find the logarithm of 1999. Ans. 3.30085.
450. To Find the Logarithm of a I>ecimal Fraction.
'RxTLE^'—Take out the logarithm of a whole number consist
i^ of the same figures, and prefix to it the proper negative
characteristic. (Art. 444.)
Note.— If the number consists of an integer and a decimal, find the
logarithm in the same manner as if oA the figures were integers, and
prefix the characteristic which belongs to the integral part. (Art. 443.)
6. What is the log. of 0.95 ? Ans. 1.97772.
7. What is the log. of 0.0125 ? Ans. 2.09691.
8. What is the log. of 0.0075? ^^^' 3«27So6
9. What is the log. of 16.45 ? ^^^' 1*2 16 16.
10. What is the log. of 185.3 ? Ans. 2.26787.
448. How find the logarithm of a nnmber fW>m i to zo? 449. From xo to xooo ?
450. How And the log. of a decimal? Note. Of an integer and a decimal f
236 LOGARITHMS.
451. To Find the Number belonging to a given Logarithm.
Bulb. — Look for the decimal figures of the given logarithm
in the toMe under the column marked ; and if all of them
are 7iot found in that column, look in the other columns on
the right till you find them exactly, or very nearly ; directly
opposite, in the column marked JV", will he found the first
iwo figures, and at the top, over the logarithm, the third
figure of the given number.
Make this number correspond to the characteristic of the
given logarithm, by pointing off decimals, or by adding
ciphers, if necessary, and it will be the mimber required.
Note. — If the characteristic of a logarithm is negative, the nnmber
belonging to it is a. fraction, and as many ciphers most be prefixed to
the number found in the table, as there are units in the characteristic
less I. (Art. 444.)
452. When the Decimal Part of the given Logarithm is
not exactly; or very nearly, found in the Table.
Rule. — From the given logarithm subtract the next less
^logarithm found in the tables ; annex ciphers to the remain
der, and divide it by the tabular difference {marked D)
as far as necessary.
To the number belonging to the less logarithm anneoi the
quotient, and make the number thus produced correspond to
the characteristic of the given logarithm, as above.
Note. — ^For every cipher annexed to the remainder, either a siff
nificant figure or a dphtr must be put in the quotient.
11. What number belongs to 2. 1 7231 ? Ans. 148.7.
12. What number belongs to 1.25261 ? Ant. 17.89.
13. What number belongs to 3.27715 ? Ans. 1893.
14. What number belongs to 2.30963 ? Ans. 204.
15. What number belongs to 4.29797 ? Ans. 19858.29.
16. What number belongs to 1.14488 ? Ans. 0.1396.
17. What number belongs to 2.29136 ? Ans. 0.01956.
18. What number belongs to 3.30928 ? Ans. 0.002038.
■* ■■■■■■ ■ I ■ ■  ■ ^^^ _^
451. How find the iinml)er belonging to a logariUun ?
LOGABITHMS. 23?
453. Gompntations by logarithms are based upon the
following principles :
1°. The sum of the hgarithms of two numbers is equal to
the logarithm of their product.
Let a and c denote any two numbers, m and n their logarithmsy
and 6 the base.
Then If = a
And J* = c
Multiplying, 6*+» = etc,
2°. The logarithm of the dividend diminished by the
logarithm of the divisor is equal to the logarithm of the
quotient of the two nutribers.
Let a and c denote any two numbers, m and n their logarithms,
and h the base.
Then 6» = a
And h* = c
Dividing, 6*» = a^e,
454. To MtUiiply by Logarithms.
a
!• Eequired the product of 35 by 23.
SOLijnoN.— The log. of 35 = 1.54407
M «
'• 23 = 1. 36173
Adding, 2.90580. (Art 453, Prin. z.)
The number belonging is 805, Ana, Hence, the
Rule — Add the logarithms of the factors; the sum will
be the logarithm of the product.
Notes. — i. If the sum of the decimal parts exceeds 9, add the tens
figure to the characteristic.
2. If either or all the characteristics are negative, they must be
added according to Art. 65. But as the mantissa is always positive,
that which is carried from the mantissa to the characteristic must be
considered positive.
2. What is the product of 109.3 by 14.17 ?
3. What is the product of 1.465 by 1.347 ?
4. What is the product of .074 by 1500 ?
453. Upon what two principles are computations by logarithms based ? 454. How
maltiply by logarithms ?
238 LOGAEITHHS.
455. To I}ivide by Logarithms.
5. Bequired the quotient of 120 by 15.
Solution.— The log. of 120 = 2.07918
•« " " 15 = 1. 17609
«  ** quotient = 0.90309. An8.%. Hence^the
Rule.— JVom the logarithm of the dividend subtract tk
logarithm of the divisor ; the difference will be the logarithm
of the quotient. (Art. 453, Prin. 2.)
Notes. — i. When either of the characteristics is negative, or when
the lower one is greater than the one above it, change the sign of the
subtrahend, and proceed as in addition.
2. When I is carried from the mantissa to the characteristic, it
must be considered positive^ and be added to the chaiacteiistic before
the sign is changed.
6. What is the quotient of 12.48 by 0.16 ?
7. What is the quotient of .045 by 1.20?
8. What is the quotient 6f 1,381 by .096 ?
456. Negative quantities are divided in the same manner
08 positive quantities.
If the sign of the divisor is the same as that of the
dividend, prefix the sign + to the quotient; but if different,
prefix the sign — .
9. Divide —128 by —47.
10. Divide —186 by —0.064.
11. Divide —0.156 by —0.86.
12. Divide —0.194 by 0.042.
457. To Involve a Number by Logarithms.
MuUiplieati&n, by logarithms is performed by addition. (Art. 453.)
Therefore, if the logarithm of a quantity is added to itself once, the
result will be the logarithm of the second power of that quantity ; if
added to itself twice, the result will be the third power of that
quantity, and so on. Hence, the
EuLE. — Multiply the logarithm of the number by the
exponent of the required poiver.
455. How divide by them 1 457 How involve a niunber by logarithmB?
LOQABITHMS. 23^
NoTE& — I. This rule depends upon the principle that logarithms
are the exponents of powers and roots, and a power or root is involved
hj multiplying its index into the index of the power required.
2. In this rule, whatever is carried from the mantissa to the
characteristic is poeUize, whether the index itself is positive or negative.
13. What is the cube of 1.246.
Solution. — The log. of 1.246 is 0.09551
Index of the required power is 3
Log. of power is 0.28053. •^^^ i<93435
14. What is the fourth power of .135 ?
15. What is the tenth power of 1.42 ?
16. What is the twentyfifth power of 1.234?
458. To Extract the Root of a Number by Logarithmt.
A quantity is resolved into any number of equal factors, by dividing
Hs index into as many equal parts. (Art. 281.) Hence, the
Rule. — Divide the logarithm of the number by the index
of the required root.
Note. — ^Thls rule depends upon the principle that the root of a
quantity is found by dividing the exponent by the number expressing
the required root. (Art. 296.)
17. What is the square root of 1.69?
Solution.— The log. of 1.69 is 0.22789
The index is 2, 2 ) .2278 9
Logaritim of root, 0.11394. Ant. 13.
18. What is the cube root of 143.2 ?
19. What is the sixth root of 1.62 ?
20. What is the eighth root of 1549 ?
21. What is the tenth root of 1876 ?
459. If the characteristic of the logarithm is fiegative,
and cannot be divided by the index of the required root
without a remainder, make it positive by adding to the
characteristic such a negative number as will make it
exactly divisible by the divisor, and prefix an equal positive
number to the decimal part of the logarithm.
_._ —  — — ■
458. How extract tbo root f
240 LOGARITHMS.
22. It is required to find the cube root of .0164.
Solution. — The log. of .0164 is 2.21484.
Preparing the log., 3)3 + 1.214 84
1.40494. Ans, 0.25406+.
23. What is the sixth root of .001624 ?
24. What is the seventh root of .01449 ?
25. What is the eighth root of .0001236 ?
460. To Calculate Cktinpimnd Interest by Logarlthmt.
BuLE. — Find the amount of 1 dollar for 1 year; multiply
its logarithm by the number ofyearsy and to the product add
the logarithm of the principal. The sum will be the logarithm
of the amount for the given time.
From the amount subtract the principal, and the remain'
der will be the interest.
Notes. — i. If the interest becomes due half yearly or quarterly ,
find the amount of one dollar for the half year or quarter, and multiply
the logarithm by the number of half years or quarters in the time.
2. This rule is based upon the principle that the several amounts in
compound interest form a geometrical series, of which the principal is
the first term, the amount of $1 for i year the ratio, and the number
of years + i the number of terms
26. What is the amount of I1565 for 40 years, at 6 per
cent compound interest ?
Solution. — The amt. of $1 for i year is $1.06 ; its log., 0.02531
The number of years, 40
Product, 1.0124c
The given principal, $1565 ; its log., 319453
Ans. $16103.78. 4.20693
27. What is the amount of $1500, at 7 per cent compound
interest, for 4 years ? Ans. I1966.05.
28. What is the amount of $370, at 5 per cent compound
interest for 33 years ? ^ W5. Ii 85 1. 2 7 + .
460. How calculate compound interest by logarithms ?
lOaAKIIHMS.
241
TABLE OF COMMON LOGARITHMS.
N.
1
.00000
2
.3oio3
8
4
5
6
.77815
7
8
9
D.
47712
1284
.60206
.69897
2119
.84510
.90309
.95424
3743
7*55
10
.00000
0432
0860
1703
5691
253 1
29,38
3342
4f6
II
4130
791 B
4532
4922
53o8
6070
6446
6619
7188
379
13
8279
8636
2??5
9342
U
t?27
♦38o
Vll
1059
349
i3
.11394
1727
2057
2711
3354
3672
6732
9590
3988
43o2
322
14
4613
4922
7698
5229
5534
5836
t)i37
9o3j
6435
7026
7319
3oi
i5
7609
8184
8469.
8752
93i3
9866
♦140
281
i6
.20412
o683
0952
3553
1219
38o5
1484
1748
4304
201 1
2272
253 1
528?
264
\l
3045
33oo
4o55
455 1
4797
7184
5o42
249
235
5527
7875
5768
6007
6245
6482
6711
9003
6951
7416
7646
19
8io3
833o
8556
8780
9226
9447
9667
9885
223
20
.3oio3
0320
o535
0750
2838
0963
1175
1387
3443
1597
1806
201 5
212
21
2222
2428
2634
3o4i
3244
3646
3846
4044
2o3
22
4242
4439
4635
483 1
5o25
5218
541 1
56o3
^m
5o84
7840
I^
23
6173
636i
6549
6736
856i
6922
J 107
7291
7475
24
8021
8202
8382
8739
♦483
2160
8917
9094
9270
9445
9620
178
25
26
9794
r/^i
♦140
i83o
♦3l2
3??6
♦654
2325
♦824
2488
^,
1162
2814
i33o
456o
III
U
3297
3457
m
3933
4001
5637
4248
44o5
1 58
4716
4871
5o25
5119
6687
5485
5788
5939
6090
1 53
29
6240
6389
6538
6835
6982
7129
7276
7422
7567
147
3o
•477»2
7857
8001
8144
8287
96oi
8430
8572
8714
8855
8096
♦379
143
3i
9136
'J.t
9416
9554
9831
9069
l322
♦ !06
♦243
1 38
32
.5o5i5
0786
0920
io55
1188
1455
1 587
1720
1 33
33
i85i
1983
2114
2244
2375
3656
25o5
2634
2763
2892
4i58
3o2o
i3o
34
3148
3275
34o3
352Q
4778
3782
3908
4o33
4263
126
35
4407
453 1
4654
4900
5o23
5145
5267
5388
55oo
123
36
563o
5751
5871
5991
6110
6229
7403
8546
6348
6467
6585
670 J II 9
ll
6820
7978
8093
7054
8206
7I7I
8320
^433
365?
7634
8^71
9879
m
7864
8995
1,6
ii3
39
9107
9218
9329
9439
9550
9660
9770
9q88
♦097
no
4o
.60206
o3i4
0423
o53i
o638
0746
i8o5
o853
0959
1066
1172
]o8
41
1278
i384
1490
1595
1700
1909
2014
2118
2221
io5
42
2325
2428
253 1
2634
2737
2839
2941
3o43
3i44
3246
102
43
3347
434d
3448
3548
3649
3749
4738
3849
3949
4048
4147
5128
4247
5223
100
44
4444
4542
4640
4836
4934
5o3i
98
45
5321
5418
55i4
56io
5706
58oi
5897
5992
6087
6181
95
46
6276
6370
6464
6558
6652
6745
68^9
69J2
7025
7117
93
S
7210
8124
7302
8215
7394
83o5
7486
8395
9285
^4^5
2^
8574
^4
7852
8753
^l
8o34
8o3i
49
9020
9108
1
9197
9373
9401
6
9^8
9636
7
9723
9(10
8d
9.
2
8
4
6
8
9
B.
MSt
tiOOAStlHHS.
H.
1
9984
0842
2
♦070
0927
8
♦167
1012
4
♦243
1096
6
♦329
1 181
6
♦41 5
1265
7
♦5oi
1 349
8
♦586
1433
9
♦672
i5i7
5o
5i
.69807
•70757
1600
52
1684
1767
2591
i85o
1933
2016
2099
2181
2263
2346
53
2428
2010
2673
3480
2754
3D60
2835
2917
2997
3078
3i5q
54
323o
•74036
3320
3400
; 3640
45o8
4586
3878
4663
3957
55
4ii5
4194
4273
5o6i
435i
4420
5200
4741
55ii
56
4819
5588
4896
4974
5128
5282
5358
5435
u
5664
5740
58i5
5891
5967
6042
6118
6193
6268
6343
6418
6492
7232
6567
7300
6641
6716
%V,
6864
6938
7012
59
7085
7159
7379
7452
7597
7670
7743
6o
.77815
8533
6604
t^^
8o32
8104
8176
8247
8958
83i9
8390
8462
6i
8746
8817
1 8888
9029
9099
"4
62
9239
9309
9379
9449
9519
! 9588
9657
9727
9706
♦482
63
9934
.80618
♦oo3
♦072
0754
♦140
♦209
♦277
♦346
♦414
♦55o
64
0686
0821
1 558
0956
1023
lOOC
1757
ii58
1225
65
1291
1954
2608
1 358
1425
1491
2l5l
1624
1^90
1823
1880
254^
66
2020
2086
2217
2866
2282
•2347
24i3
2478
tl
2672
2737
2802
2930
3569
3^'
3o59
3i23
3187
325i
33i5
3378
3442
35o6
3696
4386
3822
69
3885
3948
4011
4073
4i36
4199
4261
4323
4448
70
.84510
4572
4634
4696
4757
4819
4881
4942
5552
5oo3
5o65
7»
5i26
5187
5248
5309
5370
543 1
5491
5612
5673
72
5733
6332
5794
5854
5914
6570
7157
6o34
6004
6688
61 53
6213
6273
6864
73
6302
6982
7564
8i39
8705
645 1
65io
6629
6747
7332
6806
74
6923
7040
7^
8253
7216
7852
8423
7390
7448
8024
8503
t
7506
8081
7622
8196
8762
S^
M
8480
m
??
8649
88r8
8874
94^
8930
8086
9542
9042
^?3
91^4
9210
9265
^l^i
9376
9487
9598
9708
79
9763
9818
9873
9927
9982
♦037
♦091
♦146
♦200
♦255
80
.90309
0363
0417
0956
0472
o526
o58o
o634
0687
0741
^v\
81
0849
0902
1009
1062
1116
1169
1698
1222
^V\
i328
82
i38i
1434
1487
1 540
1593
1645
1751
i8o3
i856
83
1908
ig6o
2012
2o65
2117
2169
2686
2221
2273
2789
2324
2376
84
2428
2480
253i
2583
2634
2737
2840
2891
85
2942
35oo
3o44
3095
3 146
3197
3247
3298
3349 3399 i
86
345o
355i
36oi
365 1
3702
3752
38o2
3852
3902
4399
U
3952
4002
4o52
4101
4i5i
4201
425o
43oo
435o
4448
4498
4547
5o^
4645
4694
5182
4743
4792
4841
4890
89
4939
4988
5o37
5i34
523 1
5279
5328
5376
90
.95424
5473
5521
5569
56i7
6093
5665
5713
5761
5809
5856
9'
5904
5952
6000
6047
6142
6190
6237
6708
6284
6332
9»
6370
6848
6426
6473
6520
6567
7o3d
6614
6661
6755
6802
93
68o5
7818
8272
6942
6988
7081
7i28
7174
7635
6091
8543
7220
7267
H
^
73i3
1772
8227
74o5
7864
83i8
745 1
8408
7543
8000
8453
8046
8498
7681
8137
8588
8182
8632
?2
8677
9123
8722
9167
8767 8811
9211 9255
8856
9300
8901
9344
8945
9388
8900
9432
9034
9476
907B
9520
99
9564
9607
1
965i
8
9695
8
9739
4
r
9782
5
9826
6
9870
7
9913
8
9957
8
H.
CHAPTER XX.
MATHEMATICAL INDUCTION AND
BUSINESS FORMULAS.
461. Mathematical Induction consists in proving
by trial that a proposition is true in a certain case ; and,
finding it true in the next case, then in the third, and so
on, we conclude it must be trve in all similar cases.
462. Many of the principles and formulas of Arithmetic
and Algebra are established by this mode of reasoning.
463. Take the familiar principle in Arithmetic :
The product of any two or more numbers is the same, in
whatever order the factors are taken.
To prove this principle of two numbers, as 5 and 3, the pnpll
represents the number 5 by as many unit marks in a
# # # # #
horizontal row, and under this places two similar rows.
# # # # #
He sees that the number of marks in the horizontal
^ ^ 4f # #
r<no taken 3 times is equal to the number of marks in
a perpendicular row taken 5 times ; that is, 3 times 5 = 5 times 3.'
He then takes three factors and finds the proposition true, and so
on. Hence, he concludes the principle is universally true,
464. Next, suppose it be asserted :
The product of the sum and difference of two quantities i2
equal to the difference of their squares. (Art. 103.)
Taking two quantities, as 4 + 3 and 4—3, or a+& and a— 6.
Multiply
4 + 3
Or
a + h
By
43
By
a — 6
4* + 4x3
a^ + ab
4x33'
a66»
Product,
4« 3«
o» 6«
461. In what does Mathematical induction coneist ? niastration.
244 MATHEMATICAL INDtJCTlOK.
He takes another example of two quantities, and finds the
4itatement true ; then another, and so on. Hence, he condades the
proposition is a universal truth.
465. Suppose this proposition were ennnciated:
The sum of any number of terms of the arithmetical series
I, 3, 5, 7, etcy to n termSy is equal to n^.
