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THOxMSON'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 OF A SERIES OF MATHEMATICS.
Lb=SSSi^ NEW YORK:
Effingham Maynard & Co.,
SUCCESSORS TO
Clark & Maynakd, Publishers,
771 Broadway and 67 & 69 Ninth St.
1889.
THOMSON'S Mathematical Series.
I. A Graded Series of Arithmetics, in three Books, viz. :
Naw Illustrated 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 Classes.) 224 pp.
New Practical Arithmetic. Adapted to a complete business education.
(For Grammar Departments.) 384 pp.
n. Independent Books,
Key to New Practical Arithmetic. Containing many valuable 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
Grammar Schools and Academies. 168 pp.
in. 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.
^lectrotyped by Smith 8c McDougal, 82 Beekman St., New Yo'k.
PREFACE.
XT has loDg 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 methods
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 h^ve 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
rales, 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 difficulties 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.
184012
iv PREFACE.
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 appHcation of Algebra to business
Formulas.
In 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.
Brookltn, N. Y., Sept, 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.
Introduction^ ....•..-9
Definitions, ---••-•- 9
Algebraic Notation, ----••-9
Algebraic Operations, -----. 14
Classification of Algebraic Quantities, - - • ^9
Force of the Signs, - - - - - - - 21
Axioms, --.---•--22
Addition, -.- 23
Subtraction, 29
Applications of the Parenthesis, • " " ' SS
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
Factoriuff, ----..--53
Prime Factors of Monomials, - - - - - 54
Greatest Common Divisor of Polynomials, ^ - (>Z
Demonstration, 6^
Least Common Multiple of Polynomials, - • 69
6 CONTENTS,
PAGE
Fractions, •..yo
Signs of Fractions, --.--..71
Reduction of Fractions, ----- . y^
Common Denominators, - - - • - - 77
Least Common Denominator, - - • - - 7^
Addition of Fractions, - - - - - 80
Subtraction of Fractions, ---.-- 82
Multiplication of Fractions, - • • - - 84
Division of Fractions, --.••- 89
Simple Equations, 95
Transposition, -----.--96
Reduction of Equations, -- - - - - 97
Simiittaneons Equations, - - - - 112
Elimination by Comparison, - - - - -113
Elimination by Substitution, - - - - -114
Elimination by Addition or Subtraction, - - - 115
Three or More Unknown Quantities, - - - - 120
Generalir^ation, 124
Formation of Rules, - - - - - - -126
Generalizing Problems in Percentage, - - - 128
Generalizing Problems in Interest, - - - - 13^
Conjunction of the Hands of a Clock, - - - 133
Involution, 134
Reciprocal Powers, - -135
Negative Exponents, - - - - - - -135
Zero Power, - - - - - - - -136
Formation of I^owers, i3<^
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, -- - - • - -149
CON^TENTS. 7
PAGE
Signs of Roots, -------- 150
Square Root of the Square of a Binomial, - - 151
Square Root of a Polynomial, 152
General Rule, 153
JRadical 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, - - - 179
Third Method of Completing a Square, - - - 180
Problems, -------- 184
Simultaneous Quadratics, - - - - - 187
Hatio, - 192
Proiwrtion, --..--- 196
Theorems, ------- 198-203
Problems, -------- 204
Arithmetical I^rof/ression, - - - - 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 JProffression^ - - - - 215
The Last Term of a Geometrical Series, - - 216
The Sum of a Geometrical Series, - - - - 2 1 7
8 CONTENTS
PAGE
Miscellaneous Formulas in Geometrical Progression, - 220
Inserting Geometrical Means, - - - - 221
Problems, -------. 221
Harmonical Progression, - - - - - 223
Infinite Series, - - - - - . . -226
})Ogarithnis, 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 ProhlemSy - - - - 261
Problem of the Couriers, - - - - - -262
Imaginary Quantities, - • • • - -265
Iiidsterminate 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
ALGEBRA
CHAPTEE I.
INTRODUCTION.
Art. 1. Algebra* is the art of computing by letters and
signs. These letters and signs are called Symbols,
2. Quantity is anything which can be measured; as
distance, weight, time, number, &c.
3. A Measure of a quantity is a unit of that quantity
established by law or custom, as the Standard Unit.
Thus, the measure of distance is the yard ; of weight, the Troy
pound; of time, the mean solar day, etc.
NOTATION.
4. Quantities in Algebra are expressed by letters, or
by a combination of letters and figures j as, a, h, c, z^,
4y, Sz, etc.
The first letters of the alphabet are used to express known
quantities; the last letters, those which are unktiown.
Questions.— I. What is algebra ? Letters and signis called ? 2. Quantity? 3. A
measure ? 4. How are quantities expressed ?
* From the Arabic al and gdbron, reduction of parts to a whole.
10 INTEODUCTION-.
5. The Letters employed have no fixed numerical
value of themselves. Any letter may represent any num-
ber, and the same letter may represent clijfere7it numbers,
subject to one limitation; the same letter must always stand
for the same number throughout the same problem.
6. The delations of quantities, and the operations to
be performed, are expressed by the same signs as in Arith-
metic.
7. The Sign of Addition is a perpendicular cross,
cdXlQdi plus ; * as, +.
Thus, a+& denotes the sum of a and &, and is read, " a plus &," or
••a added to &."
8. The Sign of Subtraction is a short, horizontal
line, called mi?ius j f as, — .
Thus, a — b shows that the quantity after the sign is to be subtract-
ed from the one before it, and is read, " a minus b," or "a less b."
9. The Sign of Multiplication is an oblique
cross; as, x.
Thus, a X 6 shows that a and b are to be multiplied together, and is
read, *'a times &," " a into 6," or "a multiplied by 6.'*
10. Multiplication is also denoted by a period be-
tween the factors ; as, a • h.
But the multiplication of letters is more commonly ex-
pressed by writing them together, the signs being omitted.
Thus, sa&c is equivalent to 5 x a x & x c.
11. The Sign of Division is a short, horizontal
iine between the points of a colon ; as, -^.
Thus, n-i-b shows that the quantity before the sign is to be divided
by the one after it, and is read, " a divided by b."
5. Value of the letters ? 6. Relations of quantities expressed ? 7. Describe the
sign of addition. 8. Subtraction. 9. Mntiplication. 10. How else denoted ?
♦ The Latin term plus, signifies more.
\ The Latin minus, signifies it«*t
DEFIN^ITIONS. 11
12. Division is also denoted by writing the divisoi
under the divide7id, with a short line between them.
Thus, T shows that a is to be divided by 6, and is equivalent to a-i-h.
13. The Sign of Equality is two short, horizontal
lines, equal and parallel; as, =.
Thus, a = 6 shows that the quantity before the sign is equal to tht
quantity after it, and is read, " a equals h" or *' a is equal to W*
14. The Sign of Inequality is an acute angle, with
the opening turned toward the greater quantity; as, ><.
Thus, a>h shows that a is greater than 6, and a<.h shows that a
IS less than h.
15. The Parenthesis ( ), or Vinculum ,
indicates that the included quantities are taken collectively,
or as one quantity.
Thus, 3 (05 + 6) and a + 6 x 3, each denote that the sum of a and &
is multiplied by 3.
16. The Double or Ambiguous Sign is a combi-
nation of the ^[gn^plus and minus; as, ±.
Thus, a±J) shows that 6 is to be added to or subtracted from a, and
is read, " a plus or minus h.'*
17. The character ,*, , denotes hejice, 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 oL
minus ; as, + « + J + c, or —x — y — z,
20. Unlike Signs include both plus and minus; as,
a — b-\-c and —x + y — z,
II. Describe the sign of division ? 12. How else denoted ? 13. The sign of
equality? 14. Of inequality? 15. Use of a parenthesis or vinculum ? 16. Double
sign ? 17. Sign for " hence," etc. ? 18. By what is every quantity preceded ? When
none is expressed, what is understood ? 10, Like signs ? 20. Unlike *
12 INTRODUCTION.
21. A Coefficient * is a number or letter prefixed to a
quantity, to show liow many times the quantity is to be
taken. Hence, a coefiicient is a multiplier ox factor.
Coefficients may be numeral, literal, or mixed.
Thus, in 5«, 5 is a numeral coefficient of a ; in he, 6 is a literal co-
efficient of c ; in 2>dx, ^d is a mixed coefficient of x.
When no numeral coefficient is expressed, i is always
understood.
Thus, a^ means lajy.
EXERCISES IN NOTATION.
22. To express a Statement by Algebraic Symbols,
It is required to express the following statement in
algebraic symbols:
1. The product of a, I, and c, divided by the sum of c and
d, is equal to the difference of x and y, increased by the
product of a multiplied by 7.
Ans. axbxc-T-{c + d) = {x — y) -{- ja.
Or —77^ = {^ — y) + 7«- Hence, the
Rule. — For the words, substitute the signs which indicate
the relations of the quantities and the oj)erations 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.
21. A (Coefficient? When no coefficient is expressed, what is understood f
22. How translate a statement from common lan^age into algebraic symbols ?
* Coefficient, Latin, con, with, and efficere, to effect ; literally, a
eo-operator.
EXEKCISES IK KOTATIOK. 13
4. The quotient of 3^ divided by 5^, increased by 4m,
equals the sum of c and 6d, diminished by the product of
7« and x.
5. If to the difference between a and I, 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
\a and h minus m, equals the product of c and d, increased
by 15 times m.
These and the following exercises should be supplemented by
dictation, until the learner becomes familiar with them.
23. To translate Algebraic Expressions Into Common
Language.
Express the following statement in common language :
Substituting words for signs, we have the sum of a and h,
divided by d, equals twice the product of «, 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 h, Ans. Hence, the
EuLE. — For the signs indicating the given relations and
operations, substitute words.
Express the following in common language:
2. \- a — b =: 1- axy — ^cd,
X c
Sa ^a — hcd
4o ax -\- be = 7,x,
5 X ^ ^
aic — x^ . cdh + x
Zd --^- ' ^^- 2a
xy a—b _x -\- y 2
Sa X a 3c
6 4Q^^y , a—b _x-\- y 2a -\- d
23. How translate algebraic exprespions into common language J
14 Il^^TRODUOTION'.
ALGEBRAIC OPERATIONS.
24. An Algebraic Operation is combining quanti-
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 ?
By Arithmetic. — 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 opebatiok.
A's share by a?, and form an -t-iGt o; = A s share,
equation by treating this letter then ^X = B's share,
as we treat the answer in proving and X -{- 2>^ ^= 12 dollars,
an operation. If x represent A's ^^^^t is,' 4:^: = 1 2 dollars,
share, 3a; will represent B's, and xt ^ i a
, ^ n .1 !> Hence, a; = 3 doL, A.
a;+3a;=i2 dollars, the sum of ' a ^ n
both. Uniting the terms, we 3^ = 9 ^oL, B.
have the equation, 4a? = 12 dollars. To remove the coefficient 4, we
24. What Is an algebraic operation ? 25. A problem ? A solution ? 26. Equality
denoted ? 27- Tlie expression of equality called ?
ALGEBRAIC OPERATIONS. " 15
divide both sides of tlie equation by it. For, if equals are divided by-
equals, the quotients are equal. Therefore, :c = 3 dollars, A's share,
and 3a; = 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 doDars + 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 both 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
B's number, 2.x will represent A's,
and X+2X, or 3a;, will represent
the number of both. Dividing
both sides by the coefficient 3, we
have a; = 5 pears, B's number, and
2X = 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. What
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 m.any as
C, 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 ?
2^. How remove a coefficieijt ?
OFBBATION,
Let
X = B's number;
then
2X = A's «
and
3ic = 15 pears.
, *,
x= s pears, B's.
22;= 10 pears, A's.
16 IN^TRODUCTION".
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, rephed
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?jghter,
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 JPower is the product of two or more equal
factors.
Thus, the product 2 x 2, is the sqttare or second power of 2 ;
a; X a; X a; is the cube or third power of x.
31. The Index or JEx2Jonent 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.
a' ** ax ax a, the cube, or third power, etc.
32. A Hoot is one of the equal factors of a quantity.
30. What is a power? 31. How denoted?
ALGEBRAIC EXPRESSIONS. 17
33. Roots are denoted by the Madical Slf/n ^
prefixed to the quantity, or by a fractional exponent placed
after it.
Thus, -\/a, 0^, or ^a denote the square root of the quantity a ;
^ a shows that the cube root of a is to be extracted, etc.
34. The Tudedc of the Hoot is the figure placed
over the radical sign. The index of the square root is
usually omitted.
(For negative indices, see Arts. 256, 258,)
Eead the following examples :
1. «2 + 3«. 7. 4(a — hf.
2. ¥ — c2. 8. «2 ^ 2ah + 52.
3. « + 52 __ ^, g^ Va + h,
4. 01^ — y + y^* 10. 's/a^ — ^.
5. 2y^ ■\- ^ — z, II. 2«^ + c^
6. 3(a2^^>). 12. 4x^ + 2y\
Write the following in algebraic language :
13. The square of a plus the square of b,
14. The square of the sum of a and i.
15. The sum of a and b, minus the square of c.
16. The square root of a, plus the square root of x,
17. The cube root of x, minus the fifth power of y.
18. The cube root of a, plus the square of i.
ALGEBRAIC EXPRESSIONS.
35. An Algebraic Expression is any quantity ex-
pressed in algebraic language ; as, 30^, 5« — 7^, etc.
36. The Terms of an algebraic expression are those
parts which are connected by the signs + and — .
Thus, ina+h, there are two terms ; in ic+y x z—a there are three.
32. A root? 33. How denoted? 34. What is the fi^ire placed over ft? rac^iQ^
$ipi caU'-l ? 35- Wbat is an alget>r8i<? expression ? 36. Xts t^riug t
18 Il^^TRODUCTION".
Note. — Letters combined by the signs x or -?- do not constitute
separate terms. Such a combination, to form a term, must have the
sign + or — prefixed to it, and the operations indicated b,v the signs
X or -f- must be performed before the terms can be added to or sub-
tracted fiom the preceding term. (Art. 36.) Thus, a + bxc has two
terms, hxc foiming 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, cib contains two factors, a and &, and is of the second degree.
a^x contains three factors, a, a, and x, and is of the third degree.
¥^ contains five factors, 6, b, 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 b}- the signs. (Art. 36.)
40. To Find the Numerical Value of an algebraic expression.
1. If a = s, ^ = 7, and x = g, what is the value of
6« + 85 + 3^ ?
Analysis. — Since a = 5, 6a must
equal 6 times 5, or 30 ; since & = 7, 8&
must equal 8 times 7, or 56 ; and since
x — g, 3X must equal 3 x g, or 27. Now
30+56 + 27=113. Therefore, the value
of the given expression is 113. Hence,
the
EuLE. — For the letters, substitute the figures which the
letters represent, and perform the operations indicated by
the signs.
2. If 5 = 3, c= s, and d=z8, what is the value of
Sb-{-7c + 6d?
Suggestion. 15 + 35 + 48 = 98, ^tw.
37. The dimensions of a term? 38. Degree? 39. Numerical valae of an alge*
bra ic expression ? 40. Hqw found?
OPERATION.
6a = 5 X 6 =
Sb = 7 xS =
30
56
SX = SX9 =
Ans,
27
"3
ALGEBKAIC QUAI^TITIES. 19
Find the numerical value of the following expressions,
when a = 2, J = 3, c = 4, J = 5, and x=z6,
3. 4a + 6ah + 5c = how many ? Ans. 64.
4. (a -\-b) X c xd — (x-^ c) = how many ?
5. (:f — a) + rt2;4- (c -j- «) = how many ?
6. 2; -f- 2 + (6/ — 6') + Jf — ic ^ how many ?
7. f^x— (« X c) 4- (« X ^) + a; = how many?
8. ^ + (:i; X a) + « — ii'' + c = how many?
CLASSIFICATION OF ALGEBRAIC QUANTITIES.
41. Quantities in Algebra are primarily divided into
hnotvn and unknown,
42. A Known 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. IdJce Quantities are those which are expressed
by the same power of the same letters; as, a and 2a,
2X^ and x^»
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 x^.
Note. — An exception must be made in cases where letters are re-
garded as coefficients. Thus, ax^ and Ix'^ are like quantities, when a
and 6 are considered coeflBlcients.
45. A Positive Qnantity is one that is to be added,
and has the sign + prefixed to it ; as, 40^ + 3 J-
46. A Negative Quantity is one that is to be sub-
iracted, and has the sign — prefixed to it; as, 4a — 3Z'.
41. How are quantities in Algebra primarily divided ? 42. A known quantity?
Unknown? 43. Like quantities? 44. Unlike? 45. A positive quantity? 46. A
negative ?
20 Il^TRODUCTION".
47. The terms Positive and JSegative denote oppo-
siteness of direction in the use of the quantities to which
they are applied. If lines running Nortli from any point
are positive, those running South are negative. If future
time \& positive, past time is negative; if credits ^tq positive,
debts are negative, etc.
48. A Simple Quantity is a single letter, or several
letters written together without the sign 4- or — ; as, a.
db, Zxy,
49. A Compound Quantity is two or more simple
quantities connected by the sign + or -— ; as 3« f 4J,
zx — y,
60. A Monomial * has but one term ; as, a, il.
61. A binomial f has two terms ; as, « -|- J, « — h.
Notes. — i. The expression a — 6 is often called a reddual, because
It denotes that which remains after a part is subtracted.
2. A binomial is sometimes called a polynomial.
62. A Trinomial % has three terms ; as, a + J — c.
63. A Polynomial \ has two or more terms ; as,
a -\- h — c ■\- X.
64. An Homogeneous Polynomial has all its
terms of the same degree.
Thus, 2ab + cd + s^V is homogeneous ; but ^ahc ■\-c^ + sx is not.
66. The Reciprocal of a quantity is a unit divided by
that quantity.
Thus, the reciprocal of a is - ; the reciprocal ofa + biB r •
47. What do the terms positive and negative denote ? 48. A simple quantity.'
49. A compound? 50. A monomial? 51. A binomial? 2^ote. The expression
0 — 6 called? 52. A trinomial? 53. A polynomial? 54. When homogeneous?
55. The reciprocal of a quantity ?
* Greek, monos, single, and nome, terra, having one term.
f Latin, bis, two, and nome, name (a hybrid), two terms.
X Greek, treis, three, and nome, name, having three terms.
\ Greek, polus^ manj^, and noim, name, having many terms.
FOECE OF THE SIGKS. 21
FORCE OF THE SIGNS.
56. JEach 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 + 5— 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.
Tims, the expression 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 -f- 2, shows that 8 is to be
divided by 2 and the quotient subtracted from 16, and is equal to 12.
58. If 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 + & or a — 6 is to be multiplied or divided by c, the oper-
ations are indicated by {a+b)x e, or c{a+b); (» — &)-*- c, or •
c
EXERCISES.
!• 50 + 5x2= what number ?
2. 50 — 5x2 = what number ?
3. ac + 4b X 2 = what ?
4. ^b — 6d -T- 3 = what ?
5- 15 + 5 X 3 + lo-^ 2 =what?
6. 18 — 2x4-^2 + 10 = what ?
7. 3X+Sy-i-4-\-axb = what ?
8. 6b — jc X X + ga -^ S = what ?
9. {b -\- c) X xy = what ?
56. By what are all algebraic terms preceded ? The force of each of these signs ?
57. Of the figns x and n- ? 58. Of the parenthesis and vinculuia ?
22 H^TTEODUCTIOS".
la ^x X S^ -^ 2Z -{- a = what ?
11. {b — a) -T- xy -]- 2Z = what ?
12. 3a; + a;^ + 22 X 32/ = what ?
13. The difference of x and y, multiph'ed by a less b, and
the product divided 'bjd = what ?
Find the value of the following expressions, in which
a = s> ^ = 4} (^ = 2, X = 6, y = 8, and z= 10:
14. a-\-(axx)-^C'\-yxz = what ?
15. 2^ ~ (a; — Z») 4- « X ^ X y + 2;^ = what ?
AXIOMS.
59. An Aociom, 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 bj 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 t^^^^^^e 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 oi all its parts.
10. Like powers and like roots of equal quantities, are
equal.
Note. — The importance of tliorougUy understanding the defini-
tions 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 them.
CHAPTER II.
ADDITION.
60. Addition in Algebra is uniting two or more quan
fcities and reducing them to the simplest form.
61. The Mesult is called the Sum or Amount.
62. Quantities expressed by letters are regarded as
concrete quantities. Hence, their coeflBcients may be added,
subtracted, multiplied, and divided like concrete numbers.
Thus, ja and 4« are ya, 46 and 5& are gb, as truly as 3 apples and
4 apples are 7 apples, or as 4 bushels and 5 bushels are 9 bushels.
PRIi\!CIPLES.*
63. 1°. Like quantities only can he united in one term.
2°. Tlie sum of two or more quantities ii the same in
whatever order they are added.
CASE I.
64. 10 Add like Monomials which have like signs.
I. What is the sum of i$ab + isao + igab ?
Analysis.— These terms are like quantities operation.
and have like signs. (Art. 19.) We therefore + ^S^^
add the coefficients, to the sum annex the com- _j- i^ab
mon letters, and prefix the common sign. The 1 tq^A
result, + 4706, is the answer required. (Ax. g.)
47 « J, Ans.
60. What is addition ? 61. The result called ? 62. How are quantities expressed
by letters regarded ? 63. First principle ? Second ?
*■ The expressions 1°, 2°, 3°. etc.. denote first, second, third, etc.
24 ADDITION.
2. What is the sum of — 142^^, — i6a;y, and — i8a:y ?
Analysis. — Since these terms are like quan- — 14^^
tities, and have like signs, we add them as i6xil
before, and prefix the sign — to the result, for r>
the reason that all the quantities have the —
sign — . Hence, the — A^xy, Ans.
KuLE. — Add the coefficients; to the sum annex thecoma
man letters, and prefix the common sign.
(3.)
(4.)
(S)
(6.)
(7.)
sab
s«y
7a'
— 'jhcd
-4^f
sal)
8x1/
3a»
— Zbcd
-3^^^
6ai
xy
4^2
— Sbcd
- a^y^
yah
3«y
a^
— Ucd
-Sa^y^
8. Add 5«&2 ^ i>ja¥+ i^al^.
9. Add — Sabx^y^ — saix^y^— 2Sabx^y\
10. Add s^dm^ + Wdm^ + 9l^dm^ 4- W^dm^
11. If 3a -f 5« + « + 7a = 48, to what is a equal ?
Solution. 3a + sa + a+'ja=i6a ; hence, a=4S-7-i6, or 3. Ans.
12. If 4bc + ghc + 2hc + ^bc = 80, to what is he equal?
13. If xy + s^y + s^y + 4xy = 65, to what is xy equai ?
CASE II.
65. To Add like Monomials which have Unlike signs.
14. What is the sum of ^ab — ^ab — yah + gab + tab
^Sab?
Analysis, — For convenience in operation.
adding, we write the negative terma Sab — S^b
one under another in the right- gab — jab
hand column, with the sign — be- ^^j 3^J
fore each, and the positive terms ; T"
in the next column on the left. ^^^^ " ^^^^ = ^^^^ ^^^^^
We then find the sum of the coefficients of the positive and negative
64. How add lT!o^omia^• which have like signs ?
ADDITION. *Z5
terms separately ; and taking the less sum from the greater, the result
2ab, is the answer. Hence, the
KuLE. — I. Write the positive and negative terms in sepa-
rate columns with their proper signs, and find the sum of the
coefficiefits of each column separately,
II. From the greater subtract the less j to the remainder
prefix the sign of the greater, and annex the common letters.
Note — If two equal quantities have opposite signs, they balance
each other, and may be omitted.
15. Add 4d + ^d — ^d -\- 6d — 2d. Ans. 6d.
16. Add — ^x -}- 6x -\- 8x — ^x -\- gx — yx.
17. Add 3«Jc + i2aic — 6abc + s^^^o — loaic — saic.
18. Add 2J — 5^ + 45 — 65 — yb.
19. Add —6g-\-4y^Sy — gg-\-Sy — y.
20. Add 4m -\- i6m — 8m — 9m + 5m — 10m.
21. If 6ab + i4ab — yab + i^ab — i2ab + i6ab = 32, to
what is ab equal ?
22. To what is bed equal, if bed — $bcd + 4bcd + 4bcd
-Sbcd=7s^
Eemark. — The sum in Arithmetic is always greater than any of its
parts. But, in Algebra, it will be observed, the sum of a positive
and negative quantity is always less than the positive 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 yg and 3(? is neither log nor lod, any more than
7 guineas and 3 dollars are 10 guineas or 10 dollars. Their sum is
79 + 3^- (Art. 63, Prin. i.)
67. Poh/no^nials are added by uniting like quantities,
as in adding monomials.
65. How add monomials having unlike Bigne. ? Bern. What is true of the Bura in
Arithmetic ? In Algebra ? 66. How add unlike quantities ? 67. Polynomials ?
26 ADDITION.
23. What is the sum of the polynomial ^ah — 35 + 46?
— 3^5 — 5^^ -\- ^x — c— 2d; and bg ■\- d + zab -f d ?
Al^ALYSIS. For OPERATION.
convenience, we 3^^ — 3^ + 4^ — 3^ — C
write tlie quanti- — 5«5+ 5 — 2^+ 4a;
ties so that like ^ab + d + J^
tenns shall stand — —
one under another, - 2^> + 3^/ + x + bg - C, Ans.
and uniting those which are alike, the result is —2h + 2,d-{-x-\-l}g—c.
68. From the preceding illustrations and principles we
deduce the following
GENERAL RULE.
I. Write the given quantities so that like terms shall stand
one under another,
II. Unite the terms which are alihe, and to the result
annex the unlike terrns with their proper signs. (Art. 65.)
1. Add $a — ^a -{- 6a + ya -{- ga -\- 2b — ^d,
2. Add ^mn -}- ^mn — ^mn + gmn — xy ■\- be.
3. Add 2,bc — "jbc -\- xy — mn + 1 1 Jc + gbc.
^ Add $ab — 37W?^ — ab -{• ^ab -\- 2Z — ^ab + ab.
5. Add 2t^y — 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-{-b) + 5 (a + b) + 7 {a-\-b)?
Solution. 3 (« + 6) and 5 (a + &) and yia + b) are 1 5 (a + b). An».
f. Add 13 (« 4- J) -1- 15 (a + 5) - 7 (« + h).
8. Add 2>c{x — y) + ic^—y) — So{x—tj) -\-yc{x—y).
9. Add saVxy + saVxy — jaVxy + SaV^^y-
10. Add sVa + 3a/« — sVa + 9 V« — 3 V^.
11. Add S^x — y — 3^/^ — y + sVx — y.
68. The general rule for addition ? 69. How aid quaniities included in a paren-
thesif 'i
PROBLEMS. 27
70. The sum of unlike quantities haying a common letter
or letters, may be expressed by inclosing the other 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 5^2; + ^ix — 4cx?
Solution. 5«ic + 3bx-4cx = (5« + 3&— 4^) aj, or x (5a + 3b— 4c]. Ans.
13. Add ja — 6ba + sda — ^ma.
14. Add ahy + sy — 2cy — ^tny.
15. Add 9m + abm — ^jcm ■\- ^dm,
16. Add iT^ax — 2fix -{-ex — ^dx + mx.
1 7. Add axy + 5a;^ — cxy,
PROBLEMS.
71. Problems requiring equal quantities to be added to each
side of the equation^
1. A has 3 times as many marbles as B, lacking 6 ; and
both together have 58. How many has each ?
Analysis. — If x represents operation.
B's number, then will 3a;-6 Let X = B's No. ;
represent A's, and 37 + 32;— 6 ^Yien ^X — 6 = A's "
= 58, the sum of both. To , /- o v. ix.
. j^ , X 4- 'XX — 6 = ^S, both,
remove —6, we add. an eqiml
positive quantity to each side a; + 3a; — 6 + 6 = 58 + 6
of the equation. (Axiom 2.) 4^ ^^ 64
Uniting the terms, we have x= 16, B's NOo
4X = 64, and x = 16, B's, and ox 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 do 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 ?
70. How may the sum of unlike quantities which have a common letter be ex-
^8 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. If 44a; + 6sx — 24 = 85, what is the value of x?
7. If jx — $ -{- 2X = 60, what is the value ofx?
8. If 4?/ + 2?/ + 5?/ — 7 = 70, what is the value of y?
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?
1 1. What is the value of ?/, if 41/ + 32/ + 2?/ — 1 2 = 60 ?
12. A lad bought a top and a ball for 32 cents ; the price
of the ball was 3 times that 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; -f 6a; + 4a; -f 52; — 8 = 154. Ans. 9.
16. 2X -\- sx ■{- sx — 10 = 130.
17. 4X + 3.r + 7a: — 12 = S6.
18. lox — 4X + gx — 2S = 155.
19. 15a; — 7a; — 2a; — 60 = 300.
fio. 18a: — 4a; -f- a;— 75 = 225.
OHAPTEE III.
SUBTRACTION.
73. Subtraction 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 = 4rtj ; 8& — 5& = 36.
PRINCIPLES.
75. I®. Like quantities only can le subtracted one from
another.
2°. Tlie 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, leaves 6.
Adding —4 to 6 + 4, we have 6 + 4—4.
But (Ax. 7) 6 + 4—4 is equal to 6.
4°. Subtracting a negative quantity is the same as adding
an equal positive one.
73. Define subtraction. The Minuend. Subtrahend. Difference. 75. Name the
Qrpt principle, Second. Illijstmte Prtn. 3 upon the blackhoar^. IlJustmte Prin. 4.
30
SUBTRACTION".
Thus, let it be required to subtract —4 f jom 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
yalue of the estate will be $500— $300 = $200. Taking $50 from the
issets has the same effect on the balance as adding I50 to the liabilities
in like manner, taking $50 from the liabilities has the same effect as
adding $50 to the assets.
76. To Find the Difference between two like Quantities.
This proposition includes three classes of examples, as
will be seen in the following illustrations:
1. From 25a subtract ija.
Remark. — i. In this example the signs are
alike, and the subtrahend is less than the min-
uend. Subtracting a positive quantity is
equivalent to adding an equal negative one.
(Prin. 3.) We therefore change the sign of
the subtrahend, and then unite the terms
25a— 17« = Sa,
2. From 4a subtract ja.
Remark. — 2. In this example the signs
are alike, but the subtrahend is greater than
the minuend. Changing the sign of the sub-
trahend, and uniting the terms as before, the
subtrahend cancels the minuend, and has
—3a left. (Prin. 3.)
3. From 45 «5 subtract — 2gah.
Remark. — 3. In this example the signs
aie unlike. Subtracting a negative quantity
is the same as adding an equal positive one.
(Prin. 4.) Changing the sign of the sub-
trahend and proceeding as before, we have
iSab + 290* = 7406. Ans.
OPERATION.
2$a Minuend.
— lya Subtrahend.
Set Difference.
in addition. Thus.
OPERATION.
4(1 Minuend.
— 'JCt Subtrahend,
— 3a Difference.
OPERATION.
4$ab Minuend.
•^ 2gab Subtrahend.
74«^, Ans,
7^. How And the difference between two like quantities ?
SUBTRACTION. '31
4. From ^hc + 7^ — S^r, take ^bc -{- 2d — 4X.
Analysis. — In subtraction of operation.
polynomials, for convenience, we g^c -}- jd — ^X
place like terms under each other. - j^ 2d -\- dX
Then, changing the signs of all the r
terms in the subtrahend, we unite ^^^ + 5^ ^f ^^^*
them as before.
77. From the preceding illustrations and principles we
deduce the following
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 quantities 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.
2. From 49a:, take 23:^ + 3.
3. From 2Sxgz, take i4xgz.
4. From — 43ab, take + igab.
5. From 4ab, take — i^al).
6. From 4s^y, take + i6xy.
Ans. 1 8c.
. 262; — 3.
(7.) (8.) (9.) (lo.)
From 2oaG 42aa^ 370^25 — 290:^
Take — 23ac ^ax^ — i4«^J + 15^^
77. General rule for subtraction ? Note. How subtract unlike quantitiee.
32 SUBTRACTION.
(II.) (I2.) (13,) (14.)
From 3 1 a^J I gabx^ — ssm^x 4 1 x^y
Take — ya^b igaba^ + 44m^ ^ i2a^y
15. A is worth $100, and B owes I50 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 ic
degrees below ?
17. By speculation, A gained on a certain day I275, and
B lost $145. What was the difference in the results of their
operations ?
(18.) (19.) (20.)
From 'jxy — Sa 8J2 -f jam 13^^ — jy^
Take ^xy — 2a — 5^ — gam — s^^ — ^y^ — ^^
21. From i^ab -{- d — x, subtract ^m — ^n,
22. From gcd — ab, take 2m — 37^ — 4^.
23. From 13m — 15, take — 5m -|- 8.
24. From jx^ — ^x + 15, take — ^x^ ■\-^x-\- 15.
25. From igab — 2c — 7<:7, take ^ab — 15c ~ Zd,
26. From a, take b — c, and prove the work.
27. From II (a -f b), take 5 (<^ + b).
28. From ij (a — b ■\' x), take 8 (« — J + a;).
29. Subtract — 18 (oj + Z>) from — 13 (« -|- b),
30. Subtract 21 {x^ — 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 — 1500; 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?
Remaek. — The svhtraJiend, in Algebra, is often greater than the
minuend, and the difference between a positive and negative quantity
greater than either of them. It js thence called Algebraic Dif-
ference,
SUBTRACTION, 33
78. The Difference of unlike quantities which have a
common letter or letters may be indicated by e?idosing all
the other letters, with their coefficients and signs, in a
parenthesis, and annexing, qy prefixing the- common letter
or letters to the result.
SS' From $am, take 2bm,
Analysis. 3am = 3^ times m, and 2bm = 26 times m ; therefore,
Sam—2bm = {3a— 2h) m, or m (3a— 2&). Ans.
34. From 2l)x^, take cx^ — dx\
35. From ahy, take cy ■\- dy — xy,
36. From ic^, take W — ca\
37. From ahx, take ^cx -\- dx ■\- mx.
38. From ^xy, take alxy — cxy -\- dxy,
39. From 5a<? + J/w<?, take 3«c — Jc.
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 signs of the included
terms remain unchanged.
Thus, a— 6+(c— (Z + e) = a— 6+c— e?+e. Hence,
81. Any number of terms may be inclosed in a parenthe-
sis and the sign + placed before it, if the signs of the
inclosed terms remain unchanged.
Thus, a + &— c + (? = « + (&— c + <f), or <? + & + (— c + <f).
Note. — This principle affords a convenient method of indicating
the addition of polynomials. (Art. 67.)
■ ttt; •
rS.^How subtract unlike quantities having a cw&mon letter or letters ? 79. What
is the object of a parenthesis? 80. How removed when the sign + is prefixed to U.
34 SUBTRACTIOI'T.
82. A parenthesis having the sign — prefixed to it, may
be removed by changing the signs of all the inclosed terms
from + to — and — to +.
Thus, removing it from tlie equal expressions,
1 ,, . ')-=a — o + c — a. Hence,
« — 6 — (c2 — c) j
83. Any number of terms may be inclosed by a paren-
vhesis, and the sign — placed before it, if all the signs of
the inclosed terms are changed.
Thus, a—l-^c—d = a—(b—c+d), or a—'b—{—c+d)y etc
Note. — This principle enables us to express a polynomial in diffei*
ent forms without changing its value.
1. How express a — x -\- c, using a parenthesis?
Ans. a — X •\- c =z a — {x — c), or a — ( — c + x).
2. How express a — h — x — y-\-z, using a parenthesis ?
Ans. a — h — {x -\- y — z), or
a — b — (^ -\- X — z), or
a — h—{—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—\h—c-{d + x)-\-e\=a—(h—c—d—x+e)=a—'b+c + d + x—6.
Note. — Quantities may be included in more than one parenthesis,
by observing the preceding rules.
Remo^^^ the parentheses from the following expressions-.
3. ah — {he — d). Ans, ah — hc + d.
4. h — (c — d -{- m),
5. c^x — {— y -\- ab ^ /^d).
6. 2« — [& + c — (a; + «/) — ^.
7. a — (b — c) — (a ~ c) ■\- c — (a — h).
Jg^ 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 inclose terms in a parenthesis
With — prefixed to it?
OHAPTEE IV.
MULTIPLICATION.
85. Multiplication is finding the amount of a qnaii'
tity taken or added to itself, a given number of times.
Thus, 3 times 4 are 12, and 4 taken 3 times (4 + 4 + 4) = 12.
The Multiplicand is the quantity to be multiplied.
The 3Iultiplier 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
quantity,
2°. The 'product is of the same nature as the multiplicand;
for, repeating a quantity does not alter its nature,
3°. T/ie product of two or more factors is the same in
whatever order they are multiplied.
CASE I.
88. To Multiply a Motiotnial by a MonotniaZ,
I. What is the product of « multiplied by c?
Ans. a X c, 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 multiplication. The multiplicand. Multiplier, Product. 86, Fao
tors. 87. Name Prin. x. Prin. 3. Prin. 3.
36 MULTIPLICATIOir.
2. If I ton of iron costs a dollars, what will x tons cost ?
Analysis, x tons will cost x times as much as 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 25 ?
Analysis. — Since each coefficient and each letter opbbation.
in the multiplier and multiplicand is a factor, it fol- 4^
lows that the answer must be the product of the 2 J
coefficients with all the letters of both factors an- . r~T
nexed. Hence, the
EuLE. — Multiply the coefficients together, and prefix the
product to the product of the literal factors.
Multiply the following quantities :
4. 4ab by 50?, Ans, 2oabx,
5. 6bc by 7«. 9. 'jxy by 8ah»
6. ^abc by ^xy, 10. 6ac by jdx.
7. 8dm by xy, 11. gbd by 6cm,
S, gbcd by 'jxyz, 12, jxyz by gadf
SIGNS OF THE PRODUCT.
89. The investigation of the laws that govern the signs
of the product, requires attention to the following
PRINCIPLES.
1®. Repeating a quantity does not change its sign,
2°. The sign of the multiplier shows whether the repetitions
of the multiplicand are to he added, or subtracted,
90. If the Signs of the factors are alihe, the sign of the
product will be positive ; if unlike, the sign of the product
will be negative.
88. How multiply a monomial by a monora.al ? 89. Name Principle i. Prin. 2,
90. If signs of factorH are alike, what ia the sign ol the product? If unlike 1
MULTIPLIC ATIOK. 37
91. Demoksteation. — There are four points to be
proved:
First, That -f- into + produces +.
I/et +a be the multiplicand and +4 the multiplier. It is plain
that +05 taken +4 times is +4*. (Prin. i.) The sign of the multi-
plier being -r, shows that the product +4^, is to be added, which is
donf» \}j setting it down without changing its sign. (Art. 66.)
Second. That — into -f produces — .
Let —a be multiplied by +4. Now —a taken 4 times is —4a ; for
a negative quantity repeated is stiU 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 4- into — produces — .
Let +a be multiplied by —4. We have seen above that +a taken
4 times is +^a. But here the sign of the multiplier being — , shows
that the product +4a> is to be subtracted. This is done by changing
its sign from + to — , on setting it down. Thus, +a x —4 = —4a.
(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
tnat 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^. (Art. 77.) Hence, universally,
92. Factors having like signs produce -f, and unlike
signs — .
13. Multiply -I- 4ab by ~ 'jcd, Ans. — 2d>ahcd.
14. Multiply — <^xy by + ^ad.
15. Multiply + ()db by + ^dc.
16. Multiply — ^xy by — i()alc.
17. Multiply + i2>aic by — 232:^.
18. Multiply — 35^?/ by — 272*^^.
91. Prove the first point from the blackboard. The second. Third. ' Fourth
92. Rule for signs.
38 MIJLTIPLIC ATIOH.
93. When a letter is multiplied into itself, or taken twice
as a factor, the product is represented hj a x a, or aa\
when taken three times, by aaa, 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 ««?
Analysis, aaa y.axi=: aaaaa^ or a^, Atib. Now aaa = a^, and
aa = 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. aV^(? by a^c, Ans. a^^d^,
21. 2>(^Wxhj 2al^y, Ans. 6a^¥xy.
22. 32;?/2 by 5ic2, ^^. ab'^hjab^.
23. 6a^ by 4«5^. 26. $xyz by 2xy,
24. a'^x'^y by a^a^y. 27. 6a^^c by :^a^dl^c.
28. If a = 3, what is the difference between sa and a^?
29. If a; = 4, what is the difference between 4X and a^ ?
95. The preceding principles illustrating monomials may
be summed up in the following
EuLE. — Multiply the coefficients and letters of hoth factors
together; to the product prhj.^ . :.? proper sign, and give to
each letter its proper index.
Note. — It is immaterial in what or 2 the factors are taken, but it
is more convenient, and therefore cusix>mary, to arrange the letters in
alphabetical order. (Art. 87, Prin. 3.)
30. Multiply — z'^y ^y — 2a;.
31. Multiply 6^2^ by — ^c^lc.
94. How multiply powers of the same letter together? 05. What is the rule for
multiplying moaomifids ?
MULTIPLICATIOK. 39
(32.) {33) (34.) (35)
Multiply 4x7/
By x^y
7«^
5^y
(36.)
Multiply s^y
By - -xf
(37.)
7a5c3
CASE II.
(38.)
— 7ac
(39.)
xyz
96. To Multiply a Polynomial by a Monomial,
1. What is the product of a + 5 multiplied by J ?
Analysis.— Multiplying each term of the operation.
multiplicand by the multiplier, we have axb Ct -\- 0
= fl*, and & X & = &2. The result, «d + &^ is the ^
product required. Hence, the Ans, ah A- V^
Rule. — Multiply each term of the multiplicand ly the
multiplier; giving each jtartial product its proper sign, and
each letter its proper index.
Multiply the following quantities :
2. he — adhj ah, Ans, al^c — a%d,
3. 3«a^ -f 4cd by 2c.
4. 5«^ — 2cd -\- xhj lax,
5. 4«2 — 3^5 _|- ^2 ]3y __ 2.hd.
6. 3^2 — 4W — 2C^ by — $a^c,
CASE III.
97. To Multiply a Polynomial by a Polynomial,
7. What is the product oi a + h into « -f- Z* ?
Analysis. — Since the multiplicand is to operation.
be taken as many times as there are units in a -j- h
the multiplier, the product must be equal to a -{- h
a times a+b added to 6timesa + &. *Now ~2~T k
a times a+b = a^ + ab, and b time^ a+b ^
= +ab+b^. Hence, a+b times g;+5 must -{- ah -^ i^
be equal to a'' + 2ab+¥. Ans. a^ + 2^5. -f W-
96. How multip'.y a r;o'»\TioTninl l)v 1 monomi:;.
40 MULTIPLICATIO:^.
8. Multiply J 4- 2e? — 3c by 05 4- 5.
Analysis. — We multi- operation.
ply each term in the multi- 2a -jr i — $0
plicand by each term in the a -{- b
multiplier, giving to each 2^2+ ab -^ sac
product the proper sign.
