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English Classics, 

Classes in English liferatitre, /heading. Grammar, etc. 
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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, - - - - - - - 198-203 

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 -j-6 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'T-ic + 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 
eo-operator. 



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^) + ai-f (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 — 2x4-7-2 + 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'{x--i) + a X d X y -j- 2z = what ? 

AXIOMS. 

59. An Axiom is a self-evident 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^ o-f-o 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 =4-40. (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^-i-a^b 

a + b 4a%— 3aJ 




a^+Sa^+saU^+l^, Ans. i2fl»J2_2^j2_3^j2^ ^^j^. 

20. Multiply a^ — ab + l^hj a-i- b. 

21. Multiply a^ — ab + ff^hja^-^ab-^ l^. 

22. Multiply a^-i-x+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)W-yy 


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 i8-i-3 = 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 iSab-i-6a = 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^c-T- 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__^ r-M^ ^^ ^"^^-^^ 

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*+2ai-f 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 three-fourths 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. 268-270.) 



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.'5-A 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)(a-6). (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){x--y). 

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?y-f 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'—^' = (fl-fft) 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« 
lei-m? 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 6-1-3, 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 a-l ^ \i- 
^r 12' 1 ' 2T. J '2 > Divides the fraction. 
Multiplying the denommatory ) 

,-« 2-t-2 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 6-i-2 3 

4°. Multiplying and dividing a ) Does not change its 
fraction by the same quantity j value., 

2X2-S-2 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 

aS-3» 


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 thirty-fifths. 

_ __ 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. 



xyz-T-x =yz 
xyz -T-xy = z 
xyz -i-yz =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 j-xx? 



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(i-t.). 

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. (i-n2)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- r-T-^^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. (i-fl2)^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 ' 
x--y 



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(a-nyi) -- 

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. 2-5.) 

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, 

4H-2 = 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 
ft-actionB 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-[- c-r- 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 ^ — 6-f-7- = T-f^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 r-T = -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 one-fourth 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 one-half 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 
4-J 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. One-tenth 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) 


aJ-y= 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 war-steamer 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. One-third 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 left-hand 
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 
8-y = 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 -\-x-i-y + 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 

s-hd 
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), p-f-6 = 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 hour-hand arrives at I, the minute-hand has passed 
12 hour spaces, and made an entire circuit. But since the hour-hand 
has moved over one space, the minute-hand has gained only 1 1 spaces. 
Now, if it takes the minute-hand 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 minute-hand 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 

(fl-k-by = a» + 5a^+ioa»&»4-ioa26a4-5a&*+&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, 

{a-k-h+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 + \ aj-i 

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 2X4-1 , 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^-i-ofi = 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- = a-i. 
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 = ^'^•ssaass-i- , 

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 4-a 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 
= 4-a' ; and —a x —a = 4-a'. 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^^(a-x) 
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 = iX-J- = ^ 

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 V5-V^. 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 

e-t-b, 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 one-third of a namber multiplied b; 
one-fourth 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 -t-Cy 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 fore-wheel of a carriage makes sixty revolutions 
more than the hind-wheel 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» = 25--iqy+^ (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 ar-y= '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 84-4 = 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 ' A-L ± \ 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 72-1- 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 = bc-i-d, b = tul+c, c = ad-i-b, 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), a-k-b :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 (152-5-2) 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 3-5, 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 (n-i)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-(n-i)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 3-5, 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(5-0 

^ _ log[g + (r-i)g]-logfl 

log r 

_ logZ-log[/r-(r-i)^] ^^ 

logr 



» = 



+1 



fl 



. 8 8 

r» r = I « 

a ' a 



r» + 



8 



^rft— 1 — -. 



1--S' "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 



a-f- 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 324-1-27 =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, i-f-3 = .333333, and so on» to infinity. 

Or, n-3 = 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 (i-f-i)^, 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 twenty-fifth 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 

Logarit|im 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., 3-19453 

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 


4-3 


By 


a — 6 




4* + 4x3 




a^ + ab 




-4x3-3' 




-a6-6» 


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 ^baXp-t-e 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. 242-246.] 

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(i-f 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 semi-annually, - 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 semi-annually ? 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 (i-i-r)»— 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^a-r- - — ■ — • 

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 court-house ; 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* = 8x-8 = -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^ V--i, 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 school-room 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+yf-i- B[a+yy -i- 0{a+p) = D; 
or u4(a'+3d»y4-3fly*+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^'i-C"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's-eye 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-= = tstt-s — 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;4-8« = 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 fore-wheel of a steam-engine makes 40 revolutions 
more than the hind-wheel 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 river-boatmen 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 amiU-pond, 
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*4-30*+ 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» ^ i-a/ • (i-a^)^ 



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 
two-thirds and B three-fourths 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 one-third 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;8y-j ^ ^ '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 + ^4--jV + 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 _ lt-2ia gb^iia 

7-^^^ Sja^b^ SJa-iY 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. 

1-3. 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. 

1-3. Given. 
4. S— c+d — w. 
5- S^+y—aif 
+4d. 

6. 2a — b — c+x 

+y+d, 

7. a—b + c—a 

•j-c+c—a+b. 



MULTIPLICATION. 



rage 36. 

1-4. 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. 

19-21. 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+4bxr-cx 
—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*4-2a:y + 2a;2? 

^y^j^2yz-\-^. 

Page 43. 

1. a^+2a+i. 

2. 4a2+4a-f-i. 

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. 14-1^ 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. — 6-1:2:. 