We see by inspection that the sum of the first two terms, 1 + 3 = 4,
or 2* ; that the sum of the first three terms, i + 3 + 5 = 9, or 3* ; that
the sum of four terms, i + 3 + 5 + 7 = 16, or 4*, and so on. Hence, we
may conclude that the proposition w true if the series be extended
indefinitely. Or,
Since we know the proposition is time when n denotes a small num
ber of terms, and that the value of any terin in the series, as the 5th,
7th, 9th, etc., is equal to 2/1—1, we may suppose for this value of n,
that 1 + 3 + 5 + 7 .... +(2H—i) = w*. (i)
Adding 272.+ 1 to both members, we have
1 + 3+5 + 7.... +(2»i)+(2»+i) = »*+2n+i. (2)
Factoring second member of (2), w* + 2» + 1 = (« + 1)^.
Therefore, since the sum of n terms of the series = n^, it follows
that the sum of n + i terms = (71 + 1)^, and so on. Hence, the prop
osition must be unvoeraaUy true.
466. In Geometry, we have the proposition :
The sum of the three angles of a triangle is equal to two
right angles.
We find it to be true in one case ; then in another, etc.
Hence, we conclude the proposition is universally true.
Notes. ~ I. It is sometimes objected that tMs method of proof \& less
satisfactory to the learner than a more rigorous process of reasoning.
But when it is fully understood, it is believed that it will produce the
fullest conviction of the truths designed to be established.
2. In metaphysics and the natural sciences, the term in^Uietian is
applied to the assumption that certain laws are general which by
experiment have been proved to be true in certain cases. But we
cannot be sv/re that these laws hold for any cases except those Tirliich
have been examined, and can never arrive at the conclusion that they
are necessary truths.
BUSIKESS FOKMULAS. 245
BUSINESS FORMULAS.
467. The principles of Algebra are not confined to the
demonstration of theorems and the solution of abstruse
equations. They are equally applicable to the development
of formulas and for business calculations.
Note. — In reciting formulas, the student should first state the
proposition, then write the formula upon the hlackhoard, explaining
the several steps by which it is derived as he proceeds. He should
then translate the formula from (dgebraic into common langv4xge.
PROFIT AND loss:
468. Profit and Loss are computed by the principles
of Percentage.
469. To Find the Profit or Loss, the Cost and the per
cent Profit or Loss being given.
Let c denote the cost, r the per cent profit or loss, and p th«
percentage, or sum gained or lost.
Since per cent means hundredt?is, r per cent of a number must be t
huindredths of that number. (Art. 237.) Therefore,
c dollars x r =:^ p, the sum gained. Hence, the
FOBMULA. p r= CV.
EuLE. — Multiply the cost by the rate per cent, and the
product will be the profit or loss, as the case may require.
(Art. 237.)
1. Suppose c = $3560, and r = 12 per cent. Required
the profit.
Solution. $3560 x .12 = $427.20, Ans.
2. If a house costing $4370 were sold at 8 per cent lesn
than cost, what would be the loss ?
470. To Find the per cent Profit or Loss, the Cost and
the Sum Gained or Lost being given.
Let c denote the cost ; p the percentage, or sum gahiea or lost ; and
f the per cent*
246 BUSIKESS FOBMULAS.
Since the eost multiplied hj the late gives p, the given profit ot
looB, it follows ^baXpte must be the rate. (Art. 469.) Hence, the
Formula. r = — •
c
BnL& — Divide the gain or loss by the cost, and the quo
tient will be the per cent profit or loss.
3. A farm costing $2500 was sold for I500 adyance.
Bequired ihe per cent profit. Afis. 20 per cent.
4j a teacher's salary being liSoo a year, was raised $300.
What per cent was the increase ?
471. To Find the Cosi^ the Profit or Low and the per cent
Profit or Loss being given.
Let ji doiote the sum gained or lost, r the per cent, and c the cost.
Since the sum gained is equal to the cost multiplied by the rate per
cent (Art. 469), it follows that p dollars the sum gained, divided by r
the rate, wiU 4)e the cost. Hence, the
Formula. c = —•
r
BuLE. — Divide the profit or loss by the rate per cent.
5. The gain on a bill of goods was $67.48. At 25 per
cent profit, what was the cost ? Ar^. I269.92.
6. An operator in stocks lost $15759 which was 12^ per
cent of his investment. What was the investment?
472 To Find the SeUing Price, the Cost and per cent
Profit or Loss being given.
Let e denote the cost, r the per cent of profit or loss, and 9 the
selling price.
When there is a gain^ the amount of $1 = i + r ; when there is a
lou, the amount of $1 = i—r ; and the amount of e dollars = 6(1 ± r).
Therefore, we have the general
Formula. « = c(i ± r).
BuLE. — Multiply the cost by i plus or minus the rate, as
the case may require. (Art. 240.)
7. A man paid $750 for a piano. For what must he sell
it to gain 15 per cent? Ans, $862.50.
8. A man bought a carriage for $960, and sold it at a loss
of 1 2 J per cent. What did be receive for it ?
BUSINESS FORMULAS. 247
473. To Find the Coatf the Selling Price and the per cent
Profit or Loss being given.
Let 8 denote the selling price, r the per cent profit or loes, and c the
cost required.
When there is a gain, the selling price equals the cost plus r per
cent of itself, that is, i +r times the cost; when there is a l<w, the
selling price equals the cost minus r per cent of itself, that is, i— r
times the cost ; therefore the cost equals the selling price divided by
1 ±r. Hence, we have the general
Formula. c = — : — •
I ±r
Rule. — Divide the selling price by i plus or minus the
rate, as the case way require ; the quotient will be the cost.
9. A goldsmith sold a watch for $175 and made 20 per
cent profit. What was the cost ?
SoLTTTiON. 1 + 20 per cent = 1.20 ; and
$175 H 1.20 = $145.83^. 4IW.
10. A jockey sold a horse for $540^ and thereby lost
zo per cent. What did the horse cost him ?
SIMPLE INTEREST.
474. The Elements or Factors involved in calcula
tions of interest are the same as those in percentage^ with the
addition of time.
475. Interest is of two kinds, simple and compound. By
the former y interest is derived from the principal only; by
the lattery it is derived both from the principal and the
interest itself as soon as it becomes dne.
476. To Find the Time in whioh any Sum at Simple Interest
will Double itself, at any given Rate Per Cent.
Let p denote the principal, r the rate per cent, t the given interest,
and t the time in years.
Then % = prt. (Art. 242.)
Making i equal to p, p = prt.
Dividing by pr,  = t. Hence, the
248 BUSINESS FOBHULAS.
Formula. f = ~.
r
BuLE. — Divide i by the rate, and the quotient will be the
time required.
11. How long will it take $1500, at 5 per cent^ to doable
itself?
Solution. t =  = — = 20. Ans. 20 years.
r .05 ^
12. How long will it take i6So, at 6 per cent, to double
itself?
13. How long will it take $8475, ^^ ^^ V^^ ^^^ ^ double
itself?
477. To Find the Rate at which any Principal, at Simple
interest, will Double itself in a Given Time.
By the preceding f ormnla, t = •
T
Maltiplymg by  , we have the .
Formula. r = ^
BuLE. — Divide i by the time j the quotient mil be the rate
per cent required.
[For other formulas m Simple Interest, see Arts. 242246.]
14. If I1700 doubles itself in 8 years, what is the rate ?
Ans. 12^ per cent.
15. At what rate per cent will $5000 double itself in
40 years ?
COMPOUND INTEREST.
478. Interest may be compounded annudHy, semi
annually, quarterly, etc. It is understood to be com
pounded annually, unless otherwise mentioned.
479. To Find the Atnount of a given Principal at Compound
Interest, the Rate and Time being given«
Let p denote the principal, r the rate, n the number of years, and
a the amount
BUSINESS FORMULAS. 249
Since the amount equals the principal plus the interest, it follows
that the amount of $i for i year equals i+r ; therefore, p(i+r)
equals the amount of p dollars for i year, which is the principal for
the second year.
Again, the amount of this new principal p(if r) for i year =
p(i+r)(i+r)=j?(i+r)', which is the amount of p dollars for two
years.
In like manner, p (i + r)» is the amount of p dollars for three years
and so on, forming a geometrical series, of which the principal p is
the first term, i +r the ratio, and the number of years + i, the num
ber of termis. The terms of the series are
The last term, p(i + r)*, is the amount of p dollars for n yeans.
Hence, the
Formula. a =p(t + r)\
EuLB. — Multiply the principal by the amount of i for
I year, raised to a power denoted by the number of years;
the product will be the amount.
480. To Find the Compound Interest for the given Time
and Rate.
Subtract the principal from the amount, and the remainder
will be the compound interest.
Note. — When the number of years or periods is large, the oper
ations are shortened by using logarithms.
1 6. What is the amount of $842, at 6 per cent compound
interest, for 4 years ?
Solution. $842 x (1.06)* = $1063, Ans,
17. What is the amount of $1500, at 5 per cent for 6 yrs.,
compound interest ?
r
481. If the interest is compounded semiannually,  will
denote the interest of $1 for a half year. Then, at com
pound interest, the amount of p dollars for n years is
3n
I .
250 BUSINESS FORMULAS.
482. If the interest is compounded qvartertyy then 
will denote the interest of ti for a quarter. Then, at
compound interest, the amount of p dollars for n years is
JO (i 4 J , etc.
i8. What is the amount of I2000, for 3 years at 6 per
cent, compounded semiannually ? Arts. $2388.05.
19. What is the amount of I5000 for 2 years, at 4 per ct.,
compounded quarterly ?
483. By transposing, &ctoring, etc., the formula in
Art. 479, we have,
The first term, p =
The nmnber of tenuB, n =
(i + r)«
log. a — log. p
log. (i+r)
DISCOUNT.
484. Discount is an allowance made for the payment
of money before it is due.
485. The Present Worth of a debt payable at a future
time is the sum which, if put at legal interest, will amount
to the debt in the given time.
486. To Find the Present Wofth of 2i Sum at Simple Interest,
tiie Time of Payment and the Rate being given.
Let s denote the gnm dne, n the number of years, and r the mterest
of $1 for I year.
Since r is the interest of $1 for i year, nr must be the interest for n
years, and i + nr the amount of $1 for n years. Therefore, «s(i + 7ir)
is the present worth of the given sum.
Putting P for present worth, we have the
Formula. P = — ;
I +nr
Rule. — Divide the sum due hy the amount of $1 for the
given time and rate; the quotient toill be the present worth.
'yAns.
BUSINESS FORMULAS. 261
487. To Find the Discount, the Present Worth beiiig given.
Subtract the present worth from the debt.
20. What is the present worth of I2500 payable in 4 yra,
interest being 7 per cent ? What the discount ?
Solution. P =. f = 1^ = $1953.125, pres. worth
$2500 — I1953.125 = I546.875, disooont,
21. What is the present worth of I3600 due in 5 years, at
6 per cent ? What is the discount?
22. Find the present worth of I7800 due in 6 years,
interest 5 per cent ? What the discount ?
COMPOUND DISCOUNT.
488. To Find the Bresent Worth of a Sum at Compound
Interest, the Time and the Rate being given.
Let B denote the sum due, n the number of years, r the rate per ct.
Since r is the rate, i + r is the amount of $1 for i year ; then the
amount for n years compound interest is (i +r)". (Art. 479.) That is,
$1 is the present worth of (i + r)» due in n years. Therefore, «<(i + ry
must be the present worth of the given sum.
Putting P for the present worth, we have the
FOBMULA. F =
Rule. — Divide the sum due by the amount of li at com
pound interest, for the given time and rate ; the quotient
will be the present worth.
23. What is the present worth of $1000 due in 4 years, at
5 per cent compound interest ?
Q ^1000
SOLITTIOK. P = ; = ? r = 822.7I» Ans.
(i+r)» (1.05)*
24. What is the present worth of $2300 due in 5 years, at
6 per cent compound interest ?
253 BUSINESS FORMULAS.
COMMERCIAL DISCOUNT.
489. Cofnfn€rci4U IHscaunt is a per cent taken
from the face of bills^ the marked price of goods^ etc.,
without regard to time.
490. To Find the Cknnmercial Discount on a Bill of Goods,
the Face of the Bill and the Per Cent Discount being given.
Let b denote the base or face of the bill, r the rate, and d the
discount or percentage.
Then 6 x r will be the discount. Hence, the
Formula. d = br.
Rule. — Multiply the face of the bill by the given rate, and
the product will be the commercial discount.
491. To Find the Cash VtUue or Ifet Proceeds of a Bill.
Subtract the commercial discount from the face of the hiU.
25. Bequired discount and net value of a bill of goods
amounting to $960, on 90 days, at 12^ per cent off for cash ?
Solution. $q6o x .125 = $120, discount ; $960 — $120= $840, An»,
26. Required the cash value of a bill amounting to $2500,
the discount being 10 per cent, and 5 per cent off for cash.
492. To Mark Goods so as to allow a Discount, and make
any proposed per cent Profit.
Let e denote the cost, r the per cent profit, and d the per cent disc
Since ris the per cent profit, i + r is the selling price of $1 cost, and
e(i +r) the selling price of c dollars cost.
Again, since d is the per cent discount from the marked price, and
the marked price is 100 per cent of itself, i^d must be the net value
of $1 marked price. Therefore,
c(i + r) i (i—d) = the marked price.
Putting m for the marked price, we have the
^ c(i +r)
Formula. m = — ^ — ^^
Rule. — Multiply the cost by i phis the per cent gaiuy and
divide the product by i minus the proposed discount.
BUSINESS FORMULAS. 253
27. A trader paid $25 for a package of goods ; at what
price must it be marked that he may deduct 5 per cent, and
yet make a profit of 10 per cent ? 
« c(i+r) $25x1.10 ^ „ .
Solution, m = ^^ — ^ = ^^ = $28.94+, Ans,
I— a .95
28. A merchant buys a case of silks at $1.75 a yard;
what must he mark them that he may deduct 10 per cent,
and yet make 20 per cent ?
29. A grocer bought flour at $6^ a bjprel ; what price
must he mark it that he may fall 8 per cent, and leave a
profit of 25 per cent ?
INVESTMENTS.
493. The Value of an Investment in National and
State securities, Railroad Bonds, etc., depends upon their
market value, the rate of interest they bear, and the cer
tainty of payment.
494. The Dividends of stocks and bonds are reckoned
at a certain per cent on the par value of their shares, which
is commonly $100.
495. To Find the Per Cent which an Investment will pay, the
Cost of a Share and the Rate of Dividend being given.
Let c denote the cost or market value of i share of stock, p its par
valuej and r the annual rate of dividend.
Since r is the rate of dividend and p the par value, pr must be the
dividend on i share for i year. Therefore, pr^e will be the per cent
received on the cost of i share. (Art. 238.)
PuttiDg R for the per cent received on the cost of i share, we have
the
Formula. U = ^'
c
EuLE. — Find the dividend on the given shares at the given
rate, and divide this by the cost ; the quotient will he the per
cent received on the investment.
254 BUSINESS FOBKULAS.
NOTB. — ^WheD the stock is above or bdow par, the premium or
ducauiU must be added to or sabtracted from its par value to give the
cost
30. What per cent ioterest does a man receive on an
inyestment of $5000 in the Bank of Commerce^ its dividends
being 10 per cent, and the shares 5 per cent above par?
Solution. — ^The premium on the etock = $5000 x x)S = 125a
Therefore, the cost = $5000+ $250 = $5250.
Again, the dividend on stock = $5000 x .10 = $5oa Therefora,
$50o^$525o = g^{ per cent, Ans.
31. A invested $6000 in New York 6 per cent bonds, at
3 per cent premium. What per cent did he receive on his
investment ?
32. A man lays out $1000 in Alabama 10 per cents, at a
discount of 20 per cent. What per cent did he receive on
his investment ?
33. What per cent will a man receive on 50 shares of
Pennsylvania Eailroad stock, the premium being 4 per ct,
and the dividend 10 per cent ?
34. Which are preferable, Massachusetts 6 per cent bonds
at par, or Ohio 8 per cent bonds at 2 per cent premium ?
496. To Find the Amount of a given Remittance which a
Factor can Invest and Reserve a Specified Per Cent for his
Commission.
Let s denote the snm remitted, and r the per cent commission.
The sum remitted includes both the sum invested and the commis
sion. Now 1 1 remitted is 100 per cent, or once itself ; and adding the
per cent to it, we have i + r, the cost of $1 invested. Therefore,
» I (i + r) must be the amount invested.
Putting^ a for the amount invested, we have the
POEMULA. a =
1 +r
Rule. — Divide the retnittance hy i plus the per cent
commission ; the quotient will be the amount to invests
BUSINESS FORMULAS. 255
35. A clergyman remitted to his agent $2500 to purchase
books. After deducting 4 per cent commission^ how much
does he lay out in books ?
SoLimoN. a = — ^ = ??55? = $2403.85, Ana.
i+r 1.04 ^—^''
36. A gentleman remitted $25000 to a broker, to be
invested in stocks. After deducting i^ per cent, how much
did he invest, and what was his commission ?
SINKING FUNDS.
497. Sinking Funds are sums of money set apart or
deposited annually, for the payment of public debts, and for
other purposes.
CASE I
498. To Find the Afno^mt of an Annual Depooit at Compound
Interest, the Rate and Time being given*
Let s denote the annual deposit or som set apart, r the rate per
cent, n the number of years, and a the amount required.
Since the same sum is deposited at the end of each year, and put
at compound interest, it follows that the deposit at the end of the
ist year = s
2d *' =« + «{i+r)
3d " = 8 + 8(i + r) + 8{i\rf
nth. " = « + «(i+r) + «(i+ry .... + ^(i+r)"',
forming a geometrical series ; the annual deposit being the first term,
the amount of $1 for i year the ratio, the number of years the num
ber of terms, and the annual deposit multiplied by the amount of $1
for I year, raised to that power whose index is i less than the numbei
of years, the last term ; and the amount is equal to the sum of the
series. (Art. 402.) Hence, we have the
POBMULA. a = ^ ■ 8.
r
EuLE. — Multiply the amount of $1 annual deposit for the
given time and rate hy the given annual deposit ; the product
will be the amount required.
256 BU8IKE8S FORMULAS.
37. A clerk annnallj deposited $150 in a savings bank
which pays 6 per cent compound interest What amonnt
will be due him in 5 years ?
o (iir)»— I (i.o6)« — i^ _„ .
SoLunoK. a = ^ 9 = ^ — ^ — $150 = $845.75, ^w
T .00
38. A man agrees to give I5300 annually to build a church.
What will his subscription amount to in 4 years, at 7 per
cent compound interest ?
39. If a teacher lays up $500 annually, and puts it at
5 per cent compound interest for 10 years, how much will
he be worth ?
CASE II.
499. To Find the Annual JDeposii required to produce a
given Amount at Compound interest, tlie Rate and Time being given.