(Art. 89.) Finally, we add
+ 2ab + ^ — 3&g
the partial products, and the Ans, 2a^ + ^ab — ^ac -{- b^ — 2,bc
result is the answer required.
98. The various principles developed in the preceding
cases, may be summed up in one
GENERAL RULE.
Multiply each term of the multiplicand by each term of
the multiplier, giving each product its proper sigti, and each
letter its proper exponent.
The sum of the partial products will be the true ^jroduct.
Note. — For convenience in adding the partial products, like terms
should be placed under each other.
Multiply the following quantities:
1. 2a-\- bhj s^ + y' 5« sa + 4b — chjx—y.
2. 3^ + 4y^J a — b. 6. ^x + sy +^^ya + b.
3. /^b — c by 2>d — a, 7. jcdx — ^ab by 2m — $n.
4. 6xy — 2a hj b -{- c. 8. Sabc + 4m by $x — 4?/.
9. Multiply ^ab"" by Sa^. Ans. 24a^"+\
10. Multiply — 'jacff" by — Sa^af. Ans. 56a^a;"*+".
11. Multiply 3a Jc" by xyz"^.
12. Multiply acd"^ by iibcd\
13. Multiply — ax^ by — ax\
14. Multiply x{a-\- by hy c{a + by. (Art. 15.)
15. Multiply c{a^ bf by 5 (a — bf.
16. Multiply a{x-\- y)"" by be {x + y)".
17. Multiply 3X {a + by by — (« + Z>).
98. fiow multiply a polynomial by a polynomial?
MULTIPLICATIOiq'. 41
99. When the polynomials contain different powers of the
same letter, the terms should be arranged so that the first
term shall contain the highest power of that letter, the
second term the next highest power, and so on to the last
term. This letter is called the leading letter,
(i8.) (19.)
a ■)- b 4a^—$ab
a^^2a^+ ab^ i2a^b^-\-4a'^b^
a^^^a^ + sal^-^l^, Ans. i2aW—sa^b^—zaW, Am.
20. Multiply a^ — ab -{- b^ hj a -^ b.
21. Multiply a? — ab + W by a^ -^ ab -^ ^.
22. Multiply x^ ■\- X ■\- I by 0^ — x -\- i.
23. Multiply ^x^ — 2xy + 5 by a;^ + 2xy — 6.
24. Multiply ^ax — 2ay by ^ax + 30!^.
25. Multiply d -{- bx \)j 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 + 6) is equivalent to {a + h)x{a + h).
Note. — Algebraic Expressiotis are said to be developed or
expanded, wlien the operations indicated by their signs and exponents
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) (« + i).
29. Expand {x^ + 2xy + y'^) {x + y).
30. Expand (a"^ + b"") (a"^ + 5").
31. Expand [x -{- y -\- z) {x -^ y -{- z).
99. How arrange different powers of the same letter ? 100. How indicate the mul-
tiplication of polynomials ? ''
42 KTJLTIPLICATION".
THEOREMS AND FORMULAS.
101. Theorem i.—The Square of the Sum of two quan-
tities is equal to the square of the first, plus twice their
product, plus the square of the second,
1. Let it be required to multiply
a-\- b into itself.
Analysis. — Each term of the multipli-
cand being multiplied by each term of the
multiplier, we have a times a + 6 and b times
a + b, the sum of which is a^ + 2ab + &*.
Hence, the A71S.
Formula. (a + W = a^ -\- 2ab + l^.
102. Theorem 2.— The Square of the Difference of two
quantities is equal to the square of the first, minus tivice
their product, plus the square of the second,
2. Let it be required to multiply
a — hhya — b. a — b
Analysts. — Reasoning as before, the re- ^
a-Jf-b
a-\-b
a^-\- ab
+ ab+1^
a^ + 2a6 + ^>2
suit is the same, except the sign of the mid- ^ — ^^
die term 2ab, which has the sign — instead — ab ■\- 1^
of -I-. Hence, the A7IS. a^ — 2ab + ^
Formula. (a — bf = a? — 2ah-\- b\
103. Theorem ^.—The Product of the Sum and Differ-
ence of two quantities is equal to the difference of their
squares,
3. Let it be required tp multiply
a -\-bhj a — b. a + h
Analysis. — This operation is similar to ^ "
the last two ; but the terms -f- ab and —ab, <^ + ^^
In the partial products, being equal, balance ^b fr
each other. Hence, the Ans. cf — W'
Formula. {a + h) (a — h) = a? — l^.
loi. What is Theorem i ? 102. Thoorom 2 ? 103. Theorem 3 ?
MULTIP-LI CATION".
43
104. TJie product of the sum of hvo quantities into a third,
is equal to the sum f^f their products.
4. Let X and y be two quantities, whose sum is to be
multiplied ^y a. Thus,
Tl\e product of the sum (a; + y) x a —ax-\-ay
The sum of the products ofa;xa+yxa=:aa;+ay
And aic + ay = aaj + ay. Hence, the
Formula. a (a? + 2/) = aoc + 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 (a;— y) x a = ax— ay
The difference of the products of a; x a—y x a = ax— ay
And ax— ay — ax— ay. Hence, the
Formula. a(x — y) = ax — ay.
Remark. — 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:
{^x — i) (4.r — 1).
(55+i)(5J+i).
(i _ x) (i - x).
(i +2:z;)(i + 2x).
{Sb — 3a) {Sb - 3a).
{ah + cd) {ah -f cd).
(3« — 2y) (3« + 2y).
{x^^y){x^-y).
{x - y^) {x — y^).
{20^ + X) (2«2 _ ^,
I.
(a+ i)(«+ i).
II.
2.
(26? 4- I) (265+ l).
12.
3.
i2a — h) {2a — h).
13.
4.
{x + y) {x + y).
14.
5-
(^ -y){x- y).
15-
6.
{i + x){i-x).
16.
7.
W-y){iy'-y)'
17.
8.
(4m — 3n) {4m + 3^)-
18.
9-
(^2_^)(^2 + ^).
19.
10.
(i _ 7:^,) (i _|_ 7:^;).
20.
104. What is the product of the sum of two quantities into
105. Of the difference?
third equal to?
Let x=.
: No. apples ;
X
: pears.
'<-
24
^x + x =
:72
4X =
,', X =
:72
: 18 apples.
X _
: 6 pears.
44 MULTIPLIC ATIONo
PROBLEMS.
106. Problems requiring each side of the equatKon 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-
ber of apples, then - will represent the
number of pears, and x + - will equal
24, the number of both. Tbe denomi-
nator of X is removed by multiplying
each term on both sides of the equation
by 3. (Ax. 6.) The result is 3a; + x, or
4a; = 72. Hence, cc=i8, the apples,
and 18-7-3 = 6, the pears. Hence,
3
107. When a term on either side of the equation has a
deiiominator, that denominator is removed hy multiplying
every term on hoth sides of the equation hy it. (Ax. 4.)
2. What number is that, i seventh of which is 9 ?
Ans. 6^,
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^^^ ^ __ ^^^
X X
will - + - = 21, by the conditions. Multiplying x X
34 -+-=21
each term on both sides by the denominators 3 3 4
and 4 separately, we have /\x+2)X=- 252. (Ax. 4.) 4^ "f" 3^ ^^ 252
Uniting the terms, 7a; = 252, and x — 36, Ans. ,' . X = 36
Proof. ^ of 36 = 12, and ^ of 36 = 9. Now, 12 + 9 = 21.
107. When a quantity on cither s'.de .of an equation has a denominator, how re-
move it ?
MULTIPLICATION. 45
6. What number is that, 2 thirds of which exceed i half
of it by 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
l^hole worth ?
A.C(y
9. If -- equals 20, to what is x equal ?
10. If — is equal to 20, to what is x equal ?
4
11. If — is equal to 24, to what is x equal ?
AX
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 i 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 v.
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. 1°. When the divisor is a quantity of the same hind
as the divide7id, the quotient is an abstract number.
2°. When the divisor is a number, the qiiotient is a quan-
tity of the same Tcind 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 IVIonomial.
I. What is the quotient of abed divided by cd?
Analysis.— The divisor cd is & factor of the divi- operation.
dend ; therefore, if we cancel this factor, the other cd ) abcd
factor ab, will be the quotient. (Prin. 4.) Ans. ab.
108. Define division. The dividend. Divisor. Quotient. Remainder. 109. Oi
what is division the reverse? Explain, no. Name Uie first principle. The second.
Third. Fourth.
DIVISIOK. 4:7
2. What is the quotient of i2>al) divided by 6a ?
Analysis.— DividiDg the coefficient of the divi- opbbation.
dend by that of the divisor, and cancelling the com- 6a ) l?>ab
mon factor a, we have T8a6-j-6a = 3&, the quotient A^lS. $b,
required. (Prin. i.) Hence, the
EuLE. — Divide one coefficient by the other, and to the re^
suit annex the quotient of the literal parts.
Divide the following quantities :
(3-) (4.) (5-) (6.)
2c) 4abc 4b) 2obxy 8xy ) 4oxy 16b) S2ai
(7.) (8.) (9-)
gm ) 4sabm 2omn ) 6obcmn 24xy ) g6mnxy
SIGNS OF THE QUOTIENT.
112. The rule for the signs in division is the same as
ihat in multiplication. That is,
If the divisor and dividend have lihe signs, the sign of the
quotient will be + ; if unlike, the sign of the quotient
will be — .
Thus, +a X +b = +ab; hence, +ab -r- +b = +a.
—a X +b = —ab ; hence, —ab -r- +b = —a.
+ a X —b = —ab; hence, —ab -. b = +a.
—a X — & = +ab; hence, +ab -i b = —a.
Divide the following quantities :
10. — ^2abc by — 4ab. Ans. Sc.
11. iSabxhj—$z. Ans.
12. 2iabc by — yab. 15. 4Sabc by — Sac.
13. — 2Sbcd by — 4cd. 16. 6$bdfx by gbx.
14. $$cdm by 7cm. 17. — 'j2acgm by Scm.
III. How divide a monomial by a monomial? 112. What is the rale for the
signs?
48 DIVISlOITo
113. To Divide Powers of the same letter,
1 8. Let it be required to divide a^ by a\
Analysis. — The term a^ — aaaaa, and a^ — aaa. Rejecting tho
factors aaa from the dividend, the result aa, or d^, 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^ -j- a^ _ ^2
(Arts. 31, no. Prin. 4.) Hence, the
EuLE. — SuUraci the index of the divisor from that of the
dividend.
Divide the following quantities :
19. ^ by 6^. 22. xyz^^hj xyz^.
20. x^^ by ^, 23. i6ah^ by ^ab.
21. ac^ by ac^, 24. 6xy by 3^2^
114. The preceding principles may be summed up in
the following
Rule. — Divide the coefficient of the dividend hy that of the
divisor ; to the result annex the quotient of the literal factors^
prefixing the proper sign and giving each letter its proper
exponent.
Proof. — Multiply the divisor and quotient together, as in
arithmetic.
Note. — If the letters of the divisor are not in the dividend, the
division is expressed by wanting the divisor under the dividend, in the
form of a fraction.
25. What is the quotient of ^x divided by 3?/? Ans. — •
26. — 24aWc^ -. 2>^ib. 32. 2>'^x^y^^ -^ dp?yz.
27. — 2i^x^y^^ -^ ^^y^' 33* <)6aWc-^ i2ah.
28. $aW -^ ah. 34. 84d^:i^y^ -7- jd^xy.
29. — jofiy^ -. xy. 35. loSaix^ -i- ga^a^.
30. a^b^c^ -T- aWc. 36. i^2X*y!^ -r- iix^yzK
31. 1 6aWc^ -7- Sa^^c^. 37. 121 m^n^a^ -r- 1 1 mhia^.
113. ITow divide powers of the same letter? 114. Rule for division of moiia
jnials ? Proof? If *^e Jitters of the divit-or are not in the dividend, what i.- done ?
DIVISIO.TST, 49
CASE II.
115. To Divide a Polynottiial by a Monomial.
1. Divide ah ■\- ac + ad by a.
Analysis.— Since the factor a enters into oPERATioif.
each term of the dividend, it is plain that a) ab -\r dC A- ad
each term of the dividend must be divisible Ans h 4- C -i- d
by this factor. Hence, the
EuLE. — Divide each term of the dividend ly the divisor,
and connect the results hy their proper signs.
Note. — If a polynomial which contains the same factor in every
term, be divided by the other quantities connected by their signs, the
:iuotient will be that factor.
Divide the following quantities:
2. 6a^ + 10^2 __ i^a by 2a, Ans, sa^ 4- 5«5 — 7.
3. 40^ — Sa^ + i2«2 by — 2«2. Ans. — 2^2 4. 4^5 _ 6,
4. at/^ -i- a(^ -{- ad^ by a.
5. isx^y + 250:^ by s^y.
6. 6adc — 2a-i-Sah by 2a,
7. —i6by^'{-4y^hj—Sy,
8. i4X^y — yxy^ by — yx^^
9. xy^ -f xz — xhj X,
10. 35fl5 -i- 28J— 42 by — 7.
11. i$a^ — 15^2 by 5a.
12. i6x^ac + i2acd^ — 4xa^c by — 4CC.
13. 4a* — 20^2 + Sab by 4«.
14. 3^5 -f i^a^ — 2^a^)d by 3flr5.
15. ZaV)c — xdatJ^c — 2oabc^ by ^abc,
16. 6a; {a + ^»)2 -+- 9a;2 {a -{- 5)2 by 3ir.
17. ^S{^-y)-V2>o{x — y)hYS'
18. «a;2 (5 — c) — aH {b — c) by ax.
19. i8«4 (a + by — 12«3 (^ _|. ^,)2 by 6«2 (« 4. by.
20. a"+i — «"+2 4. a«+3 by «".
115. Eow divide a polynomial by n -lonomial ?
60 Divisioiy.
CASE III.
116. To Divide a Polynomial by a Polynomial.
I. Divide a^ + Za% + ^aV^ + ^ by a2 + 2ab + J*.
Analysis. — For conve- opbbation.
nience, we arrange tlie terms a^ + sa^b + 3«^ + Z*^ a^ -|- 2rt J 4 5^
so that the first or leading a^^2a^+ aW a + b Quot
.letter of the divisor shall be ^7~~ TTTTs
the first letter of the divi- ^ 0 + 2afr'+d
dend. The powers of this a^-{-2ai^+I)^
letter should be arranged in
order, both in the divisor and dividend, the highest power standing
Jirst, the neoct 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% +b times. Multiplying the divisor by +& and
subtracting the product, the dividend is exhausted ; therefore a + & ia
the quotient. Hence, the
Rule. — 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.
IL Multiply the whole divisor by the term placed in the
quotient ; subtract the product from the dividend, and tc the
remainder bring doivn as many terms of the dividend as the
case may require.
Repeat the operation till all the terms of the dividend are
divided.
Note. — If there is a remainder after all the terms of the dividend
ore brought down, vlace U over the divisor, and annex it to the quotient
ii6. How divide a polynomial by a polynomial ? Jf tba*o is a remainder, what la
done with it?
PROBLEMS. 61
2. Divide 4^2 4- ^ab + i^ hj 2a +bo Ans. 2a + b,
3. Divide x^ + 2xy -\- y^ hj x -{- t/,
4. Divide a^ — 2ab + l^ by a — b.
5. Divide a^ -- sa^b + sab^ — i,^ hy a ^ b.
6. Divide ac ■\- be -{- ad -\- bd hy a -{• b,
7. Divide ax -{• bx — ad — bd by a + b,
8. Divide 22^ -f 'jxy 4- 6^2 by a; + 2^
9/ Divide a^ — V^hya + b.
10. Divide a? — y^ by a; — y.
11. Divide a^ — l^ by o — J.
12. Divide 6^3 4. 13^5^ 4. 552 by 205 -f 35.
13. Divide «2 __ ^ __ 5 by a — 3.
14. Divide a^ -- 7,a^x + 3aic2 _ ^^js by a — a;
15. Divide 6ar* — 96 by 3a; — 6.
16. Divide x^ ■\- "jx -\- 10 by x -\- 2.
17. Divide x^ — 5^? + 6 by x — 3.
18. Divide c^ — 2cx -\- 7? hy c — a?.
19. Divide a^ + 2ab ■{- b^ hy a -{- h
20. Divide 22 (« — Z>)2 by 1 1 (a — h).
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 5 6 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 ?
53 DIVISION
6. A farmer has 9 times as many sheepas cows, and the
number of both is 2oo„ How many of each ?
7. Divide 57 into two ?uch parts that the greater shall be
3 times the less^ plus 3. What are the numbers ?
8. Given 2X -{- 4X -\- .4; — ^ = 60, to find x,
9. A and B are 35 miles apart, and travel toward each,
other, A at the rate of 4 miles an hour, and B, 3 miles. In
bow many hours will they meet ?
lOo Given a -}- $a + 6a + 2a + y = iig^ to find a.
11. Given 85 -f- 5^ -f 7^ — 10 = 130., to find b,
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 a?,
15. What number is that, to ths half of which if 3 be
added, the sum will be 8 ?
160 Three boys have 42 marbles ; B has twice as many as
As and C three times as many as A. How many has each ?
17. If A has 2x dollars, and B twice as many as A, and
C twice as many as B, how many have all ?
r8. 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 0 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
oe added, the sum will be 23 ?
OHAPTEE VI.
FACTORING.
117. Factors are quantities which multiplied togethei
produce another quantity. (Art. 86.)
118. A Composite Quantity is the product of two
or more integral factors, each of which is greater than a
unit.
Thus, 3a, 5&, also x^y^, are composite quantities.
119. Factoring is resolving a composite quantity
into its factors. It is the converse of multiplication.
120. An Exact Divisor of a quantity is one that will
divide it without a remainder. Hence,
Note. — The Factors of a quantity are always exact 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 6, are prime quantities. Hence,
Note. — The least divisor 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 6c, are prime to each other.
123. A Multiple is a quantity which can be divided
by another quantity without a remainder. Hence,
A multiple is a product of two or more factors.
•117. What are factors? ii8. A composite quantity? 119. Wliat is factoring?
120 An exact divisor? 121. A prime quantity? 122. When prime to each other?
ii-3. A multioie?
54 FACTORING.
PRINCIPLES.
124. 1^. If one quantity is an exact divisor of another ^
the former is also an exact divisor of any multiple of the
latter.
Thus, 3 is a divisor 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 their product.
Tims, 3 is a divisor of 9 and 15, 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 ly each of its prime
factors, by the product of two or more of them, and by no
other quantity.
Thus, the prime factors of 30 are 2, 3, and 5. Now 30 is divisible
hy 2, by 3, and by 2 x 3 ; by 2 x 5 ; by 3 x 5 ; by 2x3x5, and by no
other number.
CASE I.
125. To Find the Prime Factors of Monomials.
I. What are tlie prime factors of i2«2j?
Analysis. — The coefficient 12 = 2x2x3, and a^h — adb. There-
fore the prime factors of i2a^6 are 2 x 2 x '^aab. Hence, the
KcTLE. — Find the prime 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.
Resolve the following quantities into their prime factors :
2. is^f' Ans. zy^s^^y y y-
3. i2,d^b\ 7. i-jx^z.
4. lobo?]^, 8. 2$al^c3?,
5. $$aWc^, 9. Tja^c^d.
6. 2ixy^^. 10. 6$m^n^x.
124. Name Principle i. Principle 2. Principle 3. 125. How find the prime fee-
tors of monomials ?
FACTORING. 56
CASE II.
126. To Factor a Polynomial.
1. Kesolve ^a^h + ^ab — Sac into two factors.
Analysis. — By inspection, we per- operation.
ceive the factor 2a is common to 2a ) /^a% + 8fl^^ — 6ac
each term ; dividing by it, the quo- 2ah + ^h — 3 c
tient 2a& + 4&-3/5 is the other factor. ^^^^ ^a izah + 4J - 3c)
For convenience, we enclose this fac-
tor in a parenthesis, and prefix to it the factor 2a, as a coejfident.
Proof. — The factor (206 + 4&— 3c) x 2a=^a^l) + %ab—6ac. Hence, the
EuLE. — Divide the polynomial ly the greatest common
monomial factor ; the divisor will he one factor, the quotient
the other, (Art. 115.)
Note. — 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^ + alt^ into two factors, one of which shall
be a monomial. Ans. db (a + ^), a (ah -{-¥), or h (a^ -f ah),
3. Factor a + ah + ac. Ans, a (i -^ h -\- c),
4. Factor by -\- he -\- t^x,
5. Factor 2ax •\- 2ay — 4az,
6. Factor ^hcx — Ghcx — ^ah(k
7. Factor Mmn — 2^dm^
8. Factor 35^7/2 -f- 14^0:.
9. Factor 2^hdx^^^dmy,
10. Factor 6a^ -f ga^c.
11. Factor 2iao[^y -f- 35«a;y.
12. Factor 25 + 152:2 — 200^2^.
13. Factor x -\- x^ -\- ^,
14. Factor 3a; -f 6 — 9^.
15. Factor 19^52; —19^5,
126. How factor a polynomial ?
56 rACTORiiir®.
CASE III.
127. To Resolve a Trinomial into two equal Binomial Factors.
1. Resolve x^ + 2xy -f y"^ into two equal binomial factors.
Analysis. — Since tlie square of a * operation.
quantity is the product of two equal /^xi ==: X, V^ = 2/>
factors (Art. 30), it follows tliat the ^ c^ 2
square root of a quantity is one of the " ' " '" •2'
two equal factors which produce it. (^H"^) (^ + 2/)? Ans*
(Art. 32.) Therefore the square root of
Q? is ar, that of y^ is y. And since the middle term 2Ty is twice the
product of these two terms, x^->r2xy+y'^ must be the square of the
binomial x+y. Consequently, x + y is one of the two equal binomial
factors.
2. Resolve x^ — 2xy -f- y"^ into two equal binomial factors.
ANAiiTSis. — Reasoning as before, the operation.
quantity x^—2Ty+y^ is the square of ^x^ z=x, V^ = ^
the residual x—y. Therefore, the two ^ „ j_ 2
equal factors must be x—y and x—y. ' ' if '^ if
Hence, the (^— ^) (^ — y)> ^^S.
Rule. — Find the square root of each of the square terms,
and connect these roots hy the sign of the middle term.
Note. — A trinomial, in order to be resolved into equal binomial
factors, must have two of its terms squares, and the other term timce
the product of their square roots. (Art. loi.)
Resolve the following into two equal binomials:
3-
^2 + 2a'b + J2.
9.
f+2y+i.
4.
x^—2xy-\- y\
10.
I _ 2C2 + C^.
5.
m^ -j- 4mn + 4^2.
II.
^2m ^ 2^myn ^ y^^
6.
i6a^ + 8« + I.
12.
4^2" _ 4a» 4- I.
7-
49 + 70 + 25.
13.
a^ + 2aW + IK
8.
4^2 _ i2ah + 9J2.
14.
a^x^ + 2ax^y + y\
197. How resolve a trinomial into equal binomial fectors f
PACTOBING. 67
CASE IV,
2.28. To Factor a Binomial consisting of the Difference of
two Squares.
I. Resolve ^a^ — 9J2 into two binomial factors.
Analysis.— Both of these terms operation.
are squares; the root of the first is ^ ^0? = 2fl5
2a, that of the second is 36. But the /~w 7
difference of the squares of two quan- ^
tities is equal to the product ot their * ' ^^ 9^
sum and difference. (Art. 103.) Now (2«4-35) (2a— 3^), ^W5.
the sum of these two quantities is
2a + 3&, and the difference is 2a — 3& ; therefore, j\a? — 96^ =
(2a + 3&)(2a— 3&). Hence, the
Rule. — Find the square root of each term. The sum of
these roots will he one factor, and their difference the other.
Note. — This rule is one of the numerous applications of the for-
mula contained in Art. 103.
2. Resolve a^ — x^ into two binomial factors.
3. Resolve ^x^ — i6y^ into two binomial factors.
4. Resolve y^ — 4 into two binomial factors.
5. Resolve g — x^ into two binomial factors.
6. Resolve a^ — i into two binomial factors.
7. Resolve i — ^ into two binomial factors.
8. Resolve 25^^ — 16^2 into two binomial factors.
9. Resolve ^x^ — y^ into two binomial factors.
10. Resolve 1 — 16^2 into two binomial factors.
II. Resolve 25 — i into two binomial factors.
12. Resolve x^ — y* into two binomial factors.
13. Resolve a^x^ — b^y^ into two binomial factors.
14. Resolve m^ — # into two binomial factors.
15. Resolve a^'"" — h'^" into two binomial factors.
xaS, How fector a binomial consisting of the difference of two squares ?
58 1^ AGIO KING.
CASE V.
129. Various classes of examples of higher powers may be
factored by means of the following
PRINCIPLES.
1°. The difference of any tivo powers of tlie same degree is
divisible by the difference of their roots.
Thus, (x^—f) -J- {x—y) — x+y.
{^—f) -4- {x-y) = a^+xy+^.
(oe*—y^) H- {x—y) — a^ + x^y+xy^+y\
(a^— y5) -r- (x—y) = x^ + Q^y + x'^y'^ + xy^+y^.
2°. The differ e7ice of two even poivers of the same degree is
divisible by the sum of their roots.
Thus, {Q?—y^) -i- (x+y) — x—y.
(aj4_2^) ^ (x+y) = x^—x^y+xy'^—y^.
(afi—y^) 4- (x+y) = x^—x^y + a^y^—x^y^ + xy*—^,
3°. TJie Slim of two odd poicers of the same degree is divi-
sible by the sum of their roots.
Thus, i^+y^) -^ {x+y) = o^-xy+yK
(a^+y^) -r- (x+y) = xl^—a^y + x-y'^—xy^+y*.
(x'^+y'') ■+■ (x+y) — afi—x^y+xY—a^y^+xY—xy^+y\ etc.
Note.— The indices and signs of the quotient follow regular laws :
I St. The index of the first letter regularly deoreoMS by i, while that
of the following letter increases by i.
2d. When the difference of two powers is divided by the difference
of their roots, the signs of all the terms in the quotient are pltis. When
their sum or difference is divided by the sum of their roots, the odd
terms of the quotient are plus, and the even terms minus.
1^ 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.)
129. Kecite Prin. i. Prin. 2. Prin. 3. Note. What is the index of the first let
ter f Of the following letter ? What is said of the si^uB ?
FACTORING. 59
130. To Factor the Difference of any two Powers of the
same Degree.
I. Eesolve a? — y'^ into two factors.
Solution.— The binomial (a^— y^) -^ {x—y) = x^ + xi/+^. .*. x—y
md x^ + xy+y^ are the factors. (Prin. i.) Hence, the
'RuL^.— Divide the difference of the powers hy the differ-
ence of the roots; the divisor will he one factor, the quotient
the other,
Eesolve the following into two factors :
3' ^ — y^' 5. I —
131. To Factor the Difference of two even Powers of the
same Degree.
6. Eesolve a^ — Z>* into two factors.
Solution.— By Prin. 2,a^—¥ia divisible hy a + b. Thus, (a*— 6^)
^ (a + b) = a^—a^b + c^—b^, the divisor being one factor, the quotient
the other. Hence, the
EuLE. — Divide the difference of the given powers ly the
sum of their roots ; the divisor ivill he one factor, the quo-
tient the other, (Art. 129, Prin. 2.)
Eesolve the following quantities into two factors :
7. J2 _ x\ 10. iC^ — 1.
8. d^ — ^, II. 1—^6,
9. a^ — h^, 12. a^ — I.
132. To Factor the Sum of two odd Powers of the same
Degree.
13. Eesolve a} + h^ into two factors.
Solution. — Dividing a^ + V^ by a + b, the factors are a + b and
a^—ab+y"* (Prin. 3.) Hence, the
EuLE. — Divide the sum of the powers hy the sum of the
roots; the divisor and quotient are the factors.
130. How factor the diflference of any two powers of the same degree ? 131. How
factor the difference of two even powers of the same degree ? 132. The sum of two
odd powers of the same degree t
60 FACTOEIKG.
Resolve the following quantities into two factors :
14. a^-{-f. 17. i + y\
15. a^ + I. 18. I + a\
16. a^ + I. 19. I + W.
133. It will be observed that in the preceding examples
of this Case, binomials have been resolved into tivo factors.
These factors may or may not be prime factors.
Thus, in Ex. 6, d^-h^ = {a + 'b){a^-a'^h-\- (0)^-1)% But the /'actor
{a^—a?h-\-ah'^—¥') is a composite quantity = {a—b)(a^ + ¥).
134. When a binomial is to be resolved into prime fac-
tors, it should first be resolved into two factors, on<> of
which is prhne ; then the composite factor should be
treated in like manner.
20. Let it be required to find the prime factors of «^ - ¥.
Solution.— The ^/a^ = a\ and /y/ft^^ft^ (Art. 128.)
Now a^-¥- = (a2-&2) (^2 + jf.y But a^-h^ = («+&) {a-V). (Art. i^j.)
Therefore the prime factors of a^—¥ are {a^ + ¥){a + l){a—h).
Resolve the following quantities into their prime factors '
21. «^— I. Arts. {dJ^ ■\-i){a-\- i){a—i).
22. i—f. Ans. (i+y^){i-{-y){i-yy
23. ofi — y^.
Ans. {x^ -xy + y^) {a^ + ccy + y^) {x •\-y){x — y).
24. x^ — 2x^y^ 4- y^.
Ans, {x^ - y^y = {x + y) {x -h y) (^ - «/) (^ ~ V)-
25. a^ — I.
Ans. (x 4- i) {x- i) {x^ -\-x^-i){xi-x-\- i).
26. a« + 2a%^ + 2'^.
^?^5. {a -\-h){a + I) {a^ - a2> + ^) {a^ - ah -]r ^),
27. «2 _^ 9a + 18. Ans. {a + 6) (rt + 3}.
28. 4^2 ~ i2«& + 9&2. ^ws. (2a — 2fi) (20 - 3^)
(See Appendix, p. 284.)
OHAPTEE YII
DIVISORS AND MULTIPLES.
135. A Common I>ivisor is one that will divide two
or more quantities without a remainder.
136. Commensurable Quantities are those which
haye a common divisor.
Thus, aW and dbc are commensurable by ah.
137. Incommensurable Quantities are those
which have no common divisor. (Art. 122 )
Tims, ab and xyz are incommensurabla.
138. To Find a Common Divisor of two or more Quantities.
1. Find a common divisor of abx, act/, and adz.
Analysis. — Resolving the given quanti- operation.
ties into factors, we perceive the factor a, is abx = a xb XX
common to each quantity, and is therefore a act/ = axcXV
common divisor of tbem. (Art. 119,) Hence, ^^^i ^ axdxz
*^^ A71S. a.
Rule. — Resolve each of the given quantities into factors,
one of ivhich is common to all.
Find a common divisor of the following quantities :
2. -^alcd and gahm. Ans. ^ah.
3. x^yz and lahx. 6. 2ax, dlx, 14CX.
4. a% bed, ah^xy. 7. 35^^, "jm"^, 42m^x.
5. 2adc, acx% a^cy, 8. 24a^, i2al>^, 6aW.
135. What is a common divisor ? 136. Commensurable quantities ? 137. Incom-
mensunible quantities ? 138. How find a common divisor of two or more quantities ?
0x5 DIYIS0R8.
139. The Greatest 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 foMor of those quantities, and the g. c. fZ.* is their greatest
common factor.
2. A common divisor is often called a common measure, and the
greatest common divisor, the greatest comm/m measure.
PRINCIPLES.
140. 1°. TJie greatest common divisor of two or more
quantities is the product of all their common prime factors.
2°. A commoji divisor of tioo quantities is 7iot altered ty
multiplying or dividing either of them hy any factor not
found in the other.
Thus, 3 is a common divisor of i8 and 6 ; it is also a common divisor
of 1 8, and of (6 x 5) or 30.
3°. The signs of a 2^olynomial may he changed hy divid-
ing it hy — I.
Thus, (— 3a + 4&— 5c)h 1 = 3a— 46 + 5c. (Art. 112.) Hence,
4°. Tlie signs of the divisor, or of the dividend, or of hoth,
may he changed without changing the common divisor.
141. To Find the Greatest Common Divisor of Monomials by
JPrune Factors,
I. What is the g, c, d, of 35aca;, 2Sahc, and 2iay?
Analysis. — Resolving the operation.
given quantities into their prime 3SCICX := <^XT XaXCXX
factors, 7 and a only, are com- ^SaJc = 2X2X7X«X*XC
mon to each ; therefore their
product 7 X a, IS the g,c,a,re- ^ -» / ./
quired. (Prin. i.) •'• 7Xfl5 = 7«- ^ns.
139* What is the greatest common divipor of two or more quantities ? Note i.
What is true of a common divisor of two or more quantities? Of the g. c. d.?
140. Name Principle i. Principle 2. Principle 3.
* The initials g, c, d, are usod for the greatest common divisor.
jivisoRS. 63
2. Find the g, c, d. of 4a^b% loaW, and i^abdx.
Analysis. — Resolving these quan- operation.
titles into their prime factors, the ^aWc = 2 x 2 X oaahhc
factor 2 is common to the coefficients; locfil^ = 2 X 5 X CLobbb
also, a and 6 are common to the lit- ^o^aMx = 2 X 7 X obdx
eral parts Now multiplying these ^^^^ 2^a^h = 2ab
common factors together, we have
ly.ay.'b — 2ab, which is the g. c, d. required. (Prin. i.) Hence, the
Rule. — Resolve the given quantities into their prime fac-
tors ; and the product of the factors common to all, 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 6aV and ^dbc,
4. Of i6a^xy and i^acx^y.
5. Of i2aWa^^ and i6a^x^zK
6. Of daV^x^^, i2a^xh^, ana i2>a^xh\
142. To Find the Greatest Common Divisor of Quantities by
Continued Division,
I. Required the greatest common divisor of 30a; and 422;.
Analysis. — If we divide the greater operation.
quantity by the less, the quotient is i, 30a; ) 42:?; ( i
and 12a; remainder. Next, dividing the 302;
first divis'or 30a;, by the first remainder \ /
^1, J.- \ ■ J XI -J I2iC)302;(2
12a;, the quotient is 2 and the remainder ^ ^ ^
6x. Again, dividing the second divisor ^4^
by the second remainder, the quotient 6x ) 1 2:?; ( 2
is 2 and no remainder. The last divisor,
tx, is the ff, c, d.
12a;
Demonstration. — Two points are required to be proved :
ist. That 6x is a common divisor of the given quantities.
2d. That 6a! is their greatest common divisor.
First. We are to prove that 6x is a common divisor of 30a; and 42a;.
By the last division, 6a; is contained in 123*, 2 times. Now as 6a; k a
141. How find the g.c. d. of monomials by prime factors? Note. In finding the
prime factors of the literal part, what exponents are given ?
64 DIVISORS,
divisor of i2aJ, it is also a divisor of the product of i2aj into 2, or 24a;.
(Art. 124, Prin. i.) Next, since 6x is a divisor of itself and 24a*, it
must be a divisor of the sum of tx-\-2^, or 30a!, which is the smaller
quantity. For the same reason, since ()X is a divisor of 12a; and 30a;, it
must also be a divisor of the sum of 12a; + 302;, or 42a;, which is the
larger quantity. Hence, tx is a common divisor of 30a; and 42a;.
Second. We are to prove that 6x is the greatest common divisor of
30a; and 42a?.
If the greatest common divisor is not bx, it must be either greater
or less than bx. But we have shown that tx is a common divisor of
the given quantities ; therefore, no quantity less than tx can be the
greatest common divisor of them. The assumed quantity must there-
fore be greater than bx. By supposition, this assumed quantity is a
divisor of 30a? and 42a! ; hence, it must be a divisor of their difference,
42aj— 30a;, or 12a;. And as it is a divisor of 12a;, it must also divide the
product of 12a; into 2, or 24a;.
Again, since the assumed quantity is a divisor of 3oaJ and 24a;, it
must also be a divisor of their difference, which is bx ; that is, a greater
quantity will divide a less without a remainder, which is impossible.
Therefore, 6a! must be the greatest common divisor of 3oaj and 42aJ,
the second point to be proved. Hence, the
B,XJL^.— Divide the greater quantity by the less, then divide
the first divisor dy 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, fifid the g. c» d,
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 48^5, 72 a, and io8a?
Suggestion. — The </, c, d, of 48a and 72a is 24a ; and that of 24a
and loSa is 12a. Therefore, 12a is the (/. c, d, required.
142, «. The Greatest Common Divisor of Poly-
nomials is found by the above rule, as illustrated in the
following examples:
142. Show upon the blackboard the truth of this rule? r.-p a How find the
g. ^. d. of polynomials ?
DIVISORS
65
3. What is the g. c. d. of /^a^ — 2ia2 4. 15a 4. 20 and
<j3 __ 6a + 8 ?
OPKBATION.
^a^ — 2ia^ + 150^ +20 a^ — 6« + 8 istdiviBor.
\a^ — 24^2 4. 32«
4^ + 3
zst quotient.
-f 3a^ — 1 7a + 20
f 3^2 _ i8a -f 24
a« — 6« + 8
a^ — /^a
— 2a + 8
— 2a + 8
a — 4 'Pt remainder and 2d divisor.
a — 2 ad quotient.
Ans. a — 4.
Analysis. — 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,ofa^— 2xy + y^ and 01? — y^"^
Analysis.— Dividing the
greater by the less, the first
remainder is — 2xy + 2y^.
Cancelling from it the com-
mon factor 2y, we have for
the second divisor —x + y.
Changing the signs, it be-
comes a;— y. (Art. 140, Prin. 4.)
Dividing as before, the quo-
tient \^ x + y, and no remainder,
common divisor.
OPKBATION
0^ — 2Xy-\- y^
7? — ^2
2y) —2xy+2f'
Divisor, — ^ + y
or, ^—y
Quotient, ^-\-y
X^ — y^ Divisor,
I Quotient.
a? — y"^
■t
X?-
The last divisor, x—y, is the greatest
143. How does it affect the g. c. d. if a factor is introduced into eitlier orbotli the
given quantities ? How if one is cancelled ? What is true of the si^jne ?
66
DIVISORS.
5. What is the g. c, d. of 4^3 — 6a;2 — 4a; + 3 and
2X^ + X^ + X—1?
dividend, 4^ — 6x^ — 4X + S
43:^ -\- 2X^ -{- 2X — 2
2d divisor, — 8,7^ — 6a; -f 5
2d quotient,
OPERATION.
2C(^ -^ X^ -^ X — I ist divisor.
2 igt quotient.
— X
3d dividend, — Sx^ — 6x -{- 5
— Sx^ + S^X — 16
4th divisor,
4tli quotient.
— 21 ) — 42a; +21
, 2ir — I
— a; + 4
8a.'^ + 4X^ -{- 4X — 4 2d dividend
8'Ji^ + 6a;2 — 5ic
— 22:^ -i- 9a; — 4 3d divisor.
4 3d quotient.
— 23^^+93; — 4 4th dividend.
~ 2X^ -{- X
2X — I is the g, c, d.
8a; — 4
8a; — 4
Analysis. — Dividing the greater by the less, the first term of the
first remainder, — Sa*-, is not contained in 2a:^> the first term of the
second dividend. We therefore multiply this dividend by 4, and it
becomes Sx^ + 4X^ + 4X— 4, and dividing this by the second divisor, the
second remainder is —22;'^ + 90?— 4. Dividing the preceding divisor by
this remainder, we see that the third remainder, —422; + 21, is not
contained in the next dividend. Cancelling the factor —21, the fourth
divisor becomes 2X — i, the greatest common divisor required.
Find the g, c. d. of the following quantities:
6. a^ — y^ and a^ — 2x2/ + ^.
7. «3 _f_ ^z an^ ^2 _j_ 2ab + ^.
8. h^ — 4 and J2 ^ 45 + 4.
9. a:^ — 9 and x^ -{- 6x -\- 9.
10. a^ — 3(1 + 2 and a^ — a — 2.
11. «^ + 3^2 -f 4flj -|- 12 and a^ -f 4^2 _{- 4^5 _{-. 3,
12. a^ -{- I and x^ + mx^ -f 772a; + 1.
13. d^ — Ifi and a^ — W,
14. fl^2 _ 2^j _j_ 4^2 and c^ — a^h + 3fl!j2 _ ^js
15. 3^ — 102;^ + 15^ + 8 and x^ — 2a;^ — 6ar^ + 4a:2
4- 13a; -f 6.
MULTIPLES. ;37
MULTIPLES.
144. A Multiple is a quantity which can be divided
by another quantity without a remainder. (Art. 123.)
145. A Coimnon Multijde is a quantity which can
be divided by two or more quantities without a remainder.
Thus, 1 8a 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. 1°. ^ multiple of a quantity must contain all the
prime factors of that 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 the given quantities.
Thus, 42, a common multiple of 14 and 21, contains all the prime
factors of those quantities ; viz., 2, 3, and 7.
3°. The least com7non multiple of tiuo or more quantities
is the least quantity which contains all their prime factors^
each factor heing tahen the greatest numderof times it occurs
in either of the given quantities.
Thus, 30 is the least common multiple of 6 and to, and contains all
the prime factors of these quantities ; viz., 2, 3, and 5.
144. What is a multiple ? 145, A common multiple ? 146. The least common
multiple ? 147. Name Principle i. Principle 2. Principle 3.
68 MULTIPLES.
148. To Find the Least Common Multiple of Monomials by
Prime Factors,
1. Find the I, cm,* of i5«V, ^l^cx, and 9&A
Analysis. — The prime factors operation.
of the coefficients are 5, 3, and 3. l^^a'^OC^ = 3X5X«*Xa?
The prime factors of the letter a t^W-cx =1 ^xl^ XCXX
are a, a, a, a, which are denoted by qbc^z = ^X^X^X<^X^
a*. In like manner, the prime fac- j^^^^ A^a^l^C^Z.
tors of X are denoted by a;', those of
b by b'^, and those of c by c^ ; z is prime. Taking each of these factors
the greatest number of times it occurs in either of the given quanti-
ties, the product, 45a^&-c%-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, tahing each the greatest num-
her of times it occurs in either of the given quantities. Tlie
product is the I, c, m, required.
Or, Find the least common multiple of the coefficieyits, and
annex to it all the letters, giving ecLch letter the exponent of
its highest poiver in either of the quantities.
Note. — Tn finding the I. c, in, of algebraic quantities, it is often
more expeditious to arrange them in a horizontal line, then divide,
«tc., as in arithmetic.
Eequired the I, c, Vfl, of the following quantities :
2. ^a\ i2a^A and 24^2:2^. Ans. yza^x^g,
3. ^ab^ 2Sbc^, and ^6a^M,
4. iSoc^'ifz, 2oy% and Sxyz^.