28. 5a& 

29. 72;^. 
abc. 
2abc 

Sa^c. 



30 
31 
32 

33 
34 



35- 



a 



36. 122^2?. 

37. IIWl^. 

Pagre 4:9. 

1-3. 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?— 2flKc-|-a;'. 

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. 










1-3. 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)(i-S). 


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)(i-J + i«~68 

20-28. 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. 

1-5. Given. 
6. a; — y. 

MULTIPLES. 

7- 315^^^- 
8. S^m^n^. 

rage 69. 

9-11. 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:. 



1-3. 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. 




1-4. 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 

S-6' 

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 

1-4. 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^ 
aj-hy' 

232:2? 
viyja-hby 



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. 28|feet. 




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; 

•93-7S> o*^®r- 

Page 108. 

37- ^336, P' one. 
$280, " other 

38. i4yrs.,y'ngest; 
16 " next; 
18 " eldest. 

39. i6|day8. 

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. 

9-1 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 130-133. 

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. 

$i339-29- 

$2*22.22+. 

2jyrs. 

Given. 

3h. i6-i^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^+wflT-iJ+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. 

20-23. 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- 


C-2. 


6. 0"*+". 


16. 


x*y~^^. 


7. 0-755. 


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''. 
22-25. 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. 

1-5. 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. \^4-c^, ^^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. 

1-3. 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. 

1-3. 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\/jf4-y. 



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«». 



1-3. 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. 

1-3. Given. 

X 

c 
x+2Vxy+y 
x-y 

x{Va -j- V c) 
V3 — X 



7 
8 



V3 + 



RADICAL EQUATIONS. 



1-3. 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. 


"^--^a-i 
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. 


1-5. 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. 

1-3. 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, fore-w. ; 
IS ft., hind-w. 

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- I-973- 

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- 33-335- 

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 245-260. 

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 265-268. 

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. 
b-i 



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^ 

+ 9-6J2 + 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 



v--?>-^ 




^t^'-iK 



^ri 



-J 




V^- (k 'J^i^^ 








^(V^^^too^'"^ 










/>744 1 /•^' 











^' /rut fi/JPL ^yi^>^'^ d^n^iAAO' ^i/iU/ir^^ le^ 



uAe 







•^ 



LM?. / 



'^ .^ 





Text-Book oh Bhetoeio; 



Supplementing the Development of the Science with 
Exhaustive Practice in Composition, 

A COrilSB OP PRACTICAL LESSONS ADAPTED POR USE IN HIGH-SCHOOLS 
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 
write a ))ractica1 text-book for Iligh-Scliools, Aeademies, and the lower 
classes lif Colleges, base d upon the science rather than an exhaustive 
treatise upon the science itself. 

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 
which has worked its way down mto his tongue and fingers, enabling 
him to speak and write the belter for Jiavin^ studied it. The author be- 
lieves that the aim of the study should be to put Uie pupil in possession 
of an art, and that this can be done not by forcing the science into him 
through eye and ear, but by drawing it out of him, in products, through 
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 





Anderson's Historical Serie^ 



A Junior Class History cf t!io Uaitfcd States. Illustrated with 
hundreds of portraits, views, maps, etc. 272 pisiges. i6mo. 

A Grammar School History cf the United Ststtes. Annotated; 
and illustrated with numerous portraits' and views, and with'more than fort/ 
maps, many of which are colored. 340 pp. lOmo. 

A Pictorial Gchod History cf the United States. TmV.j iHvlz- 
trat^ with maps, portraits, vignettes, etc 420 pp. Z2mo. 

A Popular School History of the United States, in which are in- 
serted as a part of the narrative selections from the writings of eminent 
American historians and other American writers cf note. Fully illustrated 
with maps, colored and plain ; portraits, views, etc. 356 pp. i2mo. 

A Manual of General History. Illustrated with numerous en- 
gravings and with beautifully colored maps showing the changes in the 
political divisions of tli3 v/orld, and giving the locatioa of important places. 
484 pp. Z2mo. 

A New ManncI cf General History, with particular attention to 
Ancient and Modem Civilization. With numerous engravings and colored 
maps. 600 pp. i2mo. Also, in two parts. Parti. Ancient HiSToav: 
300 pp. Part. II. MoDi!aL>i History: 300 pp. 

A School History of England. Illustrated with numerous engrav- 
ings and with colored maps showing the geographical changes ia the coun* 
try at different periods. 333 pp. i2mo. 

A School History of France. Illustrated with numerous engravings, 
colored and nncolored maps. 373 pp. i2mo. 

A History of Rome. Amply illustrated with maps, plans, and engrav- 
ings, 543 pp. By R. F. Leiguton, Ph. D. (Lips.), 

A School History of Greece. In preparation. 

Anderson's Bloss's Ancient History. Llustrated with engravmgs, 
colored maps, and a chart. 443 pp. i2mo. 

The Historical Reader, embracing selections in prose and verse, from 
standard writers of Ancient and Modem History ; wiUi a Vocabulary of 
Difficult Words, and Biographical and Geographical Indexes. 544 pp. X2mo. 

The United States Reader, embracing selections from eminent Ameri- 
can historians, orators, statesmen, and poets, with explanatory observations, 
notes, etc. Arranged so as to form a Class-manual of United States His- 
tory. Illustrated v/ith colored historical maps. 414 pp. xsmo. 

CLARK & MAYNARD, Publisliers, 

734 BROADWAY, MEW YORK.