B7 the formula in the preceding article, we have
(i+r)»— I
i^ ~ 8 = a.
r
Dividing by coefficient of «, we have the
■CI (i + !•)» — 1
Formula. s^ar  — ■ — •
r
Rule. — Divide the amount to he raised by the amount of
1 1 annual deposit for the given time and rate; the quotient
will be the annual deposit required.
Note. — To cancel the debt at maturity, the sum set apart as a
sinking fund is supposed to be put at compound interest for the given
time and rate.
40. A father promises to gives his daughter $5000 as a
wedding present. Suppose the event to occur in 5 years,
what sum must he annually deposit in a Trust Company, at
5 per cent compound interest, to meet his engagement ?
(i + r> — I A (1.05)6 — I $250
Solution. 9 = a* ^I—L = $5000 h * — ^ = ^^ =
r JOS 276
$905.80, Arts.
BUSINESS FORMULAS. 267
41. A man having lost his patrimony of $20000, wishes
to know how much must be annually deposited at 10 per
cent, to recover it in 5 years ?
42. A county borrows I30000, at 6 per cent compound
interest, to build a courthouse ; what sum must be set
apart annually as a sinking fund to cancel the debt in
10 years?
ANNUITIES.
500. Annuities are sums of money payable annually,
or at regular intervals of time. They are computed accord
ing to the principles of compound interest.
CASE I,
501. To Find the Anwunt of an Unpaid Annuity at Compound
Interest, the Time and Rate per cent being given.
Let a be the annuity, i + r the amount of i for i year, and n th«
number of years.
The amount due at the end of the
ist year = a,
2d " = a + a(i+r;
3d " = a + a(i+r) + a(i^r)\
4th " = a + a(i+r) + a(i+r)' + a(i+r)',
«
nth year = a + a (i + r) + a (i +r)* + a (i + r)» . . . a(i +r)»i.
forming a geometrical progression, the annuity being the first term, the
amount of $1 for i year the ratio, the number of years the number of
terms, and the annuity multiplied by the amount of $1 for i year
raised to that power whose index is i less than the number of years,
the last term. Therefore, the amount is equal to the sum of the
series.
Putting 8 for the amount (Art. 498), we have the
Formula.
g __ (i + r)"  I ^
iS = a.
BuLE. — Multiply the amount of ii annuity, for the given
time and rate, by the given annuity.
258 BUSINESS FORMULAS.
NoTE.^LogaiithmB may be nsed to advantage in some of the
following examples.
43. What is due on an annuity of $650, unpaid for
4 years, at 7 per cent compound interest ?
Solution. 8 = (iJtlt.'ZJL a = (^^7)'  J ^^^ _ 2886, Am,
r .OT
44. An annual pension of I880 was unpaid for 6 years ;
what did it amount to at 6 per cent compound interest ?
45. An annual tax of I340 was unpaid for 7 years ; what
was due on it at 5 per cent compound interest ?
CASE II.
502. To Find the Bresent Worth of ^n Annuity atCompound
Interest, the Time of Continuance and the Rate being given.
Let P denote the present worth ; then the amount of P in n years
will be equal to the amount of the annuity for the same time.
Therefore,
r
Dividing each member by (i + r)» (Art. 279), we have the
I (i I t\^
Formula. F = ^^ — ' — '— a.
r
Note. — In applying the formula, the negative exponent may be
made positive by transferririg the quantity which it affects from the
numeraioT to the denominator (Art. 279).
1
Thus. P = l^±J[n a ^ LJIl^a.
r r
EuLB. — Multiply the present worth of an annuity of $1
for the given time hy the given annuity.
46. What is the present worth of an annuity of $375 for
6 years, at 7 per cent compound interest ?
o n I — (i+r)» I — (1.07)*^ I — .66^
Solution. P = ^ '— a = ^ ^ $375 = ~— — $375
r .07 .07
= $1785.71, Ans,
47. What is the present worth of an annual pension of
I525 for 5 years, at 4 per cent compound interest ?
BUSINESS FOBKULAS. 259
CASE III.
503. To Find the Present Worth of a Perpetual Annuity, the
Rate being given.
Let n denote infinity, then redndng the formula in Art. 502, we
have this
POBMULA. ^—y (Art. 435.)
Rule. — Divide the annuity by the interest of ii for
1 yeavy at the given rate.
48. What is the present worth of a perpetual scholarship
that pays 1 150 annually, at 7 per cent compound interest?
Solution. P= = ?i5? = $2142.86, Ans.
r .07
49. What is the present worth of a perpetual ground rent
of I850 a year, at 6 per cent ?
CASE IV.
504. To Find the Fi^esent Worth of an Annuity, commencing
in a given Number of Years, the Rate and Time of Continuance
being given.
Let n be the number of years before it will commence, and N the
zinmher of years it is to continue. Then,
„ I — (i + r)(»+^> I — (i + r)*
P = a ^ ^ a ^^ ~ — .
r r
Performing the subtraction indicated, we have the
Formula. P =  [(i + !•)— — (i + r)""^.
EuLE. — Mnd the present worth of (he given annuity to
the time it terminates ; from this subtract its present worth
to the time it commences.
50. What is the present worth of an annuity of $600, to
commence in 4 years and to continue 12 years, at 7 per
cent interest ?
Solution. P = ^[{i + r)« — (i + r)"^.
r
p = ^ [(1.07H  (i^)^'»].
.07
260 BUSINESS FORMULAS.
51. A father left an annual rent of $2500 to his son for
6 years, and the reversion of it to his daughter for 12 years.
What is the present worth of her legacy at 6 per cent
interest ?
CASE V.
505. To Find the Annuity, the Present Worth, the Time,
and Rate being given.
By the formula in Article 502,
r
DiTiding by the coefficient of a, we have the
Formula. a =
I  (i + r)^
BcLE. — Divide the present worth hy the present worth of
an annuity of%i for the given time and rate,
52. The present worth of a pension, to continue 20 years
at 6 per cent interest, is $668. Required the pension.
Pr $668 X. 06 $668 X. 06 ^ . .
Solution, a = — ; — = ^ — zr*, = ^^o — = $58.23^11*
I— (i+r)» I— (1.06)^ .6882 ^ ^
53. The present worth of an annuity, to continue 30 years
at 5 per cent interest, is $3840. What is the annuity?
Note. — The process of constructing fonnulas or rules, it will be
seen, is based upon the principles of generalization combined with
those of algebraic notation. The student will find it a profitable
exercise to form others applicable to different claases of problema.
CHAPTER XXI.
DISCUSSION OF PROBLEMS.
506. The Discussion of a Problem consists in assign
ing all the different values possible to the arbitrary quanti
ties which it contains, and interpreting the results.
507. An Arbitrary Quantity is one to which any
value may be assigned at pleasure.
Problem. — If h is subtracted from a, by what number
must the remainder be multiplied that the product may be
equal to c ?
Let X = the number.
Then (a — 6) a; = c.
I*
Therefore, x =
508. The result thus obtained mav have five different
forms, depending on the relative values of a, J, and c. To
represent these forms, let m denote the multiplier.
I. Suppose a is greater than J. In this case a  6 is pontive,
and c being pontive^ the quotient ia positive. (Art. ii2.) Consequently,
the required multiplier must be positive, and the value of x will be of
the form of + m,
II. Suppose a is less than b. In this case a — h is negative,
and e divided by a —6 is negative. (Art. 112.) Hence, the required
multiplier must be negative, and the value of a; is of the form of — m.
III. Suppose a is equal to b. In this case a — 6 = o. There
fore, the value of a; is of the form of — , or a; =  = oo. (Art. 434.)
506. In wliat does tbe discoMion of probtoma consiat i 507. What it an arbitrary
quantity Y
262 DISOVSSIOK OF PB0BLBM8.
lY. Suppose ciBo, and a is either greater or less than b.
In this case the yalue of a; haa the form — , or 2B = o.
m
Y. Suppose c equals o, and a equals b. In this case iha
▼aloe of X haa the form  •
o •
Note. — The stiident can easUy t^st these prindples hj substituting
nombers for a, b, and c
509. The Discussion pf Problems may be further illus
trated by the solution of the celebrated
PROBLEM OF THE COURIERS*
Two couriers A and B, were traveling along the same
road in the same direction, from C toward Q; A going at the
rate of m miles an hour, and B n miles an hour. At
12 o'clock A was at a certain point F; and B d miles in
advance of A, in the direction of Q. Wh^n and where were
they together ?
P d Q
This problem is general ; we do not know from the statement
whether the couriers were together before or after 12 o'clock, nor
whether the place of meeting was on the right or the left of P.
Suppose the required time to be after 12 o'clock. Then the time after
12 is positive, and the time before 12 is negative; also, the distance
reckoned from P toward Q is positive^ and from P toward C is negative.
Let t = time of meeting in hours after 12 o'clock ; then mt — dis
tance from P to the point of meeting.
Since A traveled at the rate of m miles an hour, and B n nules an
hour, we have
971^ = the distance A traveled.
And nt ** " B
Again, since A and B were d miles apart at 12 o'clock,
mt — n^ = dL
Factoring and dividing we have the
* Originally proposed by Clairaut, an eminent French mathemati
cian, born in 1 7 13.
DISCUSSION OF PROBLEMS. 263
FOBMULA. t =
The problem may now be discussed in relation to the
time ty and the distance mt, the two unknown elements.
I. Suppose m > w.
Upon this supposition the values of t and mt \vill both be positive;
because their denominator m — nis positive. Now since t is positive,
it is evident the two couriers came together after 12 o'clock : and as
mt is positive, the point of meeting was somewhere on the right of P.
These conclusions agree with each other, and correspond to the
conditions of the problem. For, the supposition that m>n Implies
that A was traveling faster than B. A would therefore gain upon B,
and overtake him some time after 12 o'clock, and at a point in the
direction of Q.
Let (2 = 24 miles, m = 8 miles, and n = 6 miles.
d 2A.
By the formula, t = = . V. = 12 hoars.
m — u — 6
wi^ = 8 X 12 = 96 miles A traveled.
Tii = 6 X 12 = 72 •* B "
Now, 96 — 72 = 24 771. their distance apart at noon, as gfiven above.
These values show that the couriers were together in 12 hours past
noon, or at midnight, and at a point Q, 96 miles from P and 72 miles
from d.
II. Suppose m <in.
Then In the formula, the denominator m — n\s negative, therefore
both t and mt are negative.
Hence, both t and mt must be taken in a sense contrary to that
which they had in supposition (I), where they were positive ; that is,
the time the couriers were together was before 12 o'clock, and the
place of meeting on the left of P.
This interpretation Is also Id accordance with the conditions of the
problem under the present supposition. For, if w < 7^ then B was
traveling faster than A ; and as 6 was in advance of A at 12 o'clock,
he must have passed A before that time, somewhere on the left of P,
in the direction of C.
Let (? = 24 miles, r» = 5 miles, and 71 = 8 miles.
By the formula, t = = ^ ^ = — 8 hours. 
^ m—n 5—3
And m^3:5x— 8 = — 40 miles A traveled
n* = 8x8 = 64 "B "
264 BISCUSSIOX OF PKOBLEMS.
These Talues show that the oooriers were together 8 hoars before
noon, or at 4 o'clock A. m., and at a point C, 40 miles from P and
64 miles from d,
III. Suppose m = n«
Upon this supposition we have m — n = o, and
. d  . md
t =  =Qo, also mt = — = »•
o o
According to these results, t the time to elapse before the couriers are
together, is infinity (Art. 434) : consequently they can never be together.
In like manner 7nt, the distance from P of the supposed point of
meeting, is infinity ; hence, there can be no such point.
This interpretation agrees with the supposed conditions of the
problem. For, at 12 o'clock the two couriers were d miles apart, and
it m = n they were traveling at equal rates, and therefore could
never meet.
IV. Suppose J = o, and m either greater or less than n.
We then have t = = o, and mt = o,
m — n
That is, both the time and distance are nothing. These results show
that the couriers were together at 12 o'clock at the point P, and at no
other time or place.
This interpretation is confirmed by the conditions of the problem.
For, if <f = o, then at 12 o'clock B must have been with A at the point
P. And if m is greater than w, or m is less than n, the couriers were
traveling at different rates, and must either approach or recede from
each other at all times, except at the moment of passing ; therefore
they can be together only at a single point.
V. Suppose J = o, and m = w.
Then we have ^ =  , and m^ =  •
o o
These results must be interpreted to mean that the time and the
distance may be anything whatever, and that the couriers must be
together at all times, and at any distance from P.
This conclusion also corresponds to the conditions of the problem.
For, if <f = o, the couriers were together at 12 o'clock, and if in = n,
they were traveUng at equal rates, and therefore would never part
IMAG1J5(ABY QUAKXIIIES. 266
IMAGINARY QUANTITIES.
510. An Imnginary Quantity is an indicated even
root of a w«gra/m quantity ; as^ Vi, V— a, ^—7.
Notes. — i. Imaginary quantities are a species of radiecUSf and are
called imaginary, because they denote operations which it is impossi
ble to perform. (Art. 294.)
2. Though the operations indicated are in themselves impo^Me,
these imaginary expressions are often useful in mathematical analyses,
and when subjected to certain modifications, lead to important results.
511. Imaginary quantities are added and subtracted like
other radicals. (Arts. 310, 311.)
But to multiply and divide them, some modifications in
the rules of radicals are required. (Arts. 312, 313.)
512. To Prepare an Imaginary Qaanttty for Multiplication
and Division.
BuLE. — Resolve the given quantity into two factors, one
of which is a real quantity, and the other the imaginary
expression V— 1.
NoTBS.— I. This modification is based upon the principle that any
negative quantity may be regarded as the product of two quantities,
one of which is — i. Thus, —a = a x — i ; —ft* = 6* x — i.
2. The real factor is often called the coefficient of the imaginary
expression, 'y/— i.
I. Multiply V—a hy V^J.
Solution, y"— a = y^ x y^^, and y'— & = y^ x y^i*
Now y^ X y^^ X y^ x y^^ = y^ x — i = — y^a&, Ans,
2. Multiply + V — a? by — V— y.
3. Multiply a/— 9 by V— 4.
:,ia What are imag^imary qoAQtities? 5x1. How. added and Babtractod?
5x8. How prepare them for multiplication and division ?
12
266 IMAGINARY QUANTITIES.
513. It will be seen from the preceding examples:
First, That the product of two imaginary quantities is a
real quantity.
Second, That the sign before the product is the opposite
of that required by the common rule for signs. (Art 92.)
For, while the sign to be prefixed to an even root is ambiguooi),
this ambiguity is removed wiien we know whether the quantity whose
root is to be taken has been produced from positive or negative
quantities. (Art. 293.)
4. Multiply V^^ by VTs,
5. Multiply V^^ by Vy.
Note. — i. From these examples it will be seen that the product of
a Teal quantity and an imaginary expression, is itself imaginary.
6. Divide V^ by V—y.
SOLUTIOK. ^^ = ^ ^^=z y Zf ^^
7. Divide V— a: by V^a?,
NOTB. — %. Hence, the quotient of one imaginary quantity divide ^
another, is a real quantity ; and the sign before the radical is the same
as that prescribed by the rule. (Art. 92.)
8. Divide V^x by VJ/y
9. Divide Vx by V—y,
Note. — 3. Hence, the quotient of an imaginary quantity dlvidadtij
a real one, is itself imaginary, and vice verw.
10. Divide 10 V— 14 by 2 a/— 7,
1 1. Divide c V^^ by d V— i,
514. The development of the different powers ot V— i,
12. (\/^)2 = — I. _ IS iV^y = + V^.
13 (V^)^ = — V^i. 16. (a/^)« = — I.
14. {V^y= +1. 17. {V'^iy= V"^.
Hence, the even powers are aUematdy — z and +1, and the odd
powers — \/— I and +/y/— x.
IKDETEBMIKATB PROBLEMS. 267
INDETERMINATE PROBLEMS.
515. An Indeterminate Problem is one which does
not admit of a definite answer. (Art. 220.)
Note. — ^Among the more common indeterminate problems, are
1st. Those whose conditions are satisfied by different values of the
same unknown quantity. (Art 220.)
2d. Those which produce identical equations. (Art. 200.)
3d. Those which have a less number of independent simultaneous
equations than there are unknown quanti ies to be determined.
4th. Those whose conditions are inconsistent with each other.
1. Given the equation a: f y = 9, to find the value of x.
Solution. — Transposing, a? = 9 — y. Ana, This result can be
verified by assigning any values to a; or ^.
2, What number is that, f of which minus i half of itself
is equal to its 12th part plus its sixth part?
Let X = the number.
Then 3«_«^«.^.«
4 2 12 6
Clearing of fraetions, etc, 90; = 92
Transposing and factoring, (9—9) x = o
o
/. aj = .
o
IMPOSSIBLE PROBLEMS.
516. An Impossible Problem^^a one, the conditions
of which are contradictory or impossible.
I. Given a; + y = 10, a: — y = 2, and xy = 38.
OFSBATION.
Solution. — By combining equations (i) ic + y = 10 (i)
and (2), we find as = 6 and y = 4. Again, x — y z= 2 (2)
X X y = 6 X 4 = 24, But the third condition
requires the product of x and ^ to be 38,
which is impossible. • '• ^ = "
= 4
5x5. What is an indetennlDate problem ? $x6. Wbat ie an impoBBible problem f
B68 KSOATIYE SOLUTIONS.
2. What number is that whose 5th part exceeds its 4th
part by 15 ?
3. DiTide 8 into two such parts that their product shall
be 18.
NEGATIVE SOLUTIONS.
517. A Negative Solution is one whose result is a
minus quantity.
518. An odd root of a quantity has the same sign as the
quantity. An even root of a positive quantity is either .
positive or negative, both being numerically the same.
(Ari 293.)
But the results of problems in Simple Equations, it is
understood, are positive; when otherwise it is presumed
there is an error in the data, which being corrected, the
result will be positive.
1. A schoolroom is 30 feet long and 20 feet wide. How
many feet must be added to its width that the room may
contain 510 square feet ?
Solution. — Let x = the number of feet,
Then (20 + a;) 30 = area.
B7 conditions, 600 + yyx = 510
Transposing,. ^ox = — 90
.*. oj = —3 ft., Ans,
Notes. — i. It will be observed that tbls is a problem in Suuple
Equations. The steps in the solution are legitimate and the result
satisfies the conditions of the problem algebraically, but not arith
metically. Hence, the negatwe remit indicates some miHake or
inconsistency in the conditions of the problem.
If we subtract 3 ft. from its width, the result will be a positiu
quantity.
2. Were it asked how much must be added to the width that the
room may contain 690 square feet, the result would be + 3 feet.
5x7. What to a negative 80luti<ni f
horwer's method. 269
3. In Buch cases, by changing some of the data, a similar problem
may be easily foand whose conditions are consistent with a possible
resnlt.
2. What number must be subtracted from 5 that the
remainder may be 8 ?