5. isa^l^c, ga¥c% and iSa^c^.
6. 2Sab\ i4«2J*, 35«^^, and 42a*h.
7. 2ix^yh^, 35^y^z^, and d^xyH.
8. ^mVy, i2mhiy'^, 2i>Yi^ T^mn^y^.
148. How find the I. c. tn. of monomials by prime factors ? What other method
* The initials I, c, tn, are used for the least common multiple.
MULTIPLES. 69
149. To Find the Least Common Multiple of Polynomials.
9. Kequired the l. c. ni» of ^^ + ^ and a^ — 1^.
Analysis. — Resolving operation.
tlie quantities into their tt^— Z*^ = (« + d) X («— J)
prime factors, as in the «3+^ = (« + J) X (tt^— ffJ + ^)
margin, (« + &) is common (^^ j) >< (^_y^ ^ (cii_ah + m =
to both, and is their </. c. d. a dz , xq xi ^ '
(A.rt. 139.) Now multiply-
ing these factors together, taking each the greatest number of times
it occurs in either of the given quantities, the product a*—a^b + ab^—b*
is the Lcm, required. (Art. 148.)
Second Method.
^nce the g. 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
g, c. d. and the quotient multiplied by the other, the product will
be the I, c, tn. Hence, the
EuLE. — Resolve the quantities into their prime factors
and multiply these factors together, taking each the greatest
number of times it occurs in either of the given quantities.
Their product is the I. e, m. reqtiired.
Or, Find the greatest common divisor of the given quanti'
ties^ and divide one of them hy it. The quotient, multiplied
ly the other, will he their I, c, m*
10. Find the I, c. m. of 2a — i and 4^2 _ i.
Solution.— The g, c. d, is 2a— 1. Now(4a2— i)-^(2a— i)=(2ffi+ 1) ;
and (2a + 1) X (2a— i) = 4a^— I, Ans.
Find the I, c, m, of the following quantities :
11. x^ — y"^ and x^ — 2xy + i/^.
Ans. u? — x^y — xy"^ + y\
12. «2_j2 and a^ — tfi.
13. a? — I and x^ ■}- 2X -\- i.
14. 2^2 ^ ^a — 2 and 6a^ — a — i.
15. m^ -j- m — 2 and m^ — i.
(See AppS'Uclix, p. 284.)
CHAPTER VIII.
FRACTIONS.
150. A Fraction is one or more of the equal parts into
which a unit is divided.
151. Fractions 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 helow the
line, and shows into Jiow many equal parts the unit is
divided.
153. The Numerator is the quantity above the line,
and shows how many parts are taken.
Tlius, the expression - shows that the quantity is divided into h
equal parts, and that a of those parts are taken.
154. The Unit or Base of a fi-action is the quantity
divided into equal parts.
155. The Ter^ns 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, 305, 5, 7.
157. A Mixed Quantity is one which contains an
integer and a fraction.
Thus, a + - is a mixed quantity.
150. What is a fraction ? 151. How exprensed? 152. What does the denomina-
tor show? 153. The numerator? 154. What is the base of a fraction? 155. Tlia
terms of a fraction ? 156. An integer ? isz* ^ mixed quantity ?
FKACTIOKS. 71
158. Fractions arise from division, the numerator 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 3m.
SIGNS OF FRACTIONS.
X60. JEvery Fraction has the sign + or — , expresse(J
or understood, before the dividing line.
161. The Dividing Line has the force of a viiicu^
luM or parenthesis, and the sign before it shows that the
value of the ivhole fraction is to be added or subtracted.
162. Every Numerator and Denominator is
pi-^ceded by the sign + or — , expressed or understood.
In this case, the sign affects only the single term to which
it is prefixed.
163. If the Sign before the Dividing Line is
changed from -f to — , or from — to +, the value of the
fraction is changed from + to — , or from — to -f .
^, hx — ex — dx , -, . X hx — cx — dx
Thus, a -\ = a + o — c — d; but a =
x X
a — b + e + 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 + 6 ; but
158. From what do fractions arise ? 159. What is the value of a fraction •
160. What is prefixed to the dividing line of a fraction ? i6r. What is the force of
the dividing line ? 162. By what is the numerator and denominator pi'eceded ?
How far does the force of this sign extend? 163. If the sign hefore the dividing
line is changed, what is the efi"ect ? 164. If ,all the signs of the numerator are
changed ?
72 FRACTIONS.
165. If all the Signs of the Denominator are
changed, the value is also changed in a corresponding
manner.
Thus, — =+a; but — = -a. Hence,
+x —X
166. If any two of these changes are made at the same
time, thej will balance each other, and the value of the
fraction wiU not be altered,
_- ab — ab —ah ab
^-^^ -j = -ITi = - -J- = - — 6 = + «•
.. ab — ah ab — ab
PRINCIPLES.
167. The principles for the treatment of fractions in
A-lgebra are the same as those in Arithmetic.
1°. Multiplying the numerator, or \ Multiplies the
Dividing the denomi^iator, j fraction.
,p, 2X2 4 2 . , 2 2
Thus, - = ^ = -. And - = - •
6 63 t-i-2 3
2°. Dividing the numerator, or ) r^ • • 7 .-, ^
nr 1^' 1 • A-L 7 ' ± \ Divides the fraction.
Multiplying the denominator, ) -^
»^ 2^2 I . , 2 21
6 6 6x2 12 6 •
3°. Multiplying, or dividing loth \ Does not change its
terms by the same quantity ) value.
-«, 2x2 4 2 I . , 2-T-2 I
Thus, = -t = = _. And = -
6x2 12 6 3 6-5-2 3
4°. Multiplying and dividing a \ Does not change its
fraction by the same quantity j value. •
-^ 2x2-5-2 2
Thus, = - .
6x2-7-2 o
165. If all the signs of the denominator arc changed ? i66. If both are changed?
167. Name Principle i. Principle 2. Principle 3. Principle 4.
BEDUCTION OF FRACTIONS. 73
REDUCTION OF FRACTIONS.
168. Heduction of Fractions is changing their
terms without altering the value of the fractions.
CASE I.
169. To Reduce a Fraction to its Lowest Terms.
Def. — The JLoii^est Terms of a fraction are the
smallest terms in which its numerator and denominator
can be expressed. (Art. 122.)
1. Reduce - — =-^ to its lowest terms.
i$abcx
Analysis.— By inspection, we perceive the operation.
factors S, a, b, and x are common to both terms. 5^ (rax aoa^
Cancelling these common factors, the fraction l^abcx 3c
becomes . Now since both terms have been divided by the
same quantity, the value 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.)
Note. — It will be observed that the factors 5, a, &, and x are prime ;
therefore, the product e^abx is the g, c. d. of the numerator and
denominator. (Art. 121.) Hence, the
Rule. — Cancel all the factors common to the numerator
and denominato'*
Or, Divide both terms of the fraction by their greatest
common divisor. (Art. 167.)
2. Reduce f- to its lowest terms. Ans. -•
i2abc 3
3. Reduce to its lowest terms. Ans. —
2,ac c •
168. What is reduction effractions? 169. What are the lowest terms of a frac-
tion ? How reduce fractions to the lowest terms ?
14: REBTTCTIOK OP FRACTIOKS.
Reduce the following fractions to the lowest terms:
3^y . lo. 3^—3^^
ga^y^ ' 2X^y — 2xyz
i2a^l)<? a -{-he
^' 4abcd ' ' (a -{- be) X x
I'jh^cxy x^ — y^
^it^cxy ' x^ — y/^
^AaWc^ ax — x^
7. — • 13. •
loSabx^y^ cfi — a^
c^ — IP' a — I
c? 4- 2ab + 6^ c^ — 2a -\- \
^—y^ ^ + 1
^* x^ — 2xy + ^' ^' a? -\^ 2xy + y^'
CASE II.
170. To Reduce a Fraction to a JVIiole or Mixed Quantity,
I. Reduce to a whole or mixed quantity.
Analysis.— Since the value operation.
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 46, 2& times. Placing the remainder
e over the denominator, we have a + 26 + -, the mixed quantity
required. Hence, the
Rule. — Divide the nunurator hy the denominator, and
placing the remainder over the divisor, annex it to the
quotient.
Note. — This rule is based upon the principle that both terms are
divided by the same quantity. (Art. 167, Prin. 3.)
170. How reduce a fraction to a whole or mixed quantity? Note. Upon what
principle is this rule baeed?
KEDTJCTIOK OF FRACTIO^-S. 75
Reduce the following to whole or mixed quantities :
ax — x^
X
3.
al--^
a
4-
V^-^
1) ■\-c
f
^ + C2
a — h
a^ -{- a^ — aa^
c? — ax
\2^ ■\- /i^x — zy
^' h — c ^' 4X
CASE III.
171. To Reduce a Mixed Quantity to an Improper Fraction.
1. 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« b 3^^ b
times 3 thirds, or 3^; and 3? + -^=^^, "" 3~7 3
3 3 3 3 3a h 3a-{-b
the fraction required. Hence, the 1 — = —
3 3 3
Rule. — Midtiply the integer hy 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 :
, cd . ahd — cd ,
2. at) i- Ans, 3 = ao ^ c,
d d
xy —h , . I — ^
3. 3X + -^— 6. a; — I +
\ •\- X
T ^ a — c a — t
S. a + l + ^y S. 8. + 3^.
OPEBATIOH.
3« =
3«
I
I
X 5 _
X5
isa
5
Hence, the
76 EEDUCTION OF FKACTI0N8.
CASE IV.
172. To Reduce an Integer to a Fraction having any required
Denominator.
1. Reduce 3« to fifths.
Analysis. — Since in i« there are 5 fifths, in
3a there must be 3 times 5 fifths, or -^.
Or, reducing the integer 3a to the form of a
fraction, it becomes — ; multiplying both
terms by the required denominator, we have -^,
Rule. — Multiply the integer hy the required denominator,
and place the product over it.
2. Reduce 2x to a fraction having 6m for its denominator.
3. Reduce 6ax to a fraction having 4ab for its denominator.
4. Reduce ^a + 4b to a fraction having 6c^ for its
denominator.
5. Reduce x — y to a fraction having x -i- y for its
denominator.
6. Reduce 2x^y to a fraction having ^a^ — 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- ofebatiok.
nominator, by the given denominator 3, the ^ 2 -^ 3 = 4
quotient is 4. Multiplying both terms of a X 4 4«
the given fraction by the quotient 4, the * 2X4 1 2 '
result, — , is the fraction required. Hence, the
Rule. — Divide the required denominator hy the denomi-
nator of the given fraction, and multiply loth terms hy the
quotient.
172. How reduce an integer to a fraction having any required denominator?
REDUCTION OF FRACTION'S. 77
Reduce the following to the required denominators :
2. Reduce — to thirty-fifths.
Solution. 35 -5- 7 = 5. ^o\^ = — . Ans.
7-^5 35
3. Reduce - to the denominator ac,
c
4. Reduce — to the denominator 49a.
7
5. Reduce ^ to the denominator a? — 2xy -f ^.
X — y
6. Reduce -— — to the denominator M^(x + ti)\
x + y \ JJ
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
Coinmon IJenoitiinaior.
fh C X
I. Reduce t> ;t> and -, to equivalent fractions having
a common denominator.
173. How reduce a fraction to any required denominator ? 174. What is a com-
jnon denominator ? Principle i ? Principle a ?
78 REDUCTION OF FRACTIONS.
Solution.— Multiplying the denominators h, d, and y together, we
have bdy, which is a common denominator.
b X d X i/ = hdy The common denominator.
a X d X ^ = ady )
ex i X y = hey V The new numerators.
X X l X d =z hdx )
J a ady c hey x hdx
h~ hdy' d~~hdy' y^ hdy'
To reduce the given fractions to this denominator, we multiply-
each numerator by all the denominators except its own, and place the
results over the common denominator. Hence, the
Rule, — Multiply all the denominators together for a com-
mon denominator, and each numerator into all the denom-
inators, except its oion.for the neio numerators.
Notes. — i. It is advisable to reduce the fractions to their loicest
terms, before the rule is applied. (Art. 169.)
2. This rule is based on the principle, that the Dolue of a fraction is
not changed by multiplying both its terms by the same quantity.
(Art. 167, Prin. 3.)
Reduce the following to equivalent fractions having a
common denominator:
2.
a X ^
~c' y' 4*
Ans. 4^^ 4^^ ''^-.
4cy 4cy 4cy
3.
c h 2d
d' x' T
8.
2a c -\- I
4.
a h X
2x' ~?' ~y
9.
2 a c -\- d
3' ^' c-d
5.
2a X
10.
xy I 2«
Sh' a + b
z' 2' h'
6.
x — y x^y^
x-\-y' x — y'
II.
2a x^ + ^^
7.
a + h 5a— 1
3 ' a '
12.
a — x a + X
a -{- x' a — x
175. How reduce fractions to equivalent fractions having a common denomina-
tor r JV^oie. Upon what principle is this rule based ?
EEDUCTIOK OF FRACTIONS.
79
CASE VI.
176. To Reduce Fractions to the Least Common Denominator.
I. Eeduce -
X' xy yz
Solution. — Tlie I, c, m, of the denominators U
, and — to the I, c,d.
xyz = L c, d.
xyz -r-x = yz
xyz -T-xy =i z
xyz -i- yz ^x
The
multipliers.
a X yz
X X yz
b X z
xy X z
d X X
xyz. (Art. >j8.
_ ^?/^
~xyz
_bz^
~ xyz
dx
The
fractions
required.
yz X X xyz
To change the given fractions to others whose denominator is xyz, we
multiply each numerator by the quotient arising from dividing this
multiple by its corresponding denominator. Hence, the
Rule. — I. Find the least common multiple of all the
denominators for the least common denominator.
II. Divide this multiple by the denominator of each frac-
tion, and multiply its numerator iy the quotient.
Note. — AH the fractions must be reduced to their lowest terms
before the rule is applied.
Reduce the folloAving fractions to the I, c. d, :
a Ic y
2.
2^' a: ' 4C
cd 2X xy
3-
ab^ 2,a^ ac
a i c X
4-
~ ' 'Z' T ' T.'
2 3 4 y
a^c 2cd x^y
S-
ah' Wc' ^U
6.
2db 3 a; I
Zac' 4' OL^c' 8
2« cd x^y
h
45 ' Ic' lex
8.
10.
II. —
12. -^
a^l a — I a^+ly^
a — y a + V a^-lj''
2(x-\-y) a ah
zix-^tjY xy' 'ejx'-^y)'
d X
a^' Wb'
X m y
ac Wc' &d
X a ■\-h d
' xz
y
2'
xy
13-
m-\-n m—n
m*
W
2ax^ ' ^cx
176. How reduce fractions to the least common denominator!
so ADDITION OF FRACTIOKS
ADDITION OF FRACTIONS.
177. When fractions have a common denominator, their
numerators express like parts of the same unit or base, and
are like quantities, (Art. 43.)
178. To Add Fractions which have a Common DenominatoPc
1. What is the sum of f , |, and | ?
Solution, f and | are V-, and f are ^i, Ans. Hence, the
EuLE.— ^flf^ the numerators, and place the sum over the
common denominator,
. ,, 75 45,85 . loj
2. Add — , — , and — . Ans, -^—
m m m m
3. Add ^ — , , , and ^
2xy 2xy 2xy 2xy
. ,, 7c?a:2! i7(?a;2! iidxz , 4<7a;;?
4. Add '—r , , , — 7- , and ^^ •
$aoc ^aoc $aoc ^aoc
5. Add ?^+f to 3i-rf . 6. Add ^^±^ to 4^=^.
179. To Add Fractions which have Different Denominators.
7. What is the sum of t^ -7> and — ?
h d' X
bxdxx = hdx, c. d,
a adx c bcx
Analysis, — Since these fractions
have different denominators, their
numerators cannot be added in their 5 odx d udx
present form. (Art, 66.) We there- m hdm
fore reduce them to a common denomr ^ ~^x
inator, then add the numerators. ^^^ _^ j^^. ^ j^^
(Art. 175.) Hence, the j^ -, Ans^
Rule. — Reduce the fractions to a common denominator,
and place the sum of the nuriierators over it.
Note.— All answers should be reduced to the lowest terms.
177. When fractions have a common denominator, what is true of the numerators!
178. How add such fi:3,ction8 ? 179. How, whea they have different denominators ?
ADDITIOI^ OF FRACTIONS. 81
Find the sum of the following fractions :
8. --f--^ +
V X m
2X 2y 2xy . 2X^m -\- lyhn -f 2X^y^
mxy
cd Ay Ix
^ ZX^ 2d^ S
9-
3^ + ^ + 1
4 5 3
lO.
^ + J- + M.
3 26? 4
a . X
II.
I ■\-c ' l — c
12.
x+y x—y
2xy xy
I-?-
2 +x 3 -{-ax
y ay
^ A.
a ah
i6.
a 271 -\- d
d sh
a d
17. - H
y -m
,8. :^+ -^
y m — n
— 4 — 16
19. — ^ +
2 7 — 3
4« 6c xm
20. -^^ — — •
X -^ y x — y b d 2>x
180. To Add Mixed Quantities.
Ji vn
1. What is the sum of « + - and d ?
c n
BOLunoiS'. — Adding the integral and fractional parts separately^
\06 result S&a^r d ^ , tlie sum required. Hence, the
en
Efle. — Add the integral and fractional parts separately,
and unite the results. (Art. 179.)
Note. — Mixed quantities may be reduced to improper fractions, and
tlien be added by the rule. (Art. 171.)
2. What is the sum of « + - and c + - ?
2 X
3. What is the sum oi x -\- ^ and ?
0 m — y
4. What is the sum of 3d and a -\ ?
2 I
a ~— 1/
5. What is the sum of 5a; + ^ and — ^ ?
0 2
180. How add mixed quantities ? iVofe. How else may they be added ?
82 SUBTRACTION OF FRACTIONS.
181. To Incorporate an Integer with a Fraction.
c ^
6. Incorporate the integer ah with the fraction •
Solution. — Reducing db to the denominator of tlie fraction, we
have ah = '^^^^M . and 3_«&^^ + -£z:i = 'i^x^^y^;C-d ^^^
2,x+y 2>^+y o^x+y y^^y
Hence, the
EuLE. — Reduce the integer to the denominator of the frac-
tion, and place the sum of the numerators over the given
denominator, (Art. 172.)
za
7. Incorporate the integer 3^ with the fraction -r-
8. Incorporate — 4^ with «
9. Incorporate — a with ^«
a — "b
10. Incorporate z^ + V with
x — y
II. Incorporate — a -\- ^h with ^•
12. Incorporate 2a; + 2y with
iC
SUBTRACTION OF FRACTIONS.
182. The numerators of fractions which have a common
denominator, we have seen, are like quantities, (Art. 177.)
Hence, they may be subtracted one from another as integers.
1. Subtract | from |.
Solution. l-l^T^^^l or -. Am, - •
00004 4
2. From T subtract y ^ns. t — 7 = — 7 — •
6 J III
181. How incorporate an integer with a fraction? 182. What is true of the
numerators of fractions having a common denominator ? How suhtract euch frac-
tions ?
3
X 2 =
6, c.
.c?.
7«_
3
14a
6
2
ga
6
I4«
6
9«_
6 "
6'
An
SUBTRACTION OF FRACTIONS.- 83
3. From —7— subtract -^-^
4. From — -^ subtract - — •
^ a a
183. To Subtract Fractions which have Different
Denominators.
q. From -- subtract — •
^ 3 2
Analysis. — Since these fractions
have different denominators, they can-
not be subtracted one from the other
in their present form. We therefore
reduce them to a common denominator ,
which is 6, and place the difference of
the numerators over it. Hence, the
RcTLE. — 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^^^. Ans, 5^:iJ£^.
X ' y xy
7. From — , take •
' m' y
8. From ~~ , take •
m y
^ a -\- -Kd ^ ^ m — 2d
9. From — '-^j take ^
4 3
.•83. How when they have different denominators ? Note i. How subtract mixed
quantities ? Note 2. How a fraction from an integer, or an integer from a fraction T
84 MULTIPLICATION^ OF FEACTIONS,
10. From — , take — - »
m y
11. From -, take m,
y
12. From 4« + -, take 3a — :^'
c d
2 3
14. From T-^—, take , ^ ■.
15. From « — -, take ^.
f 2
16. From "^ - ^ take "^-^
10 ' i^ + y
n. From a; ^""'', take ^""^
-a.
MULTIPLICATION OF FRACTIONS.
CASE I.
J84. To Multiply a Fraction by an Integer,
1. Multiply -7 by m.
A-{^ALysis.— Multiplying the numerator of the ofkiutioi.
fhftCtiun by the integer, the product is am. ^ ^ ^ ^^^
(Art. i6;, Piin. i.) b ~ b '
2. "W^at :s the product of 7- x a;?
Analysis- 'A fraction is multiplied by ^ y^ x ^
dividing its denominator ; therefore, if we bx bx -r- X
divide hx by gc, the re.sult will be the product a a
required. (Art. 167, Knn. 1.) Hence, the ^^"ZZTJ ^^ 5*
EuLE. — Multivly i?a numerator by the integer.
Or, Divide the detwminaior by it.
Notes. — i. A fi'action is multiplied by a quantity equal to its
denominator, by cancelling the denominator. (Art. 1 10, Prin. 4.)
184. How multiply fi frdctioo by on integer?
MULTIPLIOATIOK OF FRACTIONS. 85
2. A fraction is also multiplied by dJij factor in its denominator, by
cancelling that factor. (Art. no, Prin. 4.)
Find the product of the following quantities:
3-
4
5
6
7.
8.
9
10
II
^2/
— — T X (a — - J). -4?^5. 3a;,
—y X d.
cd
—7-^ X (3 + m).
«^
— X 6.
24
2a; — XII , _
v^~i X (3c + 2fZ). 14.
ax — X 6x.
SX
a + h
; X ^x.
20X 4- 2^xy ^
a -\- ab
he -\- c
2X-\- s
X 2ac.
X 2or2;.
5 a^ — z^
4 «^
Ans, — •
c
12.
a — i ,
^ X (12a: + 18).
13-
-J X (a — x),
d — x ^ '
14.
a + b
15.
40Z — 10 ^ ^
16.
3c — d
20 >^'5.
17.
18.
-;^ X X 12; + «).
CASE II.
185. To Multiply a Fraction by a Fraction^
I. What is the product of - by —
Analysis. — Multiplying the numerator of operation.
the fraction - by d, the numerator of the axd_ad
^ ad C c
multiplier, we have — , But the multiplier is -j -,
— ; hence the product — is JW times too lar^e. C X ffl cm
^ d a
To correct this, multiply the denominator — X —-=:--
by m. (Art. 167, Prin. 2.) ^ ^- ^^
Note I. How is a fraction multiplied by a quantity equal to its denominator J
Note 2. How by any factor in its denominator ?
e6 MULTIPLICATION OF FRACTIOITS.
2. Required the product of — -^ multiplied by — •
Analysis. — The factors 2, r, and e are opebation.
common to each term of the given frac- 2al) cm 2abcm
tions. Cancelling these common factors, 6cd ax 6acdx
the result is -^ , the product required 2aocm om .
ddcdj'jc 1(1 cr
(Art. 167, Prin. 3.) Hence, the ^
Rule. — Cancel tlie common factors; then multiply the
numerators together for the new numerator, and the denom-
inators for the new denominator.
Notes. — i. Mixed quantities should be reduced to improper fractions,
and then be multiplied as above.
Or, i^Q 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 lowest 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, |off = fxf = ^.
Find the products of the following fractions:
2X xxii 2dy ^13
3. — X — I X — ^« 6. — ; X I-
^ y 2d X a + 3a: 8
he X d (a + m)y.h aii
4. — X T- X — 7. ^^ — X -r-~--{ —
a by c IX {a-{-m)xc
x-^y ^x + y a±h cd
5» X ; — • o. 5 — X — •
yz y + z (? X
9. What is the product of -^— by — ^^^ ?
Solution. — Factor and cancel. Ans. —(x+y).
185. How multiply a fraction by a fraction ? Note i. How mixed quantities !
Vote 2. How shorten the operation ? Note 3. What is the force of the word 0/ ia
compound fractiouB ?
MULTIPLICATION OF FEACTI0:N^S. 87
HT li.- 1 2a , x^ — y^ . 2a (x + y)
10. Multiply by ^- Ans. — —j-^'
11. Multiply -^ by \~ *
^ "^ 4a: "* y^ — 2xy
TIT 1 J.- 1 4^ — 2h , 2a — h
.2. Multiply ^—--^ by -^.
13. Multiply « + -^- by ^.
14. Multiply ^ + 1^ by -t/.
15. Multiply a; — — by - + ^.
16. Multiply « 4- -T- by — y
CASE III.
186. To Multiply an Integer by a Fraction*
dx
I. Multiply the integer a by —
Analysis. — Changing the integer to the
a « =
a
form of a fraction, we have - to be multiplied I
. dx ... . adx^ ^ ^. a dx _adx
by — , which equals . Hence, the 7 ^ 77; — "777 *
^ cy* ^ 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 denom,inator.
Notes. — i. Multiplying an integer by b. fraction is the same as find-
hig B. fractional part of a quantity. Thus, a; x | is the same as finding
f 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.
186. How multiply an integer by a fraction ? Note. To what is this operation
similar?
88 MULTIPLICATION OF FRACTION'S.
Find the product of the following quantities :
9. {x'-f)x ""'
2.
, dx
abc X —
cy
3.
, b -\-c
ad X — ' —
xy
4.
ax X '
4«
5.
(« + ^)xf
3 (^ + «/)
^ 2 (« + /^)
II. (:z:2 + i) X ^^^
3(^—0
12. 2a;^(flj — Z*) X
;2 — Z»2
6- (3« — 2/) X ^- 13. 3a (2; — i) X ~^^
y *. ^ V ' ^'^ _ I
7. (-^+0X^. 14. (2.. + .2)x^^.
8. (i~«^)x-^. 15. (I-^^2)x '
I + « ^ ' n-\- \
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 15a;.
2. Multiply ^^ by y^ — i.
3. Multiply ^ + — by —^.
,, ,,. , 2ax ^ab , xac
4. Multiply — X — - by --^.
^ "^ a ac '' 2ab
5. Multiply ^- by ^.
^ ^ -^ loy ^ ga
6. Multiply X 7 by
a a -\- b '' a
187. What Ib the general rule for multiplying fractions ?
DIVISION OF FRACTIONS. 85
7. Multiply ^— — by x + z.
8. Multiply ^ by 2y\
if
9. Multiply by a;^ — 2xy + «/2.
a; — y
10. Multiply - — —-^ by 8;? — 2.
^ -^ 402; — 10 -^
11. Multiply a; — -^ by - + ^»
^ -^ X -^ y X
2a^ 2ah
12. Multiply a + ^ by — ^ •
., Multiply (i±l)ib,^^.
14. Multiply ^ by -5 ^'
^ -^ 4a; y^ — 2xy
15. Multiply & + ^^^ by Z» - ^^^.
2// ^2 ^2
16. Multiply by —-
^ -^ x — 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?
71/
Analysis.— One is i third of 3; therefore, opeeation.
oa 9^ . _ 3^
I orange will cost i third of — dollars, and "^ "^ 3 — ~
- of — dollars is — dollars. (Art. 167, Prin. 2.)
2. Divide - by m.
c ^
Analysis. — Since we cannot divide
OPBEATION.
the numerator of the fraction by m, — -i- m rz= ^ z=. — •
we multiply the denominator by it. C ' C X m cm
90 DIVISION OP FRACTIONS.
The result is — . For, in eacli of the fractions - and — , the same
cm c cm
number of parts is taken ; but. since the unit is divided into m times
as many parts in the loiter as in the former, it follows that each part
in the latter is only — th of each part in the former. Hence, the
'' m
Rule. — Divide the numerator by the integer.
Or, Multiply the denomhiator hy it.
Note. — If the dividend is a mixed quantity, it should be reduced
to an improper fraction before the rule is applied. (Ex. 3.)
Divide the following quantities
3.
a -\ by d,^
X
4.
^ + l^hjxy.
5.
/ by s^y-
9-
6.
2«^ 1 X
— by 5.
sac -^
10.
7.
« H hy a,
c
II.
8.
ax H by xK
12.
. ax -\- be
Ans. ::
dx
. abx 4- 11
Ans, — J — -•
abxy
c^ -\- ax ,
-^ — by a-\- X,
2b ''
a^ — <^ ,
^— - — by a — c.
7? + 2xy + if .
— — — ^—^-^~ by x-\-y.
— ■ — f by « + J.
a — b^
CASE II.
189. To Divide a Fraction by a Fraction,
This case embraces two classes of examples :
First. Those in which the fractions have a common
denominator.
Second. Those in which they have different denominators.
188. How divide a fraction by an integer? NoU. If the dividend is a mixed
quantity, how proceed ?
DIVISION^ OF FRACTIONS. 91
1. At — dollars apiece, how many kites can a lad buy for
dollars ?
m
Analysis.— Since these fractions have a com- opbbation.
mon denominator, tlieir numerators are like 12a ^ ^a
quantities, and one may be divided by the ni ' m
other, as integers. (Art. 177.) Now yi is con- Ans, 4 kites.
tained in 12a, 4 times. (Art. no, Prin. i.)
2. How many times is - — contained in -^— ?
'' X X
3. What is the quotient of ^^ divided by — ^-
n c
4. It is required to divide - by -•
X y
Analysis. — Since these frac-
tions have different denomina-
tors, their numerators are unlike
quantities ; consequently, one
cannot be divided by the other
in this form. (Art. 114, no^e.) We
therefore reduce them to a com-
mon denominator ; then dividing
one numerator by the other, the
result is the quotient.
Or, more briefly, if we invert the divisor, and multiply the dividend
by it, we have the 8am£ combinations and the same result as before.
(Art. 185.) Hence, the
EuLE. — Multiply the dividend hy the divisor inverted.
Or, Reduee the fractions to a eommon denominator, and
divide the numerator of the dividend hy that of the divisor.
Notes. — i. A fraction is inverted, when its terms are made to
change places. Thus, ^ inverted, becomes - •
0 a
2. The object of inverting the divisor is convenience in multiplying.
3. After the divisor is inverted, the common factors should be can^
celled before the multiplication is performed.
189. How divide a fraction by a fraction when they have a common denominator ?
When the denominators are different, how ?
rmST OPERATION.
a _
x~
_ay c _cx
~ xy' y~~ xy'
ay ^ ex _ay
xy ' xy~ ex
SECOKD OPEBATIOK.
a
X
y X c ex
92 DIVISION OF FBACTIONS.
Divide the following fractions :
^ cd ^ ay ^2* 2¥c^ ^ 6bc
6. ^-r- by -^. 14. — by ^— .
taoy ''ax 4cd '' 2d
7. 3^ by a ,5. J^by^f"?.
8. -^Lby^ t6. A^by3^^-?.
a;— 1*^2 a -\- h '' 2y
g— I 1^ _^ a:* , «a?
,0. 5^ by -'-5^. ,8. 36«f!i, £8aS_
„. iil£±^) byii^±^. 10. -SLbyl^.
12. -^-r-i by 20. -—4 by -f— ^.
a + 0 "^ a iSab *^ 36^6?
CASE III.
190. To Divide an Integer by a Fraction.
I. Divide the integer ydc by ^'
Analysis.— Having reduced the opbbation.
integer to a fraction, and inverted ndc -^ ^ =
the divisor, we cancel the common ^
factor d, and proceed as in the last jdc h ']hG .
case. Hence, the ~i ^ ^ ~ 3« ' '
Rule. — Reduce the integer to a fraction, and multiply it
hy the divisor inverted.
Divide the following quantities :
^ ' ■ jx
190, How divide au integer by a fraction ?
2.
, ex
'^y ■■ dm
3-
ax: ^ .
m -{- n
4.
J/
DIVISION OF FRACTION'S.
93
191. Complex JFractions are reduced to simple ones,
by performing the division indicated.
8. Keduce — to a simple fraction.
3
Analysis. — The given fraction is equiv- a
alent to
« . 3
. Performing the division
b
, the dividend.
indicated, we have ^, the simple fraction
required. (Art. 189.)
-, the divisor.
4
^ y 4 _ 4a .
b 3 3b
Reduce the following fractions to simple ones :
10.
II.
b
Id"
«4-i
« — I
a— I
a+ I
a-b
^ + .V
x — y
a + b
Ans.
12.
13-
bed
a? — y'^
a-b
x + y
x-\-y
a-^b
a + b
x — y
192. The various principles developed in the preceding
cases may be summed up in the following
GEN ERAL 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 reduce complex fractions ? 192. What is the general rule for dividing
fractionB?
94
DIVISION OF FKACTIONS,
1. Divide — — by sic
4xyz
2. Divide ^^ ^ by gxy.
2'jy
^. ., i6xy , 2cd
-?. Divide by —
^ fyn. '' inn.
^. ., 42ah , 145
4. Divide -^ 5 by
5. Divide -^^^-^ by -^.
, TA. ., x'^ — 2xy + ifi , ^ — y
6. Divide / -^ by ^.
7. Divide -X ; — -„ by
Solution. — Factoring and cancelling, we have,
a^—m* _ (a® + m^) (a + m){a—m) ^ a? -{■am _ a(a + m) ^
a^—2am + m^~ {a—m)(a—m) ' a—m ~ a—m
(a^-\-m-)(a + m)(a—m) a—m a'^ + m^ m^ .
' ^-^, — -\ X -7 ^ = , or a + — » Ans,
(a—m) {a—m) a{a + m) a a
8. Divide -^^ by -^—.
cfi — x^ ''a — X
-p. . . , 4^2 — 8c , c2 — 4
0. Divide by -*
^ x + y '' X -\- y
10. Divide ; by ^-73-- — (•
ac -^ ax -^ A^\d -^ X)
11. Divide -=— — r bv — , —
a^ -\- & " a •\- c
12. Divide ■— by ^-^^ — -—^^
X "^ a — x
,3. Divide ^qr-,^jq.-^ by -^-^.
14. Divide ^— by ^--^.
iq. Divide by — ^^ -" (See Appendix, p. 285.)
■^ x-^- ax -^ I ■\- 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 First Member is the part on the left of the
sign =.
The Second Manber is the 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,
CuMc, etc.
197. A Simple Equation is one which contains
only ihQ first power of the unknown quantity, and is of the
first degree ; as, ax = d.
198. A Quadratic Equation is one in which the
highest power of the unknown quantity is a square, and is
of the second degree ; as, ax^ -\- ex ^ d.
199. A Cubic Equation is one in which the highest
power of the unknown quantity is a cuie, and is of the
third degree ; as, aa^ + dx^ — ex = d.
193. What is an equation? 194. How many parts? 195. Which is the flri«t
member ? The second ? 196. How are equations divided ? What other divisions ?
197. What is a simple equation ? 198. A quadratic ? 199. Cubic ?
96 SIMPLE EQUATIONS.
200. An Identical Equation is one in which both
members have the same form^ or may be reduced to the
same form.
Thus, ab—c = ab—c, and 8a;— 3a? = 5a;, are identicaL
Note. — Sucli an equation is often called an identity/.
201. The Transformation of an equation is chang-
ing its form without destroying the equality of its members.
Note. — The members of an equation will retain their equality, so
long as they are equally 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 equahty of its members.
203. Unknown Quantities may be combined with
known quantities by addition, subtraction, multiplication,
or division.
Note. — The object of transposition is 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 from one Member of an
Equation to the other.
1. Given x -{-1) z=a, to find the value of x.
Solution.— By the problem, x+b = a
Adding —b to each side (Ax. 2), x+b—b = a—b
Cancelling ( + &—&) (Ax. 7), .*. x = a—b
This result is the same as changing the sign of b from + to — in
ihe first equation, and transposing it to the other side.
2. Given x — d z= c, to find the value of x.
Solution. — By the problem, x—d = c
Adding +dto 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. Equality.
202. What is transposition of terms? 203. How combine unknown quantities T
Note. Object of transposition ?
ONE UNKNOWN QUANTITY. 97
This result is also tlie same as changing the sign of d from — io 4- ,
and transposing it to the other side. Hence, the
^vJjE.— Transpose the term from one member of the equa-
tion to the other, and 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 subtraction.
3. Given b — c-{-x = a — d, to find x,
4. Given x-\-al) — c=a-\-l), 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 05+6 = <?, it follows that x+c—d = o.
REDUCTION OF EQUATIONS.
207. Tlie Heduction 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 dimin-
ished by the same quantity ivithout destroying the equality.
204. How transpose a term from one member of an equation to the other?
205. What is the effect of changing all the signs ? 206. Of transposing all the terms f
207. In what does the reduction of an equation consist ? 208. What is the value of
an nnkno\ATi quantity ? What sometimes called ? 209. Upon what principle does
the reduction of equations depend ?
98
SIMPLE EQUATIONS
210. This principle may be illustrated by a pair ol
scales. If 4 balls, each weighing i lb., are placed in each
scale, they balance each other.
Adding 2 lbs. to each scale,
4 + 2 = 4 + 2
Subtracting 2 lbs. from each,
4-2 = 4-2
Multiplying each by 2,
4x2 = 4x2
Dividing each by 2,
4-5-2 = 4-7-2
211. To Reduce an Equation containing One Unlcnown
Quantity by Transposition,
5. Given 2a; — 3^? -f 7 = a; -f 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.x = 2S + 3a
Therefore, 28 + 3a is the value of x required. Hence, the
EuLE. — Transpose the unhnoivn quantities to one sid6,
and the known 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 reduction of equations by addition or svbtractiovu
2. If the unknown quantity has the sign — before it, change the
Bigns of all the terms. (Art. 205.)
212. When the same term, having the same sign, is on
opposite sides of the equation, it may be cancelled. (Ax. 3.)
6. Reduce 7,x ■\- a ^ 6 = b — 4 -^ 2X.
7. Reduce x— ■^■^c=z2X'\-a — d.
8. Reduce 21/ -{- he — ad = y + 2m — 8.
9. Reduce sab — y -}- d = — 2y-f 17.
10. Reduce 4cd -f 27 — 4ic -|- J = 28 — 3a: -f 3M.
11. Reduce 5 -f c — 4a; = 32 -f & — $x + d.
12. Reduce a: -f 4 — 2a; — 3 = 3a; -f 4 -f 8 — ^x.
aio. Illustrate this principle? 211. What is the rule for reducing equations- ?
Vote. To what is transposition equivalent ? 212. When the same term, having the
fam^ sign, is on opposite sides, what may be done ?
ONE unk:n^own quantity. 99
CLEARING OF FRACTIONS.
213. To Reduce an Equation containing Fractions.
1. Given - + -— = 27, to find the value of x.
Solution. — By the problem, ? + ??== 2?
2 6
Multiplying each term by 6, the I, c. ni, )
of the denominators (Art. 148), f 3^+ 22? = 162
Uniting the terms, 52? = 162
Dividing each side by the coefficient, Ans. x = 32I
2. Given - -\- - = —, to find the value of x,
23 4
Solution.— By the problem, ? 4. ? = ^
234
Mult, by 12, the I, c, in. of denominators, 6a; +42; = 90
Uniting terms, and dividing (Art. 211), Ans. x= g
Therefore, the value of x is 9. Hence, the
KuLE. — Multiply each term of the equation ly the least
common multiple of the denominators ; then, transposing
and imiting the terms, divide each member hy the coefficient
of the unknown quantity.
Notes. — i. An equation may also be cleared of fractions, 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 fnultipUed by the same
quantity. (Ax, 4.)
3. A fraction is multiplied by its denominator by cancelling the
denominator. (Art. 184, Wote 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 multiplication and division. Hence, the
operation is often called, reduction of equations by multiplication and
division.
213. Rule for clearing an equation of fractions ? Notes, i. In what other way may
fractions be removed ? 2. Why does not this process destroy the equation ? 3. What
:8 the effect of cancelling a denominator? 4. Effect of removing a coefficient'
5. If a numerator is a multiple of its denominator, how proceed ?
100 SIMPLE EQUATIONS.
3. Eeduce — + 12 = — + i.
5 3 ;
4. Eeduce - = 6x — 66»
3 6
5. Reduce — 4--= 'zk —x.
10 5
214. When the sign — is prefixed to a fraction and the
denominator is removed, the sigjis of all the terms in the
numerator must be changed from -f to — , or — to +.
X — 2
6. Eeduce 30; = 20,
Solution.— By tlie problem, 3a; — ^^ = 20
Removing tlie denominator 5, 15a;— a; + 2 = 100
Uniting and transposing the terms, 14a; = 98
Dividing by the coeflacient, Ans. x— 7
7. Given 4^^::^^ =_ 3^, to find ^.
X d'
8. Given ^x — — = a —^ , to find x,
5 10
9. Given — a;H [-^ = -^, to find x.
3 4 24'
10. Given «-|-5 + c = --i [-^ + ^+<?-^5^ to find rzr.
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. (Art. 213.)
II. Transpose and unite tlie terms. (Art. 204.)
III. Divide loth sides hy the coefficient of the unknown
quantity, (Art. 213, Note 4.)
Proof. — For the unknoivn quantity substitute its value,
and if it satisfies the equation, the worlc is riglit.
214. If sign — is prefixed to a fraction ? 315, What is the general rule for simple
equations ? How proved ?
OlJE UNKNOWN QUANTITY, 101
EXAMPLES.
1. Given x ■] 1- - = 14, to find x,
2 4
2. Given - 4- a; = — + 40, to find x,
2 10
3. Given — -f 10 = — + 13, to find x.
^ 5 10 ^^
4. Given — - — \- 6 = S, to find x,
^ X — 2
2X -t- I
5. Given x -\ =10, to find x.
6. Given 2x + ^^±1L = 18 + 9£r:29 ^^ ^^^ ^
5 4
7. Given - H ^ - = 78, to find x.
234
8. Given ^^^-8=^-^ + 5, to find x.
0 4
9. Given x + ^^~^ + '^^~ = 12, to find a;.
26
10. Given 2a; — 16 = ^^ , to find x.
3
^. 2a; — 8 , a; + 32 , ic . £ ;i
11. Given 5—^- + - = 30, to find x,
12. Given - + -^=16 + -, to find a:.
20 o
rt- 3^+1 2a; , a;— I , ^ ,
13. Given — 10 = 1 — , to find x.
2 30
14. Given f- v = — — i A^ to find x.
^ 10 6 15 ^^
15. Given Sx + 6i — - = Si- — + •^, to find x.
-^ ^2 ^72
Note. — Sometimes there is an advantage in uniting similar terms,
before clearing of fractions. Tims, uniting 6^ witli 8^ ; also witli
— , we nave, 8a; = 2 + 8a; ; /. a; = 7, Ans.
102 SIMPLE EQUATIONS.
i6. Given f -6 +^ = 1+ 2, to find a;.
^ . AX ZX
17. Given — = — + 15 — 12, to find x.
5 4
18. Given 20; — 4 = - + 2, to find x,
2
19. Given ^+ ^_ i|::^3^ ^ ,^j^ j^ g^^ ^
45 5
X
20. Given - = 5 -}- c, to find x,
(t
fix
21. Given — = ^Z, to find x.
^. ax hx , n -,
22. Given = c, to find x.
23
^. 2ax -{- h ex + d ^ „ ^
2$. Given -^— = — 3_ to find x.
a c
24. Given h- = - + -, to find x.
a X 2 a
25. Given |+| + f = f + ^f-^ to find x.
26. Gi,en ^ +'-^ = '1 -^'J-^, to nnd X,
2 3 5 3
^. 3flf + a; 6 , ^ -
27. Given - — ■ 5 =-, to find x,
X ^ x'
X — I I
28. Given — ; h i = -, to find x.
ic + I a
r^. X , X a , n -,
29. Given - H = - + «. to find x.
a c —■ a c '
a?
30. Given x-{-h=z , to find x.
31. Given x — a = — -^— - , to find x.
X — a
32. Given 3 (^*) + p^-*) = 4 (^*), to find ^.
3:> Given ^-~ = x --, to find x.
OKE UKKNOWK QUANTITY. 103
34. Given ~-\- x = 25, to find x.
4 2
X X
35. Given 80 = 4a; , to find x.
2 6
,^. 2iC+I a;^^
36. Given ! — = 2X -^-^ , to find x,
3 4
37. Given 10 -> 22; = 5^+^ - £1Z136 ^^ g^^ ^^
3 3
38. Given a; — 3 = 15 _ ^Jli, to find x.
39. Given a; + 2 = 3a; + ^-±-? - ?-±-^, to find a?.
4 3
^. Z^ , X — 4 a; — 10
40. Given ^ + 5 _ ^ _ 5^^ g^^
422
^. iia;— I ex— 11 x—i
41. Given — - — = ^ to find x.
12 4 10
42. Given ^ _ ?? = 120, to find x.
5 10
43. Given ar — 20 = — ^^+1 to find x,
5
44. Given ~^ = ^—^ ■\-t2-x, to find x,
^ 3
45. Given ^-^ + 10 = — _ 1^=^, to find x.
46. Given — ^ - -^- = 5, to find a:.
I* -j- I Q, I
r\' X 2 + X C
47. Given ^-^ - ^_ =.____, to find x.
48. Given ^^=5+^, to find ^.
49- Given — = ^ — — , to find a?.
2 3
50. Given 8«= ^1^, to find x,
1 -\- X
51. Given -~^-^- = ~^yto find a;.
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 deduction of this equation.
217. To Solve Problems in Simple Equations containing
one unknown Quantity.
I. A farmer 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 ?
I. Formation— Let x = A'a number.
By the conditions, - = B's **
2
4
Therefore, by Ax. 0, x + - + — — 2= 52, the whole.
24
2c Reduction— 4a;+2a;+3«— 8 = 208
Transposing, etc., gx = 216
Removing the coefficient, x = 24, A's number.
Ans. A had 24, B had 12, and C had 18—2 = 16.
From this illustration we derive the following
GENERAL RULE.
I. Represent the unknown quantity ly a letter, then state
in algelraic language the operations necessary to satisfy the
wnditions of the prohlem.
II. Clear the equation of fractions ; then, transposing and
uniting the terms, divide each member by the coefficient of the
unhnown 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
letter, and the method of forming the equation.
tx6. What is the Bolution of a problem ? Of what does it coneist ? 217. What is
the general rale ?
OKE TJKKKOWiq- QTJAI^TITY. 105
2. The bill for a coat and vest is I40 ; 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 C 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 1 20 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 1 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 I of my money, and then J of the
remainder, I have $72. What sum had I at first ?
10. Divide $300 between A, B, and 0, so that A may
have twice as much as B, and C as much as both the others.
1 1 . 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
1 1 260 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, I ; the third day,
I; the fourth day, ^; and he then has $300 left. How
much had he at first ?
106 SIMPLE EQUATIOKS.
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 J 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 f 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 lioo, the cows at $45, and the sheep
at I5, 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.
How many should each receive ?
24. Four geese, three turkeys, and ten chickens cost lio ;
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 1 2 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 \, f , and f of it amount
to $1222 ?
OKE UHKHOWN QUANTITY. 107
29. What number is that, |, J, J, and J 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 ?
^^. At an election, the successful candidate had a
majority of 150 votes out of 2500. What was his number
of votes ?
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 0, giving A $100
more than B, and $200 less than 0. What is the share of
each ?
36. A prize of $150 is to be divided between two pupils,
and one is to have | as much as the other. What are the
shares ?
* When the conditions of the problem contain fractional expressions,
as ^, J, \, etc. , we can avoid these fractions, and greatly abridge the
operation, by representing the quantity sought by such a number of
ic's 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 \2X = the number.
Then will 6x — i half.
« " 4a; = I third.
** " 3iC = I fourth.
" '* 225—1 sixth.
Hence, 6a; + 4a! + 3a; + 2a? = 60
.'. a; = 4
Finally, a;x 12 or 12a! = 48, the number required
108 SIMPLE EQUATIONS.
37. Two horses cost 16 16, 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's as 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 1 2 be
added to the product, the sum will be 66.
47. A man bought sheep for $94 ; having lost 7 of them,
he sold ^ 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
$100 in debt. What is their income ?
* When the quantities sought have a given ratio to each other, the
solution may be abridged 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, 2X will
represent the second part ; then 5 j; + 3a; = 48, etc.
ONE UKKITOWK QUAKTITY. 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 vi and n
separately, the difference of their products shall be c? ?
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 h, are to each
other as « 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 his head and tail together. What was
the weight of the fish ?
54. An express messenger travels 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
\ of the nitre. What was the amount of gunpowder ? *
* The operation will be ^ortened by the following artifice :
Let 42a; + 48 = the number of pounds of powder.
Then 28a; + 42 = nitre ; jx+s^ = sulphur ; 4X + 4 = charcoal.
Hence, 390; + 49 ^ = 42a; + 48.
.'. X
110 • SIMPLE EQUATIONS.
59. Divide $6 into 3 parts, sucli that J of the first, J of
the second, and ^ of the third are all equal to each other.
60. Divide a line 21 inches long into two parts, such that
one may be | of the other.
61. A milliner j)aid $5 a month foi rent, and at the end
of each month added to that part of 1 er 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?
6^. Divide $4200 between two persons, so that for every
I3 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 ^ of his annual income for
board, -| for clothing, -^ in charity, and saves I318. What
is his income ?
66. A certain sum is divided so that A has $30 less than ^,
B lio less than |, and 0 $8 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 principle :
If four quantities are proportional, the product of the extremes 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 into an
equation at this stage of his progress.
ONE UNKNOWN^ QUANTITY-. Ill
equal to the product of fhe 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 3a; be equal to their respective ages.
Then 2a;+4 : 3a' + 4 - 5 : 7-
Making the product of the extremes equal to the product of the
mean^ we have,
140;+ 28 = 150;+ 20.
Transposing, uniting terms, etc., a; = 8.
.'. 2X= 16, the younger ; and 3a; = 24, the older.
68. 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 pulLat
the rate of 9 miles an hour, twice as long to come up the
river as to go down. At what rate does the river flow ?
71. One-tenth of a rod is colored red, ^ orange, -^ yellow,
-}^ green, -f^ blue, ^^ indigo, and the remainder, 302 inches,
violet. What is its length ?
72. Of a certain dynasty, | of the kings were of the same
name, J of another, J of another, ^^ 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.)
CHAPTEE X.
SIMULTANEOUS EQUATIONS.
TWO UNKNOWN QUANTITIES.
218. Simultaneous * Equations consist of two oi
more equations, each containing two or more unk7iown
quantities. They are so called because they are satisfied by
the same values.
Thus, x + y^'j and 5a?— 4y = 8 are simultaneous equations, for in
each aj = 4 and y = 3.
219. Indepe^ident Equations are those which
express different conditions, so that one cannot be reduced
to the same form as the other.
Thus, 6aj— 4^=14 and 23^ + 3^=: 22 are independent equations.
But the equations ic + y = 5 and 3a; + 3^ = 15 are not independent, for
one is directly obtained from the other. Such equations are termed
dependent.
Note. — Simultaneous equations are usuaWj 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 a?+y=7 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 mani/ simultaneous equations as there are
unknown quantities.
If there are more equations than unknown quantities,
some of them will be superfluous or contradictory.
218. What are Bimultaneous equations ? 219. Independent equations ? 220. Ho^
many equations must each problem have ?
* From the Latin ^i'n^id, at the same time.
TWO UNKN^OWN QUAl^TITIES. 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. JSllmination* 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,
CASE I.
222. To Eliminate an Unknown Quantity by Comparison,
I. Given x -\- y =z i6, and a; — «/ = 4, to find x and y.
Solution.— By the problem,
« **
Transposing the ^ in (i),
" the 7/ in (2),
By Axiom i.
Transposing and uniting terms.
Substituting the value of y in (4),
x+p= 16
(I)
x-y= 4
(2)
X = it-y
(3)
x= A+y
(4)
4+y= it—y
(5)
2y= 12
(6)
.-. y= 6
aj= 10
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 he eliminated in terms of the other quantities.
II. Form a new equation from these equal values, and
reduce it ly 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.)
For convenience of reference, the equations are numbered (i),
(2), (3), (4), etc.
•
221. What is elimination? Name the methods. 222. How eliminate an unknown
ttuantity by comparison ? Note. Upon what principle does this rule depend ?
* From the Latin eliminare, to cast out.
114 SIMPLE EQUATIONS.
2. Given x -\- y = 12, and x—y-ir^ z= 8, to find x and y
3. Given 3X-]-2y = 48, and 2x—$y = 6, to find x and y.
4. Given x-\-y = 20, and 2:^+3?/ = 42, to find x and y.
5. Given 4^+32/ = i3> and 3^^ + 21/ = 9, to find x and ?/.
6. Given 30; +2?/ =118, and a; +5?/= 191, to find a; and ^^'.
7. Given 4x-\-^y = 22, and 72:4-32^=27, to find x and 2/.
CASE II.
223. To Eliminate an Unknown Quantity by Substitution.
8. Given x-{-2y = 10, and ^x-^2y z= 18, to find x and y.
Solution. — B7 the problem, x+2i/=io (i)
" 3a;+22/ = i8 (2)
Transposing 2y in (i), . x = 10— 2y (3)
Substituting tlie value of x in (2), 30—4^ = 18 (4)
Transposing and uniting terms (Art. 211), 42/ = 12 (5)
'- y= 3
Substituting the value of 2^ in (i), a; = 4
5^" For convenience, we first find the value of the letter which is
least involved. Hence, the
Rule. — I. From one of the equations find the value of the
unknown quantity to he eliminated, 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 x + ^y = 19, and $x — 2y = 10, to find x and y.
10. Given - + ^ = 7, and - -|- ^ = 8, to find x and 11.
23' 3^2' ^
11. Given 2x-{-^y =z 28, and 30;+ 2^ = 27, to find x and y.
12. Given 4X-{-y = 43, and 5^+2^ = 56, to find x and y.
13. Given s^+^ = 7^> and 5?/ + 32 = 'jx, to find x and y.
14. Given 4X-]-sy= 22, and yx+^y = 2^, to find x and y.
223. How eliminate an unknown quantity by substitution? If^ote Upon what
principle docs this method depend?
TWO UNKNOWN QUANTITIES. 115
CASE III.
224. To Eliminate an Unknown Quantity by Addition or
Subtraction,
15. Given 4a; + 3^ =18, and 5a:— 2?/=: 11, to find x and y.
Solution. — By tlie problem, 4a; + 3^=18 (i)
" " e,x—2y~ II (2}
Multiplying (i) by 2, the coef. of y in (2), 8a5+ 6y = 36 (3)
Multiplying (2) by 3, the coef. of y in (i), isx—6y = 33 (4)
Adding (3) and (4) cancels 6y, 23a; = 69 (5)
,', X =3
Substituting the value of a; in (i), 12 + 3^ = 18
y= 2
^° In the preceding solution, y is eliminated by addition,
16. Given 6a: + 5?/ =2 8, and 8a; + 3?/— 30, to find x and y.
Solution. — By the problem, 6a; + SV = 28 (1)
" '* 8a; + 3y= 30 (2)
Multiplying (i) by 8, the coef. of a; in (2), 48a; + 402/ = 224 (3)
Multiplying (2) by 6, the coef. of x in (i), 48a; + i8y = 180 (4)
Subtracting (4) from (3), 22^ = 44
.'. y= 2
Substituting the value of y in (2) 8a;+6 = 30
.'. x= 3
B^ In this solution, x is eliminated by subtraction. Hence, the
EuLE. — I. Select the letter to he eliminated; then multiply
or divide one or doth equations hy such a number as ivill
make the coefficients of this letter the same in loth, (Ax. 4, 5.)
II. If the signs of these coefficients are alilce, subtract one
equation from the other ; if unlihe, add the two equations
together. (Ax. 2, 3.)
Notes. — i. The object of multiplying 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 ^rime to each other, multiply each equation by the coefficient
of this letter in the other equation, as in Ex. 15.
224. What is the rule for elimination by addition or snbtraction ? Notes. — i. The
object of multiplying or dividing the equalion ? 2. If coefficients are prime ?
116 SIMPLE EQUATIONS.
3. If not prime, divide the I. c, in* 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 and 8, the coefficients of a; in Ex. 16, is 24;
hence, the multipliers would 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 I, c, in. is avoided.
17. Given 3a; +4?/ =2 9, and 'jx-\-iiy=j6,to find a: and y.
18. Given 9a;— 4^=8, and 13^-1- 72/== loi, to find ir and ^.
19. Given ^x—'jy='jy and 12a: + 5^/^94, to find x and y.
20. Given 3^^+21/= 118, and 0:4-5^=191, to find xandy.
21. Given 4a; -{-52/= 22, and 7a; + 32/= 27, to find x and y.
Rem. — The preceding 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
frax^tions.
The second is convenient when the coefficient of one of the unknown
quantities is i ; if more than i, it is liable to produce /7•ac^^07^«.
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. 2a; + 3^= 23, 5. 5a; + 7^= 43,
5a; — 2^= 10. \ix -\- 9?/= 69.
2. 4a; + y— Z^, 6. 8a; — 2iy= n,
4y+ x= 16. 6a: + 357/= 177.
3. 30a; + 4oy = 270, 7. 21?^ + 20a; = 165,
50a; 4- 30^ = 340. 772^ — 30^ = 295.
4. 2a; + jyzzz 34, 8. I la; — loy = 14,
SX+ gy= 51. s^+ iy= 41.
Notes. — 3. If not prime, how proceed ? 4. If the coefficients have comuion feo-
tors, how shorten the operation ?
TWO UNKNOWK QUANTITIES. ll?
lO.
II.
12.
13-
14.
6y~
2a; = 208,
15.
Sx +
y =
'42,
loa; —
42/ = 156.
2X +
42/ =
18.
4X +
Sy= 22,
16.
2X +
42/ =
20,
5^-
72/= 6.
4X +
52/ =
28.
SX-
5«/= i3>
17.
4X-\-
32/ =
50,
2X +
7^= 81.
3«-
32/ =
6c
5^-
72/= 33j
18.
3^ +
52/ =
57,
iia: +
121/ = 100.
5« +
32/ =
47.
i*
|=.8,
19.
; +
2( _
3
7>
re
2
^ = 21.
4
i-
4
5.
162; +
17?/ = 500,
20.
2a; +
2/ =
50,
17a; —
3«/= no.
i*
^ _
7
5.
PROBLEMS.
1. Required two numbers whose sum is 70, and whose
difference is 16.
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 ; 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 I570. If A's share were
three times and B's five times as great as each really is, then
both would have I2350. 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 j^^ou lend me J 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 I8500, 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 f of one part,
added to f of the other, shall be 40.
16. Divide I1280 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 liorse ; 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. Kequired 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 far 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 I the less be taken from | 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 ^ 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 1 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 ?
120 SIMPLE EQUATION".
THREE OR MORE UNKNOWN QUANTITIES.
225. The preceding methods of elimination of two
unknown quantities are applicable to equations containing
three or more unknown quantities. (Arts. 222-224.)
226. To Solve Equations containing three or more Unknown
Quantities.
I. Given 3a; + 2^ — 52? = 8, 2X -{- $y -\- 4Z = 16, and
$x — 6y ■}- ^z = 6, to find x, y, and z.
-.UTlON.— By the problem.
30!+ 2y- sz =
8
(I)
K ft
2X+ 3y+ 4^ =
16
(2)
tt U
SX— 6y+ 3Z =
6
(3)
Multiplying (i) by 2,
6x+ 4y—ioz =
16
(4)
(2) by 3,
6x+ gy+i2Z =
48
(5)
Subtracting (4) from (5),
5.?/ + 22S =
32
(6)
Multiplying (2) by 5,
ioa; + 1 5^ + 202 =
80
(7)
(3) by 2,
1005— i2y+ 6s =
12
(8)
Subtracting (8) from (7),
27y + i4Z =
68
(9)
Multiplying (6) by 27,
1352^ + 5942 =
864
(10)
(9) by 5,
1352/ + 70s =
340
(II)
Subtracting (11) from (10),
5242 =
524
T
(12)
Substituting the value of 2 in (6), y =
1
2
Substituting the value of y
and z in (2), x =
3
Ana. x = 3, y = 2, 2=1.
Hence, the
Rule. — I. From thegiveti equations eliminate one unknown
quantity, by combining one equation with another,
II. From the resulting equations eli?ninate another unknown
quantity in a similar manner. Continue the operation ufitil
a single equation is obtained, with but one unknown quan-
tity, and reduce this by the preceding rules.
Note. — The letter having the smallest coefiScients should be elimi-
nated first ; and if each letter is not found in all the given equations,
begin witli that which is in the least number of the equations.
226. What iB the rale for solving equation? having three or more unknown quan-
tities ? Note. Which letter should be eliminated flret ?
THEEE OR MORE UNKNOWN QUANTITIES. 121
Eeduce the following equations :
2. 5^ — 3«/ + 2^ = 28, 5. 5ic + 2?/ 4- 42; = 46,
SX + 2y — 4Z = 15, 32; + 2?/ + z = 23,
3^ + 42; — a; = 24. loa; + 5?/ + 4^ = 75.
3. 2a; 4- 5«/ — 32^ = 4, 6. a; + ?/ + 2; z= 53,
4^ — 3y + 22; = 9, a; + 2y 4- 32; = 105,
5x + 6y—2z=ziS. ic + 32^ + 40 = 134.
4. 2a; -\-sy — 4Z = 20, 7. 33: + 42; = 57,
^—2?/ + 3^= 6, 2y— z = ii,
SX — 2y-}-sz = 26. . 5a: + 3?/ =: 65.
234 345 450
9. Required the value of w, x, y, and z in the following
equations :
w+x-\-y-\-z=i4 (i)
2W 4- a; + 2/ — 2? = 6 (2)
2w -\- z^ — y + z = 14 (3)
«^ — ^ + 32/ + 4^ = 31 (4)
SOLUTION.
Adding (i) and (2), 2)'^+ 2X-\- 2y = 20 (5)
" (2) and (3), 410+ 4X =20 (6)
Multiply (3) by 4, Sw+i2X— 4^+42= 56 (7)
Subtract (4) from (7), 7z^+i3a;— 7^ = 25 (8)
Multiply (5) by 7, 2it^+i4a; + i4y =140 (9)
" (8) by 2, 14^^ + 26a!— i4y = 50 (10)
Add (9) and (10), 35?5 + 4ac =190 (11)
Multiply (6) by 10, 40^0 + 401? =200 (12)
Subtract (11) from (12), 5«/J = 10
.'. w = 2
Substituting value of w in (6), 8 + 4aj = 20
.'. «— 3
Substituting value of w and a? in (5), etc.,
y = 4, and 2 = 5.
6
L;i2
SIMPLE EQUATIONS.
227. The solution of equations containing many unkno\;\'t]
quantities may often be shortened by substituting a single
letter for several.
lo. Eequired the value of w, x, y,
and z in the adjoining equations.
io-^x-\-y= 13
w-\-x-\-z = 17
iv-\-y-i-z = 18
x+y-{-z = 2i
(I)
(2)
IS)
(4)
Note. — Substituting s for the sum of the four quantities, we have,
8 = w + x+y + z.
Equatien (i) contains all the letters but z, s—z = 13 (5)
(2) « « " p, s-y = 17 (6)
(3) " " '* X, s-x =18 (7)
(4) " " " w, s-w = 21 (8)
Adding the last four equa- )
tions together.
or 4S—{z + y-{-x + w) j^ = 69
or 4s— 8
That is.
3s = 6g
.-. 8= 23
Substituting 23 for s in each of the four equations, we have,
w = 2, x= s, y = (>, z= 10.
II. Eequired the value of v,
Wf X, y, and z, in the adjoining
equations.
(9)
(10)
V -\-w-\-x-\-y := 10 (1)
V -\-W-\-X-\-Z = II (2)
V -\-w-\-y-\-z=i2 (3?
V +x -\-yi-z = is (4)
lw + x+y-{-z = i4 (5)
Note. — Adding these equations, 4v + 4w + 4X + 4y+4Z = 60 (6)
Dividing (6) by 4, v+ w+ x+ y+ z = 15 (7;
Subtracting each equation from (7), we have,
s = 5, y = 4, x = 3, w = 2, and v = 1.
I£. W + X -\- Z z= 10,
X -}- y + z = 12,
w +x -\-y= 9,
w -\- y -\- z = II.
13-
I I
X y
5
6'
y z 12'
X z 4
(See Appendix, p. 286.)
THREE OR MORE UNKNOWN QUANTITIES. 123
PROBLEMS.
1. A man has 3 sons ; the sum 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 32. 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 I300; and
of a third, 12 sheep and 20 lambs for $140, 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: J the first
with J in the second, is 60 less than in the third ; | of the
third with | the first, is 165 less than the second. How
many are in each ?
5. A grocer has three kinds of tea: 12 lbs. of the firsts
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
II 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 0 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 $90 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; As 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 C's. What are the respective distances ?
9. A, B, and C purchase a horse for $100. The payment
would require the whole of A's money, with half of B's ; or
the whole of B's with i of C's; or the whole of C's with J
of A's. How much money has each ?
OHAPTEE XI.
GENERALIZATION.
228. Generalisation is the process of finding a
formula, or general rule, by which all the problems x)f 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 I^roMems 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.*
SuGGESTiON.^If & 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
General Solution. — ^Let x = the required factor.
By the conditions, xxb,OThx = a,the product. Hence, the
FOKMULA. X = ^'
h
Translating this Formula into common language, we have the
following
EuLE. — Divide the product ly the given factor ; the quo-
tient is the factor required.
228. What is generalization ? 129. When is a problem generalized ? 230. Hov»
are quantities expressed in general problems ?
* New Practical Arithmetic, Article 93.
GENERALIZATIOIS'. 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 C ?
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.
Suggestion. —Substituting a for the product, h for one factor, and
e for the other, we have the
General Solution. — Let x = the required factor.
By the conditions, xxbxc, or bcx = a, the product.
Removing the coeflBcient, we have the
FOKMULA. X = i —
be
Rule. — Divide the given product hy 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 $392, and the
difference in their prices was 1 1 8. What was the price of
each ?
Note. — From the items given, this problem may be generalized
as follows :
231. When the product of two factors and one of the factors are given, how find
the other factor ? 232. When the product of three fiactors and two of them are
given, how And the other foctor ?
126 ge:n^eralization".
233. The Sum and Difference of two Quantities being given,
to Find the Quantities.
Suggestion. — Since the sum of two quantities equals the greater
filns the less ; and the less plus the difference equals the greater ;
it follows that the sum plus the difference equals twice the greater.
Substituting .s for the sum, d for the difference, g for the greater,
9xid I for the less, we have the following
General Solution. — Let ^ = the greater number,
and 1= '* less "
Adding, gr-f-? = «, the sum.
Subtracting, g—l — d, the difference.
Adding sum and difference, 2g = s+d
8A-d
Removing coefficient, g = , greater.
Subtracting difference from sum, 2I = s—d
g ^
Removing coefficient, I = — — , less. Hence, the
Formulas.
This problem may be solved by one unknown quantity.
7 — ^ — ^
2
FORMATION OF RULES.
234. Many of the more important rules of Arithmetic
are formed by translating Algehraic 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.
II. To find the less, suUract half the difference from half
the sum.
8. Divide I1575 between A and B in such a manner that
A may have $347 more than B. What will each receive ?
233. When the stun and difference of two quantities are given, how find the
quantities ? 234. Give the rule derived from the last two formulas.
GENERALIZATION". 137
9. At an election there were 2150 vot^s 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 0 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. Substituting a
for 8 days, and h for 12 days, we have the
General Solution. — Let x = the time required.
Dividing x by a, we have - = part done by B,
a
X
** X by 6, we have, h~ ** ** ^'
By Axiom q, - + - = i, the work done.
a J)
Clearing of fractions, bx+ax = db
Uniting the terms, (a + b)x = ab,B and C*s time.
Removing tlie coefficient, we have the
Formula. x = i--
HvLiE.— Divide the product of the numbers denoting the
time required ly each force, hy the su7n of these numbers ;
the quotient is the time required by the united forces.
11. A cistern has two pipes; the first will fill it in
9 hours, the second in 15 hours. In what time will both
fill it, running together ?
235. The time beiug given in which two or more forces can produce a result, how
find the time reauired for the united forces to produce it ?
128 GEN^EKALIZ ATIOl!f.
1 2. A can plant a field in 40 hours, and B can plant it in
50 hours. How long will it take both to plant it, if they
work together ?
236. In Generalizing Problems 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.
Note. — The elements 0Yfact(y,'s in percentage are,
ist. The Base, or number on wliicli percentage is calculated.
2d. The Rate per cent, which shows how many hundredths of
the base are taken.
3d. The Per^centage, or portion of the base indicated by the rate.
4th. The Amount, or the hsiseplus or minus the percentage. Thus,
Let b = 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 ?
Note. — The items in this problem may be generalized as follows :
237. The Base and Rate being given, to Find the
JPerceutage,*
Suggestion. — Per cent signifies hundredths; hence, any given
per cent of a quantity denotes so many hundredths of that quantity.
But finding a fractional part is the same as multiplying the quantity
by the given fraction. Substituting h for the cost or base, and /• for
the number denoting the rate per cent, we have the
General Solution. — Let p = the percentage.
Multiplying the base by the rate, br = p. Hence, the
FoKMUiA. p = hr.
Rule. — Multiply the base hy 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 factors in percentage? 237. When base
and rate are given, how find the percentage ?
♦ New Practical Arithmetic, Arts. 336 — 340.
GENERALIZATION^. 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 follows :
238. The Base and Percentage being given, to Find the Rate,
Suggestion. — Percentage, we have seen, is a product, and the hase
is one of its factors (Art. 237, note) ; tlierefore, we have the product
and one factor given, to find the other factor, (Art. 231.) Substituting
2> 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-r-h = r. Hence, the
Formula. r = ^'
o
Rule. — Divide the percentage ly the lase; 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 $750 by a speculation, which was*
25 per cent, of the money invested. What sum did hb
invest ?
Note. — The particular statement of this problem may be tran».
formed into the following general proposition :
238. When the hase and percentage are given, how find the rate?
130 GENERALIZ ATIOIT.
239. The PeVcentage and Rate being given, to Find the Base,
Suggestion. — We have the product and one of its factors given,
to find the otJier factor. (Art. 237, note.) Substituting p for the
percentage, and r for the rate per cent gained, we have the
General Solution. — Let 6 = the base.
Then (Art. 237), _p-r-7' = 6. Hence, the
Formula. b = —'
r
KuLE. — Divide the percentage by the rate, and the quotient
is the base.
21. A paid a tax of I750, 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 $2500 a year. What is his income ?
23. xit 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 Amount,
Suggestion. — Since B laid up 8 per cent., he was worth his origi-
nal stock plus 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 will 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 ± /•)•
KuLE. — Multiply the base by i i the rate, as the case may
require, and the result ivill be the amount.
239. When percentage and rate are given, bow find the base ? 240, How find the
amount when the base and rate are given ?
GENEKALIZATION. 13i
Note. — When, from the nature of the problem, the amount is to he
greater than the base, the multiplier is i plus 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 = 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 sum of principal and interest.
26. What IS the interest of $465 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, pr x t
must be the interest of p dollars for t years. Hence, the
Formula. i = prt.
EuLE. — Multiply the principal hy the interest of %i for
the given time, and the result is the interest.
27. What is the interest of I1586 for i yr. and 6 m., at
8 per cent ?
28. What is the int. of $3580 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 amount is j^reater or less than the base, what is the multiplier?
241. What are the elements or factors which enter into computations of interest?
242. When the principal, rate, and time are given, how find the interest ?
132 GEN^ERALIZATION.
243. The Principal, Rate, and Time being given, to Find the
Amount,
General Solution. — Reasoning as in the preceding article,
the interest = prt.
But the amount is the sum of the principal and interest.
/. p+prt = a. Hence, the
FoBMULA. a = p-{- prt.
Rule. — Add the interest to the principal, and the result is
the amount.
30. Find the amount of $4375 for 2 years and 6 months,
at 8 per cent.
31. Find the amt. of $2863.60 for 5 years, at 7 per cent.
244. The delation between the four elements in the
Formula, a = p ^ prt,
is such, that if any three of them are given, the fourth may
be readily found. (Art. 243.)
245. The Amount, the Rate, and Time being given, to Find
the Principal.
Transposing the members and factoring, we have the
Formula. p
1 -\- rt
32. What principal will amount to I1500 in 2 years, at
6 per cent ?
2,3' What sum must be invested at 7 per cent to amount
to $300 in 5 years ?
246. The Amount, the Principal, and the Rate being given,
to Find the Time.
Transposing 2> and dividing hy pi', (Art. 244), we have the
Formula. t = ^'
pr
34. In what time will $3500, at 6 per cent, yield $525
interest ?
243. The amount f 244. What is the relation between the four elements in the
preceding formula. 245. When the amount, rate, and time are given, state the
formula, 246. When the amount, principal, and rate* are given, state the formula.
GENERALIZATIOK.
133
247. The Hour and Minute Hands of a Clock being together
at I2!VI., to Find the Time of their Cottjuncfioii 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.
Analysis. — 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 11 spaces.
Now, if it takes the minute-hand * hour to gain 1 1 spaces, to gain
I space will take -^j of an hour, and to gain 12 spaces it will take 12
times as long, and 12 times -^j hr. = ^f hr. = Tj\ 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 jJy hr., or i hr. 5y\ 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, h 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
Formula. t = ah.
Rule. — Multiply the time required to gain 12 spaces by
the given hour past 1 2 o^dock ; the product will de 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 XII.
INVOLUTION *
248. Involution is finding a power of a quantity.
249. A Power is the product of two or more equal
factors.
Thus, 3x3 = 9; axaxa = a^', 9 and a^ are powers.
250. Powers are divided into differe^it degrees ; as first,
second, third, fourth, etc., the name corresponding with the
number of times the quantity is taken as 2i factor to produce
the power.
251. The First Power is the quantity itself.
The Second Poiver 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 cuhe, etc.
Note. — The quantity caUed the first 'power is, strictly speaking, not
a power, but a root. Thus, <2^ or a, is not the product of any two equal
factors, but is a quantity or root from which its powers arise.
252. The Index or Exponent f of a power is a, figure
or letter placed at the right, above the quantity. Its object
is to show hoto many times the quantity is taken as a factor
to produce the power.
Thus, «' = a, and is called the first power.
a'^ = axa, the second power, or square,
a^ = axaxa, the third power, or cube,
<35* = ax ax ax a, the fourth power, etc.
248. What is involution? 249. A power? 250. How divided? 251. The first
power ? Second power 1 Third ? 252. What is the index or exponent ? Its object ?
* Involution, from the Latin involvere, to roll up.
f Index (plural, indices), Latin indicare, to indicate.
Exponent, from the Latin exponere, to set forth.
IKVOLUTIOK. 135
Notes. — i. The index of the first power being i, is commonly
omitted.
2. The expression «* ig read " a fourth," "the fourth power of a"
or " a raised to the fourth power ;" x^ is read, " x nt\v" or " the
wth power of x."
Read the following: a\ d^, x\ y^% %^, IT, cC".
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^, etc.
255. Reciprocal Powers are those which arise from
the continued division of a unit by the direct powers of that
4uantity. (Art. 55.)
Thus, the continued division of a unit by the direct powers of a
^ves the series,
I I I I I I
a' ~a^* ^3' "^' ¥5' ¥'
256. Reciprocal Powers are commonly denoted by
prefixing the sign — to the exponents of direct powers of
the same degree.
Thus, -^a~\ 4; = «-^ — , = a-', -^ = a"^, etc.
257. The difference in the notation of direct and recip-
rocal powers may be seen from the following series :
(I.) a\ a\ a\ a\ a\ i, \, -^, ^, -^„ ^, etc.
(2.) a^, aS a% a\ a\ a^, a-\ a'"^, a"^, aS a~^ etc.
Note. — The first half of each of the above expressions is a series
of direct powers ; the loM half, a series of reciprocal powers.
258. Negative Exponents are the same as the
exponents of direct powers, with the sign — prefixed to
them.
Note. The index i ? 253. How else are powers divided? 254. Direct powers?
255. Reciprocal? 256. How is a reciprocal power denoted ?
136 INVOLUTION.
N0TES.-~i. This notation of reciprocal powers is derived from the
continued division 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 poicers are often called positive, and reciprocal powers,
negative. But the student must not confound the quantities whose
exponents have the sign + or -, with those whose coeJicientshaiYe the
sign + or — . This ambiguity will be avoided, by applying the term
direct, to powers with, positive exponents, and reciprocal, to those with
negative exponents.
259. The Zero I^oiver of a quantity is one whose
exponent is o ; as, a° ; read, *^ the zero power of a."
Every quantity with the index o, is equal to a unit or i.
For, — r= «"-» = ao (Art. 113) ; but — = i ; hence, «« = i.
SIGNS OF POWERS.
260. When a quantity is positive, all its poivers 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. All JPowers of a quantity may be formed by
multiplying the quantity into itself. (Art. 249.)
262. To Raise a Monomial 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 jwoduct 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 tahen. (Art. 87, Prin. 3.)
1. Given 3^^ to be raised to the third power.
SOLUTION.
{UhJ = 3a¥ X 3«^ X sah^ (Art. 261),
or, sxsxsxaxaxaxb^xb^xIP (Prin. 2),
.-. (3ab^y = 27«3&6, Ans.
Involving each of tliese factors separately, we have, (3)3 = 27 ;
(af = a^ ; and {¥f = ¥ ; and 27xa^x¥ = 2']a^¥, Ans. Hence, the
EuLE. — Eaise the coefficient to the poiuer required, and
multiply the index of each letter by the index of the power,
prefixing the proper sign to the result, (Art. 90.)
Notes.— T. A single letter is involved by giving it the index of the
required power.
2. A quantity which is already a power is involved by multiplying
its index by the index of the required power.
3. The learner must observe the distinction between an index
and a coefficient. The latter is simply a multiplier, the former shows
how many times the quantity is taken as a, 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^^ ?
8. What is the fourth power of $aM^c?
9. What is the sixth power of 2a^c^'i
10. What is the eighth power of abcd'i
11. What is the nih power of xyz ?
262. How raise a monoinial to any power? Note. A pingle letter ? A" quantity
already a power ? Distinction between index and coefficient ?
138 INVOLUTION.
12. Find the fifth power of {a + hy.
13. Find the second power of {a -h hY",
14. Find the nth power of {x — «/)*".
15. Find the 7it\\ power of {x + yy.
16. Find the second power of (a^ + If).
17. Find the third power of {aW¥).
263. To Involve a Fraction to any required Power*
18. What is the square of — ^?
EuLE. — Raise both the numerator and denominator to the
required power.
'i(it^
10. Find the cube of ^ — •
2a
20. Find the fourth power of ^
21. Find the square of ^— gTn*
2
22. Find the wth power of -•
23. Find the wth power of — ^ •
<y
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 + Z>2. Ans. ^a^-^-SaU^-^hK
25. What is the square of « + 5 + c ?
Ans. 0? + 2ab + 2ac + 5^4- 2bc + &;
26. What is the cube of a; + 2?/ + 2 ?
265. It is sometimes sufficient to express the power of a
compound quantity by exponents.
Thus, the square of a + & = (« + &)2 ; the nth power of a& + c + 3«?^ -
363. How involve a fraction? 264. How involve a compound quantity «
INVOLUTION-
139
L
—
:i
-
'
FORMATION OF BINOMIAL SQUARES.
266. To Find the Square of a Binomial in the Terms of
its Parts.
1. Given two numbers, 3 and 2, to find the square of
their sum in the terms of its parts.
Illustration.— Let the shaded part of the diagram represent the
square of 3 ; — each side being divided into
3 inches, its contents are equal to 3 x 3, or
9 sq. in.
To preserve the form of the square, it is
plain equal additions must be made to two
adjacent sides ; for, if made on one side, or on
opposite sides, the figure will no longer be a
square.
Since 5 is 2 more than 3, it follows that
two rows of 3 squares each must be added at
the top, and 2 rows on one of the adjacent sides, to make its length
and breadth each equal to 5. Now 2'into 3 plus 2 into 3 are 12 squares,
or tvyice the product of the two parts 2 and 3.
But the diagram wants two times 2 small squares, to fill the comer
on the right, and 2 times 2, or 4, is the square of the second part. We
have then 9 (the square of the first part), 12 (twice the product of the
two parts 3 and 2), and 4 (the square of the second part). Therefore,
(3 + 2)2 = 32 + 2 X (3 X 2) + 2^.
2. Required the square oix-^y, Ans. x^-{-2 xxy-\-y\
Hence, universally.
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.
Note. — The square of a Unomial always has three terms, and con-
sequently is a trinomial. Hence,
No binomial can be a perfect square. (Art. loi.)
266. To what is the square of the sum of two quantities equal ? How illustrate
the square of the sum of two quautities in the terms of its parts ?
140 BII^OMIAL 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 Bi7iomial Theorem.*
268. The Binomial Hieorem is a general formula
by which any power of a binomial may be found without
recourse to continued multiplication.
To illustrate this tlieorem, let us raise the binomials « + & and a—h
to the second, third, fourth, and fifth powers, by continued multipli-
cations :
{a + hf = a^ + 2(0) + ¥.
(a + hf = a^ + sa^b + saf^ + ¥.
{a + Vf = a* + ^a% + ta%'^ + 40*3 + j^^
{a + yf = a^ + 5a*& + loaW + loarW + 506* + }fi.
(a - bf = a? - 2ab + b^.
{a - bf ^ «3 _ 3^2^ + 2>ab'' — b\
(a - bf = a'^ - 4a^b + 6a?b^ - ^ab^ + b\
{a - bf -oJ> - sa^ft + \oaW - loa^&a + 5^54 _ y,^
269. Analyzing these operations, the learner will discover
the following laws which govern the expansion of Unomials :
1. The number of terms in any power is one more than the
index of the power.
2. The i7idex of the first term or leading letter is the
index of the required power, which decreases regularly by i
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.
268. What is the Binomial Theorem ? 269. What is the law respecting the num-
ber of terms in a power ? The indices of each quantity ? The sum of the indice*
iu each term ?
* This method was discovered bj Sir Isaac Newton, in 1666.
BINOMIAL THEOEEM. 141
4. The coefficient oi the first and last term of every power
is I ; of the second and next to the last, it is the tJidex of
the power; and, universally, the coefficieuts of any two
terms equidistant from the extremes, are equal to each other.
Again, the coefficients regularly increase in the first half
of the terms, and decrease at the same rate in the last half.
5. The signs follow the same rule as in multiplication
270. The preceding principles may be summed up in the
following
GENERAL RULE.
I. In'DICES. — Give the first term or leading letter the index
if the required power, and diminish it regularly hy i through
the other terms.
TJie index of the following letter in the second term is i,
and increases regularly hy i through the other terms.
II. Coefficients. — The coefficient of the first term is i.
To the second term give the index of the poiuer ; and,
universally, multiplying the coefficient of any term hy the
index of the leading letter in that term, and dividing the
product by the index of the following letter increased hy i,
the result will he the coefficient of the succeeding term.
III. Signs. — If hoth terms are positive, mahe all the terms
positive; if the second term is negative, make all the odd
terms, counting from the left, positive, a?id all the even terms
the binomial formula.
n—- I
(a + hy = a'' -\-n x a""-^ h + n x a^'-^h^ etc.
Note. — The preceding rule is based upon the supposition that the
index is a positive whole number ; but it is equally true when the index
is e\ih.Qr positi've or negative, integral or fractional.
The coeflacients of the first and last terms ? The law of the signs ? 270. What
is the general rule ?
142 BINOMIAL THEOKEM.
Expand the following binomials :
1. {a + by. 6. (^-\-zy^.
2. (a — by. 7. (a — by.
3. {c + ay, 8. (m + 7iy\
4. {x + yy, g. (x—y)^.
5- {^—yy* lo. {a+by.
2T1.. 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 the opera-
tion is completed, the value of the terms must be restored.
11. Required the fifth power of i^ + 3^2
Solution.— Substitute a for x'^, and h for 32^2 . then
(a + If = a^ + 5a^& + loaW + ioa?l^ + soft* + 6^.
Restoring the values of a and &,
(a;2 + 3^2)5 _ ajio + 1 53,8^2 ^ goa?^^ + 2-joxh^ + 4052;^ + 2432/^®.
12. Expand (x^ — 2,by.
A71S. 7^ — i2^b + s^b^ — io8ic2^»3 4. 8iJ4.
272. Every power of i is i, and when a factor it has no
effect upon the quantity with whicli 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 mm of the two exponents in each term
is equal to the index of the power.
13. Expand (ic + i)^. 15. Expand (\ ~ ay.
14. Expand \b — i)*. 16. Expand (i -f of,
(See Appendix, p. 287.)
271. When the terms have coefficients or exponents, how proceed ?
POWERS or POLYNOMIALS. 143
273. A Polynomial may be raised to any power by actual
multiplication, taking the given quantity as a factor as many
times as indicated by the exyonent of the required power.
But the operation may often be shortened by reducing the
several terms to two, by substitution, arid then applying the
Binomial Formula.
17. Kequired the cube of x -^ y -\- z.
Solution. — Substituting a for (y + e), we have aj + (y +2) = aj + a.
By formula, {x-vaf = q^-\- yi?a + ym^ + a^
Restoring the value of a,
274. To Square 2i Polynomial without Recourse to
Multiplication.