Sqltttion.— Let (P = the nunber.
Then 5 — « = 8
Transposing, x = ^ 2, Ana.
3. A man at the time of his marriage was 36 years old
and his wife 20 years. How many years before he was twice
as old as his wife P Ans. — 4 year&
HORNER'S METHOD OF APPROXIMATION*
519. This method consists in transforming the given
equation into another whose root shall be less than that of
the given equation by the first figure of the root, and
repeating the operation till the desired approximation is
found.
The process may be illustrated in the following manner:
Let it be required to find the approximate value of a; in the general
equation,
A(x* + Ba? + Cx = D, (i)
Having found the firat figure of the root by trial, l^t it be denoted
by a, the second figure oy b, the third by Cy and so on.
Substituting a for a; in equation (i), we have,
Aa* + Ba^ + Oa — D, nearly.
Factoring and dividing,
"^ " CTBa + Aa^ <*)
* So caUed from the name of its author, an English mathematician,
who communicated it to the Royal Society in 1819.
270 hokneb's method.
'By patting y for the sum of all the figures of the root except the
fint, we have x = a^jf, and subRtitnting this value for x in equation
(i), we hare,
Aia+yfi B[a+yy i 0{a+p) = D;
or u4(a'+3d»y43fly*+y») + B{a*+2ay+$/^) + C(a+y) = D.
Factoring and arranging the terms according to the powers of p,
we obtain
To simplify this equation, let us denote the coefficient of i/* hy B^,
that of phj C, and the second member by D' ; then.
Ay' + By^ + C'y = D'. (4)
It will be seen that equation (4) has the same form as (i). It is the
first transformed equation, and its root is less bj a than the root of
equation (i).
By repeating the operation, a second transformed equation may be
obtained. Denoting the second figure in the root by h, and reducing
as before, we find,
, _ ly ..
C { Bh ■\ AJ^' ^5^
Putting s foi the sum of all the remaining figures in the root, we
have y =zb^z; and substituting this value in equation (4), we obtain
a new equation of the same general form, which may be written,
Ai^hB"t^'iC"B = D". (6)
This process should be continued till the desired accuracy is attained.
The first figure of the root is found by trial, the second figure from
equation (5), and the remaining figures can be found from similar
equations.
But it may be observed that the second member of equation (5)
involves the quantity &, whose value is sought. That is, the value of
b is given in terms of b, and that of e would be given in terms of e, and
so on. For this reason, equations such as (5) might appear at first
sight to be of little use in practice. This, however, is not the case ;
for after the root has been found to several decimal places, the value
of the second and third terms, as B'b+Ab^ and B"c+Ai^ in the
denominators, will be very small compared with C and C, oonae
horneh's method. 271
qaentl3r as & is very nearly equal to D' divided hj C\ they may be
neglected. Therefore the successive figures in the root may be
approximately found by dividing D' by C\ D" by 0'\ and so on,
regarding C\C'\ etc., as approximate divisors.
In transforming equation (i) into (4), the second member Z)' and
the coefficients C ' and B' of the transformed equation may be thus
obtained.
• Multiplying the first coefficient A by a, the first figure of the root.
and adding the product to B, the second coefficient, we have,
B + Aa (7)
Agidn, multiplying this expression by a, and adding the product to
0, the third coefficient, we have,
C + Ba + Aa\ (8)
Finally, multiplying these terms by a, and subtracting the product
from D, we have
D(Ca + Ba^ k Aa^ = !>,
which is the same as the expression for 2^ in equation (4).
Now to obtain C, we return to the first coefficient, multiply it by «,
add the product to expression (7), and thus have the sum
B + 2Aa, (9),
which we multiply by a, and adding the product to expression (8)
obtain,
C + 2Ba + 3^a' = C',
which is the desired coefficient of ^ in equation (4).
Finally, to obtain B'y we multiply the first coeffident by a, and add
the product to expression (9), and thus obtain,
B + zAa = B',
In this way the coefficients of the first transformed equation are
discovered ; and by a similar process the coefficients of the second,
third, and of all subsequent transformed equations may be found.
520. This method of approximation is applicable to
equations of every degree. For the solution of cubic equa
tions^ it may be summed up in the following
Rule. — ^I. Detach the coefficients of the given equation,
and denote them by A, B, 0, and the second member by D.
JFKnd the first figure of the root by trial, and represent it by a.
272 HORNEB'S METHOD.
MvUiply A by a, and add the product to B. Multiply
the sum by a and add the product to C. Multiply this sum
by a and subtract the product from D. ITie remainder is
the first dividend, or D\
II. Multiply A by a and add the product to the last sum
under B. Multiply this sum by a and add the product to
the last sum under C. The result thus obtained is the first
divisor, or C
III. Multiply A by a and add the product to the last sum
under B» The result is the second coefficient, or B'.
IV. Divide the first dividend by the first divisor. The
quotient is the second figure of the root, or b.
V. Proceed in like manner to find the subsequent figures
of the root
IS^omL — I. In finding the second figure of the root, some allowance
should be made for the terms in the divisor which are disregarded ;
otherwise the quotient will furnish a result too large to be subtracted
fromlX.
EXAMPLES.
I. Given a^ + ax^ + 3a; = 24, to find x.
SOLmON.
AS C D a be
I +2 +3 =24 « = ( 2.08, u4n£
2 _8 22
4 II 2 = ly
3 12 I.891712
6 23 = 0' .108288 = D"
2 .6464
8 = ^ 23.6464
.08 .6528
8.08 24.2992 = (7"
.08
8.16
.08
9.24 = B"
Horner's method. 278
Note. — 2. In the following example, the last figures of the root are
found by the contracted method of division of decimals, an expedient
which may always be used to advantage after a few places of decimals
have been obtained. (See Higher Arithmetic.)
2. Given a^ + 12a? — iSx = 216, to fiud x.
SOLUTION.
A
B
D a be
I
+ 12
18
= 216 (4.24264+.
_4
+64
184
16
+46
32=1>
J:
Jo
26.168
20
126 = 0"
5.832 = D"
^
4.84
5.468224
24 =
B
130.8 4
.363776 = zy'
24.2
4.88
275385
24.4
24.6 =
24.64
24.68
z B"
135.7 2 = 0"
.9856
136.7 056
.9872
137.0 928 =
88391
82615
5776
5508
X s= 4.24264+, Atu,
3. Given afi + $3^ + sx = lyS, to find z.
X = 4.5388, Ana.
4. Given 52^ + 9a;* — 70; = 2200, to find x,
X = 7.1073536, Ans,
5. Given a:® + a;* + a? = 100, to find x,
i2; = 4.264429+, Ans,
i« •
274 TEST EXAMPLES FOB BEYIEW.
TEST EXAMPLES FOR REVIEW.
1. Beqnired the valne of
6a H 4a X s + 8a J 2 — 3fl + 12a X 4.
2. Beqnired the yalne of
(8a? + 3a:) s + 4a; + 7  (sa; + gx) ^ 7.
3* Required the yalue of
Scue — flj + /^cd — (200? — 4aJ + 2cd).
4* Required the value of
4be + [scd — (2a;y — mn) 5 + 3&c].
5. Show that subtracting anegatiyequautity is equivalent
to ftdding a positive one.
6. Explain by an example why a positive quantity
multiplied by a negative one produces a negative quantity ?
7. Explain why a minus quantity multiplied by a minus
quantity produces a positive quantity.
8. Given —  (a: + 8) = ^ + —  1 7f , to find a?.
3 9 7
o. Given ^ H [ 2a? = ^ x ^ to find as.
^ 5 5 3 ^
10. Resolve 38^^ — 6V(? — (?d into two factors.
1 1. Resolve $afif/ — ga^z — iSa^yz into two factors.
12. Resolve a^ — J^* into two factors.
13. Resolve 8a — 4 into prime factors.
14. Resolve a* — i into prime factors.
15. Divide 31 in to two such parts that 5 times one of them
shall exceed 9 times the other by i.
16. Make an algebraic formula by which any two numbers
may be found, their sum and difference being given.
17. Two sportsmen at Creedmoor shoot alternately at a
target ; A hits the bull'seye 2 out of 3 shots, and B 3 out
of 4 shots; both together hit it 34 times. How many shots
did each fire ?
TEST EXAMPLES FOR REVIEW. 275
1 8. Find two quantities the product of which is a and Obe
quotient b.
lo. Beduce r — r to its lowest terms.
ao —b
20. Beduce 5 ?= to its lowest terms.
fl2 — ^
21. Besolre gx^y^ + i2xyz + 42? into two factors.
22. Besolve 96^ — 6bc + (? into two factors.
23. Make a formula by which the width of a rectangular
surfiace may be found, the area and length being given ?
24. A square tract of land contains \ as many acres as
there are rods in the fence inclosing it What is the length
of the fence ?
^5. A student walked to the top of Mt. Washington at
the rate of i^ miles an hour, and returned the same day at
the rate of 4J miles an hour ; the time occupied in traveling
being 13 hours. How far did he walk?
26. Given b ^^ = o, to find x.
1 —x
27. Prove that the product of the sum and difference of
two quantities, is equal to the difference of their squares.
28. Prove that the product of the sum of two quantities
into a third quantity, is equal to the sum of their products.
29. Beduce t^ ox / — — r to its lowest terms.
{p? + 2xy + y^)(x — y)
a* — J*
30. Beduce 7= = tstts — ior to its lowest terms.
31. Beduce . ^ to a single fraction having
the least common denominator.
32. Find a number to which if its fourth and fifth part
be added, the sum will exceed its sixth part by 154.
33. Two persons had equal sums of money; the first
spent $30, the second $40 : the former then had twice as
much as the latter. What sura did each haye at first ?
276 TEST EXAMPLES FOB REVIEW.
*34. A French privateer discovers a ship 24 kilometers
distant, sailing at the rate of 8 kilometers an hour, and
porsnes her at the rate of 12 kilometers an hoar. How
long will the chase last ?
35. Given ?i±^ = 7 and lllZ^^y^o, to find
X andy.
36. Oiven z^^^ — h 5 and 4y ^ — = 3* to find
X and y.
37. Make a rule to find when any two bodies moving
toward each other will meet, the distance between them and
the rate each moves being given ?
38. A steamer whose speed in still water is 12 miles an
honr, descended a river whose velocity is 4 miles an hour,
and was gone 8 honrs. How far did she go in the trip ?
39. Find a fraction from which if 6 be subtracted from
both its terms it becomes , and if 6 be added to both, it
becomes J.
40. Eeqnired two numbers whose sum is to the less as 8
is to 3, and the difference of whose squares is 49.
41. Given ioa:+6y = 76, 4y— 2j? = 8, and 6a;48« = 88,
to find X, y, and z.
42. Given 2X + $y + z = 24, 3a? + y + 2« = 26, and
a? J 2y + 3z = 34, to find x, y, and z,
43. Three persons. A, B, and 0, counting their money,
found they had I180. B said if his money were taken from
the sum of the other two, the remainder would be $60;
C said if his were taken from the sum of the other two, the
remainder would be J of his money. How much money
had each?
44. The forewheel of a steamengine makes 40 revolutions
more than the hindwheel in going 240 meters, and the
circumference of the latter is 3 meters greater than that of
the former. What is the circumference of each ?
45. A man has two cubical piles of wood; the side of one
TEST EXAMPLES FOK REVIEW. 277
is two feet longer than the side of the other, and the differ
ence of their contents is 488 cubic feet. Required the side
of each.
46. Eequired a formula by which the height of a rectan
gular solid may be found, the contents and base being given.
47. Divide 126 into two such parts that one shall be a
multiple of 7, the other a multiple of 11.
48. A tailor paid i 20 for French cloths ; if he had bought
8 meters less for the same money, each meter would have
cost 50 cents more. How many meters did he buy ?
49. A shopkeeper paid $175 for 89 meters of silk. At
what must he sell it a meter to make 25 per cent ?
50. Make a formula to find the commercial discount, the
marked price and the rate of discount being given.
51. A man pays $100 more for his carriage than for his
horse, and the price of the former is to that of the latter as
the price of the latter is to 50. What is the price of each ?
52. Make a formula to find at what time the hour and
minute hands of a watch are together between any two
consecutive hours?
53. A father bequeathed 165 hektars of land to his two
sons, so that the elder had 35 hektars more than the younger.
How many hektars did each receive ?
54. What number is that, the triple of which exceeds 40
by as much as its half is less than 5 1 P
55. A butcher buys 6 sheep and 7 lambs for ^71 ; and, at
the same price, 4 sheep and 8 lambs for I64. What was the
price of each ?
56. At a certain election, 1425 persons voted, and the
successful candidate had a majority of 271 votes. How
many voted for each ?
57. A's age is double B's, and B's is three times C's; the
sum of all their ^es is 150. What is the age of each ?
58. Eeduoe the ^243 to its simplest form.
39. Beduce V^ 4 oy^ to its simplest form.
:i78 TEST EXAMPLES FOR REVIEW.
60. Bednce x^ and ^ to the common index ^.
61. Beduee 3 (a — J} to the form of the cube root.
62. A farmer sold 13 bushels of com at a certain price ;
and afterward 17 bnshels at the same rate^ when he received
(3.60 more than at the first sale. What was the price per
bushel ?
63. A sold two stoves. On the first he lost $8 more than
on the second; and his whole loss was I2 less than triple
the amount lost on the second. How much did he lose on
each?
64. A number of men had done J of a piece of work in
6 days, when 12 more men were added^ and the job was
completed in 10 days. How many men were at first
employed ?
65. A company discharged their bill at a hotel by paying
$8 each; if there had been 4 more to share in the payment,
they would only have paid I7 apiece. How many were
there in the party ?
66. In one factory 8 women and 6 boys work for $72 a
week; and in another, at the same rates, 6 women and
1 1 boys work for 8o a week. How much does each receive
per week?
67. What factor can be removed from ViSZ^ ?
68. Given Vx + 12 = Vfl + 12, to find x.
69. Given — ~ = ^""_^ , to find y.
y Vy ^
70. Given VaJ® — 4^^ = a — ft, to find x,
71. From a cask of molasses ( of which had leaked out,
40 liters were drawn, leaving the cask half full. How many
liters did it hold ?
72. Make a formula to find the per cent commission a
factor receives, the amount invested and the commission
being given.
73. Divide 20 into two parts, the squares pf which shall
be in the ratio of 4 to 9.
TEST EXAMPLES FOR REVIEW. 279
74. After paying out ^ of my money and then  of the
remainder, I had I140 left. How much had I at first?
75. If I be added to both terms of a fraction, its valne
will be J; and if the denominator be doubled and then
increased by 2, the value of the fraction will be 4. Required
the fraction.
76. Tiffany & Co. sold a gold watch for $171, and the per
cent gained was equal to the number of dollars the watch
cost. Required the cost of the watch.
77. Two Chinamen receive the same sum for their labor;
but if one bad received $15 more and the other $9 less, then
one would have had 3 times as much as the other. What
did each receive ?
78. A drover bought a flock of sheep for $120, and if he
had bought 6 more for the same sum, the price per head
would have been $1 less. Required the number of sheep
and the price of each.
79. A certain number which has two digits is equal to
9 times the sum of its digits, and if 6^ be subtracted from
the number, its digits will be inverted. What is the
nnmber ?
80. Two riverboatmen at the distance of 150 miles apart,
start to meet each other ; one rows 3 miles while the other
rows 7. How far does each go ?
81. A and B buy farms, each paying I2800. A pays $5 an
acre less than B, and so gets 10 acres more land. How
many acres does each purchase ?
82. Find a factor that will rationalize Vx f V7,
83. Find a factor that will rationalize V3X — VJy
84. Given Vi^+^x= /"^^ _. , to find x.
85. The salaries of a mayor and his clerk amount to
$13200 ; the former receives 10 times as much as the latter.
Required the pay of each.
S6. What two numbers are those whose sum is to their
280 TEST EXAMPLES FOR BEVIEW.
difference as 8 to 6^ and whose difference is to their prodnci
as I to 36 ?
87. What two numbers are those whose product is 48, and
the difference of their cubes is to the cube of their difference
as 37 to I ?
SS. Find the price of apples per dozen, when 2 less for
12 cents raises the price i cent per dozeu.
89. Two pedestrians set out at the same time from Troy
and New York, whose distance apart is 150 miles ; one goes
at the rate of 24 m. in 3 days, and the other 14 m. in 2 oajs.
When will they meet ?
90. The income of A and B for one month was $1876,
and B's income was 3 times A's. Kequired that of each ?
91. A farmer bought a cow and a horse for $250, paying
4 times as much for the horse as for the cow. Find the
cost of each.
92. A man rode 24 miles, going at a certain rate ; he then
walked back at the rate of 3 miles per hour and consumed
12 hours in making the trip. At what rate did he ride ?
93. It costs $6000 to furnish a church, or $1 for every
square foot in its floor. How large is the building, pro
vided the perimeter be 320 feet?
94. Find 5 arithmetical means between 3 and 31.
95. Find the sum of 50 terms of the series i, 1, i^, 2, 2^,
3> Sh 4, 4h etc.
96. A dealer bought a box of shoes for lioo. He sold all
but 5 pair for $135, at a profit of ti a pair. How many
pair were there in the box ?
97. Two numbers are to each other as 7 to 9, and tht*
difference of their squares is 128. Required the numbers.
98. In a pile of scantling there are 2400 pieces, and the
number in the length of the pile exceeds that in the height
by 43 : required the ntmber in its height and length.
99. Bertha is ^ as old as her mother, but in 20 years she
will be f as old. What is the age of each ?
100. Fifteen persons engage a car for an excursion; but
TEST EXAMPLES FOR REVIEW. 281
before starting 3 of the company decline going, by which
the expense of each is increased by $1.75. What do they
pay. for the car ?
loi. When the hour and minute hands of a clock are
together between 8 and 9 o'clock, what is the time of day ?
102. A and B wrote a book of 570 pages; if A had
written 3 times and B 5 times as much as each actually
did write, they would together have written 2350 pages.
How many pages did each write ?
103. A man and his wife drink a pound of tea in 12 days.
When the man is absent, it lasts the woman 30 days. How
long will it last the man alone ?
104. Find the time in which any sum of money will
double itself at 7 per cent simple interest.
105. A purse contains a certain sum, in the proportion of
$3 of gold to $2 of silver ; if $24 in gold be added, there will
then be $7 of gold for every $2 of silver. Bequired the sum
in the purse.
106. A and B in partnership gain $3000. A owns 1^ of
the stock, lacking I200, and gains I1600. Bequired the
whole stock and each man's share of it.
107. In the choice of a Chief Magistrate, 369 electoral
votes were cast for two men. The successful candidate
received a majority of one over his rival : how many votes
were cast for each ?
108. Two ladies can do a piece of sewing in 16 days ; after
working together 4 days, one leaves, and the other finishes
the work alone in 36 days more. How long would it take
each to do the work ?