18. Eequired the square oi a -\- h -{• c.
Solution. — By actual multiplication, we have,
{a + h^-cf = a'^ + 2ab+2ac + ¥ + 2bc+c^.
Or, changing the order of terms,
a'^ + ¥ + c^ + 2<zb + 2ac+2bc.
Or, factoring, we have, a'^ + 2aib + c) + ¥ + 2bc + cK
19. Eequired the square ot a -{- b -\- c + d.
Solution. — By actual multiphcation, we have,
a^ + 1^ + d^ + d^ + 2db + 2ac + 2ad + 2bc + 2bd + 2cd.
Or, changing the order of the terms, and factoring, we have,
a? + 2a(b+c-\-d)->r ¥ 4- 2& (c + cZ) -4- c^ + 2cd + d^. Hence, the
KuLE. — 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 twice its product into
the sum of all the terms which folloio it.
20. Eequired the square oix -\- y -{- z.
21. Eequired the square of a — h -{- c.
22. Eequired the square oi a -\- x -\- y ■\- z.
273. How may a polynomial be raised to any required power ? 274. What is the
rule foi- squaring a polynomial ?^ _ -
a
OFT..E -^X
144 ADDITION OF POWERS.
275. When one of the terms of a binomial is 2^ fraction,
it may be involved by actual multiplication, or by reducing
the mixed quantity to an improper fraction, and then
involving the fraction. (Art. 171.)
22t. Kequired the square of x -{• ^\ and x — J.
« + i a; — i
a- + i a!- i
+ ja; + i -ia; + i
aj'+ X + \ a^— X + \
Or, reduce the mixed quantities to improper fractions. Thus,
a; + - = -= — ; and x = , (Art. 171.)
22 22^''
/2a!+i\2 4a;-+4a;+i . /2aj— iV
4a;2— 4a;+i
Expand the following mixed quantities :
24. (fl^ + J)^. 26. (— f + 2al)cf.
25
276. Powers are added and subtracted hke 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 7^2 + 5 (« 4- J)3 _ 6a; + 3^ + «'
Add — 30^3 + 4 (^ + ^f -f 4a; 4- 43^2 — fl<
Ans. 4(f + g (a -^ by — 2X + $0^ -{- 43^ -^ a^ — a^
29. From 3^3 + 5 J2 — 4(^8 4. ^^^a _ ^5
Take — 40^^ + 3^ + 3^^ — 5^ + a*
Ans. 7^8 + 22>2 _ 7^8 4- 5^:3 4. 4^2 _ ^^s _ ^4
275. How involve a binomial, when one term is a fraction ? 276. How are power*
added and subtracted ? Why ?
DIVISION OF POWERS. 145
MULTIPLICATION OF POWERS.
277. To Multiply Powers of the Same Moot.
1. What is the product of 2>tt'^b^ multiplied by aWi
Solution. — Adding the exponents of each letter, we have yi^
and 6^ Now 3«^ x &^ = 3a66% Ans. (Art. 94.)
2. Multiply ^a^¥ by a~^l)~\
Solution. — Adding the exponents of each letter, as before, we
have 3a'^62, Ans. Hence, the
EuLE. — Add the exponents of the given quantities, and the
result will be the product. (Art. 94.)
Notes. — i. This rule is applicable to positive and negative exponents.
2. Powers of different roots are multiplied hy writing them one
after another.
Multiply the following powers:
3. fl^^ by a^. 7. a~*b by a~^bK
4. x~^ by ar\ 8. a~*cd by a'^c^d^.
5. b-^ hj bK 9. ¥ c-^ y-^ hj b-^ (^y\
6. a"* by a\ 10. a^y^z^ by a~^y-^^.
DIVISION OF POWERS.
278. To Divide Powers of the Same Root,
II. Divide a^ by a\
Solution. — Subtracting one exponent from the other, we have
a^-T-a^ = a?, the quotient sought. (Art. 113.) Hence, the
Rule. — 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 exponents.
••"•t. How multiply powers of the same root ? Note. Of different roots 1 278. What
is tne rule for dividing powers of the same root ?
146 TRAKSFEREING FACTORS.
Divide the following powers :
12. a^ by ar\ i6. oc^yz^ by ar^y^z~\
13. x~^ by ic^. 17. i2aPh~^c by 3flf2Z'"^c^
14. ¥ by 52. 18. 6a;4^V ^y 22^-2 ?/;22.
15. c~^ by c~s. 19. 6oa%^(^ by 5«~2Z>%-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 vice versa, by changing the
5i^;i of its exponent from + to — , or — to +. (Art. 256.)
20. Transfer the denominator of —3 to the numerator.
q6 I
Solution. ^ = «^ ^ "T^ia ~ ^^ ^ ^~^ = a^a;-^, Ans.
21. Transfer the denominator of --^ to the numerator.
x~^
Solution. — -=— = 61^-^ — - = axaj^ = aofi, Ans.
■^
» ^5
22. Transfer the numerator of — to the denominator.
01 a^ I . I I I . I .
Solution. — =- x a5__^___i.^-5_ — ^ ^;^5^
2^ 2^ y o^ y or^y
2^. Transfer the numerator of — to the denominator.
y
Solution. — = — ^ x - = -^ , Ans.
y a^ y a^y
ax~^
24. Transfer ar'^ to the denominator of •
y
25. Transfer y^ to the numerator of j—^'
26. Transfer d~^ to the denominator of —q—
27. Transfer af* to the numerator of — -• -
279. What inference may be drawn from the method of denoting rec;pri>4.Al
powers ? How transfer a factor ?
OHAPTEE XIIL
EVOLUTION *
280. Evolution is finding a root of a quantity. It is
often called the Extraction of roots.
281. A Hoot is one of the equal factors of a quantity.
Notes. — i. Powers and roois are correlative terms. If one quantity
is a power of another, tlie 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 evolution a quantity is resolved into equal factors. 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. Boots, like powers, are divided into degrees ; as, the
square, or second root; the cube, or third root; the fourth
root, etc.
283. The Square Hoot 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 Hoot is one of the three equal factors
of a quantity.
Thus, 3 X 3 X 3 = 27, and axaxa = a^; therefore, 3 js the cube
root of 27, and a is the cube root of a^.
280, What is evolution? 281. A root? Note. Of what is evolution the reverse?
283. What is the square root ? 284. Cube root ?
* From t^ie Latin evolvere, to unfold.
148 EVOLUTION.
285. Eoots 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, ^/a and a* denote the square root of a.
y\/a and a^ denote the cube root of a, etc.
Notes.— I. The figure placed over the radical sign, is called the
Index of the Root, because it denotes the name of the root.
Thus, ^/a, and \^a, denote the square and cube root of a.
2. In expressing the square root, it is customary to use simply the
radical sign ^ , the 2 being understood.
Thus, the expression /Y/25 = 5, is read, ''the square root of 25 = 5."
3. The method of expressing roots hj fractional exponents is derived
irom. 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
power, and the denominator the root.
Thus, a^ denotes the cube root of the first power of a; and a* denotes
the fourth root of the third power of a, or the third power of the
fourth root, etc.
Read the following expressions :
1. «i 4. bk 7. dh 10. af.
2. «1 5. c^. 8. m"^. II. «»*.
3. a^, 6. x^. 9. n^. 12. x§,
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 JPerfect Power is one whose exact root can
be found. This root is called a rational quantity.
285. How are roots denoted? 286. What docs the numerator of a fractional
exponent denote ? The denominator? 287. What is a perfect power ?
* From the Latin radix, a root.
The si^n /v/" is a corruption of the letter r, the initial of ra4i^*
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 poicers of i, are i. For, a root is a
factor, wliich multiplied into itself produces a power ; but no number
except I multiplied into itself can produce i. (Art. 272.)
Thus, r, i^ I", and y^i, /y/i, /y/i, etc., are all equal.
290. Negative Exponents are used in expressing roots as
well as j902(;er5. (Arts. 255, 257.)
Thus, — T = a-i ; -T = a-i ; -r = a-n ^
a* a* ^t
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 x^, etc, Hence,
'i92. A fractional exponent maybe expressed in decimals.
'fhus, 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 a^ in decimals. 19. Write h'^ in decimals.
1 .
17. Write a^ in decimals. 20. Write x^ in decimals.
18. Write a^ in decimals. 21. Write y^ in decimals.
22. Express a^ in decimals. Ans. a^ = ^0.333333+ ^
23. Express x^ in decimals. Ans. x^ = 2^0.66666+ ^
24. Express y^ in decimals. Ans. y^ = y-^.
«5. Write a^ in decimals. Ans. a~s = a^'^.
Note. — In many cases, & fractional: exponent can only be expressed
approximately by decimals.
288. An imperfecl power ? 289. A rard ? 290. Are negative exponents used in
expressing roots 1 292. How are fractional exDonents sometimes expressed ?
150 EVOLUTION.
293. The Signs of Hoots are gc remed by the following
PRINCIPLES.
1°, An odd root of a quantity has the same sign as the
quantity.
■ 2°. A71 even root of a positive quantity is either positive
or negative, and has the double sigti, ±.
Thus, the square of +a\aa^, and the square of —a is a^ ; therefore
the square root of a^ may be either +a or —a\ that is, ^/a^ = ± 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
knqwn whether the power arises from a positive or a negative quantity.
2. It should also be observed that the two square roots of a positive
quantity are numerically equal, but have contrary signs.
294. An Even Hoot of a 7iegative quantity cannot be
found. It is therefore said to be impossible.
Thus, the square root of —a^ is neither +a nor —a. For, +ax +a
= +a^ ; and —a x —a = +a^. Hence,
295. An even root of a negative quantity is called an
Imaginary Quantity.
Thus, \/— 4, V'— a^ 'V^— a^ are imaginary quantities.
296. To Find the B^oot of a Monomial.
I. What is the square root of a^?
Analysis. — Since a'^ — axa, it follows that one opeeation.
of the equal factors of «? is a; therefore, a is its ^/a? = a
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 a^ by 2, we have a' or a, which
is the square root of a^.
293. What principles eovem the sign?, of roots ? When is the doable sisu used ?
Illustrate this. JVote. When is the ambiguity removed ? 294. What is an even root
of a negative quantity ? Illustrate. 295. What is it called ?
EVOLUTION. 151
2. What is the square root of 9^^^ ?
Analysis. — Since 9 = 3x3, the index of operation.
jj* = 2 X 2, and the index of &2 = i x 2, it ^/oaf^ := 2t^^b
follows ^shat the square root of 9 is 3, that of
a^ is a-.; and that of ¥ is 6^ or &. Therefore, ^^a?¥ = ^a^h. Hence, the
Rule. — Divide the index of each letter hy the index of
the required root; to the result prefix the root of the
coefficient with the proper sign. (Art. 293.)
Note. — This rule is based upon Principle 3. If a quantity is an
imperfect power, its root can only be indicated.
296, a. Tlie root of a Fraction is found hy extracting the
root of ^ach of its terms.
Find the required roots of the following quantities:
3. Va^. 10. 'v/36«^Z>2.
"s/a^ or a. 11. ^/^^y\
3 /-
^/ AfXy. 12. v64a%
^2>a%\ 13. ^i^xy.
V^^ahc, 14. A/49^y.
V^i6^. 15. ^/r^.
297. To Extract the Square Moot of ihe Square of a
Binomial.
I. Required tb€ square root of a^ + 2a5 4- l\
Analysis. — Arrange the terms operation.
according to the powers of the letter O^ ■\- 2ab + Z>2 ( fl^ -f d
a ; the square root of the first term ^2
is a, which is the first term of the \_ 'h\ h A- Jvi
root. Next, subtracting its square ' / ■, j^
from the given quantity, bring down 2^0 + (r
the remainder, 2ab + W.
296. How find the root of a monomial ? iVb/«.— Upon what principle is this rulf
based ? 296 a. How find the root of a fraction ?
152 EVOLUTION.
Divide the ist term of this rem. by 2a, double the root Oms found, the
quotient h is the other term of the root. Place & 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 + h^, there is no remainder. Therefore, a + & is the root required.
The square root of a:-—2db+l^ is found in the same manner, the
terms of the root being connected by the sign — . Hence, the
HuLE. — Find the square roots of the first and third ter^n^
and connect them hy the sign of the middle term,
2. What is the square root of a^ -{- 4X -\- 4?
3. What is the square root of «2 _ 2^5 4- i ?
4. What is the square root of i + 2X -\- a^?
5. What is the square root oi x^ -\- ^x -\- ^ ?
6. What is the square root of «^ — a + } ?
7. What is the square root of x^ -{- ix -] ?
4
298. To Extract the Square Root of a Polynomial.
8. Required the square root of 4«* — i2a^-\-s^^+6a-{-i,
OPBRATION.
40* — i2fl^3 + 5^2 _f- 6« 4- I ( 2a^ — $a — i
4a*
^2-
3«) -
I2fl3
+ 9«2
+ 6a+i
4^2
-6a-
-0
-4«2
-4«^
+ 6«+ 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
md bring down the remaining terms. Divide the remainder by
louble the root thus found ; the quotient —3a is the next term 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.
297. How extract the square root of the square of a binomial?
EVOLUTIOK. 153
Finally, multiply the divisor, 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"^—3«—i. Hence, the
KuLE. — I. A rra?ige the terms according to the jjowers of
some letter, beginning ivith 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.
IL Divide the first term of the ''remainder by double the
root already found, and place the quotient both i7i the root
and on the right of the divisor,
III. 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.
Proof. — 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. x^ + 2xy + y"^ -{- 2XZ + 2yz + z\
10.
a^ — 4ab -\- 2a -{- 4J2 _ 4J _}. i.
II.
a^ + ^a^ + 4^2 _ 4^2 _ 85 -f 4.
12.
I — 4^2 _|_ 4^4 4- 22; — 4^2^; + x\
13.
4a^ — i6a^ + 24^2 _ i6a + 4.
14.
a^-ab+ ibK
X^ 1/2
15-
2-1- — •
299. The fourth root of a quantity may be found by
extracting the square root twice j that is, by extracting the
square root of the square root.
Thus, y^iea* = 4a^, and /^4a^ = 2a. Therefore, 2a is the fourth
root of i6a\
Proof. 2ax2a = 4a^ ; 4a^ x 40^ = 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 quantity ? The eighth ?
CHAPTER 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 letters placed before radicals are coefficients.
2. In the following investigations, all quantities placed under tlie
radical sign, or having a fractional exponent, whether perfect or
imperfect 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 + Vf, are radicals of the same degree.
302. Lihe Madicals are those which express the
same root of the same quantity. Hence, like radicals are
like quantities. (Art. 43.)
Thus, s^y/a'^— & and 3/y/«^— &, etc., are like radicals.
REDUCTION OF RADICALS.
303. Reduction of Madicals is changing their
form without altering their value.
304. The Simjylcst 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
poivers of the same name a« the root must be removed from
under the radical sign.
300. What is a radical ? 301. How is tlie decree of a radical denoted? 302. What
are like radicals ? 303. Define reduction of radicals. 304. What is the simplest form
}t radicals ?
EEDUCTIOK OF RADICALS. 155
CASE I.
305. To Reduce a Radical to its Simplest Form.
I. Reduce ^/\Wx to its simplest form.
Analysis. — By inspection, we ofebation.
perceive that tlie given radical is ^ \Wx = ^ ^0? X 2X
composed of two factors, <^a? and ^^ / — g / —
IX, the first being a perfect square "
and the second a surd. (Art. 289.) .'. ^/ l^a^X = ^aV2X
Removing ga^ from under the rad-
ical sign and extracting its square root, we have 3a, which prefixed
to the other factor gives s^y'z^, the simplest form required.
2. Reduce 4^/a^ — a^x to its simplest form.
Analysis. — Factoring operation.
the radical part, wejiave 4^^.^^ = 4^a^x{a-x)
the two factors, \/a^ and 3,-— »,
^a—x, the first being a
perfect cube, and the sec- .'. ^^ a^—a^X = /^a\^a—X
ond a surd. Remove a^
from under the radical sign, and its root is a, which multiplied by
the coeflBcient 4, and prefixed to the radical part, gives ^a^a—x, the
simplest form. Hence, the
Rule. — I. Resolve ilie radical into two factors, one of
which is the greatest power of the same name as the root.
II. Extract the root of this power, and multiplying it ly
the coefficient, prefix the result to the other factor, with the
radical sign letween 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 small, the greatest exact power they
contain may be readily found by inspection.
3. Reduce 3V5o«^^^ to its simplest form. Ans. isav^x.
4. Reduce 6\/s4^l/ to its simplest form. Ans. iSxV 2y.
305. Becite the rule. Ifote. Upon what based ?
156 EEDUCTIOK OF EADICALS.
306. To Find in large Radicals the Greatest Power
corresponding to the indicated Root.
5. Eeduce ViSya to its simplest form.
OPERATION.
4 ) 1872 ^1872 = ^4x4x9 X V13
4 ) 468 " = A/144 X V13
9)117(13 V1872 = 12V135 ^'ns.
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. Th% smallest power of the same
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 144, we have i2/y/i3, the simplest form
required. Hence, the
'Rv'LE,— Divide the radical by the smallest power of the
same degree ivhich 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 will
be the greatest poiver required.
Note. — This rule is founded on the principle that the 'product of
any two or more square numbers is a square, the product of any two
or more cuhic numbers is a cuhe, etc.
Thus, 2' X 32 z= 36 = 62 ; and 28x38 = 216 = e^.
Eeduce the following radicals to their simplest form :
\^54a^c.
yVga^ — 2'ja^b,
'\/64a^y.
</8i^.
V46Sa^.
306. How find the greatest power corresponding to the indicated root, in large
radicals ? Jffote. On what principle is this rule founded ?
6.
Va^.
12.
7-
A/8a2^.
13-
8.
2 Vgxy.
14.
9-
3^^.
15.
10.
SV^iSS-
16.
II.
6\/252«2J.
17.
EEDUCTION OF RADICALS. 157
CASE II.
307. To Reduce a Rational Quantity to the Form of a
Radical.
I. Keduce ^a^ to the form of the cube root.
Analysis.— The cube root of a quantity, we operation.
have seen, is one of its three equal factors. (3^^)^ = 2'ja^
(Art. 284.) Now 3a'' raised to the third power . ^ „2 \/^2T^cfi
is 27a*. Therefore 3^^ = ^i']a^. Hence, the
Rule. — Raise the quantity to tlie power denoted hy the
given root, and to the result prefix the corresponding radical
sign.
Note. — The coeflBcient of a radical, or any factor of it, may be
placed under the sign, by raising it to the corresponding power, and
placing it as a factor under the radical sign.
2. Reduce 2a^ to the form of the cube root.
3. Reduce (2a + d) to the form of the square root.
4. Reduce {a — 2b) to the form of the square root.
5. Place the coefficient of saVb under the radical sign.
6. Place the coefficient of 2a^\^ab under the radical sign.
7. Reduce 2x^y^z^ to the form of the fourth root.
8. Reduce ^abc to the form of the cube root.
9. Reduce ^{a — b) to the form of the cube root.
10. Reduce a^ 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. *
I I. Reduce ah^ to the form of the fourth root.
12. Reduce a — b to the form of the square root.
13. Reduce a'^ to the form of the nth root.
307. How reduce a rational quantity to the form of a radical? Note. How place
a coefficient under the radical sign ? Note. How raise a power to the form of *
required root ?
158 REDUCTIOK OF EADICALS,
CASE III.
308. To Reduce Radicals of different Degrees to others of
equal Value, having a Common Index.
1. Eeduce a^ and b^ to equivalent radicals of a common
index.
Analysis. — The fractional opebation.
indices i and i, reduced to a J = -^ and i = -rj
common denominator become ^.^ a^ = a^ and b^ = ^t\
j\ and f But af' = {a^)^, 4 , ,. i , , 3 ,„, 1
and 5^^ = (63p. (Art. 174.) ^" "= ^^ ^^ ^^^ ^" ^ ^^^^
Therefore (a^)^'^ and (ft^)" are the radicals required. Hence, the
EuLE. — I. Eeduce the indices to a common denominator.
II. Raise each quantity to the power expressed hy the
numerator of the new index, and indicate the root expressed
ly the common denominator. (Art. 174.)
Eeduce the following radicals to a common index :
2. a^ and (hc)^. 7. V4S and \^2aK
3. 3^ and 5^. 8. a^ and 5«.
4. a^ and 6^. 9. l^ and c».
5' Vs^, V^3, and v^. 10. (« + h)^ and (a- <^)i
6. '\/2x^ and Vs^^. 11. (^ — ^)^ and (a? 4 «/)i
CASE IV.
309. To Reduce a Quantity to any Hequired Index^
I. Eeduce a^ to the index ^.
Analysis. — Divide the index | byl; operation.
wo have | or ^. Place this index over a; l""~^"f^=i^XT^^I
it becomes <z', and setting the required i" "^ 1 ^^^ F ^^ 7
index over this, the result, (a*)^, is the /, (^v^j -<4ws.
answer. Hence, the
30^. 5ow reduce radicals to a coininon iude? ?
ADDITION" OF EADICALS, 159
Rule. — Divide the index of the given quantity hy the
required index, and 'placing the quotie7it over the quantity,
set the required index over the whole.
Note. — This operation is the same as resolving the original index
into two factors, one of which is the required index. (Art. 126, note.)
2. Eeduce a^ and ifi to the index J.
Solution. J-^i = |xf = |, tlie first index.
|-i-i = f X f = f , the second index.
Therefore, (a^)^ and (&")* are the quantities required.
3. Reduce 3^ and 43 to the common index ^,
4. Reduce a^ and i^ to the common index \,
5. Reduce a^ and h^ to the common index J.
6. Reduce a^ and ifi to the common index -f.
7. Reduce w^ and 5^ to the common index J-.
ADDITION OF RADICALS.
310. To Find the Sutn of two or more Radicals.
1. Wliat is the sum of 3 V» and 5 Va ?
Analysis. — Since these radicals are opekation.
of the same degree and have the same ^^/a -f- S^^ = SV^
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 4^/18 ?
Analysis.— These radicals operation. ^
are of the same degree, but 3^8 = 3V4 X ^2 ■=. 6^/2
the radical parts are unlike ; /—x r~ r~ /~
^, . XI . -, 4Vi8 = 4VQ X V2 = I2V2
therefore, they cannot be ~t ^ y » "^
united in their present form. /. 6y2 -j- 12 V 2 = 18 y 2
Reducing them to their sim-
plest form, we have 3V^8 = 6-^/2, and 4y^i8 = 12-^/2, which are
like radicals. (Arts. 302, 305.) Now 6y'2 and 12-^/2 = 18-^/2, Ans.
309. How reduce quantities to any required index? Note. To what is this oper-
ation simiJar ?
160 SUBTRACTION OF RADICALS.
3. What is the sum of 3A/1S and 4'V^24.
Analysis. — Reducing the operation.
radicals to tlieir simplest form, 3\/l8 = '^'V/o X a/z = qV^
we have 3 V^ = 9^2, and ^^- ^ ^^g x 'v/3 = S^v^S
4/^24 = 8/^3, whicli are un- ,— 3 ,-
like quantities, and can only ^^^- 9V 2 + Ws
be added by writing tliem one after the other, with their proper signs.
(Arts. 43, 67.) Hence, the
Rule. — I. Reduce the radicals to their simplest form.
II. If the radical parts are aliJcCf add the coefficieiits, and
to the sum annex the common radical.
If the radicals are unliTce, write them one after another,
with their proper signs.
Note. — To determine whether radicals are alike, it is generally
necessary to reduce them to their simplest form. (Art. 305.)
Find the sura of the following radicals :
4. '\/i2 and \/27. 9. 3V^ and 4V^i28.
5. V20 and V48. 10. 7^/243 and 5 A/363.
6. 2\^¥ and ^\^a^. 11. «V8i^ and 3«'v/49^.
7. a^/2,a% and cA/276/A 12. dV^S^.smd Vs^^c,
8. 3Vi8«a;2 and 2^/32^^. j^, 41^/^ and 5^/^.
SUBTRACTION OF RADICALS.
311. To Find the Difference between two Radicals.
I. From 3V45 subtract 2 A/20.
Analysis. — Reducing to the operation.
simplest form, we have gy^s 3'\/45 = 3^9 X V^ = gVs
and 4 a/ 5. which are like /— r r r
.^ ^^ r r 2V20Z11 2V4 X V5 =4V5
quantities. Now 9 y 5 — 4^/5 ■
= 5a/5, the difference re- .•.3^45-2^20 = 5^/5
quired. Hence, the
310. How ad4 radioals ? iVpfe. How determine whether tUey are like quantities I
MULTIPLICATION OF RADICALS. 161
EuLE. — Reduce the radicals to their simplest form ; change
the sign of the subtrahend, and proceed as in addition of
radicals, (Art. 310.)
(2.) (30_ (4.)_
From 4 a/772 V480 4 A/320
Take V44^ 4 A/63 — 5-A/80
5. From 3a/49«S take 2A/25fit^.
6. From ^^/a + ^ take 3V^« + 5.
7. From 3a/^ take — 4a/^.
8. From 3V^25o^% take 2'V^54J%.
9. From — a~^ take — 2^"^.
10. From 5a/J take 2a/J.
MULTIPLICATION OF RADICALS.
312. To Multiply Radical Quantities.
1. What is the product of 3a/« by 2 a/^.
Analysis. — Since these radicals are operation.
of the same degree, we multiply the 3a/^ X 2\^b ■= 6\^ab
radical parts together, like rational
quantities, and to the result prefix the product of the coefficients.
2. Multiply 3a/^ by 2'\/c.
AiiTALYSTS. — As these radicals are of differ- opbbation.
ent degrees, they cannot he multiplied together 3 ^/a =z 3 (ai)
in their present form. We therefore reduce 3/— / 2\
2 A/ C '~~ 2 ( C^ /
them to a common index, and then, multiply- ^ ^
ing as before, we have 6^a^c\ ^^«- ^ {a^cP)^
3. Multiply «3 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. ^ + 1 = f . Therefore, a^ x a^ = a^. Hence, the
311. How subtract radicals f
162 MULTIPLICATION OF RADICALS.
Rule. — I. 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 numer-
ator of the index can be divided by its denominator without a
remainder.
4. If rational quantities are connected with radicals by the signs +
or — , each term in the multiplicand must be multiplied by each term
in the multiplier. (Art. 98.)
Multiply the folkwing radicals:
4. 5a/i8 by 3V20. 10. OP- by q>,
5. a\/x by h^fx, 11. 7V^4 by 3^4.
6. ^/a + i by V« — ^. 12. V9« by Vi6a.
7. ^fax by ^fcy, 13. \/i8 by ^/ 2.
8. c^ by (^, 14. "^Zax by ^fwx,
(9-) (X5-)
Multiply a -f- V? Mult. « + \/x
By g-f- V5 By i + l\/x
ac + cVi a + Va;
+ aVd + v^ -t- ahVx -f Ja:
J/i5. «c + cA/^-f «A/^-f V^ Ans. a-{-V^-\-ahVx-^bx
16. 2\/f by 2\/J. 18. (m-]-n)^hj{m-\'n)^
17. 4V|by3V|. ^^^ J^^^J~^_.
313. How multiply radicals? iVbfe*. How arc roots of like quantities multi-
plied? Upon what principle is this rule based? When does the product of
radicalB become r'vtional? If radicals are connected with rational quantities, how
multiply them?
DIVISION" OF RADICALS. 163
DIVISION OF RADICALS.
313. To Divide Radical Quantities.
1. Diyide 4^/^400 by 2\/Sa.
Analysis. — Since the given radicals are operation.
of the same degree, one may be divided by 4\/T4ac / —
the other, like rational quantities, the quo- /^— ^^ 2V3C
2 V o<^
tient being Y 3c. (Art. iii.) To this result
prefixing the quotient of one coeflBcient divided by the other, we have
2 -v/sc, the quotient required.
2. Divide 4 Vac by 2 Vet.
Analysis. — Since these radicals are aVoc a (ac)^
of different degrees, they cannot be — rp- = ^
divided in their present form. We 2\/a 2 [a)^
therefore reduce them to a common 4\/ttC 4 (aW^
index, then divide one by the other, and •*• s/~~ ^^ , „. 1
2 A/ (l 2 \ (I I ^
to the result prefix the quotient of the ^ ^ '
coefficients. Tlie answer is 2^0^. or 2 (a(?)^, Ans.
. 3. Divide a^ by a^
Analysis. — These radicals are of different opbeation.
degrees, but have the same radical part or root. ^h — ^1
We therefore divide them by subtracting the 1. 2
fractional exponent of the divisor from that of
the dividend. (Art. 113.) Reducing the expo- a^ — a^ z=: a^
nents to a common denominator, a^ = a^, and . ^^ . /^^ — ■ rA
as— a«. Now a^-i-a^=a^, Ans. Hence, the
Rule.— I. Reduce the radical parts to a common index.
11. Divide one radical part hy the other, and placing the
quotient under the common index, prefix to the result the
quotient of their coefficients.
Note. — Boots of like quantities arfe divided Tyy subtracting thefrat-
tional exponent of the divisor from that of the dividend. (Art. T13.)
313. How divide radicals ? 2^ote. How divide roots of like quantities ?
164 IKVOLUTIOK OF KADICALS.
Divide the following radicals :
4. ^/\2a^c by ^/ ^c. 10. 14a V^ by 7a/^.
9
d^Mx^ by 2^/dx. 11. (« + 5)t by (a + Z>)«.
(a^ -f fl^ic)^ by a^, 12. 3^500;^ by v^.
12 (ay)T by («y)i 13. ^/x^ — f- by ^/x-\-y,
2^1^/ ax by S-v/flf. 14. 16^/32 by 2^/4.
iS^cs/^rc by 2cV^. 15. 8^512 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 i 2
Ct^ X d^ "^^^ (X^
equal factors, we multiply the given index by
the index of the required power. Hence, the .*. 0^, Ans.
EuLE. — Multiply the index of tlie root 'by the index of tlie
required power, and to the result prefix the required poivei
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 ^/a + J. Ans. a ■\-b.
3. Find the cube of a^,
4. What is the square of 3a/2^.
5. What is the cube of 2\^.
X ■
6. Eequired the cube of -V2X,
7. Required the cube of 4\ / — •
V 4
8. Find the fourth power of 3A / —
9. What is the square of « + \/y ?
314. How Involve radicals to any required power ? Note. How raise a root to a
yower of the same name?
BVOLUTIOK OF RADICALS. 165
EVOLUTION OF RADICALS.
315. To Extract the Boot of a RadicaL
I. Find the cube root of a^'i/W.
Analysis. — Finding the root of a radi- operatiok.
cal is the same in principle as finding the V^a^V^^ = v ^3^1
root of a rational quantity. (Art. 296.) 3
Reducing the index of the radical to an V^^^t =: ab^, Ans.
equivalent fractional exponent, we extract
the cube root by dividing it by 3. The result is a&^. Hence, the
Rule. — 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 the fractional exponent by that number.
Thus, /^a = aK Multiplying the former by 2, and dividing the
latter by 2, we have ^a = a".
2. If the coefficient is not a perfect power, it should be placed under
the radical sign and be reduced to its simplest form. (Art. 305.)
2. Required the square root of gV^,
3. Required the square root of 4\^^.
4. Find the cube root of 3a/^.
5. Find the cube root of 2bV2b.
6. What is the cube root of a (bc)^ ?
7. What is the fourth root of ^^f ?
8. What is the fourth root of Va^ V^^ ?
9. Find the seventh root of i28a/«.
10. Find the fourth root of Vayby,
II. Find the fifth root of 4a^V2a.
12. Find the nth root of aVbc.
315. How extract the root of a radical ? Notes. To what is multiplyinff the index
Of a radical equivalent ? If the coefficient is not a perfect power, what is dope ?
166 CHAI^GIKG RADICALS TO llATIOKALS.
CHANGING A
RADICAL TO
QUANTITY.
A RATIONAL
CASE I.
316. To Change a Radical Monomial to a Rational Quantity
1. Change V^ ^^ ^ rational quantity.
Analysis. — Since multiplying a root of a
quantity into itself produces the quantity, it
follows that /y/a x ^ a = a, which is a rational
quantity. (Art. 287.)
2. It is required to rationalize c^.
OPERATION.
OPERATION.
Analysis. — A root is multiplied by another
root of the same quantity by adding the expo- ^i y^ ^1 -^^ ^
nents ; therefore we add to the index \ such a
fraction as will make it equal to i. (Art. 94.)
Thus, as X as = as+ s = fl^i = «^ the rational quantity required.
3. It is required to rationalize ^.
Solution.— Multiplying a^ by ic«, the result is
a?, which is a rational quantity. Hence, the
OPERATION.
Rule. — Multiply the radical hy the same quantity having
such a fractional exponent as, when added to the given
expo7ient, the sum shall be equal to a unit, or i.
4. Required a factor which will rationalize a^.
5. What factor will rational
6. What factor will rational
7. What factor will rational
8. What factor will rational
9. What factor will rational
xo. What factor will rational
ze Va^c?
ze '^(a + by.
ze Va^?
ze ^^+jY?
ze \/(a+ by?
ze V{a + b + c)?
316. How reduce ft rftdicftl monomifll to » r^Uonal quantity?
CHANGING RADICALS TO RATIONALS. 167
CASE II.
317. To Change a Radical Binomial to a Rational Quantity.
1. Ifc is required to rationalize ^s/a + ^/}).
Analysis, — The product of the sum and opbratiok.
difference of two quantities is equal to the /y/^ _i- ^J})
difference of their squares (Art. 103); there- r- ,j-
fore (/;/«+ \/&) multiplied by {'s/a—^h)
= a—b, which is a rational quantity. a + ^/ob
Therefore, the factor to employ as a multi- — yai — b
plier is V«- V^- a — b, AnsT
2. What factor will rationalize Vx — Vy ?
Analysis.— If the binomial ^x — y'y is operation.^
multiplied by the same terms with the sign of Vcc — Wy
the latter changed to + , we have w'~ . ^T.
(\/x - Y^y) X {^^x+ ^y) = x-y. ~~~ ~
_ _ JO *""• y
(Art. 103.) Therefore, y\/x + ^/y is the fac- ,- .-
tor required. Hence, the ^^^- V^+Vy
Rule. — Multiply the binomial radical by the correspond-
ing binomial with its connecting sign changed,
3. What factor will rationalize x -\- 4 V9 ?
4. Rationalize A/9 — V^.
5. What factor will rationalize VT + Vet ?
6. Rationalize 6 — V^.
7. What factor will rationalize Vsa — V^b ?
8. Rationalize Vci' — Vs.
9. What factor will rationalize 3 's/a + a/S ?
10. What factor will rationalize 4 V^a — 5 V^?
317. How reduce a radical binomial to a rational quantity ?
168 RADICAL FRACTIONS.
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 operation.
fraction by the denominator A^h, and the a y^ ^^ aVb
result is —x— , whose denominator is V ^ X yb ^
rational. (Art. 167, Prin. 3, note.) Hence, the
Rule. — Multiply both terms of the fraction ly such a
factor as will malce the required term ratio7ial.
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
(y'a + -y/ft), we have a — & for the denominator. (Art. 103.)
2. Rationalize the denominator of -~z» Ans. - — ^.
Va ^ ^ a
3. Rationalize the numerator of ---• Ans,
^/x Vox
4. Rationalize the denominator of -■
x
5. Rationalize the denominator of — -•
6. Rationalize the denominator of —^ —•
^x — Vy
Of
7. Rationalize the denominator of —j= —*
I
I + V3
3 — V3
8. Rationalize the denominator of
9. Rationalize the denominator of
318. How reduce a radical fraction to one whose numerator or denominator is a
rational quantity? \^'^len the fractions contain compound quantities, what prin-
ciple enters into their reduction?
BADICAL EQUATIOI^S. 169
RADICAL EQUATIONS.
319. A Madical Bquation is one in whiCh the
unknown quantity is under the radical sign.
320. To Solve a Radical Equation.
I. Given a/^ 4-2 = 7, to find x.
Analysis. — Transposing 2, we have, y5 + 2 = 7
/\/x = 5. Since 5 is equal to the >\/x, it /—
follows that the square of 5, or 25, must be V — 7 — — 5
the square of y\/x. Therefore, x = 25.
a; = 52 = 25
2. Given 2a -{- \/x = ()a, to find x.
Solution. — ^By the problem, la + ^Jx — t^n
By transposing, ^x = ya
By involution, x = 49a'
3. Given 5V^a; + i = 35, to find x.
Solution.— By the problem, 5/y/a; + i = 35
Removing coefficient, ^^x + i = 7
Involving, a? + i = 343
Transposing, X = 342. Hence, the
Rule. — Involve both sides to a power of the same name as
the root denoted by the radical sign.
Note. — Before invohing the quantities, it is generally best to clear
of fractions, and transpose the terms, so that the quantities under tJ.a
radical sign shall stand alone on one side of the equation.
Eeduce the following radical equations:
4. « H- v^ -\- c = d.
8. v^22; + 3 — 6 = 13.
5. ^/X+ 2z=S.
9. v"^ — 4 = 3-
6. sVa; — 4 4-S = 7i•
10. 2^a; — 5=4.
7• 3a/^=24.
II. 5A^ = 30.
319. ^ hat is a radical equation ? 320. flow solved ? Note. What should be done
before involving the quantities ?
8
170 RADICAL EQUATIONS,
12. Given ^ "^^ — -i— = J2, to find iR
13. Reduce a/cl^ +Vx=z ^/"^^^ .
Analysis. — By removing tlie opkkation.
denominator the first member is /~Z 7=^ 3 + ^
squared. But x is still under y -r V ^ . / v
the radical sign. This is re-
moved by involving both mem- ^ + v ^ = 3 + ^
bers again. Atis. X = ($ -\- C — fl')
14. Given ^^^ = :^, to find ST.
2^2
15. Given x + a/«^ 4- ic^ = _______ to find 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 operation, the other will also disappear.
16. Given V4 + 5.'^^ — ^z^ =-- 2, to find x.
Transposing, 'Y/4 + S-c = ^/3x + 2
Involving, 4 + 5« = 4 + 4\/3'« + 3«
/ — ^
Transiiosing and dividing by 4, ysx — -
Involving, ^x = —
4
Transposing and multiplying by 4, x'^ = I2j;
Hence, x = 12, Ana.
17. Given a/^ -f 12 r=: 2 + v^, to find x.
18. Given Vs x Vx 4-2 = 2 + Vs^, to find Xo
^. Vx X — ax , „ ,
10. Given = — — , to find x,
(See Appendix, p. 289.)
CHAPTER XV.
QUADRATIC EQUATIONS.
321. Equations are divided into different degrees, as the
fii-st, second, third, etc., according to the powers of the
unknown quantity contained in them.
An equation of the First Degree is called a Simple
Equation^ and contains only the /r*^ power of the unknown
quantity.
An equation of the Second Degree is called a Quad-
vatic Equation, and the highest power of the unknown
quantity it contains is a square.
An equation of the Third Degree is called a Cuhic
EquoMon, and the highest power of the unknown quantity
it contains is a cul:)e.
An equation of the Fourth Degree is called a
Biquadratic, etc.
322. Quadratic Equations ai-e divided into pure
and affected.
323. A Pure Quadratic contains the square only of
the unknown quantity ; as, x^ = 5.
324. An Affected Quadratic contains both the first
and second powers of t h e unkn own quantity ; as, x^-\-ax= cd.
Notes. — i. Pure quadratics are sometimes called incomplete equa-
tions ; and affected quadratics, complete equations.
321. How are equations divided ? What is an equation of the first de^^ree ? The
second? Third? Fourth? 322. How are quadratic equations divided ? 325. What
I? h pure quadratic ? 324. An afl;"ected quadratic ? Note. What are they sometimes
called ?
172 PURE QUADRATICS.
2. A Complete Equation contains every integral power of tlie un-
known quantity from that which denotes its degree down to the zero
power.
An Incomplete Equation is one which lacks one or more of these
powers.
PURE QUADRATICS.
325. Every pure quadratic may be reduced to the form
a;3 = «.
For, by transposition, etc., all the terms containing x^ can be
reduced to one term, as lyx^ ; and all the known quantities to one
term, as c. Then will
W = c.
Dividing both members by &, and substituting a for the quotient of
e-r-b, the result is the form,
x^ = a.
326. Pure quadratic equations have two roots, which are
the same numerically, but have opposite signs. (Art. 293, 71.)
Thus, the square of +a and of —a is equally a^. Hence,
V^= ±a,
327. To Solve a Pure Quadratic Equation.
I, Find the value of a; in 6 = h 2,
9 3
Solution.— Given 6 = — + 2
9 3
Clearing of fractions, 5a^ — 54 = 3^^ + l8
Transposing, etc., 2x'^ = 72
Removing coefficient, a^ = 36
Extracting sq. root, x = ±6, Ans,
Substituting b for 2, and c for 72, in the third equation, we have
the form, bx^ = c.
Removing coefficient, etc., x^ = a. Hence, the
Rule. — Reduce tlie given equation to the form x^ = a, and
extract the square root of holh members. (Art. 296.)
336. Hqw manjT roots nas a pure ooa^raticf
PURE QUADRATICS. 173
Find the value of x in the following equations :
2. 3^5^ — 5 = 70. 10. 20^ + 12 ^ 3a;2_ 27.
3. 92?^ + 8 = 3a;2 -f- 62. 11. 72^2 — 7 = 32^2 + 9.
4. 5a;2 -I- 9 = 2a;2 ^ 27. 12. ahy^ = a*.
5. 6a« + 5 = 40?^ + 55. 13. {x + 2)2 = 4a; + 5.
6. ^ + 35 = 3^+7. 14. ii^-i=^^^. -
4 4
£^±8_^-6 ^(2^+9) __ 3^-1-6
8. - = ". 16. —5 1 f- = -.
42a; 4 — a;4H-a;3
:r . 2 a; . ^ ax^ia — 2)
9. - + - = - + -• 17. — ~T = I — ic.
328. Radical equations, when cleared of radicals, often
become pure quadratics.
18. Given Vx^ + n = \/2x^ — 5, to find a:.
Solution.— Clearing of radicals, aj2 + ii = 2aJ' — 5
Transposing and extracting root, jc = ± 4
4a;
19. Given 2V^ — 5 = — , to find ar.
20. Given 2V^--4 = 4V^---i, to find a?.
21. Given ^/^"-M = , to find a?.
V a; — ^
22. Given \/- — ^^^ = Vx, to find a;
23. Given = 'v/iTT^, to find x.
24. Given ■ = \/x — lo, to find x,
vx + 10
327, What is the rule for the solution of pure quadratics? 328. What may
radical equations become ?
174 PUEB QUADKATIOS.
PROBLEMS
1. The product of one-third of a number multiplied by
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. Kequired 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 tlie
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.
Required 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.
Required 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 ?
AFFECIEB QUADBAIICS 175
AFFECTED QUADRATICS*
329. An Affected Quadratic Equation is one
which contains the^rs^ and second powers of the unknown
quantity ; as, aoi? -^ Ix z=i c,
330. Every affected quadratic may be reduced to the form,
in which a, b, and x may denote any quantity, either
positive or negative, integral ov fractional.
For, by transposition, etc., all the terms containing x^ can be
reduced to one term, as cx"^ ; also, those containing x can be reduced
to one term, as dx ; and all containing the known quantities can be
reduced to one term, as g. Then, cx^ + dx = g.
Dividing both members by c, and substituting a for the quotient of
d-r-c, and b for the quotient of g -i- c, we have,
a^ + ax = b.
Take any numerical quadratic, as-^ S = x^ + — — 4,
Clearing of fractions, Saj^ — 4a; — 48 = 6aJ* + 4a; — 24
Transposing, etc., 2af^ — Sx = 24
Removing the coefficient, a^ — 4X = 12
Substituting a for 4, and 6 for 12 in the last equation, we have,
x^ — ax = b. Hence,
All affected quadratics mag be reduced to the general form,
%^ ±^ax = b,
331. The First Member of the general form of an
affected quadratic equation, it will be seen, is a Bi?iomial,
but not a Complete Square. One term is tvanting to make
the square complete. (Art. 266, note.) The equation,
therefore, cannot be solved in its present state.
329. What is an affected quadratic equation ? 330. To what general form may
every affected quadratic be reduced? 331. What is true of the first member of the
general form of an affected quadratic ?
* Quadratic, ivGm the Latin qtiadrare^ to make square.
Affected, made up of different powers ; from the Latin ad and/acw>,
to make or join ta
176 AFFECTED Q tJ ADHATICS.
332. There are three methods of completing the square
and solving the equation.
FIRST METHOD.
I, Given a:' + 2aa; = J, to find the value of x.
Analysis. — The first opeeation.
and third terms of the 01^ -{- 2ax =. b
Bquare of a binomial are a? -j- 2ax •\- d^ =. a^ -{■ h
complete powers, and the x 4- a = -h a/^2 i a
second term is twice the
product of their roots;
or the product of one of the roots into twice the other. (Art. loi.)
In the expression, x'^ + 2ax, the first term is a perfect square, and
the second term 2ax consists of the factors 2a and x. But x is the
root of the first term x'^ ; therefore, the other factor 2a must be tmce
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, x- + 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 square root of both members, and transposing a, we
have x = ~a± ^/a'^ + b, the value sought. (Art. 297.)
2. What is the value of a; in 20^ + x = 64 — yx?
Analysis. — 'transposing — 70? operation.
and removing the coefficient of a^, 2X + x =z 64. — 'jx
we have the form x^ + 4X = 32. But 23^ -}- 8x = 64
the first member, x'^ + 4X, is an incom- a^ -{- 4X =z 22
plete square of a binomial. x^ 4- ax 4- A ^= '16
In order to complete the square,
we add to it the square of half the
coefficient of x. (Art. 266.) Now,
having added 4 to one member of *• €., X = 4 OT — 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 square ?
a?+ 2 = ± 6
X= — 2 ±6
AFFECTED QUADRATICS. 177
Notes.— I. Adding the square of half the coefficient of the second
term to both membei-s of the equation is called completing the square.
2. The first member of the fourth equation is the square of a bino-
mial ; therefore, its root is found by taking the roots of the first and
third terms, which are perfect powers. (Art. 297.) From the process
of squaring a binomial, it is obvious that the middle term (4a;) forms
no part of the root. (Art. 266.)
333. From these illustrations we derive the following
EuLE. — I. Reduce the equation to the form, x^ ±ax = b.
II. Add to each member the square of half the coefficient ofx
III. Extract the square root of each, and reduce the result
ing equation.
3. Find the value of a; in — 2:?? -f 2>ax = — 65.
Solution. — By the problem, —2a;' + ^ax = —6b
Removing coefficient of x^, —x'^ + ^ax = — 3&
Making x^ positive (Art, 140, Prin. 3), x^—4ax = 3&
Completing square, a^—^ax + ^a^ — 4a' + 3&
Extracting the root, x—2a = ± Y4a'^ + 3ft
.*. x = 2a± ^\a? + 38
4. Given ^ ■\- ax -\- bx zzz d, to find x,
OPEBATION.
ix? ■{■ ax -\- bx =z d
a^+ (a -\-b)x = d
Analysis. — Factoring the terms which contain the first power of x,
we have ax + hx = {a + b)x ; hence, (a + h) maybe 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 x is found.
333. What is the rule for the first method of solving affecj^d quadratics »
178 AFFECTED QUADRATICS,
5. Given 32; — 2ic2 _ _ ^^ ^0 ^^^ ^.^
Solution. — By the problem, 3X — 2X^ = — g
Making x^ positive, etc.,
x^-
3X_
2
.9
2
Completing square.
X'-
f-
9
16
2 16
81
~ 16
Extracting root,
X-
_3 ^
4
±9
4
/. X =
f±f.
i. e.
, X =
: 3 or —
li
6. Given $x^ — 24X = — 36, to find x.
Ans. + 6 or +2.
Note, — The two Toots of an affected quadratic may have the same
or different signs. Thus, in the 6th and 12th examples they are the
same; in the ist, 2d, 3d, 4th, and 5th, they are different.
7. Given ^x^ — 40a; = 45. to find x.
8. Given x^ — 6ax = d, to find x.
9. Given 2X^ + 2ax = 2 (J -f c), to find x.
Solution. — Completing the square, x"^ + ax -^ — = b b + c,
4 4
Extracting root, x+- = ±-i/— + b + c
2 'A
Transposing, 3?= ± y — \- b + c
10. Given 2X^ — 22a; = 120, to find x.
11. Given x^ — 140 z= 13a;, to find x.
12. Find the value of a; in' x^ — ^x ■{• 1 = k^x — 15.
Solution. — By the problem, a;^— 3a; + i = 5a;— 15
Transposing, aj*— 8a; = — 16
Completing square, a;'— 8a5+i6 = o
Extracting root, «— 4 = o
.*. a; = 4
Note. — In this equation, both the signs and the numerical values of
the two roots are alike. Such equations are said to have equal roots.
i\ro<«,— What signs have the roots of an affected quadratic f
AFFECTED QUADRATICS. 179
SECOND METHOD.
334. When an affected quadratic equation has been
reduced to the general form,
x^ -\- ax = h,
its root may be obtained without recourse to completing the
square.
I. Given x^ -{-Zx=. 65, to find x.
Analysis. — After the square operation.
of an affected quadratic is com- /y3 a_ Q/y (\c
pleted and the root extracted,
the root of the third term is
a; = — 4 ± a/65T^6
transposed to the second mem- •*• 2; = 4 i 9
her, by changing its sign. (Art. i,e.y X = ^ or — 1 3
204.)
Now, if we prefix half the coeflScient 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 saiTie combinations of the same terms, as when the
square is completed in the ordinary way. Hence, the
Rule. — Prefix half the coefficient of x, with the opposite
sign, to plus or miuus the square root of the second member,
increased hy the square of half the coefficietit of x.
Solve the following equations :
2.
3^^ — 9^ — 3 = 207-
8.
a« -f- 4ax — b.
3-
42^2+ 120; + 5=45.
9-
32)2 — 74 = 6a; + 31.
4.
ix^— 14a: -f- 15=0.
10.
a^ -\- i^ = 6x.
5.
4a;2 _ pa; __ 28.
II.
{X — 2) {X— l)= 20c
6.
2X X + 2
12.
x+ 1 X _is
X ^ x-\-\ 6 '
7.
a^ + ^--ah = d,
I?
13.
^^^-j^ch = bd.
c
334. What is the second method of solving affected quadratics ¥
180 AFFECTED QUADRATICS.
THIRD METHOD.
335. A third method of reducing an affected quadratic
equation may be illustrated in the following manner:
1. Given a'3^ -^-Ixzzl c, to find x.
Analysis. — Multiply- operation.
ing the given equation by CL^ -\- hx ^z c
a, the coefBcient of a;^ and /^a^a^ + /[ahx = ^ac
by 4, the smallest square ^^Z^s _j_ ^al)X-\- W = 4«c -f J3
number, we have ^^«. i z> i . / i — m
4«2a;2 + ^ahx = ^ac, _L
the first term of which is . /„ _ — ^±V4^g + ^
an exact square, whose 2a
root is lax. Factoring
the second term, we have /s^abx = 2 (2aajx5). (Art. 119.)
As the factor 2.ax is the square root of 4a^a;-, it is evident that ^a^^
may be regarded as the first term, and /^cibx the middle term of the
square of a binomial. Since ^ahx is twice the product of this root lax
into &, it follows that & is the second term of the binomial ; conse-
quently, 6' 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 = — — , the value of x required.
2. Given 2:1^ 4- 3a; = 27, to find x.
Solution. — By the problem, 2X^ + 3a; = 27
Multiplying by 4 times coef. of a;*, i6ar'^ + 24a; = 216
Adding square of 3, coef. of a?, i6«2 + 24a; + 9 = 225
Extracting root, 4a;+3 = ± 15
Transposing, 425 = —3 ± 15
.\ flj = 3 or — 4|.
336. From the preceding illustrations, we derive the
KuLE. — I. Reduce the equalion to the form, aa^ ±_bx = c.
II. Multiply both members hy 4 tiines the coefficient of a^.
III. Add the square of the coefficient of x to each member,
extract the root, and reduce the resulting equation.
336. What is the rale for the third method of reduciug affected qoadrhtics f
AFFECTED QUADRATICS. 181
Notes. — i. Wlien the coeflBcient of x is an even numbei it is
sufficient to multiply both members by the coefl&cient of x^, and add
to each the square of half tlie coeflScient of x.
2. The object of multiplying the equation by the coefficient of x^ is
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 fractions 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 x to each, is the same in effect as adding the square
of half the coefficient of x 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 Hindo<^
3. Given ^a^ -\- 4X = 39, to find x, Ans. 3 or — 4^.
Reduce the following equations :
4. x^ — $0=: —X. 8. 2aj2 — 6a; = 8.
5. 5^ -f 32)2 _ 2. 9. 32^2 4- 5a; = 42.
6. 4X^ — 'jx — 2 = o. 10. a^ — 15a; = — 54.
7. 50^2^ 2a; = 88. II, 92^— 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 natural, being derived from
the square of a binomial ; but it necessarily involves frac-
tio7iSy when the coefficient of x is an odd number.
The second is the shortest, and is therefore 2i favorite with
experts in algebra.
The advantage of the third is, that it always avoids
fractions in completing the square.
The student should exercise his judgment as to the method
best adapted to his purpose.
Notes. When the coefficient of x is an even number, how proceed ? Object oJ
multiplying by coefficient Qix'^'i By 4 ?
182 AFFECTED QUADRATICS.
EXAMPLES.
Find the yalue of x in the following equations:
1. x^ — ^x^—i, 17. 3:^2 _ 7;;c _ 20 = o.
2. a;2_^^_,_^^ jS. ^x^ -- ido ^ IX,
3. 2a;2 __ y^ _. _ 2. 19. 2X^— 2X=\\.
4. a;2 -{- loic = 24. 20. (ir — 2) (a; — i) = 6.
5. 6:^2— 130: + 6 = 0. 21. 4(i»2__ i) _4^__ I,
6. 14a; — a;2 = 33. 22. (2X — 3)2 := 8i?;.
7. iz;2— 3 = — _^. 23. 3ic— 2 =
6 •'^' ^- --a;-i
8. ^^ = ^-^. 24. 4a; = 14.
16 100 — ga? „ t 40?
to. - + - = -. 26. a;« 4- - = -.
X a a 22
11. a^ + 2/?za; = i^. 27. x^ — 2nx = m^ — 7A
12. a^ + f=ii. .28. 9£.:=£)^£Z13«.
'3- ^2_6^-+y-^-3-
14
4^ a; — I 9a? + 7
14 — ic 3ic ~~ X *
15. 2A/rc2 — 4^ — J __ _ ^^
16. V^^S + 6 = a; + 5.
338. An Equation which contains but two powers of the
unknown quantity, the mdex of one power being kvice that
of the other, is said to have the Quadratic Form.
The indices of these powers may be either integral or
fractional.
Thus, a^— a;2 = 12 ; aj^^ + ar* = A ; and ^x — ^x = c, are equations
of the quadratic form.
Note. — Equations of this character are sometimes called trinomial
equations.
338. When has an equation the ouadratic forinf 2fo(e. What are such equatioDB
called?
AFFECTED QUADRATICS. 183
339. Equations of the quadratic form may be solved by
the rules for affected quadratics.
I. Given a^ — 2X^ = 8, to find x,
Solution — By the problem, ic* — 2a;' = 8
Completing square, x^ — 2X^ + i = g
Extracting square root, aj^ — i = i 3
Transposing, a:^ = 4 or — 2
Extracting square root again, a; = ± 2, or ± ^y/— 2
. 2. Given 0^ — 40^ = ^2, to find x.
Solution. — By the problem, afi — ^ = 32
Completing square, afi — 4iB^ + 4 = 36
Extracting square root, ar* — 2 = ± 6
Transposing, etc., a;' =1 8 or —4
Extracting cube root, . a? = 2 or ^y^— 4
3. Given x^" — 4bx^ = a, to find x.
Solution.— By the problem, x^* — 45a?* = a
Completing square, a;-^'* — 4hx^ + 45* = a + 45*
Extracting square root, a;» — 26 = ± ya+4ljr'
Transposing, a;" = 2& ± ya+^
Extracting the Tith root, x = V 26 ± ya+4b^
4. Given x^ + S = 6x^, to find x.
5. Given x* — 2x^ = 3, to find ar.
%. Given afi — ju^ = o, to find x.
x^ X 1
/. Given {- - = -^, to find ar.
2 ^ 4 32'
8. Given v^ + ^\/x — i, to find x.
9. Given 4a; + 4\/x +2 = 7, to find x.
10. Given ^, ' _ = - — -=--, to find «.
4 + V a; VaJ
184 AFFECTED QUADRATICS.
PROBLEMS.
1. Find two numbers such that their sum is 12 and theii
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 ic^tlie cost.
Then = the per cent.
100 ^
X
We now have x — x y. — = 24, to find the value of 05. -
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 |8o ; if he had
purchased 4 more for the same sum, each sheep would have
cost %i 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. Required the
number.
6. A teacher divides 144 oranges equally among her
scholars ; if there had been 2 more pupils, each would have
received one orange less. Required 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.
IIow many does he place in rank and file ?
AFFECTED QUADRATICS. 185
11. A drover buys a number of lambs for $50 and sells
them at I5.50 each, and thus gains the cost of one lamb.
^Required the number of lambs.
12. The sura 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 arrives at his
journey's end 10 hours before him. Find the rate per hour
at which each travels.
16. A and B distribute $1200 each among a certain
number of persons. A relieves 40 persons more than B, and
B' gives to each person $5 more than A. Required the
number relieved by each.
17. Divide 48 into two such parts that their product may
be 252.
18. Two girls, A and B, bought 10 lemons for 24 ceiL%
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 QUADKATICS.
23. Find two numbers the product of which is 120, and
if 2 be added to the less and 3 subtracted from the greater,
the product of the sum and remainder will also be 120.
24. Divide 36 into 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 $6.
29. What two numbers are those whose sum is i J and the
sum of their reciprocals si ?
30. Find two numbers whose difference is 15, and half
their product is equal to the cube of the less number.
31. xi 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 ?
SIMULTAKEOUS QUADRATICS. 187
SIMULTANEOUS QUADRATIC EQUATIONS.
TWO UNKNOWN QUANTITIES.
340. A Moniogeneotis Equation is one in which
the sum of the exponents of the unknown quantities is the
same in every term which contains them.
Thus, x'^—y'^ = 7, and x^—xy + y^ = 13, are each homogeneous.
341. A Symmetrical Equation is one in which
the unknown quantities are involved to the same degree.
Thus, x''+y^ = 34, and x^y—xy^ = 34, are each symmetrical.
342. Simultaneous Quadratic Equations 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 Equations consisting of, a
Quadratic and a Simple Equation.
I. Given x^ -{- y'^=z 13, and x -{- y =z 5, to find x and y.
Solution.— By the problem, x^+y^ = 13
(i)
x+y= 5 .
(2)
By transposition, x = s—y
(3)
Squaring each side of (3) (Art. 102), x'^ = 2S — ioy+y^
(4)
Substituting (4) in (i), 2S—ioy + y'^+y'^ — 13
(5)
Uniting and transposing, 2y'^—ioy = — 12
(6)
Comp. sq. (Art. 336,710(6), 4y^—2oy + 2S = —24 + 25
(7)
Extracting root, 2^—5 = ± i
.-. y=:3 or 2.
Substituting value of y in (3), a; = 2 or 3. Hence, the
340. What is a homogeneoua equation ? 341. A symmetrical equation ?
188 ' SIMULTAN^EOUS QUADRATICS.
Rule. — Find the value of one of the unhnoivn quantities
in the simple equation ly transposition, and substitute this
value in the quadratic equation. (Arts. 221, 223.)
^olye the following equations:
2. 0:2+^2:^25, 5. a:2 + ^2:=244,
y —X = 2.
6. 3^:2 _ ^2 —25 1,
^ + Ay = 38.
4. a^J_^2_28, 7. 8a;2_j_ ^^2_ 728,
6y — x=z 15.
344. To Solve Simultaneous Equations which are both
Quadratic and Homogeneous.
8. Given x^-\-xy = 40, and y'^-^xy = 24, to find x and y.
X +?/ --
= 7-
x^-^-f--
= 74,
x + y=i
: 12.
x^-f-.
= 28,
x-y =
: 2.
jUTIon. — By the problem.
x^ + a^ = 4o
(I)
»< «c
y^ + xy = 24
(2)
Let
x = py
(3)
Substituting ^3^ in (i).
pY +pf - 40
(4)
" (2),
/+i>2/^ = 24
(5)
Factoring, etc., (4),
^ p'+p
(6)
*' (5),
^ i+P
(7)
Equating (6) and (7),
40 24
p'+p i+p
(8)
Clearing of fractions.
S + 5P = 3P' + 3P
(9)
Transposing, etc.,
3P'-2p = 5
(ID)
Comp. sq., 3d meth. (Art. 336),
gp'^—6p+i = 15 + 1 = 16
(II)
Extracting root,
3P-1 = ± 4
(12)
Transposing, 3^ = i ± 4
Dropping the negative value, p = |
Substituting value of ^ in (7), y^ = 24-^(1 +|)=9
Extracting root, y = ± 3
Substituting value of p and y in (3), a? = fx ±3 = ±5.
Hence, the
343. Rule for solution of equations coneieting of a quadratic and simple equ^
tion?
SIMULTANEOUS QUADRATICS. 189
Rule. — I, For one of the unknown quantities substitute
the product of the other into an auxiliary quantity, and then
find the value of this auxiliary quantity.
II. Find the values of the unknown quantities by substi-
tuting the value of the auxiliary quantity in one of the
equations least involved.
Note. — ^An auxiliary quantity is one introduced to aid in the sola
>,ion of d problem, as p in the above operation.
9. Given
A.nd
It^^"" o^' 1 to find a; and y.
10. Given x ^y = 2, ].«, ,
1 1. Gireu 3^ - 72/' - - I, I t„ g„3 ^ ,„^
And 4^y= 24, ) ^
1 2. Given ^ - ^y + 3^ = '9, 1 to fi^a ^ ^„3
And xy=i5,) ^
345. To Solve Simultaneous Quadratic Equations when each
Equation is Symmetrical.
13. Given x + y =zg, and xy = 20, to find x and y.
.UTiON.— By the problem, x+y= 9
(I)
xy = 20
(2)
Squaring (i), aj» + 2a'2^ + / = 8 1
(3)
Multiplying (2) by 4. 40;^ = 80
(4)
Subtracting (4) from (3), a?—2xy+y^ - I
(5)
Extracting sq. root of (5), x—y — ±1
(6)
Bringing down (i), x + p = 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 x = s, y=4 ; and when aj = 4, y = 5
344. How solve equations which are both quadratic and homogeneous ? I^ote,
What is an auxiliary quantity ?
190 SIMULTANEOUS QUADRATICS.
2. The solution of this class of problems 'caries according to the
given equations. Consequently, no specific rules can be givren that
will meet every case. But judgment and practice will readily supply
expedients. Thus,
I. When the sum and product are given. (Ex. 13, 15.)
Find the difference and combine it with the sum, (Art. 2 24.)
II. When the difference and product are given. (Ex. 1 6.)
Find the sum and combine it ivith the difference,
III. When the sum and difference of the same powers are
given. (Ex. 14, 17.)
Combine the two equations by addition a7id subtraction,
rV*. When the members of one equation are multiples of
the other. (Ex. 18.)
Divide o?ie by the other, and then reduce the resulting
equation,
14. Given a;^ -f ^^ = 5, (i) ) . « , ,
^ ^ , 1 ^1 X ^ to find a; and V.
And x» —y^ =. i, (2) )
Solution, — Adding (i) and (2), and dividing, a;^ = 3
Involving, a? = 27
Subtracting (2) from (i), etc y* = 2
Involving, y = 8
15. Given 2^ + ^=27, ],«, ,
^ . , ^^ „" ^ to findrraudy.
And xy = 180, \ ^
16. Given a? — v= 14, )./>;, ,
. , ^ > to find X and y.
And xy = 147, )
,"*"'\'^^'ltofinda:andy.
* — ^^ = 3? '
17. Given x^ + y^ = 7,
And X
18. Given o^y^ + x^f =z 12,] , ^ ^ ,
A :i 00 ^ r to fin<i ^ and y.
And a?y -{- xy^ = 6, ) ^
Note,— What is true of the solution of Bimultaneous quadratics? When the tmm
and product are ffiven,how proceed? When the difference and product? When
the sum and difference of the same powers are given ? When the members of one
tquation are multiples of the other ?
SIMULTANEOUS QUADBATIC8. 193
PROBLEMS.
1. The difference of two numbers is 4, and the difference
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. Required two numbers whose sum multiplied by the
greater is 180, and whose difference multiplied by the less
is 16.
4. In an orchard of 1000 trees, the number of rows
exceeds the number of trees in each row by 15. Required
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. Required
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. Required 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 are on a side of each
of the squares ?
11. Required two numbers whose product is 3 times their
sum, and the sum of their squares 1 60.
12. What two numbers are those whose product is 6 times
their difference, and the sum of their squares 13 ?
CSee A'^r)cnclix, p. 290.)
CHAPTER XVII.
RATIO AND PROPORTION.
346. Ratio is the relation which one quantity bearb k
pother with respect to magnitude.
347. The Ter^ns of a Ratio are the quantities
compared. The first is called the Antecedent, the second
the Consequent y^ and the two together, a Couplet
348. The HUjn of ratio is a colon : \ placed between the
two quantities compared.
Ratio is also denoted by placing the consequent under the
antecedent, in the form of 2^ fraction.
Thus, the ratio of a to 6 is written, a : &, or ^ •
349. The Measure or Value of a ratio is the quotient
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 8-7-4 = 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 t or les8 than the other.
Thus, a foot has a ratio to a yard, but not to an hour, or a pound
350. A Simple Itatio is one which has but two terms *,
fts, a\h, 8 : 4.
346. What is ratio ? 347. What are the terms of a ratio ? 348. The pign ? How
»lso is ratio denoted ? 349. The measure or value ? Note. What quantities have a
ratio to each other ? 350. What is a simple ratio ?
* Ardecedent, Latin ante, before, and cedere, to go, to preceoe.
Consequent, Latin con, and sequi, to follow.
f Tlie sign of ratio : is derived from the sign of division -f-, the
horizontal line being dropped.
RATIO. 193
351. A Compound Matio is the product of two or
more siinple ratios.
Thus, 4 • 2 [ are eacli simple But 4x9: 2x3
9:3$ ratios. is a compound ratio.
Note. — The nature of compound ratios is the same as that of sim-
ple ratios. They are so called to denote their origin, and are usually-
•expressed by writing the corresponding terms of the simple ratios one
•mder another, as above.
352. A Direct Ratio arises from dividing the ante-
cedent by the consequent.
353. An Inverse^ or Heciprocal Matio 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& = -r, or v, and that of 4 to
4 I
12 = — , or - •
12 3 .
The inverse ratio of a to a& = — , or 6 ; of 4 to 12 = — , or 3.
tt 4
It is the same as the ratio of the reciprocals, - to -^ , and - to — •
a ab 4 12
Note. — A reciprocal ratio is expressed by inverting the fraction
which expresses the direct ratio. When the colon is used, it is
expressed by inverting the m^der 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 : f is the same as 6:3.
Note. — When the fractions have different denominators, reduce
them to a common denominator; then compare their numerators.
(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.
351. What is a compound ratio? l<!ote. Why so called? 352. What is a direct
ratio ? 353. A reciprocal ? 355. What is a ratio of equality ?
* Inverse, from the Latin in and verto, to turn upside down, to invert.
9
194 RATIO.
356. A !Ratio of Greater Inequality is one whose
antecedent is greater than its consequent, and its value is
greater than i.
357. A IRatio of Less Inequality is one whose
antecedent is less than its consequent, and its value is less
than I.
358. A Duplicate Hatio is the square of a simple
ratio. It arises from multiplying a simple ratio into itself,
or into another equal ratio.
359. A Triplicate Matio is the cule of a simple ratio,
and is the product of three equal ratios.
Thus, the duplicate ratio of a to 5 is a* : 6*.
The triplicate ratio of a to & is a^ : 5^.
360. A Suhduplicate Hatio is the square root of a
simple ratio.
361. A Subtriplicate Matio is the cuhe root of a
simple ratio.
Thus, the svbduplicate ratio of a; to y is ^\/x : ^/y.
The sicbtriplicate ratio of a; to y is y^x : ^y, etc.
362. Since ratio may be expressed in the form of a
fraction, it follows that changes made in its terms have the
same effect on its value, as like cUamjes in the terms of a
fraction. (Art. 167.) Hence, the following
PRINCIPLES.
I®. Multiplying the antecedent, or ) ,^ ... ,. ^-
T^' 'f' J.1 ^ \ MuUivhes tho ratio.
Dividing the consequent, j
2**. Dividing the antecedent, or ) r.- -t ^t
!•#- 7,. 7 . ,1 ± \ Divides the ratio.
Multiplying the consequent, )
3**. Multiplying or dividing loth \ Does 7iot alter the value
terms hy the same quantity, f of the ratio.
356. Of greater inequality? 357. Of lef? inequality? 358. A duplicate ratio?
359. Triplicate? 360. Subduplicate ? 361. Subtriplicate? 363. Name Principle i.
Principle a Principle a>
EATia Idd
EXAMPLES.
1. What is the ratio of 4 yards to 4 feet?
Solution. 4 yards = 12 feet ; and the ratio of 12 ft. to 4 ft. ia y
2. What is the ratio of 6a^ to 2X ? Ans, 3a:,
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 Sa^ to 4a ?
7. What is the ratio of isaic to ^ab?
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 P
11. What is the ratio of 20,^ to 4a ?
12. What is the ratio otx^ — y^tox-^-r/?
13. What is the compound ratio of 9:12 and 8 : 15 ?
Solution. 9 x 8 = 72, and 12 x 15= i8a Now 72-5-180 = t^, Ana
Or, 9 : 12 = ■^, and 8 : 15 = ^V Now j\ x yV = t¥o = tu» ^^■
14. What is the compound ratio of 8 : 15 and 25 : 30?
15. What is the compound ratio of a:b and 2h : ^ax?
16. Eeduce the ratio of 9 to 45 to the lowest terms.
Solution. 9 : 45 = t\» and ^\ = |, Ans.
17. Reduce the ratio of 24 to 96 to the lowest terms.
18. Reduce 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 grearter, the ratio of 15 : 9, or 38 : 19?
24. Which is the greater, the ratio of 8 : 25, or V4 '- V25,
25. If 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 pBOPOBiioifir.
PROPORTION.
363. Proportion is an equality of ratios.
Thus, the ratio 8 ; 4 = 6 : 3, is a proportion. That is.
Four quantities are in proportion, when the first is the same mvlti-
pie or part of the second that the third is of the fourth.
364. The Sign of Proportion is a double colon : :,♦
or the sign =. Thus,
The equality between the ratio of a to 5 and c to (? is expressed by
a : 6 : : c : <f , or by T = 3
0 d
The former is read, " a is to & as c is to (f ;" the latter, *'& is contained
In a as many times as d is contained in c.**
Note. —Each ratio is called a couplet , and each term a proportional.
365. The Terms of a proportion are the quantities
compared. ^\iq first 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 repeatedy so as
to form two terms ; as, a\h : : J : c.
Note. — Care should be taken not to confound propoHion with ratio.
In common discourse, these terms are often used indiscriminately.
Thus, it is said, " The income of one man bears a greater proportion
to his capital than that of another,'* etc. But these are loose expressions
In a simple ratio there are but two terms, an antecedent and a
cjonsequent ; whereas, in a proportion there must at least be four
terms. (Arts. 350, 366.)
363. What is proportion? 364. The sign of proportion? Note. 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 from three quantities ?
* The sign : : is derived from the sign of equality =, the four
points being the terminations of the lines.
PBOPOETION. 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 Projjortional is the last term of a
proportion having three quantities.
Thus, in the proportion a:h :: b:e,b\ask mean proportional, and
e a third proportional.
370. A Direct Proportion is an equality between
two direct ratios ; as, a:b : : c:d, $16 : : 4:8.
371. An Inverse or Heciprocal Proportion is an
equality between a direct and reciprocal ratio ; as,
8:4 :: i'.h
372. Analogous Terms are the antecedent and con-
sequent of the same couplet.
373. Homologous Terms are either two antecedents
or two consequents.
PROPOSITIONS.
374. A Proposition is the statement of a truth to be
proved, or of an operation to be performed.
Propositions are of two kinds, theorems and proUems,
375. A TJieorem is something to be proved.
376. A Problem is something to be done.
377. A Corollary is a principle inferred from d
preceding proposition.
368. What is a mean proportional ? 369. Wliat is a third proportional ? 370. A
direct proportion? 371. An inverse or reciprocal proportion? 372. What ar«
analop:ou8 terms ? 373. Homologoni?? 374. What is a proposition ? How divided?
375. What is a theorem ? 376. A problem? 377. A .^roUary?
198 PEOPORTIOK,
378. Tho more important theorems in proportion are the
following:
Theorem L
If four quantities are 'proportional^ the product of the
extremes is equal to the product of the means.
Left a : d :: e : d
By An. 363, |=|
dearing of fractions, a<2 = &&
Verification by NuMBERa
Given, a: 4:: 8: 16; and 2x16 = 4x8.
Cor. — ^The relation of the four terms of a proportion to
each other is such, that if any three of them are given, the
fourth may be found.
Thus, since ad — be, it follows that
a = bc+dt b = ad-T-c, e = ad-r-b, and d = bc-t-a, (Ax. 5.)
Notes. — i. The rule of Simple Proportion in Arithmetic is founded
upon this principle, and its operations are easily proved by it.
2. This theorem furnishes a very simple test for determining
whether any four quantities are proportional. W© have only to
multiply the extremes together, and the means.
Theorem IL
If three quantifies are proportional, the product of the
extremes is equal to the square of the mean.
Let a : 6 : : b • t
By Art. 363, |=:-
Oearing of fractions, ac = b*. .
Again, 9:6:: 6:4, and 4x9 = 6'.
Cor. — A mean proportional between two quantities is
equal to the square root of their product
PEOPOETIO]Sr. 199
Theorem HL
ff 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
9fthe other for the means.
Let adr=lfe
DividingbjM, 1 = 1
Or, by Art. 363, a:b :: e : d,
Agoia, 4x6 = 3x8, and 4:3:: 8 :&
Theoeem IV.
If four quantities are proportional, they are proportional
when the means are inverted.
Let aih :i ei d, then a i e 11 h: d
a e
For, hj Art. 363,
h~d
Multiplying by -. * s= j
Or, aie II hi d.
Again. 3 : 6 : • 4 : 8, and 3 : 4 : : 6 : 8. (Th. i.)
Note. — This change in the order of the means \a called
** AlterTuUian,**
Theorem V.
If four quantities are proportional, they are proportional
when the terms of each couplet are inverted.
Le. a ih M e \ dt then h ', a .: d \ t
By Theorem i, ' ad = he
By Theorem 3, h : a :: d : e.
Again, 6 : 2 :: 15 : 5, then 2 : 6 :: 5 : 15. (Th. i.)
Cor. — ^If the extremes are inverted, or the order of the
terms, the quantities will be proportional.
200 PROPORTIOK.
Notes. — i. If the terms of onlj one of the couplets are inverted,
ihe proportion becomes reciprocal,
2. The change in the order of the terms of each couplet Ls called
^Inversion**
3. This proposition supposes the quantities compared to be of the
same kind. Thus, a line has no relation to weight. (Art 349, note.)
Theorem VL
If four quantities are proportional , ttvo analogous or twt^
homologous terms may be multiplied or divided by the sam4
quantity without destroying the proportion.
Let a:h :: e I d
Multiplying analogous terms, am : bm :: e : d
and a : & 11 em i dm
^' h^d
Hence. (Art 362. Prin. 3). ^ = |. and | = ^
Multiplying homologous terms, am : 5 : : cm i d
And a '. hm '.: c : dm.
.„. , . ^ am em 3 a e
Hence. (Ax 4. 5), X=T' ""^ i^=^
Dividing analogous terms, — : — : r e : d,
and « : 6 ; • — : —
m m
Dividing homologous terms, — : b :: — : d
b d
and ^^ • ;r • ' ^ • ;r
m m
Clearing of fractions (Th. i), ad=:be
Cor. — All the terms of a proportion maybe multiplied
01 divided by the same quantity without destroying the
proportion.
Notes. — i. When the homologous terms are multiplied or divided,
both ratios are equally increased or diminished.
2. When the analogoiLs terms are multiplied or divided, the ratios
are not altered.
PEOPOHTION. 201
Theokem 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 h II ei d, then fl+6 : 6 :: e-\-d : d
_ a c
*^°'' »=5
Adding I to each member, ^ + i = -% + 1. (Ax. a.)
Incorporating i, — r— = —-^
Therefore (Art. 363), a+h il :: c + d : d
Again, 4:2 :: 6:3, then 4+2 : 2 :: 6 + 3 : 3
KoTE. — This combination is sometimes called ** Composition.'*
Theorem VIII.
// four quantities are proportional, the difference of the
first and the second is to the second, as the difference of the
third and fourth is to the fourth.
Let a : 6 : : c : d, then a— & : h :: c—d : d
Subtracting i from each member, r- — i = - — x
0 d
- ,. a— 6 c—d
Incorporating —1, — v— = ^ -
Therefore, a— 6 : 6 : : c—d : <2
Again, 4:2 : : 6:3, then 4—2 12:: 6—3 : 3
Note. — This comparison is sometimes called *' Division" *
* The technical terms. Composition and Division, 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
PEOPORTION.
Theoeem IX.
If two ratios are respectively equal to a third, they are
equal to each other.
Let a : b :: m : n,
a _m
l~"n'
Then
and
and
d :: m I n
d ~" »
That is, a : b = e : d
Again, 12 : 4 = 6 : 2, and 9:3 = 6:3
A 12 : 4 = 9 : 3
Theoeem X.
When any numler of quantities are proportional, any
antecedent is to its consequent, as the sum of all the ante-
cedents is to the sum of all the consequents.
Let a I b :: e : d ::e:/, etc.
Then a : b :i a+c+e : b+d+f, etc.
For (Th. i), ad — be
And, ** af — be
Also, ab = ba
Adding (Ax, 2),
Factoring,
ab+ad+af = ba+bc+be
a(b+d+f) = b{a+c+e)
Hence, (Th. 3), a :b i: {a+e+e, etc.) : (J)+di-f, 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.
PEOPOETI02!r. 203
Theorem XL
If the corresponding terms of two or more proportions are
multiplied together, the products will he proportional.
Let a ih :: ci d, and e i f i: g : h
Then ae :hf :: eg : dh
For. -=5 '«'<1 7=f
Halt ratios together (Ax. 4), h^~^
Hence, (Th. 3), ae : bf z: eg : dh.
Theoeem xn.
If four quantities are proportional, like powers or roots
of these quantities are proportional.
Let a : b : : e : d, then t = 3
0 a
■DA a« c*
Hence (Th. 3), «» : 6» ; ; c» : d»
Extracting sq. root, a* : 6* : : c* : d*
Again, 2:3 : : 4:6, then 2« : 3' : : 4" : 6*
In like manner, /y^ : ^/g : : -y/iS : -y/is.
Note.— The index » may be either integral or fractional.
Theorem XIIL
Equimultiples of two quantities are proportional to the
ruantities themselves.
Smce 1=3, hy Art. 362, Pnn. 3, r— = ^
fid" '^ bm d
Hence, am : im ii c : d.
204 PEOPORTIOK
PROBLEMS.
1. The first three terms of a proportion are 6, 8, and 3.
What is the fourth ?
Let x = the fourth term.
Then 6:8 : : 3 : x
/. 6x = 24, and x = 4.
2. The last three terms of a proportion are 8, 6, and 12.
What is the first?
3. Eequired a third proportional to 25 and 400.
4. Required 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 cubes 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 Progression is a series of quantities which
increase or decrease according to a fixed law.
380. The Terms of a Progression 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: arithmetical,
geometrical, and Tiarmonicah
ARITHMETICAL PROGRESSION.
382. An Arithmetical Progression is a series
which increases or decreases by a constant quantity called
the common difference.
383. In an ascending series, each term is found ly adding
the common difference to the preceding term.
If the first term is «, and the common difEerence d, the series is
a, a+d, a + 2d, a + ^d, etc.
If a = 2, and d = 2, the series is 2, 5, 8, 11, 14, etc.
384. In a descending series, each term is found by
sultr acting the common difference from the preceding term.
If a is the first term, and d the common difference, the series is
a, a—d, a— 2d, a— 3d, 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
progrepsion? 382. An arithmetical i)rogre8Pion ? What is this constant quantity
•ailed ? 383. An ascending series ? 384. A descending series ?
306 AEITHMETICAL PROGKESSION.
positive one, it may therefore properly be said that each successive
term of the series is derived from the preceding by the addition of tlie
common difference. (Art. 75, Prin. 3.)
Notes.— I. The common difference was formerly called arithmetical
ratio; but this term is passing out of use.
2. An Arithmetical Progression is sometimes called an Equidifferent
Series, or a Progression by Difference. In every progression there may
be an infinite number of terms.
385. 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 + 2d, a + 3d, be the series.
Adding extremes, etc., 2a + 3d = 2a+sd.
Or, let 2, 2 + 3, 2 + 6, 2 + 9, be the series.
Then 2 + 2 + 9 = 2 + 3 + 2 + 6.
386. If three quayitities are in arithinetical 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).
Cor. — An Arithmetical Mean between two quantities
may be found by taking half their sum.
387. In Arithmetical I^rogression there are five
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 difierence,
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. What is true of four quantities in arithmetical progression ? 386. Of three
quantities? 387. Name the elements in arithmetical progression? What relation
have they to each other {
ARITHMETICAL PKOGRESSION. 207
CASE I.
388. The First Term, the Common Difference, and Number
of Terms being given, to Find the Last Term.
Each succeeding term of a progression is found by adding the
common difference to the preceding term. (Art. 384.) Therefore the
terms of an ascending series are
a, a+d, a+2d, a+sd, etc.
The terms of a descending series are
a, a—d, a — 2d, a— 3d, etc.
It will be seen that the coefficient of d in each term of both series
is one less than the number of that term in the series. Therefore,
putting I for the last or nth. term, we have
Formula I. I = a ±{n— i)d.
EuLE. — I. Multiply the common difference hy the number
of terms less one.
II. When the series is ascending, add this product to the
first term ; when descending, subtract it from the first term,
1. Given « = 3, d= 2, and n-=i, to find I.
l = a± (n—i)d = 3 + (7—1)2 = 15, Ans.
2. Given « = 25, ^ = — 2, and w = 9, to find I
3. Given az= 12, d = 4, and n = 15, to find L
4. Given a = 1, d= —^, and n =13, to find t
5. Given « = |, d = \, and n = g, to find I.
6. Given « = i, d=^ — .01, and n = 10, to find h
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 J, 4, 5I, etc.
11. Find the 25th term of the series x + ^x + ^x-^-yx, etc.
12. Find the nth term of the series 2a, sa, Sa, iia, etc.
388. What is the rule for finding the last term ?
208 AKITHMETICAL PROGEESSIOIJ",
CASE II.
389. The Extremes and Number of Terms being given, tc
Find the Sum of the Series.
Let a, a + d, a + 2d, a + 3d . . . I, be an arithmetical progression,
the sum of which is required.
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) + {I — 2d) + {I — 3d) + . . . +a
Adding, 28 = a + I + {a + I) + {a+l) + {a + l) + . . . +a+l
.'. 28 = (a + T) 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 = i — ^t_Z x n.
2
EuLE. — Multiply half the sum of the extremes ly 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.
Thus, in the series, 3, 5, 7, 9. 11, 13, the sum of the first and last
terms, of the second and fifth, etc., is the same, viz., 16.
1. Given « = 4, / = 148, and n = 15, to find 5.
Solution. 4+148 = 152, and (152-5-2) x 15 = 1140, Arts,
2. Given a = ^, I = 30, and n = 50, to find s.
3. Given a = 6, Z = 42, and n = 9, to find 5.
4. Given « = 5, ? = 75, and n = 35, to find s,
5. Given a = 2, I = 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, 63, etc.,
to 15 terms.
ARITHMETICAL PRO G R E SSI OIT. 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 /owrtfiyere/i^ quantities; the
first term, the common difference, the last term, and the
number of terms. If any three of these quantities are given,
the other may be found. (Art. 388.)
I. 1 = a ±, (n — i) ^ / a,cl, and 71 being given.
3. Given d, Z, and n, to find a, the first term.*
Transposing (n—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 {n—i),
, I — a
a = •
n — I
5. Given a, d, and ?, to find n, the number of terms.
Clearing of fractions and reducing (4),
I — a
n = —-=— + I.
d
1. Given a = 2$, d=i 3, and n = 12, to find I
2. Given a = ^S, d = 5, and n = 45, to find I
3. Given d=: 3, Z = 35, and 71 = 9, to find a.
4. Given I = ^y, d = 5, and 71 = 21, to find a,
5. Given « = 15, 1 = 85, and n=z ^1, to find d,
6. Given a = 28, 1= 7, and ?^ = 26, to find ^.
7. Given a=z 2^, d = 5, and Z = 5^38, to find w.
8. Given a= 6, t?= 6, and Z=ii52, to find w.
* For Formula 2' see Art. 389.
21(/ AEITHMETICAL PROGRESSION.
By Formula II.