109. If a certain number be divided by the product of its
two digits, the quotient is 2j; and if 9 be added to the
number, the digits will be inverted : what is the number ?
no. Find 4 geometrical means between 2 and 486.
1 1 1. A trader bought a number of hats for $80 ; if he had
bought 4 more for the same amount, he would have paid $1
less for each : how many did he buy ?
282 TEST EXAMPLES FOB KEYIEW.
XI 2. If the first term of a geometrical series is 2, the ratio
5, and the number of terms 12, what is the last term ?
113. A tree 90 feet high^ in falling broke into three
nneqoal parts ; the longest piece was 5 times the shortest,
and the other was 3 times the shortest : find the length of
each piece.
1 14. The sum of 3 numbers is 219 ; the first equals twice
the second increased by ii^ and the second equals f of the
remainder of the third diminished by 19: required the
numbers.
1 15. Bequired 3 numbers in geometrical progression, sucli
that their sum shall be 14 and the sum of their squares 84.
1 16. A pound of coffee lasts a man and wife 3 weeks, and
the man alone 4 weeks : how long will it last the wife ?
117. Two purses contain together I300. If you take $30
from the first and put into the second, each will then
contain the same amount : required the sum in each purse.
118. A clothier sells a piece of cloth for I39 and in so
doing gains a per cent equal to the cost What did he
pay for it?
119. A settler buys 100 acres of land for I2450; for a
part of the farm he pays I20 and for the other part $30 an
acre. How many acres were there in each part ?
120. What is the sum of the geometrical series 2, 6, 18,
54, etc., to 15 terms?
121. There are 300 pine and hemlock logs in amiUpond,
and the square of the number of pines is to the square of the
number of hemlocks as 25 to 49 : required the number of
each kind.
122. A ship of war, on entering a foreign port, had
sufScient bread to last 10 weeks, allowing each man 2 kilo
grams a week. But 150 of the crew deserted the first night,
and it was found that each man could now receive 3§ kilo
grams a week for the remainder of the cruise. What was
the original number of men ?
APPENDIX.
521. To Extract the Cube Root of Polynomials.*
I. Beqnired the cube root of «• + 3a* — 30* — i lo* +
6a«+ 12a — 8.
OFBBATIOH.
^•+3a"— 3a*— iia'+6a*+i2a— 8 ( »•+«— a, BootBL
g', the first subtra hend.
1st Trial Divisor, 8a* ) 3a*— 3a*— no*, etc., first remainder.
Com. D., 30* + 3a' + a ) 30*430*+ a'
2d Tr. D., 3a*+6a«+3a' ) — 6a*— 12a" + 6a* + 12a— 8, 2d remainder.
Complete Divisor,
30* + 6a'— 30'— 6a + 4 ) —6a*— 120' + 6a* + 120—8 . Hence, the
BtTLE. — I. Arrange the terms according to the potoers of
one of the letters^ take the cube root of the first term for the
first term of the rooty and subtract its cube from the given
polynomial.
n. Divide the first term of the remainder by three times
the square of the first term of the root as a trial divisor ^ and
the quotient tvill be the next term of the root.
in. Complete the divisor by adding to it three times the
product of the first term by the second^ also the square of the
second. Multiply the complete divisor by the second term of
the root, and subtract the product from the remainder.
IV. If there are more than tivo terms in the root, for the
second trial divisor, take three times the square of the part
of the root already found, a7id completing the divisor as
before, continue the operation until the root of all the terms
is found. (See Key)
* For Hpmer's Metljod of Approximation, see p. 269.
284 APPENDIX.
2. Required the cube root of a* + ^a^b + yxV + V.
3. Find the cube root of a;* + 6a;* + 12a: + 8.
4. Find the cube root of a::^ — (iQ?y + i2a;y' — Sy*.
5. What is the cube root of 8a* — 48a* + 96a — 64.
6. Find the cube root of 2']c? — S4«^ + z^ax^ — 8a;*.
7. What is the cube root of «• — 6a' f 15/1* — 20a' +
15a' — 6a + I.
8. The cube root of a:* — 3a;' + 8a^ — 6a;» — 6a;* + Zt^^
522. Factor the following PolpnotnicUs :*
1. a^ — 9a: + 20. 7' ^ + y^ H y^nsf*
2. a* + 7a — 18. 8. i2a2a; — Sa^ + 402.
3. a' — 13a + 40. 9. rf* — ^akc H 3aa:' — a;®.
4. 2aftc* — i4a^ — 6oa& 10. i — a*.
5. a^y^ — 2xy + i. 11. i + 8a^
6. 8a:® — 32^*. 12. cfi — J^a;®,
523. Find the g. c. d. of the following Polynomials:
1. 4a^ — 4aa? — isa^ and 6a^ + yax — 3a:*.
2. 4aa^y*2!*, 122;'^;?*, and iddh^.
3. i6a^ — ^, and 16a:® — 8a:^ + ^.
4. 6a® + iiaa; +^0.^, and 6a® +702; — 3a^.
5. 0* — J*, and a' — 6®a^
6. a:" — a^ and x^ — a^.
524. Required the h c. fn. of the following Polynomiafs.
1. 60^ — 4a, 4a* + 2a, and 6a^ + 40.
2. 4 (i + a®), 8(1— a), 4(1— a®), and 8 (i + a).
3. fl? — 2a + I, a* — I, and a® + 2a + i.
4. 12 (ai® — js)^ 4 (^2 ^. ^5)^ and 18 (a® — J®).
5. 4a® — I, 2a — I, and 4a® + I.
6. 4 (i + a®), 8(1+ a), 4(1 — a®), and 8 (i — a).
* The following problems are classified and may be stadied in con
nection with the subjects to wliicli thej refer, or be omitted till the
other parts of the book are finished, at the option of tlie teacher.
APPENDIX. 286
525. Unite the following Fractions:
I 2ab
'' '^ZTl^ + fl* — **'
•
a — b b — c c — a +ac
ab be ac
5. Prom a + 3A take ^a — h^ i— <
2 3
6. From sa? + t take 2a;
c
X I
7. Prom a + a; + ^ ^ ^^^® a — a? +
a^ — y^ X + y
8. Prom take
a a — I
0. Prom take «•
^ I — a; I — a?
526. Multiply the following Poiynatniala :
a — b . 2& a* a^ — «»
1. I —^ X 2 H 7. 5. — ; — X r^ •
a + b a — b X { y ab
Aa , ^x 2b ^x  „ 5*
3ar 26 3a; 4a a — I
3. a« — 2a;y + ya X — ^^. 7. 5 A_ ^ — ^
^ ^ ^ x — y' 2a 5a — 10
7ft 2fl2 — 4a^  a;y ay
2^2 — 8a* 2id ^ a; + y ^ a? — y
527. Divide the following Quantities:
, 2a 2a xy 2V
a— 3 a— 3 2^ — 2 y — I
II I ab + b^ b
a? ' a:y8 • ^ y ^ ^3 __ ^ • ^ __ j
a? • a« "• Vi+r» ^ ia/ • (ia^)^
286 JLPPENDIX
528. Simplify th« following Fractions:
2a — a a^ — y*
^ + 1
a— I ^ a;? — y»
, a? I j_
o, * lO, *
529. Solve the following Equations :
1. A house and bam cost $850^ and 5 times the price of
the house was equal to 12 times the price of the bam.
Bequired the price of each.
2. A^ B, and G, together have 145 acres of land ; A owns
twothirds and B threefourths as much as C. How many
acres has each ?
3. From a cask J full of water, 21 liters leaked out, when
I the water was left. Bequired the capacity of the cask.
4 3 12
3« + 9 72^+5 16 + 4«
23s
6. h a: = a? + 2.
4 3
. a; f 8 a? — 6
7. a; — 2 = a; I ■ •
8.
2a; +
■(=^')=
3 ^ 4
9. The hour and minute hands of a watch are together
at 12 H. At what time between the hours of 7 and 8 p. m.
will they again be in conjunction ?
10. A merchant supported himself 3 yrs. for £50 a year,
and at the end of each year added to that part of his stock
which was not thus expended, a sum equal to J of this part.
At the end of the third year his original stock was dqubled.
What wafi that stock ?
4
2
^
22
; — 2y
=
4*
4
X
2
1=
&
X
3
*\=
c.
APPENDIX. 287
530. Solve the following SitnuUaneaus EguaUon$c
1. ^— ! — ^ = 26. 3. ^ ^ =0.
3
^ = o.
a 3
2.  + ^ = 8. 4.
32
2 3
5. A man bought a horse^ baggy, and harness for $400 ;
he paid 4 times as much for the horse as for the harness,
and onethird as much for the harness as for the buggy ;
how much did he pay for each?
6. What number consisting of two figures is that to
which, if the number formed by changing the place of the
figures be added^ the sum will be 121 ; but if subtracted,
the remainder will be 9 ?
10. xy =r 600 ;
xz = 300 ;
yz = 200.
8. T + r = i; "• ^ + 4 — 1=^9;
 + ^ +  = 22 ;
2 3 4
X z
12. w + 50 = x;
X + 120 = 3y;
y + 120 = 2z;
« + 19s = 3m;.
13. A^s age added to 3 times B's and C's, is 470 yrs. ;
B's added to 4 times A's and C's, is 580 yrs. ; and C's added
to 5 times A's and B's, is 630 yrs. What age is each ?
14. What 3 numbers are those whose sum is 59 ; half the
difference of the first and second is 5, and half the differ
ence of the first and third is 9 ?
X
+ y^z=z
0;
X
+ « — y =
2;
y + Z'x =
4
X
h
+ f = i;
X
h
^1=.,
c
+ £ = ..
w
+ x + y =1
6;
w
+ X + z =
9;
w
+ y + z=z
8;
X
+ y + « =
7
288 APPENDIX.
531. Generalize the following Problems and translate the For
mulas into Rules:
1. A dishonest clerk absconded^ traveling 5 miles an
hour ; after 6 hours, a policeman pui'sued him, traveling
8 miles an hour. How long did it take the latter to over
take the former ?
»
NOTS. — Sabetitute c for clerk's rate, p for policeinan's rate, n foi
number of hours between starting, and x for the time required.
Formula. a? = •
2. A can do a piece of work in 2 days, B in 5 days, and
G in 10 days ; how long will it take all working together
to do it?
Note. — ^Let rt, 6, and t represent the numbers. Then x = dbc +
(06 + oc + ftc).
Formula. ac = — i— i— •
oft + ac + 6c
3. Divide I4400 among A, B, and C, in proportion to the
numbers 5, 7, and 10.
Note. — Put a, &, and c, for the proportions, % for the sum of the
proportions, and n for the number to be divided.
4. A father is now 9 times as old as his son ; 9 years
hence he will be only 3 times as old : what is the age
of each?
532. Expand the following by the Binomial Theorem:
1. (2fl — 3J)3. 4. {a^^ + ff.
2. (3a? + 2yY. 5. (a + ar^y.
3. (i + 3a)*. 6. {a^  2ay.
533. F'ind the Produot of the following Powers :
1. abar^ by a^. 3. a?~«* by x~^.
2. fl*J«ar8 by a«*2a;8. ^ y^hjy^.
534. Divide the following Powers:
1. 6(r*hj^arK 3. i2ar* by 43?""^
2. Sa^bc"^ hy 4a^¥(fi. 4. (a+x)^ hj (a+x)'^.
APPENDIX. 289
535. Transfer Denominators to Numerators, thus forming
entire Quantities.
0? (Mem
536. Unite the following Radicals :
1. V48 + V27 4 V243. 4 ^ y/2$a^c + V36a;*c
2. VSA^ — V'96a; + ^/ 2^. 5. VSoa^ — Vaoa^.
3. 8 v^^^^d + 2 v^. 6. 3V^i28a;8yj ^ ^ 'V^i6y«.
537. Find the Product of the following Madicals:
1. (fl + y)n X {b + h)n. 3. (a; + y)i x (a; + y)i
2. 4+2 V^ X 2 — \/2. 4. sb \^d 4 y X 4 a/««
538. Dividing one Radical by Another.
1. (M)^ ~ (aa;)^. 4 (* + y)" ^ (* + y)".
2. 24a; Vay = 6 Va. 5. 40 V«S i 2 V^c.
3. Vi6a* — i2a^ ^ 2a. 6. 70 v^ r 7 aJ^iS.
539. Required the Factors which will MationcUize the fol
lowing Radicals:
I. 2 Va + VT 4. Vs — a/S.
2. ic + Vy. 5. 4 v^ — s Vy
3 . The denom. of — p • 6. The d. of — p — ^= — •
2V3 V3+V2+I
540. Solve the following JRadiccU Equations:
1. Given Vx + i = Vn 4 x, to find x.
2. Given Va; + 18 — Vs = Va^ — 7, to find x.
3. Given Va^^ — 11 = 5. 5. (13 + V23 + y*)* = S
6
"*' a/T+x ^ ^^3 + ^ 6. 2 Va = Va? + 3a.
290 APPEKDIX.
541. Solve the following Quadratics s
2
lOO — QX
3.
s«
«3
—
22? +
i6
X
lOO —
4^
92? _
= 3
2
4 ~
I
32
1
"s/ax
+ 2
4
 Vx
4 + Vx Vx
5. Find two numbers whose difference is 12, and the sum
of their squares 1424.
6. Eequired two numbers whose sum is 6, and the sum
of their cubes 72.
7. Divide the number 56 into two such parts, that their
product shall be 640.
8. A and B started together for a place 150 miles distant
A's hourly progress was 3 miles more than B's, and he
arrived at his journey's end 8 hrs. 20 min. before B. What
was the hourly progress of each ?
9. The difference of two numbers is 6 ; and if 47 be
added to twice the square of the less, it will be equal to the
square of the greater. What are the numbers ?
10. The length added to the breadth of a rectangular
room makes 42 feet, and the room contains 432 square feet
Eequired the length and breadth.
11. A says to B, the product of our years is 120 ; if I
were 3 yrs. younger and you were 2 yrs. older, the product
of our ages would still be 120 ? How old is each ?
12. V^ 4 V^ = 6 Vx.
13. X + V^ + 6 = 2 + 3 Va? + 6.
14. A man bought 80 lbs. of pepper and 100 lbs. of
ginger for £65, at such prices that he obtained 60 lbs. mare
of ginger for £20 than he did of pepper for £10. What
did he pay per pound for each ?
COLLEGE EXAMINATION PROBLEMS.
542. I. Divide 151^^ bj x ^^.
2. Divide a* — i* by {a — b).
3. Solye the equation x + ^ = 12 ^.
4. Multiply 3 V45 — 7 V5 by Vif + 2 Vpf
5. Divide a^b^ by aiji 6. Divide xif^ by ajty"i.
7. Given 30.^ + 2a; — 9 = 76, to find a;.
8. Given ia?^ — a: + 7I = 8, to find x,
9. Find two numbers, the greater of which shall be to
the less as their sum to 42, and as their difference to 6.
10. Find the value ofi+J + ^4jV + etc. to infinity.
11. Find the third term of (a + by\
12. Expand to four terms (i + x^)~^. •
. X. Divide ^±^ +? by ^i^ 
^ + y y y « + y
2. Find the product of a^, a^, a^ and a~^.
3. Solve the equation x + \W+x^ =
4. Solve 22 ?? £7_^^
X X + I X + 2
20^
Va^ + a?»
5. Solve Vx — I = a; — I.
6. Find the value of f + i + J +, etc., to infinity.
7. Given x+Vx : x—Vx :: 3 \/aJ+6 : 2 V^*^, to find jr,
8. Expand to four terms (a^ f a;)i3.
9. Expand to four terms {a^ — y^)~^.
544. I. Find the gr. c. dJ. and the I, c. w. of (243^^055 ^ ^ )
and (8ia8j4 — i) by factoring.
T^. ., 6 Vi , 20c Vi^
2. Divide — g— by ^— •
25 ya^ 2\dby€?
3. Solve the equations 22: — y = 21, 20^ + y* = 153.
292 APPENDIX.
4. A person buys cloth for $90. If he had got two yards
more for the same sum, the price would have been 50 cts.
per yd. less. How much did he buy, and at what price ?
5. Expand (a — b)^ by the binomial theorem.
6. Factor 4^' — 9^.
7. Multiply 3 y ? by 2 y^t
8. Given x+2y = 7 and 2X+sy = 12, to find x and y.
9. Reduce a ^ ^Zd^d and i/ff to their simplest forms.
10. Given ^ +  = 12 , to find a:.
32 3 '
545. I. From xx { — = subtract x •
^20 c
2. Multiply together , ^, and i A
3. Extract the square root of Sai^ + a* — 4cfib + 46*.
4. From 2 v^32o take 3 "Vao.
5. Divide a «ji by aij~i. 6. Solve x^ + 40^ =^ 12.
7. Solve ^ — x \/3 = ic — J V3.
8. What two numbers are those whose sum is 2a, and the
sum of their squares is 2S ?
9. What two numbers are those whose difference, sum,
and product are as the numbers 2, 3, and 5 respectively ?
10. Find three geometrical means between 2 and 162.
11. Expand to four terms  — *
546. t. Divide i2ic* — 192 by 3a: — 6.
2. Divide ^ H r by r — m.
a — a —
3. Solve the equation 21 + —j^ — = Z' + ^IZlZf.
4. Find the product of a^, a*, a?, and a~i.
= 2.
5. Solve the equation — — ^r—
^ x 201^
6. Find 6 arithmetical means between i and 50.
OOLLEGE PBOBLEMS. 293
t ^
7. How many different combinations may be formed of
eight letters taken four at a time ?
8. Expand (a — J)~i to four terms.
9. Divide 150 into two such parts that the smaller may
be to the greater^ as 7 to 8.
10. Giyen Sic + 2y = 29 and 21/ — a? = — i, to find
z and y.
547. I. Solve the equation — : + 4 = a
2 +y y^2 •
3. What is the relation between a, a®, and cr^ ?
3. Find two numbers such that the sum of \ the first
and % of the second equals 1 1^ and also equals three times
the first diminished by the second.
/ J\®
4. Give the first three and last three terms of (2a j •
5. Find the g. c. d. of a^ _ 52 and a^ — 2ah + V.
6. Find the L c. tn. of {a^ —  a?), and 4 (a — a?), and
(a + x).
. Add '3^  ^9b _ lt2ia gb^iia
7^^^ Sja^b^ SJaiY 5(a¥
8. Find the value of x in Vx + a = v^ + a.
548. I. Seduce — ^ — r 3 to its lowest terms.
c^ — a?
2. Multiply cr^V by =:= ; and divide a"^S* by j:=
3. Solve the equation "" 7 — II_5_ — —
^ 35 6a? — loi 5
4. Given ^^ ^ — fa: — ^^ "^ J = 7.
5. Solve the equation f 7f = 8.
6. It is required to find three numbers such that the
product of the first and second may be 15, the product of
the first and third 21^ and the sum of the squares of the
second and third 74.