392. In this formula there are four different quantities:
the first term, 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. s = X iif a, I, and^^^ being given.
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,
a = 1.
n
7. Given a, n, and s, to find I, the last term.
Transposing in (6), we have
I = a.
n
8. Given a, I, and 5, to find n, the number of terms.
Clearing (7) of fractions, transposing, factoring, and dividing,
2.S
w = ',
a ^-l
1. Given «= 9, 7 = 41, and w= 7, to find 5.
2. Given « = \, ? = 45, and n = 50, to find s.
3. Given ^ = 50, d:= 4, and w = 12, to find a.
4. Given a= 9, Z = 41, and 5 = 150, to find 71.
5. Given d=z 7, ?=2i, and ?^ = 35, to find 05.
6. Given a = 46, 1= 24, and 5 = 455, to find n.
7. Given a = 27, w = 9, and 5 = 72, to find I
8. Given a = 72, n= 8, and s = 288, to find I
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 PROGRESSION.
211
393. The remaining twelve fonnulas 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.
GlVBN.
Required.
FOBUUXAS.
9-
lo.
II.
12.
14.
15-
16.
17.
18.
19.
20.
d n, s
d, I, s
a, I, s
I, n, s
a, n, 8
d, n, s
a, d, s
a, df s
d, I, s
a, d, n
a, d, I
d, I, n
a =
25 — dn^ + dn
271
"^^f^yO+i/"'^'
_ 2(nl — s)
w (w— i)
25 — 2an
d =
d=z
d =
n^ — n
±^(2^—^)24- Sds—2a-{-d
71=1
n=z
2d
2l+d± V(2l ^df—Ms
2d
n
5 = - [2« + (^^ — i) ^
Ij^ a Z2 _ ^2
5 = Y i—
2 2d
5 = - [2?— (7^— i)cZ]
._ Of the twenty fonnulas in Arithmetical Progression, the jirsA
two are indispensable, and should be thoroughly committed to memory ;
the next six are important in the solution of particular problems. The
remaining twelve are of less consequence, but will be fountJ
interesting to the inquisitive student.
212 ARITHMETICAL PROGRESSION.
394. By the fourth formula in x\rt. 391, any number of
arithmetical means may be inserted between two given
terms of an arithmetical progression. For, the number of
terms consists of the two extremes and aU the intermediate
terms.
Let m = the number of means to be inserted.
Then m + 2 = n, the whole number of terms.
Substituting m + 2 f or n in the fourth formula, we have
d = . Hence,
m+i
27ie 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 5 arithmetical means between 2 and 60.
5. What is the sum of 100 terms of the series -J-, f, i, f,
h 2, i, h 3» etc.
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 bhe odd numbers i, 3, 5, 7, 9, 11,
etc., continued to 76 terms?
8. Required the sum of 100 terms of the series of even
numbers 2,^4, 6, 8, 10, etc.
ARITHMETICAL PRO GRESSI 0:N^. 213
9. The extremes of a 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 implied in the
conditions. Such problems may be solved by stating the conditions
algebraically, and reducing the equations.
13. Find four numbers in arithmetical progression, whose
ium shall be 48, and the sum of their squares 656.
Let X = the second of the four numbers.
And 1/ = their common difference.
By the conditions, (a^— y) + x + {x + i/) + (x + 2y) = 48 (i)
And (ps—yf + x^ + {x+ yf + {x+ 2y* = 656 (2)
Uniting terms in (i), 42; + 2^= 48 (3)
" "(2), 4»' + 4^+62/« = 656 (4)
Transposing and dividing in (3), y = 2^ — 2x (5)
Dividing (4) by 2, 27l^ + 2xy+2f = 328 (6)
Substituting value of y, 2X^ + 20(24—20;) + 3(24 — 2xY = 328
Reducing, ic*— 2405 = —140
Completing square, etc., a; = 14 or 10
Substituting in (5) y = — 4 or 4
Hence the required numbers are 6, 10, 14, and 18.
Note. — 2. The first two values of x 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
Bum is 15, and the sum of their cubes is 495.
15. If 100 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 AKITHMETICAL PROGRESSION-.
1 6. How many strokes does a common clock strike in
?4 hours?
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 up
in 365 days?
19. The clocks of Venice go on to 24 o'clock. How many
strokes does one of them strike in a day ?
20. What will be the amount of li, 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 traveller 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 overtake 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 |, 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. Required the
amount of the debt and of the last payment
GEOMETEICAL PEO G EES SIGN . 215
GEOMETRICAL PROGRESSION.
395. A Geometrical Progj^ession is a series of
quantities which increase or decrease by a constant multiplier
called the ratio. Hence,
The ratio may be an integer or o^ fraction.
Note, — When the ratio 'ib fractional, the series will decrease. For
multiplying by a fraction is taking a certain part of the multiplicand
as many times as there are like parts of a unit in the multiplier.
396. In a geometrical series, each succeeding term is
found by multiplying the preceding one by the ratio.
Thus, if o is the first term, and r the ratio, the series is
a, or, ar^ ar*, or*, ar^y ar^, etc.
If the ratio is 3, the series is
a, ax 3, «x3*, ax3«, etc.
If the ratio is |, the series is
a, ax^, axjx^, oxjx^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, aU 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
«» — 3«» +9«» —2705, +Sia, etc.
395. Wliat l8 a geometrical progression ? 397. What Is an ascending serlM f
J98. Descending? 399. Wlxat law governs the signs?
216 GEOMETBICAL PROGRESSION.
400. In geometrical progression there are five elements,
the first term, the last term, the number of terms, the
common ratio, and the sum of the terms.
Let a = the first term,
/ = 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, w, and r are given, to find I, the last term.
The successive terms of the series are
a, ar, ar^, ar^, ar^^ etc, to ar»-^ (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 less
than the number of the terms. Therefore, the last or nth term of the
series is ar''-'^. Hence, we have
Formula I. I =. ar"^^.
Rule. — Multiply the first term ly that 'power of the ratio
vjhose index is one less than the number of terms.
CoR. — 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= ^, n ^ 6, and r = 2, to find I
2. Given a= 2, w = 8, and r = s, to find I.
3. Given a = 72, 71 = 5, and r = |, to find I,
4. Given a= 5, 7^ = 4, and r = 4, to find I,
5. Given a= 7, ?i = 5, and r = 2, to find I.
6. Given a = 10, n = 6, and r = — 5, to find I
400. Name the elements In jceometrical prog^Bsion. 401. How find the last
term?
GEOMETRICAL P ROGRESSIOlif 217
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, I, and r are given, to find 8,
Since 8 = the sum of the terms, we have
8 = a-\-a/r + ar'^ + a7^-{- .... +a7^-2 + ar»-\ (i)
Multiplying (i) by r,
rs = ar + ar^ + a7^ + ar^+ . . . . +aT^^ + ar*. (2)
Subtracting (i) from (2), ra—s = ar*—a. (3)
Factoring and dividing, a = • (4)
In equation (4), ar^ is the last term of (2), and is therefore the
product of the ratio by the last term in the given series.
Substituting Ir for ar^, we have
Ir -~a
Formula II.
r— I
EuLE. — Multiply the last term hy the ratio, from the
product subtract the first term, and divide the remainder by
the ratio less one.
PW' For the method of finding the sum of an infinite descending
aeries, see Art. 435.
1. Given a = 2, ? = 500, and r = 3, to find the sum.
_ Ir—a 500 X 3 — 2 .
Solution. « = = = 749, Ans,
2. Griven « = 3, 1=^ 9375? and r = 5, to find s,
3. Given a = 9, l^ 9000, and r = 10, to find s.
4. Given a = s^ ^ = 20480, and r ^ 4, to find s.
5. Given a =: 15, 1= 3240, and r = 6, to find s.
6. Given a = 2^, 1= 6400, and r =z 4, to find 5.
402. How find the sum of the terms ?
10
218 GEOMETRICAL PE0GRESSI02T.
403. The two preceding formulas furnish the means loi
solving all problems in geometrical progression. Thej may
be varied so as to form eighteen other formulas.
By Formula I.
404. The first formula contains /owr different quantities:
the first term, the last term, the ratiOy and the number of
terms. If any three of these quantities are given, the oiuer
may be found. By the first formula,
I. I = ar**^^; a, n, and r being given. (Art. 401.)
For formula 2, see Article 402.
3. Given Z, n, and r, to find or, the first term.
Factoring (i), and dividing by y-^
/
4. Given a, ?, and 7^, to find r, the ratio.
Dividing (i) by «, and extracting the root denoted by the index,
5* Given a, ?, and r, to find n, the number of terms.
I
a
Dividing (i) by a, r""^
By logarithms, log r (n— i) = log ? — log a
-. , log I — log a
Dividing, etc., n = —-i^-^— + ^'
Note.— Sin«e this formula contains logarithms, it may be deferred
till that subject is explained.
1. Given « = 3, n = $, and r = 10, to find I
2. Given a = 5, n = 6y and r = 5, to find /. -
3. Given Z =1 256, n = S, and r = 2, to find a.
4. Given I = 243, ^ = 5, and r = 3, to find a.
5. Given a = 2^ Z = 2592, and /i = 5, to find n
6. Given a = 4, / := 2500, and w = 5, to find n
GBOMETKICAL PRO GRESSIOl?'. 219
By Formula IL
405. This formula contains four different quantities : the
fird term, the last term, the ratio, and the sum of ths
terms. If a7iy three of them are given, the other may be
found. By the second formula,
1/p ft
2, s = — , a, I, and /• being given. (Art. 402.)
For formulas 3-5, see Article 404.
6. Given I, r, and s, to find a, the first term.
Clearing (2) of fractions, etc.,
a = Ir — s {r — i)
7. Given a, r, and 5, to find Z, the last term.
Transposing in (6),
;r = a + s (r — i).
Dividing by r,
/ — « + g(r~»i)
8. Given a, 7, and s, to find r, the ratio.
Clearing (2) of fractions,
sr — s-= Ir — a.
Transposing in the last equation,
sr — Ir = s — a.
Factoring, etc.,
8 — a
r =
8-1
1. Given a =^ 2, I z= 162, and r = 3, to find s,
2. Given / =1 54, r = 3, and s =z 80, to find a.
3. Given o^ = 4, r = 5, and s z^ 624, to find I.
4. Given a = 4, I = 12500, and s = 15624, to find n
5. Given a = 5, I = 180, and r ^ 6, to find s.
6. Given a = t, r = ^, and s = 847, to find I,
220
GEOMETRICAL PROGRESSION?-,
406. 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.
TABLE.
No
Given.
Requered.
FOKMULAS.
9
lO
II
12
13
14
15
16
17
18
19
20
n, r, s
I, n, s
a, n, s
n, r, s
a, I, s
a, r, s
h r, s
tty 71, S
I, 71, S
a, n, r
I, n, r
a, I, Tfi
a =
{r—\)s
/•" — I
a{s — ay-^ = / (5 — iy-\
l{s — ly-^ = a{s — ay-K
r" — I
71 = log ^- log Q^ ^
log(s—a)—log(s—iy
]og[a + {r—i)s] — log a
log r
_log|— log [Ir— (r— 1)5]
~" log r
n
+ 1
s s
r" r = 1
a a
/•" +
%-i
I— S 1 — 8
a(r'' — i)
r -
- 1
^/i — " v«
Of the twenty formulas in Geometrical Progression, the fird
two are fundamental, and should be thoroughly committed to memory ;
the next six are important in the solution of particular prohlems. The
remainder are less practical.
GEOMETRICAL PK OGRE SSI ON. 221
407. By the fourth, formula (Art. 404), any number of
geometrical means may be found between two given
quantities.
Let m = the number of means required.
Then m + 2 = n.
Substituting 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* = j/i^? = ^ej = 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 terms of the series, i, |, f, -jy, etc.
5. If the first term of a series is 2, the ratio 3 and the
number of terms 15, what is the last tenn ?
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 GEOMETEICAL PKOGRESSIOK.
7. Find three numbers in geometrical progression, such
that their sum shall be 2 1, and the sum of their squares 189.
Let the three numbers be x, ^/xy, and y.
By the conditions, a;+ ^/m-\-y = 21 (i)
And a;2 + a;y+y2 = 189 (2)
Transposing and sq. (i), as^ + 20;^ + ^^2 = 441 _ 42 /y/^ + xy (3)
Subtracting (2) from (3), iry = 25 2 - 42 'v/^+ xy (4)
Transposing, etc., ^/xy = 6 (5)
Involving, ^ icg^ = 36 (6)
And 32^ = 108 (7)
Subtracting (7) from (2), c^—2xy+y^ = 81
Extracting root, x—y = 9 (8)
Substituting (5) in (i), x+y = 15 (9)
Combining (8) and (9), a? = 12
y= 3
Hence the numbers, 12, 6, and 3, Ans.
8. A father gives his daughter $1 on New Year's day
t(>vrards 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. What 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. Find four numbers in geometrical progression, such
that the sum of tlie 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
1 464 1. 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 I^rogression 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
ID ; 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, 6, c, d, e, /, g, be an harmonical progression, then
a : c : a—b : b—c, etc.
Note.— When three quantities are such, that the first is to the
tJiird as the difference of the first and second is to the difference of the
second and tliird, they are said to be in Harmonical Proportion.
Thus, i, 3, and 6, are in harmonical proportion,
408. What is an harmonical progression ?
"•' If a musical string be divide.] in harmonical proportion, the
/ifferent parts will vibrate in harmony. Hence, the name.
224 HARMONICAL PROGRESSIOK.
409. To Find the Third Term of an Harmonical
Progression, the First Two being given.
Let a and h be the first two terms, and x the third term.
Then a : x :: a-l : l-x
Multiplying extremes, etc., ab—ax = ax—hx
Transposing, etc., 2ax—bx = db
Factoring, and dividing by 2a— h, we have the
ab
FORMtJLA. X
2a — h
Rule. — Divide the product of the first two tenns hy twice
the first mi?ms the second term; the quotient will le tht
third term.
Note. — This rule furnishes the means for extending an harmonic
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 1 8 are the first two terms. Ans. 2,^.
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^^, 9, 12, for two terms.
A71S. 18 and 2,^,
410. To Find a Mean or IVIiddle Term between Two Terms
of an Harmonic Progression.
Let a and c be the first and third of three consecutive terms of an
harmonic progression, and m the mean.
Then a i c :: a—m . m—e
Mult, extremes and means, am—ac = ac—cm
Transposing and uniting, am + cm = 2ac
Factoring and dividing by a + c, we have the
„ 2ac
Formula. m = •
a + c
409. How find the third term of an harmonical progression, the first two being
tfiven?
HARMOtsTICAL PROGRESSION". 225
Rule. — Divide twice the product of the first and third
terms hy their sum; the quotient will he the mean or middle
term.
6. The first and third of three consecutive terms of an
harmonic progression are 9 and 1 8. Required the mean or
middle terra.
Solution. 2x9x18 = 324, and 9 + 18 = 27,
Now 324-J-27 = 12, Ans.
7. rind an harmonic mean between 12 and 20. Ans, 15.
8. Eind 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 reciprocals of 10, 12, 15, 20, etc., viz.,
tV» tV» tV» "eV* ■^' ^^^•»
are an arithmetical progression, whose common difference is ^V
Again, let a, b, c be in harmonic progression.
Then a : c :: a—b : b—c
Mult, extremes and means, ah—ac = ac—bc
Dividing hj abc, ^ ~ ^ (^^- 3^4')
Conversely, the reciprocals of an arithmetical progression
form an harmonic progression. Thus,
The reciprocals of the arithmetical progression i, 2, 3, 4, 5, etc.,
viz., \, ^, \i\, i, etc., are in 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.,
the second will sound an octave above the first ; the third
will sound the twelfth ; the fourth the double octave, etc.
410. Kow find a mean between two terms of an harmonic progression?
411. Whai do the reciprocals of an harmonical prog; ession fonL .
INFIl^ITE SERIES,
INFINITE SERIES.
413. An Infinite Series is one in which the successive
terms are formed by some regular law, and the numher of
terms is unlimited.
414. A Converging Series is one the sura of whose
terms, however great the number, cannot numerically
exceed definite quantity.
415. A Diverging Series is one the sum of whose
terms is numerically greater than 2ii\y finite quantity.
416. To Expand a Fraction into an Infinite Series,
Remark. — Any common fraction whose exact value cannot be
expressed by decimals, may be expanded into an infinite series.
1. Expand the fraction J into an infinite series.
Solution. 1-5-3 = .333333, and so on, to infinity.
Or, I ^3 = ^3^ + ^ + ^^ + ^^, etc. Hence, the
EuLE. — Divide the numerator hy the denominator*
2. Reduce to an infinite series.
1 —X
I— aj)i (i + aj+a;2 + a^+a^, etc., the quotient. (Art. 1 70. )
+x
+ic*, etc.
Therefore, = i+x+x'^ + q? + v^-{-q?, etc., to infinity.
413. What is an infinite series? 414. A converging series? 415. Diverging?
416. How expand a fraction into an infinite fcrie??
INFINITE SERIES. 227
Let x = ^; then will = = 2 ; and tlie series will be
^ ' i — xi—i
I + i+i + i + iV + aV* ®tc., tlie sum of which = 2.
If a? = 4, then will = = |, and the series will become
** I — a; I — i ^
i+i + i+^T+^T + ^5. etc. = |.
Notes. — i. If a? is less than i, the series will be convergent .
For, when x is Icm than i, the remainder must continually decrease ;
therefore, the further the division is carried, the less will be the
quantity to be added to the last term of the quotient in order to
express the exact value of the fraction.
2. If X is greater than i, the series will be divergent.
For, when x is greater than i, the remainder must constantly
increase ; therefore, the farther the division is csLrried, the greater will
be the quantity either positive or negative to be added to the quotient.
3. Eeduce the fraction to an infinite series.
Solution. i-7-(i+a;) = i—x+^—a^+x*—x^ + , etc.
This series is the same as that in Ex. 2, except the odd powers
of X are negatice.
Let a? = 2- ; then will = f ; which is equal to the series
I -i + 1 -i + iV- aV + , etc.
4. Reduce the fraction to an infinite series.
I —X
Ans. I + 20; + 2x^ -f 2x^ -f- 2X^, etc.
417. A fraction whose denominator has more than two
terms, may also be expanded into an infinite series.
5. Expand 5 into an infinite series.
^ I — X -\- x^
I— a?+«^) I {i+x—Q^—x^+x^, etc., Ans,
x—x^
x—x^ + a?
—a?, etc
228 INFINITE SERIES.
418. To Expand a Compound Surd into an Infinite Series,
6. Keduce v i + x to an infinite series.
OPERATION.
2+-| +x
a?
+ « + —
4
X^ I 25"^
4 8 "^64
2 + a? + -Z +"5 — z~» ®tc. Hence, the
4 lo I o 04
EuLE. — Extract the square root of the given sutfl
(Art. 298.)
7. Expand ^/^^2 Ans.x-t^^^^, etc.
8. Expand \/2, or Vi + i. ^ws. i + -f — ^ + -jV? ^^c*
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 q> fraction.
9. Expand {x 4- yY into an infinite series.
Solution. — The terms without coefficients are
«*» aj~5y, a;~«y% a;~»^, a;~»2^, etc.
1 X —4
The coefficient of the second term is + ^ ; of the 3d, = — J ;
— 1 X — J
of the 4th term, — = + ^\, etc.
The series is «* + ^x'^y — lx~^y^ + ^^x~^y^» etc.
418. How expand a surd into an infinite eeriei f
INFINITE SERIES. 229
420. When the index of the required power of a binomial
is a posiiivG integer, the series will termiriate. 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 negaiive,
the series will never terminate. For, by the successive
subtractions of a unit from the index, it will never become
o ; and the series may be continued indefinitely.
10. Expand {(^+1/)^ into an infinite series, keeping the
factors of the coefiicients distinct.
^ ^ 2X 2.^ 2.4.6aj* 2.4.6.8a;'
It. Expand VJ, or (i + i)^, keeping the factors of th€>
coefficients distinct.
Ans. i+i— i-+_3 3.-J 3- 5- 7 t,.
2 2.4 2.4.6 2.4.6.8 2.4.6.8.10
422. An Infinite Series must not be confounded with an
Ivfinite Quantity,
423. An Infinite Quantity is a quantity so great
that nothing can be added to it.
424. An Infinite Series is a series m 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 + l +i + TV+"s'*> ®*^'» ^^ which each succeeding
term is half the preceding, is continued to infinity, the sum of all the
terms cannot exceed a unit.
426. When one quantity continually approximates
another without reaching it, the latter is called the Litnit
of the former.
330 INFIiflTE SERIES.
427. An Infinitesimal is a quantity whose value is
less than any assignable quantity/.
428. The Sign of Infinity^ or of an infinite
quantity, is a character resembling an horizontal figure
eight ( CO ).
The Sign of an Infinitesimal is zero ( o ).
429. One infinite series may be greater or less than
another.
Thus, the series i + i + i + ^ + A* etc., whose limit is 2, is greater
than the series i + i + g + t^ + sV* 6*^., 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 -f 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 hali the dividend.
433. If a. finite quantity is multiplied by an infinitesimal,
the product will be an infinitesimal. For, with a given
multiplicand, the less the multiplier, the less will be tlie
product. Thus, xxo =z o.
434. If a fi7iite quantity is divided by an infinitesimal^
the quotient will be infinite. Thus, ic -^ o = 00 .
If a finite quantity is divided by an infinite quantity, the
quotient will be an infinitesimal. Thus, a: -^ 00 = o.
If an infinitesimal is divided by a finite quantity, the
quotient is an infinitesimal. Thus, o -j- 2; = o.
Note.— In higher mathematics, the expression o -f- o admits of
various interpretations.
IKFIl^ITE SEEIES. 231
435. To Find the Sum of a Converging Infinite Series, the
First Term and Ratio being given.
By the second formula in geometrical progression, we liave for an
increasing series (Art. 402),
Ir — a ar^ — a
8 — , or •
r—i r— I
In a decreasing series, tlie ratio r is less tlian i ; therefore. I or a/"»-i
is less than a. (Art. 398.)
Tliat both terms of the fraction , or may be positive,
T — I r — I
we change the signs of both (Art. 166), and
a — Ir
8 = •
i — r
But, in a decreasing infinite series, I becomes an infinitesimal, or o;
therefore, Ir = o. (Art. 427.) Hence, rejecting the iuHnitesimal from
f = , we have the
Formula. s = •
I — 1*
EULE. — Divide the first term ly i minus tlie ratio,
1. Find the sum of the infinite series
I + i + i + ^7 + A^ etc.
2. Required the sum of the infinite series
I — } + i — i+, etc.
3. Find the sum of the series \ + J + J +? etc,
4. Find the sum of the infinite series i + I + I +* eta
5. Find the sum of the series f + f + 27 +> etc.
6. Find the sum of the series 3 + 2 + | -f , etc.
7. Find the sum of the series 4 + ¥ + f| +> etc.
8. Find the sum of the series .ZZZZ^ etc.
9. Find the sum of the series .ddddd, etc.
10. Find the sum of the series - H -^■\ 5 + etc.
11. 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
what space would it move ?
OHAPTEE XIX.
LOGARITHMS*
436. The LogarithTn of a number is the exponent ol
the power to which a given fixed number must be raised to
produce that number.
437. This Fixed Numler is called the ^ase of the
system.
Tlius, if 3 is 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 roots of i are i, it is obvious that other
numbers cannot be represented by its powers or roots. (Art. 289.)
439. There are tv/o 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. Wliat are logarithms ? 437. What is this fixed number called ? 439. Name
the systems in use. The base of each.
* The term logarithm is derived from two Greek words, meaning
the relation of nwrnhers.
f So called fi-om Baron Napier, of Scotland, who invented log-
arithms in 1614.
X The common system was invented by Henry Briggs, an English
mathematician, in 1624.
LOGARITHMS. 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 i (Art. 259) ;
** *' 10 is I ; for 10^ " 10 ;
** *• icx) is 2 ; for lo^ " 100 ;
** " 1000 is 3 ; for lo' " 1000, etc Hence,
The logarithm of any number between i and 10 is a
fraction; for any number between 10 and 100, 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.
Thus (Art. 256), the log. of .1 is —i ; for 10-^ equals .1 ;
" '* ,01 is —2 ; for 10-2 " .01;
*' " .001 is —3; for 10-^ ** .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 o.oi and o.ooi, the logarithm is — 3 plus a fraction,
and so on.
Hence, the logarithms of all numbers greater than 10 or
less than i, and not exact powers of 10, are composed of
two parts, an mteger and o. fraction.
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 ^^rt, the llantissa,
443. The Characteristic of the 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 is the logarithm of any number from i to 10? From 10 to 100? From
100 to 1000? 442. What is Ihe integral part of a logarithm called? The decirna.]
part ? 443. What is the characteristic 01 the logarithm of a whole numlier ?
234 LOGARITHMS.
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 ^^ or .i is — i ; that of
3^^o' or .01, is — 2 ; that of ^i^^, or .001, is —3, etc. (Art. 256.)
The logarithm of .2 is — i with a decimal added to it ; that of 05
!s — 2 with a decimal added to it, etc.
Note. — It should be observed that the characteristic only is negative,
while the mantissa, or decimal part, is always podtive. To indicate
this, the sign — is placed over the characteristic, instead of before it.
Thus, the logarithm of .2 is i. 30103,
" " *' .05 is 2.69897, eta
445. The Decimal Part of the logarithm of any
number is the same as the logarithm of the number
multiplied 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 Loffaritlmis 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 0 is often the same
for several successive numbers, but is printed only once, and is
understood to belong 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 loj]jarithm of a decimal ? 445. What is the
effect upon the decimal part of the lo<?. of a number, if the number is multiplied or
divided by 10, 100, ioop, etc. <^6, What is a table of logarithms f
LOGAKITHMS. 235
448. To Find the Logarithm of any Number from I to 10.
Rule. — Look for the given numher m the first line of the
table ; its logarithrn will he found directly below it.
1. Find the logarithm of 7. Ans. 0.84510.
2. Find the logarithm of 9. A7is, 0.95424.
449. To Find the Logarithm of any Number from 10
to 1000, inclusive.
Rule. — Looh in the column marhed N" for the first two
figures of the given numher , and for the third at the head
of one of the other columns.
Under this third fiqure* and opposite the first tioo, will
he 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 diflference 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. 2.2 ^^^\.
5. Find the logarithm of 1999. Ans. 3.30085.
450. To Find the Logarithm of a Decimal Fraction.
Rule. — Take out the logarithm of a ivhole number consist-
ing 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 all 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 ? ^^^^- 3'375o6.
9. What is the log. of 16.45 ? Ans. 1.21616.
10. What is the log. of 185.3 ? Ans. 2.26787.
448. How find the logarithm of a numher from i to 10? 449. From 10 to 1000 ?
450. How find the log. of a decimal ? Nots. Of fta integer and a decime^l ?
236 LOGARITHMS.
451. To Find the Number belonging to a given Logarithm.
Rule. — Look for the decimal figures of the given logarilhin
in the table under the column marked 0 ; a7id if all of them
are 7iot found in that column, look in the other colu?nns on
the right till you fi7id them exactly, or very nearly ; directly
opposite, in the column marked N, 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 number require L
Note. — If the characteristic of a l(9garithin is negative^ the number
belonging to it is 2, fraction, and as many ciphers must be prefixed to
the number found in the table, as there are units in the characteristic
le%s 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 annex 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 sig-
nificant figure or a cipher must be put in the quotient.
11. What number belongs to 2. 1 7231 ? Ajis. 148.7,
12. What number belongs to 1.25 261 ? A71S. 17.89.
13. What number belongs to 3.27715 ? Jns, 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 ? A?is. 0.1396.
17. What number belongs to 2.29136 ? Ans. 0.01956.
18. What number belongs to 3.30928 ? Ans. 0.002038.
451. How find the number belonging to a logarithm?
LOGARITHMS. 23?
453. Computations by logarithms are based upon the
following principles :
1°. TJie sum of the logarithms of two numbers is equal to
the logarithm of their product.
Let a and c denote any two numbers, m and n their logarithms,
and b the base.
Then b^ = a
And ft» = (J
Multiplying, 6*+» = og,
2°. TJie logarithm of the dividend diminished hy the
logarithm of the divisor is equal to the logarithm of the
quotient of the two nn7nlers.
Let a and c denote any two numbers, m and n their logarithms,
and b the base.
Then
fe"* = a
And
&• = <?
Dividing,
6«-« = a-i-c
454. To Multiply by Logarithms.
1. Required the product of 35 by 23.
Solution. — The log. of 35 = 1.54407
« «' » 23 = 1.36173
Adding, 2.90580. (Art. 453, Prin. i.)
The number belonging is 805, Ans. Hence, the
Rule — Add the logarithms of the factors; the sum ivill
he the logarithm of the product.
Notes. — i. If the sum of tlie 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 positue,
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
multiply by logarithms f
238 LOGARITHMS.
455. To Divide by Logarithms.
5. Kequired the quotient of 120 by 15.
Solution.— The log. of 120 = 2.07918
«« 4i « jg _ 1. 1 7609
" ** '* quotient = 0.90309. Ans. 8. Hence, the
Rule.— ^rom the logarithm of the dividend subtract the
logarithm of the divisor ; the difference will he 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 characteristic 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 of 1.381 by .096 ?
456. Negative quantities are divided in the same manner
as positive quantities.
If the sign of the divisor is the same as that of tlie
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.
Multiplication 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
Rule. — Multiply the logarithm of the number by the
exponent of the required power,
455. How divide by them ? 457 How involve a number by logarithms ?
LOGARITHMS. 23b
Notes. — i. This rule depends upon the principle that logarithms
are the exponents of powers and roots, and a power or root is involved
by multiplying its index into the index of the power required.
2. In this rule, whatever is carried from the mantissa to the
characteristic is positive, 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 o.28b53. Ans. 1.93435.
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 Hoot of a Number by Logarithms.
A quantity is resolved into any number of equal factors, by dividing
its index into as many equal parts. (Art. 281.) Hence, the
Rule. — Divide the logarithm of the number ly the index
oj the required root.
Note. — This 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 ) .22789
Logarithm of root, 0.11394. Ana. 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 ?
:?!. What is the tenth root of 1876 ?
459. If the characteristic of the logarithm is negative,
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.
4s8 ■^Tow extract the root ?
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 + 1.21484
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 Compound Interest by Logarithms.
Rule. — Find the amount of i dollar for i year ; multiply
its logarithm ly the numler of years, 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 1 15 65 for 40 years, at 6 per
cent compound interest ?
Solution.— The amt. of |i for i year is $1.06 ; its log., 0.02531
The number of years, 40
Product, 1. 01 240
The given principal, $1565 ; its log., 3-19453
Ans. I16103.78. 4.20693
27. Wliat is the amount of $1500, at 7 per cent compound
interest, for 4 years ? Ans. $1966.05.
28. What is the amount of I370, at 5 per cent compound
interest for 33 years ? ^/? 5. $1851.274-.
460. How calculate compound interest by logarithms ?
LOGARITHMS.
241
TABLE OF COMMON LOGARITHMS.
N.
0
1
2
3
4
5
6
7
8
9
D
.00000
.3oio3
.47712
.60206
.69897
.77815
.84510
.90309
.95424
10
0432
0860
1284
1703
2119
253 1
2938
3342
3743
4.6
4i39
4532
4922
8636
53o8
569.
6070
6446
6819
7.88
7555
379
12
7918
8279
ini
9342
&
♦037
♦38o
♦721
1059
349
i3
.11394
1727
2057
2711
3354
3672
6732
3988
43o2
322
14
46i3
4022
7898
5229
5534
5836
6137
6435
7026
7319
3oi
i5
7609
8184
8469
8752
9033
93i3
9590
9866
♦ 140
281
i6
.20412
o683
^ll
1219
38o5
1484
1748
43o4
2011
2272
253 1
lin
264
\l
3045
33oo
4o55
455 1
4797
7184
5o42
249
7875
5768
6007
6245
6482
67.7
6951
7416
7646
235
19
8io3 1 833o
8556
8780
9003
9226
9447
9667
9885
223
20
.3oio3
0320
o535
0750
0963
1175
1 387
3445
1597
1806
2oi5
212
21
2222
2428
2634
2838
3041
3244
3646
3846
4044
203
22
4242
4439
4635
483 1
5o25
5218
541 1
56o3
5704
7658
5984
7840
\tl
23
6173
636 1
6549
6736
856i
6922
7107
8917
7291
7475
24
8021
8202
8382
!IS
9094
9270
9445
9620
178
25
9794
9967
♦ 140
♦3l2
♦654
♦824
IX
1162
i33o
171
i65
26
.41497
1664
i83o
1996
2160
2325
2488
2814
2975
456o
27
3i36
3297
3457
36i6
3775
3933
tS]
4248
44o5
1 58
28
4716
4871
5o25
5179
5332
5485
5788
5939
6090
1 53
29
6240
6389
6538
6687
6835
6982
7129
7276
7422
7567
'47
3o
.47712
7857
8001
8144
8287
843o
8572
8714
8855
8996
♦379
143
3i
9i36
9276
9416
9554
io55
9831
9969
l322
♦ 106
♦243
i38
32
.5o5i5
o65i
0786
0920
1 188
1455
1 587
1720
1 33
33
i85i
1983
2114
2244
2375
3656
25o5
2634
2763
2892
3020
i3o
34
3i48
3275
34o3
3529
4778
3782
3908
4o33
4i58
4283
126
35
4407
453 1
4654
4900
5o23
5i45
5267
5388
6703
123
36
563o
5751
5871
5991
6110
6229
6348
6467
6585
116
u
6820
6937
7054
7171
7287
8433
74o3
7519
7634
iin
7864
8995
7978
8093
8206
8320
8546
8659
8771
n3
39
9107
9218
9329
9439
9550
9660
9770
9879
9988
♦097
no
40
.60206
o3i4
0423
o53i
o638
Vstt
o853
0959
1066
1172
108
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
3448
3548
3649
3749
3849
3949
4048
4147 4247
100
44
4345
4444
4542
4640
4738
4836
5o3i
5128
5225
98
45
5321
5418
55i4
56io
5706
58oi
5992
6087
6I8I
95
46
6276
6370
6464
6558
6652
6745
6839
6932
7025
7II7
47
7210
73o2
7394
7486
7578
7669
7761
7852
7943
8o34
1
48
8124
8215
83o5
8395
8485
8574
8664
8753
8842
8931
9810
49
9020
9108
9197
9285
9373
9461
9548
9636
9723
88
N.
0
1
2
3
4
5
6
7
8
9
D.
u^
LOGARITHMS.
N.
0
1
2
3
4
5
6
7
8
9
B.
5o
.69897
9984
0842
♦070
♦i57
♦243
♦329
♦41 5
♦5oi
♦586
♦672
86
5i
.70757
0927
1012
I933
1181
1265
1 349
1433
i5i7
85
52
1600
1684
1767
i85o
2016
2099
2181
2263
2346
83
53
2428
25l0
2591
2673
2754
2835
2917
2997
3078
3i59
8!
54
3289
3320
3400
3480
3560
1 3640
IVot
3799
4586
3878
3957
80
55
■74036
4ii5
4194
4273
435i
4429
52o5
4663
4741
78
56
4819
4896
4974
5o5i
5i28
5282
5358
5435
55ii
'll
^7
5588
5664
5740
58i5
5891
1 5967
6042
6118
6193
6938
7670
6268
58
6343
6418
6492
',fj,
6641
6716
6790
6864
7012
75
59
7085
7159
7232
7379
7452
7525
7597
7743
73
6o
.77815
7887
7960
8o32
8104
8176
8247
83i9
8390
8462
72
6i
8533
8604
8675
8746
8817
8888
8958
9029
9099
9169
9865
62
9239
9309
9379
9449
9519
9588
9657
9727
9796
63
9934
♦oo3
♦072
♦ 140
♦209
♦277
0956
1624
♦346
♦414
♦482
♦55o
64
.80618
0686
0754
0821
0889
1023
n?7
n58
1225
ll
65
I2gi
i358
1425
1491
2l5l
1 558
1690
1823
1889
2543
66
1954
2020
2086
2866
2282
2347
24i3
2478
65
tl
2608
2672
2737
2802
2930
2995
3o59
3i23
3187
64
325i
33i5
3378
3442
35o6
3569
3632
3696
3759
3822
63
69
3885
3948
4011
4073
4i36
4199
4261
4323
4386
4448
63
70
.84510
4572
4634
4696
4757
4819
4881
4942
5oo3
5o65
62
71
5i26
5.87
5248
5309
5370
543 1
5491
5552
56i2
5673
61
72
5733
5794
5854
5914
5974
6o34
6094
6i53
6213
6273
60
73
6332
6392
645i
65io
6570
7157
6629
6688
6747
6806
6864
59
74
6923
6982
7040
]6T
8253
7216
7274
7332
7390
7448
5o
75
7306
8081
7564
7622
8196
7737
83o9
ml
7852
8423
7910
Itl
8024
58
76
8139
8705
8480
8503
9154
57
77
8649
8762
88r8
8874
8930
8986
9042
9098
9653
56
78
9210
9265
9321
9376
9432
9487
9542
9598
9708
55
79
9763
9818
9873
9927
9982
♦037
♦091
♦ 146
♦200
♦255
55
80
.90309
o363
0417
0472
o526
o58o
o634
0687
074;
0795
54
81
0849
0902
0956
1009
1062
1116
1 698
1222
1275
i328
54
82
i38i
1434
1487
1 540
1593
1645
1751
i8o3
1 856
52
83
1908
ig6o
2012
2o65
2117
2169
2221
2273
2324
2376
52
84
2428
2480
253 1
2583
2634
2686
2737
2788
2840
2891
52
85
2942
2993
3o44
3095
3i46
3.97
3247
3298
3349
3399
5i
86
345o
3doo
355i
36oi
365i
3702
375i
38o2
3852
3902
5i
ll
3952
4002
4o52
41 01
4i5i
4201
425o
43oo
435o
4399
5o
4448
4498
4547
4596
4645
4694
4743
4792
4841
4890
8
89
4939
4988
5o37
5o85
5i34
5i82
523 1
5279
5328
5376
90
.95424
5473
5521
5569
56i7
6095
5665
57.3
5761
5809
5856
48
91
5904
5952
6000
6047
6142
6190
6237
6284
6332
47
92
6379
6426
6473
6520
6567
6614
6661
6708
6755
6802
tl
93
6848
6895
7359
6942
6988
7035
7081
7128
7'74
7220
7267
94
73.3
74o5
745i
'411
7543
7589
^',
7681
8137
8588
7727
8182
46
95
7772
8227
78i8
8272
7864
83i8
ITe^
8000
8046
45
96
8408
8453
8498
8543
8632
45
97
8677
8722
8767
881 1
8856
8901
9344
8945
9388
8990
9034
9078
45
98
9123
9167
9211 9255 1
9300
9432
9476
9520
44
99
9564
9607
965 1
9695
9739
4
9782
9826
9870
9913
9957
43
K.
0
1
2
3
5
6
7
8
9
D.
CHAPTER XX.
MATHEMATICAL INDUCTION AND
BUSINESS FORMULAS.
461. 3Iathematical 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 true 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 numlers is the same, in
whatever order the factors are taken.
To prove this principle of two numbers, as 5 and 3, the pupil
represents the number 5 by as many unit marks in a
41, 4P 4t, 4C; ^
horizontal row, and under this places two similar rows.
# # # # #
He sees that the number of marks in the horizontal
# # # # #
row 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 is
equal to the difference of their squares. (Art. 103.)
Taking two quantities, as 4 + 3 and 4—3, or a + & and a—T).
Multiply 4 + 3 Or a + 6
By 4 — 3 By <g — &
42 + 4 X 3 a^ + ah
— 4x3 — 3* — ab — W
Product, 4^ ^~3^ cf~ -¥
461. In what does Mathematical induction consist ? Illustration.
244 MATHEMATICAL INDUCTION^.
He takes another example of two quantities, and finds the
statement true ; then another, and so on. Hence, he concludes the
proposition is a universal truth.
465. Suppose this proposition were enunciated :
Tlie sum of any numher of terms of the arithmetical series
I, 3, 5, 7, etc, to n terms^ 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, 1 + 3 + 5 = 9> or 3^ ; that
the sum of four terms, i + 3 + 5 + 7 = i6, or 4-, and so on. Hence, we
may conclude that the proposition is true if the series be extended
indefinitely. Or,
Since we know the proposition is true when n denotes a small num-
ber of teims, and that the value of any term in the series, as the 5th,
7th, 9th, etc., is equal to 211—1, we may suppose for this value of n,
that 1 + 3 + 5 + 7 +(2/i— I) = w^. (i)
Adding 2n + 1 to both members, we have
1 + 3 + 5 + 7 •••• +(2ri— i) + (2?i + i) = 7i2 + 2?i + i. (2)
Factoring second member of (2), 7)? + 2n+i = {n+ if.
Therefore, since the sum of n terms of the series = 71^, it follows
that the sum of w + i terms = {n+ if, and so on. Hence, the prop-
osition must be universalis/ true.
466. In Geometry, we have the proposition :
The sum of the three angles of a triangle is equal to two
right ajigles.
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 this method of f>roof\B 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 induction 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 sure that these laws hold for any cases except those which
have been examined, and can never arrive at the conclusion that they
are necessary truths.
BUSINESS FOEMULAS. 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 blackboard, explaining
the several steps by which it is derived as he proceeds. He should
then translate the formula from algebraic into common langv/ige.
PROFIT AND LOSS.
468. Profit and Loss are computed by the principles
of Fercenlage.
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 hundredths, r per cent of a number must be r
hundredths of that number. (Art. 237.) Therefore,
c dollars x r = p, the sum gained. Hence, the
Formula. p = cr.
Rule. — MiiUipI^ the cost hy the rate per cent, and the
product will he the jjrqfit 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 lese
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 gainea or lost ; and
r the per cent.
246 BUSINESS FOKMTJLAS.
Since the cost multiplied by tlic rate gives ^, the given profit or
loss, it follows thatp-f-c must be the rate. (Art. 469.) Hence, the
Formula. r = —*
c
Rule. — Divide the gain or loss hy the cost, and the quo-
tient will he the per cent profit or loss,
3. A farm costing $2500 was sold for $500 advance.
Required the per cent profit. Ans, 20 per cent.
4. A teacher's salary being $i8oo a year, was raised $300.
What per cent was the increase ?
471. To Find the Cost, the Profit or Loss and the per cent
Profit or Loss being given.
Let p denote 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, will be the cost. Hence, the
Formula. c = — •
r
Rule. — 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 ? Ans. $269.92.
6. An operator in stocks lost $1575, which was 12I per
cent of his investment. What was the investment?
472. To Find the Selling Price, the Cost and per cent
Profit or Loss being given.
Let c denote the cost, r the per cent of profit or loss, and s the
selling price.