29*. OOLLEOE PROBLEMS.
7. Find the sum of n terms of the series i, 2, 3, 4,
S, 6, etc.
8. Expand to five terms (a^ — ff)~^.
9. Find the sum of the radicals Vy>o and VtJ.
10. Solve the quadratic  \ = ^.
7 « + S
549. I. Find the sum and difference of ViSo^J® and
Vsofl^.
2. Multiply 2 a/3 — a/^ by 4 \/3 — 2 \/^.
3. Solve the equation —^ \ ^ ~~ = 7 — *
7 5 4
4. Solve the equation ~ ^—^ = — •
* ^ a: — 2 X — I 20
5. The sum of an arithmetical progression is 198; its
first term is 2 and last term 42 ; find the common differ
ence and the number of terms.
6. Expand to four terms (a* — V)^,
7. Simplify the radical {cfi — 2a^ + ait^)K
8. A and B together can do a piece of work in 3I days,
B and C in 4f days, and and A in 6 days. Bequired tiie
time in which either can do it alone, and all together.
9. Find 3 numbers such that the prod, of the first and
second may be 15, the prod, of the first and third 21, and
the sum of the squares of the second and third 74.
550. I. Given i + i = 2, i + ^ = 3> + 7 = 3; ^^
X y X z y ^
X, y, and z.
2. Given ~ = , to find x.
8 — X 3 12
3. Find the I. c. m. and g. c. d. ot a? + 4X — 21 and
Q?'— X — 56.
4. It takes A 10 days longer to do a piece of work than
it takes B, and both together can do it in 12 days. In how
many days can each do it alone ?
5. Substitute y + ^ ioT xinx^ ^ a? + 22:^ — 3 ; simplify
and arrange the result.
t «
ANSWERS
INTRODUCTION.
Page 15>
I, 2. Given.
3. 2 cts. A, 6 cts. 0. ;
4. 18, h. ; $32, c.
5. 9 and 27.
6. itf?, 0; 8p, B; i6p, A.
7. i2y, son ; ^6y, father.
Page 16.
8. I20, B's ; $80, A's.
9 iS> 30. 45
10. I7, calf; $56, cow.
11. $5.25, bridle;
$10.50, saddle;
$110.25, horse.
12. $3000, daughter;
$6000, son ;
$27000, wife.
13 234, 702, 936.
rage 19.
13. Given.
4. 98^.
5 i8
6. 10.
7 34
8. 17.
Page 21.
1. 60.
2. 40.
3. ac + 8J.
4. 5J — 2d,
5 35
6. 24.
7. 3X + 2y + ab.
8. 6ft — 7ca; + 3a.
9. bxy + ca:^.
10. ^^^ + a.
II.
2is;
J — a
+ 2Z,
12. 3a; + a:y + 6^2;.
ax — ay — bx + by
d
13
14. 92.
15. 120,
296
8UBTBACTI0K
ADDITION.
Poire 24.
Fage 26.
2.
16 cts., b;
ij 2. Giren.
I. 24/1+28—3^.
30 cts., k.
3. 2iab.
2. i6m«— a?y+&(?
4. ii^y
3. i6Jc+a;y— mn
Bage 28.
5. ISO*
4. 4ai — 3m«+22;
3
26 peaches ;
6. — 2^lcd.
5. isxy+ab+b.
49 pears.
7. — 163^.
6.' Given.
4
15 and 70.
8. 45aS*.
7. 2i(a+J).
5
15 b> 25 g
9. — 39«^^^y'•
8. i9<;(^ — y).
6.
I.
10. 2gl^dm\
9. la^xy.
7.
7.
II. Given.
10. 6Va.
8.
•
7
12. 4.
II. 10 Va;—^.
9
356, A; 94, R
13 5
10.
36 and 141.
rage 27.
II.
8.
12. Given.
12.
9 cts., top;
Poire ^^*
13 a{T—6h+zd
23 cts., balL
14, 15. Given.
— 37W).
13
*9>b; •31^ 8
16. 82;.
14. y(ai+3— 2c
14.
26 cts.y A. M.; 
17. aic.
5w).
74 cts., p. M.
18. —1 2*.
15. m(9+aS— 7^
15.
Given.
19. — i2y.
+ 3^).
16.
14.
20. — 2m.
16. a;(i3a— 3J+C
17.
7
21. I.
— 3rf+wi).
18.
12.
22. 75
17. icy(a+S c).
19.
60. ^
23. Given.
I. Given.
SUBTRACTION.
20.
20.
Po^e SI.
6. 272;^.
Pofirc 5^.
I, 2. Given.
7« 43«^
II.
38fl8J.
3. i^yz.
8. 37fla?.
12.
0.
4. — 62a3.
9. 5ia2J.
13.
— Tjrnhc.
5. 1 90^.
10. — 442^1
14.
53a5*y.
MULTIPLICATION.
297
IS *iSo
1 6. 25°.
17. $420.
18. 4xy—6a.
19. i3^+i6am.
20. iSay^ + y^ + 6a,
21. i^ab + d—x
—Sm + sn.
22. gcd—ab — 2m
+3W+4y.
23. 18^—23.
24. 120^ — 13a;.
25. i6aS+i3c+cZ.
26. a — 5+c.
27. 6(a+b).
28. 9 (a— J I a:).
29. 5(a + S).
30. — 7(2:8— y).
31. $600, A'b.
32. 60°.
33. Given.
34. {2b'C + d)2?^.
35. (ab'—C'd
36. a^(y—b + c).
37. x{ab'3c—d
38. a:y (8— aJ + c
39. c(2a + bm + d).
rage 34.
13. Given.
4. S— c+d — w.
5 S^+y—aif
+4d.
6. 2a — b — c+x
+y+d,
7. a—b + c—a
•jc+c—a+b.
MULTIPLICATION.
rage 36.
14. Given.
5. 42abc,
6. 3^abcxy.
7. Sdmxy.
8. S^bcdxyz.
9. 56a5a:y.
10. 42acdx,
11. $4bcdm.
12. S^adfxyz.
rage 37.
13. Given.
14. — 45flJa:y.
15. 42ade;t?.
16. i$2abcxy,
17. — 4i4a5(ja;y.
18. 945J(?(?a;y.
Pagre 38.
1921. Given.
22. i^s^y^,
23. 24aW
24. O^iC^y^,
25. d^i^^\
26. 6a?yh.
27. iSa^S^c^.
28. 18.
29. 240.
30. 6x^y,
31. — i8a^5^(?.
Po^e 5d.
32. 4a?y2
33. 2iaW
34. 4oc4{r»y.
35. 28aW
36. — 6oeh^.
37. 2iaWc^»
38. — 28^8^*.
39. ai^yV.
I, 2. Given.
3. 6<icQ(?\'2^(^d.
4. i$c^lfh^—6acdx
5. —Sd^bd
+ 6dit^d — 2bdm>
6. — 150*6?
+ 2oa^J2^+ loaV
7. 8. Given.
Pagre 40.
I. 6ax+3bx
+ 2ay+by.
298
DIVISION.
Po^e 40.
a. zo^+A^y
4. dhxy — 2db
+ 6czy — 2ac
5. ^ax+4bxrcx
—Zciy—Afiy+cy.
6. 5aa?+3ay+a2;
+ 5*c+3*y+S2f.
7. i4C(?m2; — 6dbm
— 2icdnx\gdbn.
8. 24ad{^+i2m2;
— S^abcy— i6my.
11. sabtf^xysf",
12. ii/^c2^"'+n
13. a^+".
14. cx{a+by.
16. a5c(a;+y)'»^".
Page 41,
20. a*fft^.
21. a*+a2j«+^*.
22. a:*+a^+i.
— 132^2 — 4X^y^
+ 22xy—^o.
Page 47.
I, 2. Given.
3. 2ad.
24. 24a^ — 6a^^,
25. 6?+ida;+c^ic
+ bc3^.
26. Given.
27. Q^—y\
28. o* + o8+o+i.
29 2:^+3^^
+ 3a:y2+yS.
30. a*»+2a"'i»
31. ic*42a:y + 2a;2?
^y^j^2yz\^.
Page 43.
1. a^+2a+i.
2. 4a2+4afi.
3. 4a'— 4£iJ + S2.
4. a^+2xy+f,
5. ^'^2xy^yK
6. I— a^.
7 49^*— i4J/^+y^
8. 1 6m*— 9^2.
9. T^'f.
10. I — 492:*.
11. 16a:® — 8a: + i.
12. 2S&2^io5+i.
13. I— 2aj+a:2.
14. i+4a: + 4a?.
15. 64J2— 48aJ
+ 9«^
DIVISION.
4. ^xy.
5. 5
6. 29.
16. aW+2abcd
17. ga^^4y*.
18. «*— y».
19. a;^ — 2xy^+y^,
20. 4a* — a;*^
Page 45»
I, 2. Given.
3. 36.
4. 24 chickens.
5. Given,
6. 48.
7. i960.
8. $96o»
9 25.
10. 16.
11. 56.
12. 77.
13. 36 apples.
14. 70 sheep ;
100^ hoth.
15. 16 and 12.
16. 24 plnms.
17. 42.
18. 144.
19. 2i; i4.
20. 141^ bu., one ;
6^ bu., other.
7. 5«*
8. 3JC.
9. 4mn.
J
DIVISIOIS'.
290
lo^ II. Oiyen.
12. — 3(?.
13. ^i.
14 s^
15. — 6J.
16. 76^.
17. — 9«y
18. Given.
19. dK
20. «•.
«i. A
22. iS^.
23 4*'
20;
24. — •
y
25. Given.
26. 8a6c^.
27. — 61:2:.
28. 5a&
29. 72;^.
abc.
2abc
Sa^c.
30
31
32
33
34
35
a
36. 122^2?.
37. IIWl^.
Pagre 4:9.
13. Given.
4 S^+c»+#.
S 3^+5
6. 3JC— 1+4J.
7. 25y2.
8. — Zic+y.
9. ^+2J— I.
10. —505—45 + 6.
11. 3aS— 3a.
12. — 42:2 — jfP
+aa:.
13. a*— sa + 25.
14. i+sa — <)ad,
15. 2a— 4S — 5c.
16. 2(a+J)2
+3aj(«+5)2.
17. 9a:— 9y.
18. x^b—c)
—a (5— c).
19. 3^2 — 2fl.
20. a — a^+o*.
i^ 2. Given.
3 a?+y.
4* flf — i.
5. a*— 2ai+J3.
6. c+rf.
7i a:— tZ.
8. 22: + 3^.
9. a — 5.
10. a:+y.
11. a3+fljj+y.
12. 3fl + 2i.
13 «+2.
14. fl?— 2flKca;'.
15. *2a.'3+4a;2+8aj
+ 16.
16. a:+s.
17. a:— 2.
18. c — X,
19. a +5.
20. 2 (a— J).
1. 10 yrs., son ;
46 yrs., father.
2. 15, P.'s m. ;
45, J.*s m.
3. 12 and 60.
4. 12 and 45 p.
5. 31 cts., ist;
62 cts., 2d ;
97 cts.^ 3d.
6. 20 cows ;
180 sheep.
7 i3fcless;
43i greater.
8. 9.
9. 5 hours.
10. 8.
11. 7.
12. 5 of each.
13. 4 hours.
14. 33i
15. 10.
16. 7 m., A's No. I
14 m., B's ;
21 m., O'a
300
FACTOBIKG.
1 7. 12a;, A'sm.;
18.
s. ^
[5^ and
20.
20. 8; i6y 24.
t4X, B's m. ;
19.
10,
A's;
21. 24.
$Sx, G's m. ;
20,
B's;
$142;^ all.
3o»
C's.
FACTORING.
Page 6t4.
4.
(^.v)(a^y).
I, 2. Given.
5.
{m + 2n) {m + an).
3. 2, 3, 3, oaSJ.
6.
(4a + i) (4a + i).
4. 2, 2, sbxxxyy.
■
7.
(7 + s) (7 + 5).
5. 5, 7, adobhcc.
8.
(2a — 35) (2a — 35).
6. 3> 7. iryy%^«.
9
{y + 1) (y + !)•
7. x^xxyyyz.
10.
(i^)(i^)
8. 5, 5, ahhcxxx.
II.
(af» + y») (a;"» + ^).
9. 7, iiaabccd.
12.
(20*— i) (2a"— i).
lo 5, i3,wwnnwa:.
13.
14.
Pagre 55.
13. Given.
4. i{y + c + 3^)'
Bnge 67 •
5. 2a (a; + y  2i!?).
I.
Given.
6. 3&(?(a; — 2a;a).
2.
{a + x) (a — x).
7. 8tZm{w 3).
3.
(3^ + ^y) (3a? — 4y).
8. 7a (5m + 2^).
4.
(y + 2) (y — 2).
9. 2'jd{bx—/imy),
5.
(3 + ^) (3 ~ ^).
10. 3a2(2j + 3c).
6.
(a + i)(a— i).
II. iaxy{z^ + s).
7.
(i+5)(iS).
12. 5(5 + 3^ — 42?^^^
8.
(Sa + 4*) (5« — 4*).
13. x{i ^x + x^).
9.
{2X + y){2X'y).
14. 3 (a? + 2 — 3y).
10.
(i + 4«)(i — 4«).
15. 19a' (a;— i).
II.
12.
(5 + I) (5  I).
(^ + y') {^  y^).
Pteflre 56,
13.
{ax + 6y) (aa; — hy\
I, 2. Given.
14.
(m2 H w2) (wi2 — w2).
3. {a \'l)(a + I)
.
15.
(a« + fr*) (a« — J").
MULTIPLES.
301
Page 59.
I. Given.
4. {x^i){x+ i).
6. Given.
7. (S — . a;) (5 + a:).
9. {a + b) (a«  a*S + a^i^
10. (a; + i) (a^ — a;2 +3;— i).
11. (i + a) (i — a + «« — flS
+ a*  a«).
12. (rt + i) {ci?
— a* + ««
13. Given.
0^ + fl*^
a^+a i).
14. (x^y){a^^a^ + x^f
15. (a + i) (a2 — o + i).
16. {a + i) (a* — a^ + a^ ..^^
17. + y) (i  y + y^).
18. (i + o) (i — a + fl^ _ ^f
+ «*).
19. (i +J)(iJ + i«~68
2028. Given.
I, 2. Given.
3 a:.
4. }.
5. flc.
6. 2X,
7. 7w.
8. 6ad.
rage 68.
I, 2. Given.
3. s^cifly^<?d,
4. ?>o7?y^A
5. goaWc^,
6. 42oa*J*.
DIVISORS.
Pagre 65.
I, 2. Given.
3 3«^
4. 2aa;y.
5. 4ah^sP.
6. 6flKi?;?;2.
15. Given.
6. a; — y.
MULTIPLES.
7 315^^^
8. S^m^n^.
rage 69.
911. Given. .
12. a* + a^ — flS^
7. a + J.
8. d + 2.
9. x + s.
10. a — 2.
11. a + 3.
12. a; +1.
13. a — J.
14. a — *.
15 ^ + 3^+3^
+ I.
13. a^ + a?*— a;— I
14. 6a^ + 1 la*
— 3a — 2.
15. m* f 2m' — m
— 2.
302
EBDUCTION OF FEACTIOKS.
REDUCTION OF FRACTIONS.
Fage 74:.
13. Given.
I
4.
sac
6. f
'jabc^
8.
a — b
a + b
x + y
x — y
10
II
12
13
14
15
3y — 3^
2a; 22?
I
— •
I
x^^f
X
a + X
I
a— I
I
a? + y
1. Giyen.
2. a — X.
4. b'—c.
5
* + '' + /c
6.
a — S.
7.
2ad
e
a
a — x
9. 3a;+i_g.
1. 2. Given.
4a?y — ^
3. — ^^^ •
y
6.
7.
8.
loM + a — (?
2b
a^+ 2ab + l^+ 2x
 m — '
a + b
7? — X
X \ 1
i2ac — a + b
39^_+_3?.
rage 76.
I. Given.
12TnX
3
4
5
6
6m
4^^
iSggg + 24b<?
a^ — f
x + y'
6a^y — 4^0 ^
3^2 — 2b
I^age 77.
I, 2. Given.
ab
2ia^
49a
a? — y®
^' a^ — 2xy + ^
6.
32fl« (a; + y)
8a2 {x + yy
BEDUCTIOK OF FRACTIONS.
303
h
3
6.
8.
lo.
II.
12.
rage 78.
2. Given.
2CX 2hd cPx
2dx' 2dx^ 2dx '
a(?y 2hxy 2c^
2(?xy^ 2(?xy^ 2ch:y
2c? + 2ah ^hx
Zah H 36^' 3«* + 3"*^'
a:^ — 2ary + y^
a^+2xy + f
a^ + ah i$a — 3
3a ' 3« *
hdx 2ad he + i
'hd'W~bd~'
2l^c—2l^d ^ac—^ad
3^c+3ra
2f?c—i}^d
2hxy hz 4az
2hz ' 2&i?^ 25j2;
ao; + ay 6x + 6y
2X + 2y ^ 2X + 2y '
23^+21/^
— ^— ^— — — ^ .
2X + 2y
c? — 2ax + a:^
g^ + 2ax + g^
2.
Pofire 79.
I. Given.
2flca; 4^ Say
4Sca;* 4ica;' /^cx
y?d 2hcx ^hxy
Zahc^ ^abc^ ^dbc
Say ^hy ^cy 12a;
i2y* 12^* i2y* i2y
4ah^ Scd^ h^
i6a25 i8aV 243?
6.
8.
24a2c* 24a*c* 24^1% '
3«^c
^^^— "^ .
a(; 20(2 2a;v
7. — • — , — ^•
2hc 2bc 2hc
{a + hy {a^hy a^+V
4a?y(a?+ y) 6g(a;hy)
^' 6a;y(a;+y)' 6a:y(a;+y)'
dbxy
6xy{x+y)
ad hx
II.
12.
13'
Vcdx acdm aVy
W&d' WM' W&d
a^z ayz+hyz dy^
xy^z* xj^z ' xy^z
4cma^ + 4cnx^
i2a^cx^ *
6a€m — 6acn ^ahri^
304
ADDITION OF FRACTIONS.
ADDITION OF FRACTIONS.
Pag^ 80.
I, 3. Given.
3 ^ —
2xy
sgdxz
$abc
^ X
6.
ya + 2b
7. Given.
rage 81.
8. Given.
Sax + 6 + 9ay
12a
abj—ac + bx + ex
2:2;^
2a + 2flKr + 3
10.
II.
12.
13
14.
IS
16.
ay
ax — ay + abx + aby
a? — y^
Scd^ + 302:y + sb da?
i^dx
^ah — 2dn — d^
Zdh
17.
18.
19.
20.
I.
2.
3
4.
S
am — dy
^y
nx — mx — Ay
my — wy
— 6.
4arfx + 6bcx — Mw
dcte
Given.
fee + 2e?
2X
x[
am — ay ^bx +bd
bm — by
sd {a + S — c —
xy+z
5^ +
2a — by
2b
rage 82»
6. Given.
$bd + 2a
8.
a
30^ — 2a;y — y* + a — b
iJ? — y
a; — y — a^ 4 6aS — 5 J*
II. ^ 7 ^^ —
a —
2a^+2xy — 2a;— 2y+a + J
a?— I
b
a
+ b
■4cy
c
X
+ y
a»
10.
12.
MULTIPLICATION OF FRACTIONS.
305
SUBTRACTION OF FRACTIONS.
JPofire 83.
I, 2. Given.
gahc
^ IT'
4« •
a
5, 6. Given.
ay — dm + hn
7.
8.
my
by ■— dy + hm
my
12
10.
II.
Page 84.
hy + hm + dm
my
h — my
'~y
6(? + ch
12. a +
13.
cd
6a + 5 J  2^
ad + ay — be + ex
bd — cbc + by — xy
15. a —
2a; + zdy
2y
16. ^ :
loiB + loy
4a? — y + 3g + 6a ^
MULTIPLICATION OF FRACTIONS.
13. ahe.
a + i
Bage 85.
14. Given.
5. A + 3A
^ ab
6. — •
4
6ca? — t)ey + 4t?a; 
edy
15c + 4c?
8. 2a&t;.
a + b
g. J .
^ 4 + 5^
2a^ + 2^2}
J + I
II. 2>x^ + 12a;.
12, 2aa; — zbx + 3a 
3*
14.
IS
16.
5
5
9^ — 3^
17. 3a?y + 3^
18. J^.
I, 2. Givien.
3. 6a:y.
306
MULTIPLICATION OF PBACTIONS.
4. — •
ay
6.
AJhy
Zcx'
(?(g + 8)
ex
9^ 10. Given.
7
8.
II.
12.
13"
14.
2xy
h — 2a
sab
' dhu + g^ + gi
xy+2X + f+ 2y
xy
ic* — V*
16. 2J + 4.
I. Given.
2. •
y
dbd + flrcrf
mx + nx
s — d —
a? — I '
8. 7a? — joas.
acx — acy
10.
II.
12.
13
3«c
2aa:^ + 2(ix
3^ — 3
Sxhf
a + b
6am
14
a; + I
2abxy + Vxy
4a + 6
15. I — w.
gcx — 3ffe
I. •
4
2. 32? (y + 1).
ay + 2a? + y» + ay
^ ^^
4. 90^
a
S6'
6. x»
10.
a; — i?
8. 6aY'
9. a^y».
5
a;* — y*
11. — 9 — •
12. 2 J + 4.
13. 2a (c + d).
I4> •
IS ^.
a?
DIVISION OF FRACTIONS.
807
DIVISION OF FRACTIONS.
BageBO. 4. Given. ^^ A^f
14. Given.
5
6.
2X
n
2a
\ h
7. I+
8. ^±i^.
X
a
^ 20
a^ + ac\(^
IX. ^.
Page 92.
1. Given.
2. 3 times.
3. S
4. Given.
^' cdx '
7. — •
2y
8.
X — I
a^ — a
^ 2
2x^y
3
II. 6.
12.
2a + 2b
,. 35Jl32.
X — a
14.
'5
6hi? '
sax + say
rage 93.
8, 9. Given.
a^+ 2a + 1
^^' a^ — 2a+ r
17.
18.
19.
3a« + 3J«
b '
b
4dy
20.1^.
I. Given.
abdmy
ex
mx + nx
4. •
25ay»  s^y
5. y
50^ + 5«
O, .: •
X + 1
7. 3^ — 3«a?.
I.
Sa
II.
a:2 — y^'
if
ot^ — xy^ + o^ — f
12. — \
x^ — f
4a;y«
^ 35*<^^
2. '
3
4
5
24^
cd
3^
ajhy'
232:2?
viyjahby
308
SIKFLE EQUATIONS.
6.
■ ' ■^—■— ■— <
c
7. Given.
a
8.
10.
c + 2'
II.
2C
12.
a^ — ac + c^
4(fl^ 2aa; + a?)
I
2? + 2a; + I *
SIMPLE EQUATIONS.
Poflre 97.
5. 24f
12.
24.
I, 2. Given.
6. Given.
13.
14.
3. a •— J + c — rf.
5^rf — 4ad
14.
I*.
4. + J— ai+c.
15
Given.
Q loa — 2* — c
16
Page 9^.
9 7i.
5. Given.
1 6c
Page 102^
6. 2 — a + J.
10.
15
16.
12.
7. 5 + (j — a — 3.
17.
60.
8. ad — ic + 2W
18.
4
— 8.
Page 101.
19.
20.
9. 17 — 3aS — d.
I. 8.
20.
oi + ^i^
10. 4cd + d — ^ih
2. 50.
6&»
21.
— I.
3 30
a
II. 32 — {? + (Z.
4. 9.
9
6c
12. II.
5 7
^^.
3a + 2*
6. 9.
od — ic
23
7. 72.
*^
ac
Page 100.
8. 30.
2ab
I, 2. Given.
9 5
24.
ac — 2C
3. 15
10. 28.
Sa + i3<^^
4. 12.
II. 12.
25
24
ONE UKKNOWK QUAKTITY.
309
26.
24J+ I
50c — 20a
45
3° Ti
= ■
2
3a — 6
a — 1
[
27.
4
^' 2
 .
28.
I
32. 7J.
2a— I
ac)
33. 84.
29.
c2
— i..
raffe 103.
Pa^e 105.
JPage 106.
34.
20.
I. Given.
IS
13
35.
24.
2. $8, vest;
16.
30. days.
36.
I.
«
$32, coat.
17.
240 m., one ;
37
If
3. $1500, A;
180 m., other.
38.
i6i
$3000, B ;
$4500, C.
18.
12 in., one;
39
i^
4. 40 men ;
16 in., other.
40.
36.
80 boys;
19.
«2S, H. ;
41.
II.
880 women.
*i7S^ C.
42.
1200.
5. 40 miles ;
20.
8h. 24m. A. M.
43
I4f
80 miles.
21.
9 A <iays.
44.
5
6. 1 33 J barrels.
22.
32 of each.
45
4f
7. 12 p., ist;
23
30, 75, and 45.
46.
*/
(!«
'').
24 p., 2d ;
60 p., 3d.
24.
25 cts., ch. ;
2a — 25
+ c
8. 28feet.
75 cts., goose ;
47
25
9. '$120.
I1.50, turkey.
48.
3a — 6
4
10. $50, B's sh. ;
$100, A's sh.;
25
26.
8 ft. 8 in.
40 and 60.
6d
$150, C's sh.
ad .
49
5b
II. 18 yrs., w. ;
27.
—r—jy less.
c + a
50
I — 8a
I + 8a
36 yrs., m.
12. $3000.
ac ,
a
13. 16 and 41.
28.
III28.
SI
2.
4
\0 "
14. $6000.
>
310
8IHPLB BQUATI0N8.
Fage 107.
29. Given.
30. 60 lbs.) b. ;
120 Ibs.^ in*
31. iSF8^B'8 5
30
a
A's/
32. 32jyr8., C'b;
37i " B's;
40I « A's.
33. 117s Totesd.;
1325
ti
fi
a
34. 164 artillery ;
472 cavalry;
564 infantry.
35 «S33iB's;
$633i A's;
t833i C's.
36. $56.25, one;
•937S> o*^®r
Page 108.
37 ^336, P' one.
$280, " other
38. i4yrs.,y'ngest;
16 " next;
18 " eldest.
39. i6day8.
41. 30 and 18.
12a
42. — •
13
43. 225 acres, A;
31S " B
44. 2^hrs.
45. s, istpart;
8, 2d
2, 3d
24, 4th **
46. 9.
47. 47 sheeji.
48. $120.
rage 109.
49. 60 min.
d
CO. •
51. 300 leaps.
ah
52. :, one.
a — c
he
y other.
a — c
53. 72 lbs.
54. 36 honrs;
312 miles.
55. 2oyrs., s.;
40 yrs., f. ,
56. 280.
57. $324, ist;
$108, 2d ;
$144, 3d.
58. Given.
rage 110.
59. 8, istpart;
12, 2d ''
16, 3d "
60.
61.
62.
63.
64.
66.
67.
9 in. and 12 in
<>75
27 days.
$1575, one;
$2625, other.
12 days.
$720.
$384, snm ;
I162, A's sh.;
ii8,B's "
io4,C'8 "
Given.
68.
69.
70.
71.
72.
73
74'
75
Page 111.
6 and 8.
3456, one;
2304, other.
3 m. an honr.
400 in.,
or 33i ft
8 k. of one name
6 k. of another;
3 k."
2 k.''
7 and 8.
246 leaps of d.
I CO days; .
30000 m., ist;
24000 io.^ 2d ;
(€
ft
BIHPLE EQUATIONS.
311
TWO UNKNOWN QUANTITIES.
Boge 114.
1. Given.
2. a;=8, y=4.
3. a:=i2, y=6.
4. a?=i8, y=2.
5. x=i, y=3.
6. a:=i6, y=3S.
7. a?=3, y=2.
8. Given.
9. a;=4, y=5.
10. aj=6, y=i2.
12. ic=io, y=3.
13. a:=:ii, y=9.
14. a:=3, y=2.
15. 16. Given.
17. a;=3, y=s.
18. a;=4, y=7.
19. a:=7, y=:2.
20. a;=i6, y=35.
21. a:=3, y=2.
1. a;=:4, y=s.
2. a?=8, y=2.
3. x—Sy y=3.
4. a:=3, y=4.
5. x=iz> y=4.
6. ic=i2, y=3.
7. a:=3, y=s.
& a?=4, y=3.
9. a?=34, y=46.
10. a?=:4, y=2.
11. a?=i6, y=7.
12. a:=8, y=i.
13. a:=6o, y=36.
14. 3?= 10, y=2o.
15 «=S> y=2.
16. a?=:2, y=4.
17. a;=8, y=6.
18. a:=4, y=9.
19. a; = 6,
y=:I2.
20. ic==i8, y=i4.
1. ir=43, y=27.
2. 4 cts.9 lemons ;
6 cts.9 oranges.
3. 233 V. ; 142 V.
4. 21 and 54.
5. $48, cow ;
196, horse.
6. 40 1.; 50 g.
Poijre lis.
7. 3 and 2.
8. mil m.^ one;
9999 m., other.
9. 56.
10. $320, B's;
I250, A's. *
12. $900^ A's;
$2400^ B's.
13. 31 and 17.
14. $6000 h. ;
12500 g.
1 5. 30 and 20.
16. $560, B's ;
$720, A's.
17. 25 y. and 35 y.
18. I180, ist;
$115^ 2d.
Page 119.
19. 140 m.^ ship;
1 60 m.^ steamer
20. 12 and 18.
21. 108 ft.
22. 3oyrs.;
13 verses,
24. 3 oxen ;
21 colts.
25 53
26. I5000, B's cap.;
I4800, A's "
27. I21 or 63 g.
an
28. X =
n+i*
y =
a
« + I
312
OBKEfiALIZATlOK.
THREE OR MORE UNKNOWN QUANTITIES.
Bage 121.
I. Giyen.
a. ir=7, y=s, 0=4.
3. ir=2, y=3, «=s.
4. a;=8^ y=4» ^^^=2.
5 «=4, y=3> «=S
I. 12 yrs.9 ist;
IS " 2d;
17 " 3d.
$20, c.
3. s, 8, and ii.
4. 630 men, ist;
675 " 2d;
600 " 3d.
50 ct&, ist ;
60 " 2d ;
80 " 3d.
6. a;=24, y^6, ^=23.
7 a^=7> y=io^ «^=9
8. a;=24, y=6o, zr=:i2o.
91 1. Given,
12. «7=2, a;=3, y=4, 2^=5
13. 2?= 2, y=3, 2P=4.
7. 18= ist;
22 = 2d;
10 = 3d;
40 = 4th.
8. 46 m., A's;
9 " B's;
6. 105 min., A ;
210 " B ;
420 " C.
u
C's.
9. $64, A's;
$72, B's;
$84, C's.
GENERALIZATION.
Page 125\
1. 3 chicken&
2. 30 rods.
3 12
S 7 ft.
6. 34.
7. $205, $187.
Pages 127, 128.
8. $961,. A's;
$614, B's.
9. 1248, g.;
902, 1.
10. 4f days.
11. sihrs.
12. 22$ hrs.
13. $67.32.
Page 129.
14. 9077
15* $1036.12^.
16. 587.19 bu.
17. 12^ per cent.
18. 40 per cent
19. 60 per cent
20. I3600.
Pages 130133.
21. $37500.
22. $31250.
23. $2700, B's;
$2300, C's.
24.
25
26.
27.
28.
29.
30
32.
33
34.
35
36.
37.
38.
$5625.
1842^ A.
$55.80.
$190.32.
$1253.
$4i8.6a
$5250.
$3865.86.
$i33929
$2*22.22+.
2jyrs.
Given.
3h. i6i^m.p.M.
6h.32^m. P.M.
9h. 49^ni. P.M.
IKVOLUTIOK '
313
:Page 137.
3. o^J^A
4. 7?i^^.
5. ol^V^(^.
6. i6a:®y*.
7. 2i6aW
8. 6250512*8^*.
9. (i^^¥<}^.
10. a^J^c®^?.
11. a:"y"i?r.
INVOLUTION.
12. (a+J)^.
13. (a + S)^.
14. (a: — y)"*".
IS (^ + 2^)^
16. (a« + S^)^.
17. a»S%^.
19
20.
8a8
i6a8yg^«
21.
49a^y
'90*6^ '
2*
a"*
a"** J"
xn
26. 3fi'\6a^ + 62^
+ i2xy^+24xy
+ i2a:+8y2
+ 8.
1. a* + 4a«5 + 6a^b^ + ^aV + **.
2. a« — sa*S + locfiV^ — loa^J^ + saJ* — J*.
3. c' + ^(fid + 2165^2 ^ 35c*rf3 + 35c»d*+2icW+7«?+(F.
4. a;® + 6a^y + i^oiMj^ + 2oa;8y8 _j. i5a;8y4 ^ 535^6 ^ y«
5. x^ — 'jofiy + 2ia:5va_3^a4^^35a4^_2ia%« + 7a;y«y.
6. y^® + lot/^z + 45^^ + i2oyV + 2iof/^iS^ + 2^2jf^sfi
+ 2ioyV + i2oy^z^ + 45^*2;® + loya;* + i?;^.
7. a» — 9a85 + z^cl^^ — 840^*^ + i26a5J* — 1260*6*+ 84a«4*
— 36a2S'' + 90*8 _ J9,
8. wi^^ + iim% + 55/w*^^ + i65mW + 33om^w*+462m*w"
+ ^62mH^ +33omW+ i6^m^n^+^$mhi^+iimn^+n^\
9. x^ — 120;^^ + 66x^^'i^ — 2200^1^ + 495^*5^^* — 7920;'^^'
+ 9240;^^ — 'jg27^y'^ f 4952:*^ — 22oofiy'^ + 66a^^
— i2iry^ + y^.
10. a^+wflTiJ+w^^^^^a^^^ + ij^^ x^^«^8»
2
2
w— I » — 2
^ W X
X ^ — 5a«*S* + etc.
3 4
314
EVOLUTIOK.
13. 3fi+ S^+ ^X+1.
J4. S^ — 4*8 + 6^ — 4* + I.
15. I — 5« + loa^ — loa^ + 50* — a^>
16. I + na + n a^ + n x
2 23
2
a^ + etc.
rages 143, 144.
17. a^+ S3^y+ 30!^z + 3xf\6xyz\ 3xs^+y^+ 3yh + ^yz^+s^.
tS, 19. Given.
•o. ix?^ + 2x{y + z) + y^ + lyz + ^i^.
II. a' — 2a (Z> — c) + Z^ — 2^^ + c^.
22. a^ + 2a{x + J/ \ z) + x^ + 2X {y + z) + t/^ + 2yz + z^.
23. Given.
9a2 + 12a + 4
24
25
4^^ — 4ac + c^
26.
27,
36 — i6Sabc \ i^6aWc^
49
^ — 6bmxy + gmh:^y^
m^
28, 29. Given.
MULTIPLICATION AND DIVISION OF POWERS.
2023. Given.
a
10.
a^y^z''.
Bagea 145, 146.
II.
Given.
I, 2. Given.
12.
a".
3 «^.
13
x^\
4. ar®.
14.
*2.
s. s«.
15
C2.
6. 0"*+".
16.
x*y~^^.
7. 0755.
17.
4abc~\
8. a^c^tP.
18.
s^y
9. J2ej3y2,
19.
i2a^€^.
24
• a^y
25.
26.
27.
ay
4
b
a
bx~*
a
Pages 148, 149.
13. ai.
14. a?J. ^
EVOLUTION.
¥.
15 y
16. a»
17. a'.
18. a'^.
19. J'®.
20. X^'\
21. y''.
2225. Given.
RADICAL QUANTITIES.
315
Images 151, 152.
I, 2. Given.
8
3 «
4. a^.
5. 4*a;%i.
6. 2a3^2
7. 3a4^iA
8. 2ar.
9. 3*a*a:2.
10. .6a^b.
18 8
11. 2^x^y^.
12. 8at^.
13. (i3)*a;iyi
14. 7a^«3^.
15. ^aUi.
10.  —
Sy
1. Given.
2. X + 2,
3. a— I.
4. I + a;.
5 a^ + f
6. a — J.
7. ic + .
rage 153.
8. Given.
9 ^ +y ^^'
10. fl? — 2ft 4 K
11. a^ H 2^ — 2.
12. I — 2^ •:} a:.
13. 2^2 — 4a + 2.
h
14. a •
2
a; y
15. 
RADICAL QUANTITIES.
Page 156.
15. Given.
6. aVi,
7. 2a'v/2^.
8. S's/xy.
9 6^/3
10. 15^/5.
11. 36a V^.
12. 3aA/2C.
13. 2ia'\/i — 3^.
14. 4i«^A/y.
15 3«V^^
16. 6«\/i3C.
17. i2a\/ii.
Page 157.
1. Given.
2. V^^.
3. \/(2a + *y^.
4. A/(a — 2b)\
5. '\/9a^*
6. V^cn.
7. Vi6^yi2f.
27
9. V^27 (« — J)8.
10. WW.
6
11. v^ai2^.
12. V(a — *)^.
13. v^a*"".
1. Given
2. (a3)i, (54^)i
3. 9*, (125)}.
4. (a «)*, (i 296)i
5. V^i5625,
^87, ^8.
6. \^4c^, ^^725^.
7. v^64a^ '^5''4a*.
8. v^fl% v^^'».
9. (S«)«^% (c2)i.
10. v^(a + Z>)3,
11. A^ (a; ~ y )«,
I, 2. Given.
3. (3*)*, (4*)*.
5. («*)*, w
7, («»)■, (j»X
816
DIVISION OF RADICALS.
ADDITION OF RADICALS.
Fiige 160.
13. Giyen.
4. sVs
S* 2VS+4V3
7. {a^ + 3c)V3ah. 12. {sbx+62?)Vc,
8. (9X + 8a)V2a.
9. 2SV^2. •
10. 118V3.
13. 43^\/y
SUBTRACTION OF RADICALS.
Bage 161.
1. Giyen.
2. 8\/7.
3. 4\/3o — i2V7^
4. 52V5.
5. (21a: — io)Vfl^
6. 2^a + d.
7. 7^
8. gbV^bx,
9. flTt.
10. if A/3.
MULTIPLICATION OF RADICALS.
Plagre 162.
13. Giyen.
4. 90\/io.
5. dbx.
6. Va* — V.
7. ^acxy.
8. a\/ac,
9. Given.
10. v^5^
11. 42 V2.
12. 12a.
13. 6.
14. /^ax.
15. Given.
16. \/s.
17 3^5
18. (m+»)'^fn+«
DIVISION OF RADICALS.
Po^e 16^.
4. V3^, or
a\/3^
5. Z'/bx.
6. (a^ + a:)*.
7. \2(ay)^.
8. 3JV5.
9. i^V*.
2
10. 2aV^
11. (a + J)».
12. l^X"^.
13 ^/x — y.
14. 16^/2.
15 32
RADICAL EQUATIONS
INVOLUTION OF RADICALS.
Page 164.
t, 2. Given.
3. ai.
4. 182;.
5. 8a.
6. — V 22;.
4
7. Sacc^Vd'
8. 9*2.
9. a2+2a\/jf4y.
EVOLUTION OF RADICALS.
rage 165.
1. Given.
2. 3V^«
3. 2V^3ic.
4. ^^9^.
5. 'i^^.
6. '^S^.
^;
8. aci.
9. 2a^^.
10. a^ftA^.
11. Vaa.
12. an^o«».
13. Given.
4. a^.
1 2
5. a^c^.
6. (a + S)i
7. V^c.
8. \^x + y.
9. v^fl + b.
10. Vflf + b + c.
Page 167.
I, 2. Given. .
3. » — 4A/9.
4. 3
5. v7— Va.
6. 31.
7 V3« + Vs^.
8. a — 5.
9. 3 Va — Vs.
10. 41/2^ + 5 Vs.
13. Given.
X
c
x+2Vxy+y
xy
x{Va j V c)
V3 — X
7
8
V3 +
RADICAL EQUATIONS.
13. Given.
4. (c? — a — cy.
5 25.
6. 4f.
7. 256.
8. 1 1 00.
9 3^
10. 21.
11. 252.
12.
2a
13. Given.
I
14.
1 — a
15. aVi
16. Given.
17. 4.
18. ^^.
19.
I —a
318
AFP ECIED QUADBATIOS.
PURE QUADRATICS.
14.
a?= ± 2.
2.
a?= ±8.
:Page 173.
IS
a;= ±3.
3
40 rods.
I. Given.
16.
a;— • ± I.
4
30 rods.
2. z= ±s.
1 'T
X 1 '
5
12, one;
3. a? = ± 3
4. a: = ± 4.
18.
"^^ai
Given.
6.
30, other.
$6.
5. a?= ±5.
19.
a;= ±3.
7.
80.
6. 2?''= ± 4.
20.
a; = ± 2a.
8.
15, less.
7. a; = ± 6.
21.
a;=±\/c2 + d2
60, greater.
8. a; = ± 4.
22.
aj=±i.
9
27 yds. ;
9. a? = ± a/6.
23
a; = ±Va^ + ^
$1.50, price.
10. a? = ± 7.
24.
a; = ± 26.
10.
X ± i6.
II. a: = ± 2,
II.
77 ft.
12. a;= ±a.
Pofirc i74.
12.
a; = ± 16.
13 a? = ± I
I.
a: = ± 36.

AFFECTED QUADRATICS.
Pages 178, 179.
pm
a
JPofire 181.
15. Given.
7
"25
I,
3.
2. Given.
6. a; = 6 or 2.
/ X ^ «^
3 or — 4J.
7. a; — 9 or — i.
±
V^*+^+4ft«*
4.
S or — 6.
8. a? = 3a
8.
— 2a
5
6.
7
i or — 2.
9. Given.
2 or — J.
4 or — 4.
± V* + 4«^.
10. 15 or — 4.
9
7 or — 5.
8.
4 or — I.
II. 20 or — ji
10.
3±2\/— I.
9
3 or — 4f
12. Given.
II.
6 or — 3.
10.
9 or 6.
I. Given.
12.
2 or — 3.
II.
4 or — 3f
2. 10 or — 7.
^^
w
3. 2 or — 5.
13
d
reige 182.
4. 3 or if.
2C
I.
3 on.
5. 4 or — I J.
±
1 V
2.
4 or I.
6. X — 2.
V 4^
3
3ori.
AFFECTED QUADRATICS.
319
4. 2 or — 12.
5. liorf
6. 11 or 3.
7. if or — if
8.  J or  iU>
9. 4 or 2 A.
10. i±a/— a^+i.
11. — m
12. I or — i.
13. I or — 28.
14. 10 or — ^.
IS —ior — f
16. 4 or — I.
17. 4 or — if.
18. 5 or — 4f.
19. I J or — f
20. 4 or — I.
21. ij or — f
22. 4 J or f
23 3 or — If
24. 4 or — I J.
25. f orf
26. ^ or — I.
27. w ± m.
28. 3ft or 305— 3?.
Page 188.
13. Given.
4. ± 2 or ± \/2.
S« ± V3 or
6. v^ = 1.91+.
7. i or — .
8. I or — 8.
9. 4iorf
10. 4 or — 2 if
rages 184, 185.
1. 8 or 4, one;
4 or 8, other.
2. 6o or $40.
3. 6 or 4, one;
4 or 6, other.
4. 1 6s., I5 each.
5. 5 or — 6f
6. 16 scholars.
7. I30 or I20 ;
$20 or $30.
8. 60 or 40, one ;
40 or 60, other.
9. ^6 rds. length ;
28 " breadth
10. 20 in file ;
80 in rank.
11. 10 lambs.
12. 2 and 2.
13. 4 and I.
14. 121 yds. long;
120 '* wide.
15. 6 m., A's rate ;
5 m., B^s rate.
16. 120, A;
80^ B.
17. 42 and 6.
18. 4 lemons, A;
6 « B.
19. 14 ft., length;
10 " breadth.
20. 12 rows;
15 trees in each
21. 52.
22. 20 persons.
rage 186.
23. 8or— 10, less
15 or —12, gr.
24. 16 and 20.
25. 50 and 25.
26. 121 and 25.
27. 12 ft, forew. ;
IS ft., hindw.
28. 2 or — 18, one;
18 or —2, oth.
29. f and f
30. 3, less;
18, greater.
31. 16 or 36 yrs.
32. 28 rods, length;
20 " breadth.
33' 15 yrs., A's;
8 yrs., B's.
34. 20 lbs. pepper.
rages 188, 189.
1. Given.
2. a; = 4 or 3;
2^ =r 3 or 4.
3. ic = 7 or 5 ;
2^ = 5 or 7.
4. a; = 8, y = 6.
5. xz=zio or — 12;
y=zi2 or —10.
6. a; = 10;
yz=z i2Jt^or 7.
^TlG
ARITHMETICAL PROGBESSION.
7. « = 9;
y = 4 or 5.
10. a; = 5 or —3;
y = 3 or —5.
11. a; = 3; y = 2.
12. a;=: ±s;
Ba>ge 190.
15. aj= 15 or 12;
y= 12 or 15.
Pa^ge 195.
3 i
5. f
6. 20.
7 3C.
8. 10.
9 I
rage 204.
2. 4.
3. 6400.
4. 12.
16. a; = 21 or —7;
y = 7 or — 21.
17. a; = 625;
y=:i6.
18. a; = 2 or i;
y = I or 2.
Page 191.
1. 8 or —4, gr. ;
4 or —8, less.
2. 30 yrs., wife ;
31 ** man.
3. a; = 9^/1, gr. ;
y=:±\/2, less.
RATIO.
0. .
1. —
2
2. a; — y.
4. f •
2
5 —
^ 3a?
2. 9.
3. 68.
7. i
8. T^y.
PROPORTION.
5. 32 and 24.
6. 10 and 8.
7. 16 and 12.
8. 6 and 4.
9. 48 and 9f .
ARITHMETICAL PROGRESSION
4. —s
5 If.
6. .91.
8 43
4. 40 rows ;
25 trees in each
5. 40 yds, length;
24 " breadth.
6. 9 and 3.
7. 1 1 and 7.
8. 31 rds., length;
19 " width.
9. ± 7 and ± 4.
10. 25 m. and 23 m.
11. 12 and 4.
12. 3 or — 2, one;
2 or — 3, other.
19. Equality.
20. Equality.
21. Gr. inequality.
22. Less inequal'y.
23. f I > ¥•
24. i^<w
25 7.
26. 98. •
10. 430 r., length ;
320 r., breadth.
11. 20 r. ; 30 r.
12. 9 and 15.
13. 20 and 16.
Page 207.
9 15
10. 44^.
11. 49a?.
12. 3a/t — «•
GEOMETRICAL PROGRESSION".
321
:Page 208.
6.
13
2. 762^.
7.
— II.
3. 216.
8.
0.
4. 1400.
9
255
S 2Sf
10.
(i2.
6. 6x0.
II
61.
•
7. I7S
8. 8io.
Bage 212.
I.
I, 7> 13. i9> 25,
Pa^e 209.
31
1. 58.
2.
3, 7}, 12, 16J,
2. 278.
2 1> 25 J, 30,341,
3. II.
39. 43J> 48.
4. —43
I.
47.
S 2i
2.
— 6.
6. IJ.
3
102.
7. 1024.
4
2, iif, 2 4,31,
8. 192.
4of , 5oi 60.
S
i683f
JPctge 210.
6.
981^.
I 175
7.
5776.
2. 1 130.
8.
10 1 00.
3. 6.
4. 6.
Ta4je 213.
S 259
9
5
GE0ME1
rRi
CAL PROGR]
rage 216.
JPagr6« 2 17 "221.
I. 160.
2.
I1718.
2. 4374.
3
9999.
3 4f
8.
5
27305
3885.
4 320.
6.
8525.
S "2.
I.
30000.
6. — 31250.
2.
15625.
10. 6, 1 3 J, 2o, 28,
35h 42 S, 50.
57i 64f, 72.
11. 12, 21.6, 31.2,
40.8, 50, 4} 60,
69.6, 79* 2. 88.8,
98.4, 108.
12. 975.
14. 3f 5> and 7.
15. loioo yards, or
Sf mi., nearly,
rage 214.
16. 156.
17. $62.50.
18. $667.95.
19. 300.
20. $1.20, int.;«
$2.20, amt.
21. 20, 40, and 6a
22. 16.61 + days.
23 3o> 40, 5o> 60.
24. 3 days.
25. 140.
26. $178, last pay^t;
^5370, debt.
3 2.
4. 3
S 6.
6. 5.
1. 242.
2. 2.
3 500
4. 5
322
BUSINESS FORMULAS.
5 215.
6. 567.
2. 1,2,8,32,128.
I. 4371
3 7174453
4. 2Hfm.
s 9565938
6. 43046721.
8. $4095.
9. $196.83, 1. c;
$295.24, wh. c.
10. I10.23.
11. 2, 6, 18.
12. I4294967.295.
13 lOj 3^y 90> 270.
14. 1 1 20, $60, $30
15 3> i5> 75> 375
1875.
16. $108, $144,
$192, $256.
17 a, or I.I.
18. 8, 4> '2> I*
Page 231.
1. i^.
3 I
4. if
Pagre 237.
2. 1548.86.
3 I973
4. Ill
6. 78.
7 •0375
8. 14.38.
9. 2.723.
10. 2906.3.
INFINITE SERIES.
S 2. 9 I
6. 9.
7. 10.
8. J.
10.
a — I
II. 50 rods.
LOGARITHMS.
11. a8i4.
12. — 4.619.
Page 239.
14. .0003321.
15 33335
16. 191.77.
18. 5.23.
19. 1.0836.
20. 2.504.
21. 2.124.
rage 240*
23 342 + .
24. .546+.
25. .324 + 
26. Given.
BUSINESS FORMULAS.
Pages 245260.
2. $349.60.
4. i6f per cent.
6. $12600.
8. $840.
10. $600.
12. 1 6f years.
13. 10 years.
15. 2^ per cent.
17. $2010.14.
19. $5414.28.
21. $2769.23, pr.w.;
$830.77, disc.
22. $6000, pr. w.;
$1800, disc.
24. $1718.75.
26. $2125.
28. $2.33f
29. $8.83 + .
31 5Mf per ceiit
32. 12 J per cent
33 9A F^ cent.
34. 0. 8 per cents.
so. $24630.54, in.;
$369.46, com.
SS. $1332.
39. $6290.15.
TEST EXAMPLES.
323
40. $905 .Sc).
41. I3278.69.
42. $2278.48.
44. $6130.67.
45. $2767.60.
47. $2336.25.
49. $14166.67.
51. $14775
53. $249.77.
Pages 265268.
2. +Vxy.
3. —6.
4. 6a/ — I.
5. V—xy
7. I.
'• \/^y
«• v^
10. 51/2.
Pagrc ;374,
1. 75a.
2. 57a? + 7
3. 3flra;+3aJ + 2Cc?
4. 7S(? + 3C6^
— ioa;y+ 5wm.
y
TEST EXAMPLES.
18. ±Vab;
±
a + 1
2. Impossible.
3. Impossible.
5 •
6. .
8. 8.
9 31.
10. c (3^ — 6^c
— cd),
11. 3^(y — 3^
— 6y^)
12. (a" + 6")(a"— i»)
13. 2x2 (2a— i).
14. (a2+i)(a+i)
X (« ■— i).
15. 209 II.
16. .
17. 24 shots.
V wi/ ,
a/1
19.
20.
b
I
a — h
21. {z^y + 2z)
x{zxy^2z).
22. (3^>— c)(3*— c).
23 .
24. 640 rods.
25. 29J miles.
bi
26.
27.
28.
h + I
29. I.
a + t
30. .
a —
31 IT'
32. 120.
33 *50' '
34.
35
37.
38.
39
40.
41.
42.
43
Page 276.
6 hours.
^ = 5 ; 3^ = 3.
44.
45
46.
47.
48.
49.
SO
51
85 J miles.
8i g. ; 5i 1
^ = 4; y = 6;
z=S.
x=2; y = 4;
z = S.
X = $40 A,
y = $60 B,
z = *8ojD.
3 meters f. w.
6 " h. w.
8 ft. one; 10 ft
49 and 77.
48 meters.
$2.46 a meter.
$100 horse^
$200 carriage.
324
TEST EXAMPLES.
52. .
53. 65 hectares y.
100 " elder.
54 26.
55. $5 1.; $6 8.
56. 577 v.; 8487.
57. 9oyrs.A;45y.
B; i5y. 0.
58. 9 V3
59. y \/r+ a.
Piige 278.
60. (.T*) i, {f)K
61. V27 (a — i)\
62. 90 cts.
63. $10; $18.
64. 15 men.
65. 28 persons.
66. $4 b. $6 w.
67. 9.
69. •
^ I — a
70. a + J.
7 r.' 240 liters.
72. : .
73. 12 and 8.
Page 279.
74. $180.
75 A
76. $90.
77 '*2i.
78. 24 8. $5 pr.
79. 81.
80. 45 m. ; 105 m.
81. 80 A; 70 B.
82. Vx — ^7.
83. V3« + V3^
84. (^2 + 6e? — 2 J2^
+ 96J2 + J4.
85. $12000 Mayor;
$1200 Clerk.
86. 216 g. ; 3of 1.
Page 280.
87. 8 and 6.
88. 8 cts.
89. 10 days.
90. $1407 B;
$469 A.
91. $50 cow;
$200 h.
92. 6 miles.
93. 100 ft. x6oft.
94 ih i2i, 17,
2 if, 26}.
95 637i.
96. 50 pair.
97. 18 and 14.
98. 32 h.; 75 1.
99. 10 y. B ;
100. $105.
Page 281.
01. 8 : 43i2j. o'cik.
02. 250 pp. A;
320
a
B.
03. 20 days.
04. i4f yrs.
05. $30.
06. $9000 whole 8
$4800 A ;
$4200 B.
07. 184 V. and
185 V.
08. 24 d.; 48 d.
09. 45.
10. 6, 18, 54, 162.
11. 16 hats.
Page 282.
12. 97656250.
13. 10 ft.; 30 ft J
50 ft.
14. 95; 42; 82.
15. 2, 4, 8.
16. 12 weeks.
17. $180; $120.
18. $30.
19. 45 A.; 55 A.
20. 14348906.
21. 125 p. 175 h
22. 375 men.
A
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TextBook oh Bhetoeio;
Supplementing the Development of the Science with
Exhaustive Practice in Composition,
A COrilSB OP PRACTICAL LESSONS ADAPTED POR USE IN HIGHSCHOOLS
AND ACADEIOES AND IN THE LOWEB CLASSES OP COLLEGES.
By BRAINERD KELLOGG. A.M..
Ftofeuor of the EnglUh LangtMge and IjitertUure in the BrocMyn Collegiate
and Pd ytechnic Inetitute^ and one of the authors of Reed <& KeUogg*§
" Graded Leseona in Englith " and *' Higher Lenone in £ngU$h.**
■
In preparing this work upon Blictoric, tLo author's aim has been to
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classes lif Colleges, base d upon the science rather than an exhaustive
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This Avork has grown up out of the belief that the rhetoric which the
pu^il needs is not that which lodges finally in the memory, but that
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tongue and pen. Hence all explanations of principles are followed by
exhaustive practice in Composition — to this everytliirg is made tributary.
When, therefore, under the head of Invention, the author is leading
the pupil up through the construction of sentences and paragraphs,
through the analyses of subjects and the preparing of frameworks, to
tlie finding of the thought for themes ; when, under tlie head of Style,
he is familiarizing the pupil with its f^rand, cardinal (^[ualities ; and when,
under the head of Productions, he divides discourse into oral prose, writ
ten prose, and poetry, and these into their subdivisions, giving the re
?[uisues and functions of each — he is aiming in it all to keep sight of the
act that the pupil is to acquire an art, and that to attain this he must put
into almo.«t endless practice with his pen what he has learned from the
study of the theory.
276 pages, ISmo, ath^activeLy iaund in cloth.
CLJlBE & HATNIBD, 7S4 Bro&dvay, Neir York
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