When there is a gain, the amount of %i = i+r; when there is a
loss, the amount of $1 = i— r ; and the amount of c dollars = c{i± r).
Therefore, we have the general
Formula. s = c{i ±r).
Rule. — Multiply the cost hy 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 12 J per cent. What did he receive for it ?
BUSINESS FORMULAS. 247
473. To Find the Cost, the Selling Price and the per cent
Profit or Loss being given.
Let 8 denote the selling price, r the per cent profit or loss, 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 ^o««, 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 bj
I ± r. Hence, we have the general
Formula. c = — ; — •
EuLE. — Divide the selling price hy i plus or minus the
rate, as the case way require ; the quotient will he the cost.
9. A goldsmith sold a watch for $175 and made 20 per
cent profit. What was the cost ?
Solution. 1 + 20 per cent = 1.20 ; and
$175 -r- 1.20 = $i45.83i» ^ns,
10. A jockey sold a horse for $540, and thereby lost
10 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, interest is derived from the principal only ; by
the latter, it is derived both from the principal and the
interest itself, as soon as it becomes due.
476. To Find the Time in which any Sum at Simple Interest
win Double itself, at any given Rate Per Cent.
Let p denote the principal, r the rate per cent, i the given interest,
and t the ti jt e in years.
Then i = prt, (Art. 242.)
Making i equal to j?, p = prt.
Dividing by pr, - ^ t. Hence, the
248 BUSINESS FORMULAS.
Formula. t — —.
r
'Rule,— Divide i by the rate, and the q^lotient will he the
time required.
11. How long will it take $1500, at 5 per cent, to double
itself ?
Solution. t = - = — = 20. Ans. 20 years.
r .05 "^
12. How long will it take $680, at 6 per cent, to double
itself?
13. How long will it take I8475, at 10 per cent, to double
itself ?
477. To Find the Rate at which any Principal, at Simple
Interest, will Double itself in a Given time.
By the preceding formula, t = -»
T
T
Multiplying by - , we have the
t
Formula. !• = -.
Rule. — Divide i hy the time ; the quotient will he the rate
per cent required.
[For other formulas in Simple Interest, see Arts. 242-246.]
14. If $1700 doubles itself in 8 years, what is the rate?
Ans. 12J per cent
15. At what rate per cent will $5000 double itself in
40 years ?
COMPOUND INTEREST.
478. Interest may be compounded annually, semi-
annually, quarterly, etc. It is understood to be com-
pounded annually, unless otherwise mentioned.
479. To Find the Atnoiitit 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, ana
a the amount.
BUSINESS FORMULAS. 249
Since the amount equals the principal plus the interest, it follows
that the amount of $i fori 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+r) for i year =
p{i+r){i+r) = p{i+ rf, which is the amount of p dollars for two
years.
In like manner, /) {\-\-rf is the amount of p dollars for three years
i,nd 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 terms. The terms of the series are
p, p(i + r), i)(i + r)2, p(n.7.)8^ p(i + r)*, . . . . i7(i+r>».
The last term, p (i + Tf» is the amount of p dollars for n years.
Hence; the
Formula. a = p (i + ry.
Rule. — Multiply the principal by the amount of %i for
I year, raised to a power denoted by the number of years ;
the product will he the amount,
480. To Find the Compound Interest for the given Time
and Rate.
SuUract the principal from, the amount, and the remainder
will he 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, Am.
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
^\ 2n
'{
•+-,)
'('*f-
250 BUSINESS FORMULAS.
482. If the interest is compounded qiiarte7'ly, then -
4
will denote the interest of $i for a quarter. Then, at
compound interest, the amount of jj dollars for n years is
471
etc.
4/
1 8. What is the amount of $2000, for 3 years at 6 per
fent, compounded semi-annually ? Ans, I2388.05.
19. What is the amount of I5000 for 2 years, at 4 per ct.,
compounded quarterly ?
483. By transposing, factoring, etc., the formula in
A.rt. 479, we have.
The first term, « = -, •
The number of terms, n = -^ — ;- — ^~ •
log. (I + 7-)
The ratio, r = l-r — l.
DISCOUNT.
484. Discount is an allowance made for the payment
of money before it is due.
485. The JPresent 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 Worth of z Sum at Simple Interest,
the Time of Payment and the Rate being given.
Let 8 denote the sum due, n the number of years, and r the interest
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 +7?r the amount of $1 for n years. Therefore, s-5-(i + nr)
is the present worth of the given sum.
Putting P for present worth, we have the
Formula. B = — ; •
I 4- fir
Rule. — Divide the sum due hy the amount of $1 for the
given time and rate; the quotient loill he the present worth.
BUSINESS FORMULAS. 251
487. To Find the Discount, the Present Worth being given.
SuUract the present worth from the debt.
20. What is the present worth of $2500 payable in 4 yrs.,
interest being 7 per cent ? What the discount ?
Solution. P = — ?— = 1^ = $1953.125, pres. worth ,, ,
i + nr 1.28 'a' V3J D, I' M ^^^^
$2500 — $1953.125 = $546,875, discount.
1
21. What is the present worth of $3600 due in 5 years, at
6 per cent ? WTiat is the discount ?
22. Find the present worth of $7800 due in 6 years,
interest 5 per cent ? What the discount ?
COMPOUND DISCOUNT.
488. To Find the Present Worth of a Sum at Compound
Interest, the Time and the Rate being given.
Let s 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 (n-r)«. (Art. 479.) That is,
$1 is tlie present worth of (1+ r)« due in n years. Therefore, «-t-(i + r)n
must be the present worth of the given sum.
Putting F for the present worth, we have the
FOKMULA. JP = V
(I + rf
Rule. — Divide the sum due hy the amount of li at com-
potmd interest, for the given time and rate; the quotient
will he the present worth.
23. What is the present worth of $1000 due in 4 years, at
5 per cent compound interest ?
Solution, P = — ?-- = f^, = $822.71, Ans.
(i+r)" (i.o5)'»
24. What is the present worth of $2300 due in 5 years, at
6 per cent compound interest ?
252 BUSINESS FORMULAS.
COMMERCIAL DISCOUNT.
489. Commercial Discount is a per cent taken
from the face of bills, the marked price of goods, etc.,
without regard to time.
490. To Find the Cofmnercial Discount on a Bill of Goods,
the Face of the Bill and the Per Cent Discoant being given.
Let 6 denote the base or face of the bill, r the rate, and d the
discount or percentage.
Then bxr will be the discount. Hence, the '
Formula. d = br.
Rule. — Multiply the face of the Mil hy the given rate, and
the product will be the cornmercial discount,
491. To Find the Cash Value or Net Proceeds of a Bill.
Subtract the commercial discount from the face of the hill.
25. Required 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=1840, Ans,
26. Required the cash value of a bill amounting to I2500,
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 c denote the cost, r the per cent profit, and d the per cent disc.
Since ris the per cent profit, 1 +r is the selling price of $1 cost, and
c{i+t) the selling price of c dollars cost.
Again, since d is the per cent discount from the marked price, ano
the marked price is 100 per cent of itself, i—d must be the net value
of $1 marked price. Therefore,
c(i +7") -5- (i— <?) = the marked price.
Putting m for the marked price, we have the
Formula. m = -- — -—•
I — d
Rule. — Multiply the cost hy i plus the per cent gain, a7id
divide the product hy i minus the proposed discount.
BUSINESS FORMULAS. 25,3
27. A trader paid I25 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-d .95
28. A merchant buys a case of silks at I1.75 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 barrel; what price
must he mark it that he may fall 3 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 sliare of stock, p \is par
value^ 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 sliare for i year. Therefore, pr -i- c will be the per cent
received on the cost of i share. (Art. 238.)
Putting JR for the per cent received on the cost of i share, we have
the
Formula. M = -—
c
KuLE. — Find the dividend on the given shares at the given
rate, and divide this hy the cost ; the quotient will he the per
cent received on the invostment*
254 BUSINESS FOKMULAS.
Note. — When the stock is above or lelow par, the premium or
discount must be added to or subtracted from its par value to give the
cost.
30. What per cent interest does a man receive on an
investment 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 stock = $5000 x .05 = $2500
Therefore, the cost = $5000 + 1250 = $5250.
Again, the dividend on stock = $5000 x. 10 = $5oa Therefore,
$5oo-f-$525o = 9II 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 1 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 sum remitted, and r the per cent commission.
The sum remitted includes both the sum invested and tlie commis-
sion. Now $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,
« -4- (i+r) must be the amount invested.
Putting a for the amount invested, we have the
FOEMULA. a =
I+r
Rule. — Divide the remittance ly 1 plus the per cent
commission ; the quotient will he the amount to invest.
BUSINESS FOEMULAS. 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 ?
Solution, a = — ^ = ^^^^ = $2403.85, Ans.
i+r 1.04
;^6. A gentleman remitted $25000 to a broker, to be
invested in stocks. After deducting i J per cent, how much
did he invest, and what was his commission ?
SINKING FUNDS.
497. Slnkinf/ 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 A^nouitt of an Annual Deposit at Compound
Interest, the Rate and Time being given*
Let s denote the annual deposit or sum set apart, r the rate per
cent, n the number of years, and a the amount required.
Since the sam^ 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 " =s + «(i+r)
3d " = s + s{i + r) + 8{i+rf
nth. " — 8 -\- s{i+r) + s{i+rf . , . . + 8{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 number
of years, the last term ; and the amount is equal to the sum of the
series. (Art. 402.) Hence, we have the
(i + ^^r— I
FOEMULA. a = S.
r
Rule. — Multiply the amount of $i annual deposit for the
given time and rate hy the given annual deposit ; the product
tvill he the amount required.
256 BUSINESS FORMULAS.
37. A clerk annually deposited $150 in a savings bank
which pays 6 per cent compound interest. What amount
will be due him in 5 years ?
Solution, a = ^ -^— ~ a = ' ^ $150 = $845.75. Ans.
38. A man agrees to give $300 annually to build a church
What will his subscription amount to in 4 years, at 7 pei
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 Ammal Dejyosil required to produce a
given Amount at Compound Interest, the Rate and Time being given.
By the formula in the preceding article, we have
(r + r)« — I
^^ s = a.
r
Dividing by coefficient of .s, we have the
FOKMULA. S=za^ (i + vY —J_
r
Rule. — Divide the amount to he raised hy the amount of
$1 annual deposit for the given time and rate ; the quotient
will he the annual devosit rec
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 giver
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 + ^)« _ I (105)^ — I $250
Solution. » = a -*- L— !— -^ = $5000 -5- i — ^ = — ^ =
r .05 .276
$905.80, An8.
BUSINESS FORMULAS. 257
41, A man having lost his patrimony of $20000, wishes
to know how much must he annually deposited at 10 per
cent, to recover it in 5 years ?
42. A county borrows $30000, 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 Amount of an Unpaid Annuity at Compound
Interest, the Time and Rate per cent being given.
Let a be the annuity, 1 + r tlie amount of $1 for i year, and n the
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-\-rf,
4th " = a + a{i + r) + aii^rf + a{i+rf,
nth. year = a + a{i +r) + a{i+rf + a{i-\rf . . . 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 S = (^ + ^)" — i ^
r
Rule. — Multiply the amount of %i annuity, for the given
time and rate, hy the given annuity.
258 BUSINESS FORMULAS.
Note. — Logarithms may be used 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 = (L±j£_Ill « ^ (L^^ll $550 = $2886, Ans.
r .07
44. An annual pension of |88o 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 Present Worth of an Annuity atCompound
Interest, the Time of Continuance and the Rate being given.
Let P denote the present worth ; then the amount of P in ?i years
will be equal to tlie amount of the annuity for the same 'time.
Therefore^
r
Dividing each member by (i + r)" (Art. 279), we have the
Formula.
Note. — In applying the formula, the negative exponent may be
made positive by transferring the quantity which it affects from the
numerator to the denominator (Art. 279).
1
r r
EuLE. — Multiply the present worth of an annuity of $1
for the given time by the given annuity.
46. What is the present worth of an annuity of $375 for
6 years, at 7 per cent compound interest ?
Solution. P = l^Jll^ « = i-(i.o7H^ = l=Jt ^^^^
r .07 .07
r= $1785.71, Ans.
47. What is the present worth of an annual pension of
$525 for 5 years, at 4 per cent compound interest ?
BUSINESS FORMULAS. 259
CASE III.
503. To Find the Present Worth of a Perpetual Annuity, the
Rate being given.
Let n denote infinity, then reducing the formula in Art. 502, we
have this
Formula. -P = v C^^*- 435-)
Rule. — Divide the annuity hy the interest of %i for
I year, at the given rate,
48. What is the present worth of a perpetual scholarship
that pays $150 annually, at 7 per cent compound interest?
Solution. P = - = i^ = $2142.86, Ans.
r .07
49. "What is the present worth of a perpetual ground rent
of $850 a year, at 6 per cent ?
CASE IV.
504. To Find the Present 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 cottimence, and N the
number of years it is to continue. Then,
P — I — (i + r)-^"+-^) __ I — (i + r)-"
~ r r '
Performing the subtraction indicated, we have the
Formula. J» = - [(i + /»)-« — (i + r)"^^.
Rule. — Find the present worth of the 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 I600, to
commence in 4 years and to continue 12 years, at 7 per
cent interest ?
SoLxrriON. P = ? [(i + r)-« — (i + r)-«--»],
P=?^[(i.o7H-(i.o7)-^«],
p ^ $600 X .4242^^^^^^^^
260 BUSINESS FORMULAS.
51. A father left an annual rent of I2500 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,
p-'-
(I +r)-«^
r
Dividing by the coefficient of a
, we have the
Formula. a
JPr
I — (i + r)-
EtiLE. — 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. Eequired the pension.
Pr $668 X. 06 $668 X. 06 ^ ^ .
Solution, a = — = ^^^-7 — -:—r. = ^ ,_, — ■ = $58.23,47^.
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 formulas or rules, it will he
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 classes of problems.
CHAPTEE XXI.
DISCUSSION OF PROBLEMS.
506. The Discussion of a Problem consists in assign-
ing all tfie 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 — ft) a? = c.
Therefore, x = -.
a — o
508. The result thus obtained may have five different
forms, depending on the relative values of a, h, and 6*. To
represent these forms, let m denote the multiplier.
I. Suppose a is greater than h. In this case a — 6 is positive,
and c being positive, the quotient is positive. (Art. 112.) 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 5. In this case ^ — & is negative^
and c divided by a —6 is negative. (Art. 112.) Hence, the required
multiplier must be negative, and the value of x is of the form of — m.
III. Suppose a is equal to h. In this case a — 5 = o. There
fore, the value of x is of the form of — , or a; = - = 00. (Art. 434.)
506. In what does the discaeeiou of problems consist ? 507. What is an arbitrary
quantity!
262 DISCUSSION OF PROBLEMS.
IV. Suppose c is o, and a is either greater or less than h.
In this case the value of x has the form — , or a; = o.
m
V. Suppose c equals o, and a equals J. In this case the
value of a; has the form -•
o
Note. — The student can easily test these principles by substituting
lumbers for <z, &, and c.
509. The Discussion of 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 ^ miles an hour. At
12 o'clock A was at a certain point P \ and B d miles in
advance of A, in the direction of Q. When and where were
they together ?
0 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 negntive.
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 miles an
hour, we have
mt = the distance A traveled.
And nt- " '* B
Again, since A and B were d miles apart at 12 o'clock,
mt — nt = d.
Factoring and dividing we have the
* Originally proposed by Clairaut, an eminent French mathemati-
cian, born in 1713.
DISCUSSION OF PROBLEMS. 263
Formula. t = •
7n — u
The problem may now be discussed in relation to tlie
time t, and the distance 7nf, the two unknown elements.
I. Suppose m > n.
Upon this supposition the values of t and mt will both be positive;
because their denominator m — n is 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 cZ =: 24 miles, m = S miles, and n = 6 miles.
By the formula, t = = ^ ^ = 12 hours.
m—n 8—6
mt = % y. 12 — ()6 miles A traveled.
71? = 6 X 12 = 72 " B *'
Now, 96 — 72 = 24 m. their distance apart at noon, as given 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\^ 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 in accordance with the conditions of the
problem under the present supposition. For, if w < n, then B was
traveling faster than A ; and as B 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 (Z = 24 miles, m = 5 miles, and ti = 8 miles.
By the formula, t = = — — = — 8 hours.
m—n 5—8
. And mt = 5 X — 8 = — 40 miles A traveled
264 DISCUSSIOIS" OF PKOBLEMS.
These values show that the couriers were together 8 hours 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 =zn.
Upon this supposition we have m — n = o, and
^ d T . md
t = - = CO, also mt = — = 00.
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 mty 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 d = 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 d = 0, then at 12 o'clock B must have been with A at the point
P. And if m is greater than n, 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 d = Of and m = n.
Then we have < = - » and mt = -'
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 d5 = o, the couriers were together at 12 o'clock, and if m = n,
they were traveling at equal rates, and therefore would never part.
IMAGIJS^ARY QUANTITIES. 265
IMAGINARY QUANTITIESo
610. An Imaginary Quantity is an indicated even
root of a ne^a^m quantity ; as, V— i, V— «, ^—7.
Notes. — i. Imaginary quantities are a species of radicals, and are
called imaginary, because tliey denote operations wliicli it is impossi-
ble to perform. (Art. 294.)
2. Though the operations indicated are in themselves impossible ,
these imaginary expressions are often useful in mathematical analyses,
and when subjected to certain modifications, lead to important results.
Sllr 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 Quantity for Multiplication
and Division.
Rule. — Resolve tlie given quantity into two f actor s, one
of which is a real quantity, and the other the imaginary
expression V — 1.
Notes.— 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 ; —W = ¥ x — i.
2. The real factor is often called the coefficient of the imaginary
expression, -y^— i.
I. Multiply V^^ by V—'b,
Solution. \^—a = ^a x -v/— i, and ^y/— 6 = '^b x /y/— 1«
Now y'a X -y/^ X y^ X y^/^ = -y/oS x — i = —y\/ab. Ana,
2. Multiply + V--^ by — V— y.
3. Multiply V— 9 by V— ^.
51a What are Imagimary quantities? 511. How added and subtracted I
51a How prepare them for mxiltiplicatiou aud divisiou 1
1%
266 IMAGIN^ARY QUANTITIES.
513. It will be seen from the preceding examples:
First. That the product of two imaginary quantities is 3
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 ambiguous,
this ambiguity is removed when we know whether the quantity whose
root is to be taken has been produced from positive or negative
quantities. (Art. 293,)
4. Multiply V— 2 by ViS,
5. Multiply a/— ^ by \/y.
Note, — i. From these examples it will be seen that the product of
a reoyl quantity and an imaginary expression, is itself imaginary.
6. Divide V— ^ by V— ?/.
Solution.—'^ 1 = X^-I^v^^ = V -, Ans.
7. Divide V —x by V~—x,
Note. — 2. Hence, the quotient of one imaginary quantity divided by
another, is a real quantity ; and the sign before the radical is the same
as that prescribed by the rule. (Art. 92.)
8. Di^*ide ^/^x by a/.V>
9. Divide Vx by V—y,
Note. — 3. Hence, the quotient of an imaginary quantity divided by
a real one, is itself imaginary, and vice versa.
10. Divide 10 a/— 14 by 2 a/— 7,
1 1. Divide c V'— i by d V — i,
514. The development of the different powers of V — i.
12. (a/^)2 = -I. 15. (V^)5 = +V^.
13. (V^^)^ = -V-i. 16. {V-if - -I.
14. {^y~lY=: +1. 17. {^~~i)l= -V'-^.
Hence, tJie even powers are alternately — i and +1, and the odd
powers — /y/— I and +y'— i.
INDETERMIITATE PROBLEMS, 267
INDETERMINATE PROBLEMS.
515. An Indeterminate Prohlem is one which does
not admit of a definite answer. (Art. 220.)
Note. — Among the more common indeterminate problems, are
I St. 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 quantities to be determined.
4th. Those whose conditions are inconsistent with each other.
1. Given the equation a; 4- y = 9, to find the value of x.
Solution. — Transposing, x = g — y, Ans. This result can be
•verified by assigning any values to x or y.
2. What number is that, J of wliich minus i half of itself
is equal to its 12th part plus its sixth part ?
Let
X =
the number.
Then
3«
4
X _
2 ~~
12^6
Clearing of fractions, etc..
gx =
gx
Transposing and factoring.
(9-
■Ci)X =
0
••
. X =
0
0
IMPOSSIBLE PROBLEMS.
616. An Impossible I^roblemh one, the conditions
of which are contradictory or impossible.
I. Given a; -f y = 10, x — y = 2, and xy = 38.
OPERATION.
Solution. — By combining equations (i) x -^ y = 10 (i)
and (2), we find x = 6 and y = 4. Again, x y = 2 (2)
xxy = 6x4 = 24. But the third condition "
2X :i^ I 2
requires the product of x and y to be 38,
which is impossible. • *• ^ = o
1=^ 4.
515. Wb«t is au in4eterifti»ate problem ? 516. Wsat is an impossible problem ?
268 NEGATIVE SOLUTIONS.
2. What number is that whose 5th part exceeds its 4th
part by 15?
3. Divide 8 into two such parts that their product shall
be 18.
NEGATIVE SOLUTIONS.
517. A Neffative Solution is one whose result i& 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.
(Art. 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 = tlie number of feet,
Then (20 + a;) 30 = area.
By conditions, 600 + 3oaj =510
Transposing, 30a; = — 90
.-. X — — 2) ft., Ans.
Notes. — i. It will be observed that this is a problem in Simple
Equations. The steps in the solution are legitimate and the result
satisfies the conditions of the problem algebraically, but not arith-
metically. Hence, the negatke result indicates some mistake or
inconsistency in the conditions of the problem.
If we subtract 3 ft. from its width, the result will be a positive
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.
517. What is a ueeative eolution?
hor'N-er's method. 269
3. In suc« cases, by changing some of the data, a similar problem
may be easily found whose conditions are consistent with a possible
result.
2. What number must be subtracted from 5 that the
remainder may be 8 ?
Solution.— Let x = the number.
Then 5 - a? = 8
Transposing, a; = — 3, Ans.
3. A man at the time of his marriage was 36 years old
and liis wife 20 years. How many years before he was twice
as old as his wife ? A71S, — 4 years.
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 x in the general
equation,
A^ + Bx^ + Cx = D. (i)
Having found the first figure of the root by trial, let it be denoted
by a, the second figure by 6, the third by c, and so on.
Substituting a for x in equation (i), we have,
Aa^ + Ba? + Ca = B, nearly.
Factoring and dividing,
* So called from the name of its author, an English mathematician,
who communicated it to the Royal Society in 18 19.
270 hoenek's method.
By putting y for tlie sum of all the figures of the root except the
first, we have x = a^y, and substituting this value for x in equation
(i), we have,
Aia^yJ + B^a + yy + C{a+y) = J);
or A {a^ + 3aV + 3«/ +2^) + ^ («^ + 2ay+y'^) + C(a+y) = D.
or Aa^ + sAal^y + sAay^ + ^^^ + Ba'^ + 2Bay + 5y- -^-Ca-^Gy — B.
Factoring and arranging the terms according to the powers of ^.
we obtain
Ay^ + (J5 + 3-4%2 + ((7+ 25a + ^Aa})y = B-{Ga + Ba? + Aa^. (3)
To simplify this equation, let us denote the coefiicient of y'^ by B',
that of ^ by C", and the second member by D' ; then,
Ay^ + B'f + C'2^ = 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 by 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 &, and reducing
as before, we find,
, _ D' .
C' + B'h + AJ^' ^5^
Putting z for the sum of all the remaining figures in the root, we
have y = !)-{■ z; and substituting this value in equation (4), we obtain
a new equation of the same general form, which may be written,
As^ + B"z'' + C"z = D", (63
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 5, v.'hose value is sought. That is, the vahie of
6 is given in terms of h, and that of c would be given in terms of c, and
so on. For this reason, equations such as (5) might appear at first
«ight 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 + AH^ and B"c + Ac^ in the
denominators, will be very small compared with C and G ', conse-
horiter's method. 271
quently as & is very nearly equal to D' divided by C , tliey may be
neglected. Therefore the successive figures in the root may be
approximately found by dividing D' by C, D" by (7', and so on,
regarding C , C , etc., as approximate divisors.
In transfonning equation (i) into (4), the second member D' and
the coefficients G and B' of the transformed equation may be thus
obtained.
Multiplying the first coeflBcient A by a, the first figure of the root,
and adding the product to B, the second coeflBcient, we have,
B + Aa (7)
Again, multiplying this expression by a, and adding the product to
C, the third coeflicient, we have,
C + Ba + AaK (8)
Finally, multiplying these terms by a, and subtracting the product
from B, we have
l)-iCa + Ba? + Aa^) = D',
which is the same as the expression for B' in equation (4).
Now to obtain C, we return to the first coeflScient, multiply it by a,
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 + 2,Aa^ = C\
which is the desired coefficient of y in equation (4).
Finally, to obtain B' , we multiply the first coeflBcient by a, and add
the product to expression (9), and thus obtain,
B + sAa = B'.
In this way the coeflBcients of the first transformed equation are
discovered ; and by a similar process the coeflBcients of the second,
third, and of all subsequent transformed equations may be found.
520. This metliod of approximation is applicable to
equations of every degree. For the solution of cubic equa-
tions, it may be summed up in the following
KuLE. — 1. Detach the coefficients of the given equation,
and denote them hy A; B, (7, and the second member hj D.
Find the first figure of the root iy trial, and represent it by a.
272 HORNER'S METHOD.
Multiply A ly a, and add the product to B. Multiply
the sum hy a and add the product to C. Multiply this sum
by a and subtract the product from D. The remainder is
the first divideyid, or D',
II. Multiply A by a and add the product to the last sum
under B. Multiply this sum hy a and add the product to
the last sum under G, The result thus obtained is the first
divisor, or C.
III. Multiply A by a and add the product to the last sum
U7ider B. The result is the second coefficient, or B'.
IV. Divide the first dividend by the first divisor, TJie
quotient is the second figure of the root, or b.
Y. Proceed in like manner to find the subsequent figures
of the root.
Note. — 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
from ly.
EXAMPLES.
I. Given a^ + 2x^
-{- Z^ = 24,
to find X,
SOLXJTION.
A B
C
D a he
I +2
+ 3
=
24 a? = ( 2 . 0 8, An&
2
4
J
II
22
2 = jy
3
6
12
23 = (7'
1.891712
.108288 = D"
2
.6464
8 = J5'
23.6464
'
.08
.6528
8.08
24.2992 =
C"
..08
•
8.16
.08
9.24 = B"
273
Note. — 2. In the following example, tlie last figures of tlie 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 ic^ + izx^ — i2>x = 216, to find x.
SOLUTION.
A
B
0
D abe
I
+ 12
-18
= 216 (4.24264+.
_4
+64
184
x6
+46
32 = iy
A
Jo
26.168
20
126 = 0"
5.832 = D"
j4
4.84
5.468224
24 =
B
130.84
.363776 = 2)"'
24.2
4.88
275385
24.4
24.6 =
24.64
24.68
= B"
135.7 2 = G'
.9856
T36.7 056
.9872
I37.0|9|2j8 =
88391
82615
5776
5508
X = 4.24264+, Ans.
3. Given a:* + 32:2 + 5a; = 178, ^o find x.
a; = 4.5388, Ans,
4. Given $3? + 92:2 — 7^ __ 2200, to find x,
X = 7.1073536, Ans.
5. Given a^ + a^ + x= 100, to find x,
a? = 4.264429+, Ans.
274 TEST EXAMPLES FOR REVIEW.
TEST EXAMPLES FOR REVIEW.
1. Required the value of
6a + 4^ X 5 + 8«j -^ 2 — 3a + 12a X 4.
2. Required the value of
(8a; + 3^) 5 + 4^ + 7 - (53J + 9^) "^ 7-
3. Required the value of
sax — ab + 4cd — {2ax — 4«5 + 2cd),
4. Required the value of
4bc + [s^d — {2xy — mn) 5 + ^dc],
5. Show that subtracting anegative quantity is equivalent
to adding 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)=^ + — - i7f, to find a;. .
3 9 7
g. Given ~ — i h 2a; = ^^ X ^ to find x,
^ 5 5 32
10. Resolve ^d^c — 6Ir^(^ — c^d into two factors.
11. Resolve $(^y — - gxh — iSx^yz into two factors.
1 2. Resolve a^' — if^' into two factors.
13. Resolve 8a — 4 into prime factors.
14. Resolve a^ — i into prime factors.
15. Divide 31 into 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 FOE REYIEW. 275
1 8. Find two quantities the product of which is a and the
quotient b,
10. Reduce -7 7 to its lowest terms.
^ 00 — 0
20. Reduce -^ 75 to its lowest terms.
c? — W'
21. Resolve 90:^ _|. \2xyz + 42:2 jnto two factors.
22. Resolve ()W- — die + c^ into two factors.
23. Make a formula by which the width of a rectangular
surface may be found, the area and length being given ?
24. A square tract of land contains J as many acres as
there are rods in the fence inclosing it. What is the length
of the fence ?
25. A student walked to the top of Mt. Washington at
the rate of \\ miles an hour, and returned the same day at
the rate of 4^ miles an hour ; the time occupied in traveling
being 13 hours. How far did he walk?
26. Given l ^^— = 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. Reduce 7-7, \ "~ ^ —^^ — r to its lowest terms.
fj^ J4
30. Reduce j-^ r"^— low-o-ri^v *o i*s lowest terms.
31. Reduce ;; ^ to a single fraction havirisj
I — a* I -\- a?' ° *^
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 I40: the former then had twice as
much as the latter. What sum did each have at first ?
276 TEST EXAMPLES FOR REVIEW.
34. A French privateer discovers a ship 24 kilometers
distant, sailing at the rate of 8 kilometers an hour, and
pursues her at the rate of 12 kilometers an hour. How
long will the chase last ?
35. Given '^ ^ = 7 and 7^ — 3^_ ^ ^^^
7 2 *^
9: and y,
36. Given x = ^"7 f 5 and 4^ ^!^^ = 3, to find
a; 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
hour, descended a river whose velocity is 4 miles an hour,
and was gone 8 hours. 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. Required two numbers whose sum is to the less as 8
is to 3, and the difference of whose squares is 49.
41. Given iox-\-6y =. 76, 4^—20 = 8, and 6x-\-Zz=: 88,
to find X, y, and z,
42. Given 20; + 3^ -f 2; = 24, ^x ■\- y •\- 2Z z=z 26, and
a; -f 2y + 32; = 34, to find x, y, and z.
43. Three persons, A, B, and 0, counting their money,
found they had 1 180. B said if his money were taken from
the sum of the other two, the remainder would be |6o;
C said if his were taken from the sum of the other two, the
remainder would be \ 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 FOR 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. Required 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 $1 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 1 175 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 ?
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 ages is 150. What is the age of each ?
58. Reduce the V243 to its simplest form.
59. Reduce Vy^ 4- ay^ to its simplest form.
278 TEST EXAMPLES FOR REVIEW.
60. Reduce x^ and y^ to the common index |^.
61. Reduce ${a — h) to the form of the cube root.
62. A farmer sold 13 bushels of corn at a certain price ;
and afterward 17 bushels at the same rate, when he received
I3.60 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 $7 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 $80 a week. How much does each receive
per week ?
67. What factor can be removed from ViSZX?
68. Given Vx + 12 = \/a + 12, to find x.
69. Given — - = ^ ~- , to find y.
y Vy
70. Given Vz^ — ^ah = a — h, 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 of which shall
be in the ratio of 4 to 9.
TEST EXAMPLES FOR REVIEW. 279
74. After paying out ^ of my money and then J 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 yalue
will be J; and if the denominator be doubled and then
increased by 2, the value of the fraction will be ^. Kequired
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. Kequired the cost of the watch.
77. Two Chinamen receive the same sum for their labor;
but if one had received I15 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 |i 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 63 be subtracted from
the number, its digits will be inverted. What is the
number ?
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 $2800. A pays I5 an
acre less than B, and so gets 10 acres more land. How
many acres does '^ach purchase ?
82. Find a factor that will rationalize ^/x -\- V7.
83. Find a factor that will rationalize V^ — V^.
84. Given ^W- + Vx z= -_^-i— ^ to find x.
V{b^ + Vx)
85. The salaries of a mayor and his clerk amount to
I13200; the former receives 10 times as much as the latter.
Kequired the pay 01 eacn.
S6. What two numbers are those whose sum is to their
280 TEST EXAMPLES ¥0E REVIEW.
difference as 8 to 6, and whose difference is to their product
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 ?
S8. Find the price of apples per dozen, when 2 less for
12 cents raises the price i cent per dozen.
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 aays.
When will they meet ?
90. The income of A and B for one month was 1 187 6,
and B's income was 3 times A's. Required 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 |i 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 -J-, i, if, 2, 2 J,
S, Zh 4, 4i etc.
96. A dealer bought a box of shoes for $100. He sold all
but 5 pair for $135, at a profit of li a pair. How many
pair were there in the box ?
97. Two numbers are to each other as 7 to 9, and the
difference of their squares is 128. Eequired 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 number in its height and length.
99. Bertha is J 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 FOE 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
I3 of gold to $2 of silver ; if $24 in gold be added, there will
then be $7 of gold for every $2 of silver. Eequired the sum
in the purse.
106. A and B in partnership gain $3000. A owns f of
the stock, lacking $200, and gains $1600. Required 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
sach to do the work ?
1 09. If a certain number be divided by the product of its
two digits, the quotient is 2 J ; 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.
III. A trader bought a number of hats for $80 ; if he had
bought 4 more for the same amount, he would have paid |i
less for each : liow many did he buy ?
282 TEST EXAMPLES FOB KEVIEW.
112. 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
unequal 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 11, and the second equals | of the
remainder of the third diminished by 19: required the
numbers.
1 15. Required 3 numbers in geometrical progression, such
that their sum shall be 14 and the sum of their squares 84.
1 1 6. 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 $2450; for a
part of the farm he pays $20 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 a mill-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
sufficient 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 3I kilo-
grams a week for the remainder of the cruise. What was
the original number of men ?
APPENDIX.
621. To Extract the Cube Root of Polynomials*
I. Required the cube root of a^ + 3^5 — 3a* — i la' -4
6^8 4. 12a — 8.
OPERATION.
a^'+Sa"— a^"*— iia^+6a"2 + i2a— 8 ( a*+a— 2, Roots
a^, the first subtrahend,
ist Trial Divisor, 3a'' ) 3a^— 3a**— iia^ etc., first remainder.
Com. D., 2,(1^ + yi^ + a- ) 3^*^ + 3^^+ d^
2d Tr. D., 3a^ + 6«3 4-3a^ ) — 6a*— i2a3 + 6«2 + i2a— 8, 2d remainder.
Complete Divisor,
2,a^ + da^ — 3a'^ — 6a + 4 ) — 6a*— I2a^ + 6a^ + i2g— 8. Hence, the
Rule. — I. Arrange the terms according to the powers of
one of the letters^ take the cube root of the first term for the
first term of the root, and subtract its cube from the given
polynomial.
II. Divide the first term of the remainder hy three times
the square of the first term of the root as a trial divisor, and
the quotient ivill be the next term of the root.
III. 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 tioo terms in the root, for the
second trial divisor, take three times the square of the part
of the root already found, and completing the divisor as
before^ continue the operation until the root of all the terms
is found. ( See Key)
* For Horner's Method of Approximation, see p. 269.
I.
x^ ^ gx + 20.
2.
a^ ■}- ya— 18.
3-
a^ — isa + 40.
4.
2aho^ — i4aJc -
5-
icy _2xy + I
6.
8x^ — 32?/2.
7.
a;2 + yh + ?y?7?2;.
8.
i2«2a; — Sa^y + 4^2;.
9.
a^ — Sci^x 4- 3a!a;2 — a^.
10.
I -a^.
II.
I + Sa\
284 APPENDIX.
2. Required the cube root of a^ + ^a^ + 30^52 + b^.
3. Find the cube root of x^ -\- 6x^ + 122; + 8.
4. Find the cube root oi a^ — 6aPy + izxy^ — 8y^.
5. What is the cube root of Sa^ — 48«r2 _}. gSa — 64.
6. Find the cube root of 27^3 — S4a^x + :^6ax^ — S.'?;^^
7. What is the cube root of a^ — 6a^ -[- 15^* — 20^3 _j.
15^2 _ 6« + I.
8. The cube root of a;^ — 3ic8 ^ Sx^ — 6a^ — 6x^ i- Sa:^-
3^+ I.
522. Factor the following Polynomials:*
6odb,
12. a^ — h^x^.
523. Find the </. c. c?. of the following Polynomials:
1. 4^2 — 4ax — 15^ and 6a^ + 'jax — 3:r2.
2. ^ax^yh^, \2y?^, and i6«3a;3;2;2.
3. i6a;2 — 2/2, and 16:^2 _ <^xy 4- ?/2.
4. 6^2 ■\- wax +3a;2, and 6^2 +7«a; — '^'^^
5. «* — 5^, and a^ — ^2flr3.
6. jr^ — a^, and a;* — «*.
524. Required the h c. m. of the following Polynomials.
1. 6fl2 _ 4fl5, 4058 _^ 2«, and 60^2 _|_ 4^,
2. 4 (i + a^), 8(1— «), 4(1— «^); and 8 (i + «)•
3. c? — la -{• I, «'* — I, and a^ -\- 2a ■\- i.
4. 12 («J2 _ J3)^ 4 (^2 _^ ^j)^ and 18 (a2 _ J2).
5. 4(^2 — I, 2a — I, and 4^2 -|- i.
6. 4 (i 4- a2), 8(1 + a), 4 (i — a% and 8 (i — «).
* The following problems are classified and may be studied in con-
nection with the subjects to which they refer, or be omitted till the
other parts of the book are finished, at the option of the teacher.
APPENDIX. 285
525. Unite the following Fractions i
X — 2 - 3iP — 3
I 2ab
- ^3 - ^ ' ^4 _ ^^
ab be ac
5. From « + 3^ take z(^ — h-\ ^^«
2 3
6. From 5a; + t take 2:?;
0 c
' X I
7. From « + ic 4- -^ -^ take a — a; +
0^5 — 2^2 a; 4- ^
8. From take
a a — I
I ■ , 2
0. From take -„•
^ I — X I — x^
526. IWultipIy the following Polynomials:
a — b , 2b a^ x^ — ifi
1. I —7 X 2 H ^' 5. — ; — X T^ .
a -\-b a — b X -^ y ab
4a ^x 2b , s^ ^ o 5^
2. — +^x h— • 6, a^ — I X — ^
3a; 2^ 3a; 4a a — I
3. a;2 __ 2xy + y^ x —^-^- 7. -^ ^ x *^
X — y 2a 5a — 10
75 2^2 — 4<5j3 ^y ^^.
527. Divide the following Quantities:
, 2a 2a 3V 2y
«— 3 «— 3 ^ 2y — 2 y — i
II .1 ab + b^ b
X xy^ ' -^ y ^' a^ — b^ ' a — b
^* ' "^ 2; • ' x^' ^' \i+a "^ i-a/ • (i^a^
286 APPEN-DIX.
528. Simplify the following Fractions:
3^ x^
2a —
- 2
2a
a —
X +
I
X
3
y +
4
.-c^-
-f
f . .
I
a
-f '
'^ a¥
l-
-^ + 1
lO.
529. Solve the following Equations:
1. A house and barn cost $850, and 5 times the price of
the house was equal to 12 times the price of the barn.
Required the price of each.
2. A, B, and C, 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 \ full of water^ 21 liters leaked out, when
\ the water was left. Eequired the capacity of the cask.
X — 4 s^ -j- 14 I
4. IX — 4 = - — — - — —
^ 4 3 12
^3^ + 9 7^ + 5 16 + 40;
2 3 5
a; + 8 a; — 6
Y X ^=^ X ■\- 2.
4 3
a; + 8 X — 6
X — 2 ^=z X A, •
4 3
2a; -h I / — X — 3^
+ 6.
^-3\
4 /
3
9. The hour and minute hands of a watc^ are together
at 12 M. 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 \ of this part.
At the end of the third year his original stock was doubled.
What was that stock ?
APPENDIX. 287
530. Solve the following Simultaneous Equations,*
I. ^^±-'^=26. 3. ^-^ =9.
3 •'42^
6x 6y ^ 2x — 2y
-23 4
a. - + - = 8. 4. - + ^ = d.
32 23
X y ^ . y
2^3 3 4
5. A man bought a horse, buggy, and harness for I400 ;
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 ?
X + y — z =
0;
10.
xy = 600 ;
x+ z—y =
2;
xz = 300 ;
y -hz-'X =
4.
yz = 200.
II.
. y z
X . z
« 3 4
c d
a; , , z
-+y + - = 33.
w -\- X -^ y =
:6;
12
?^ -f 50 — ic;
W -{■ X -{- z =
9;
X + 120 = 3y;
IV + y -h z =
8;
y -{• 120 = 2Z',
X +y + z =
7-
^ + 195 = Z^'
13. A's age added to 3 times B's and C's, is 470 jts. ;
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 ?
288 APPEl^DIX.
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 pursued him, traveling
8 miles an hour. How long did it take the latter to over-
take the former ?
Note, — Substitute c for clerk's rate, p for policeman's rate, n foi
flumber of hours between starting, and x for the time required.
Formula. x = •
p — c
2. A can do a piece of work in 2 days, B in 5 days, and
0 in 10 days ; how long will it take all working together
to doit?
Note. — Let «, &, and c represent the numbers. Then x = dbe -*•
{ctb + ac + be).
-r, abc
Formula. x = — ; ^-«
ab + ac + be
3. Divide $4400 among A, B, and C, in proportion to the
numbers 5, 7, and 10.
Note.— Put a, b, and c, for the proportions, s 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?
632. Expand the following by the Binomial Theorem :
1. {2a--2fiY. 4. (^' + /)'.
2. {zx + 2y)\ 5. {a + a-^)\
3. (i + 3«)'. 6. (fl2 _ 2a)\
533. Find the Product of the following Powers :
1. dbor^ by a^. 3. x~'^ by x~^,
2. a^h-^x-^}ijarWx-\ 4. y-'^hjy\
634. Divide the following Powers:
1. 6a~» by 3«-2. 3. 1 2x-^ by 4X~^.
2. Sa-*bc-^ by 4a^¥c^, 4. {a + a;)-» by {a + a;)-»».
APPEi^^DIX. 289
535. Transfer Denominators to Numerators, thus forming
entire Quantities.
J ^, ^, -^ — •
536. Unite the following Radicals :
1. V48 + V27 + '\/243. 4. X A/25^ + "^/z^xf^c,
2. A/54^ — '\/9(>x 4- A/24a;. 5. V^ofi^^ — '\/2oa^x.
3. 8 V^^^jJ + 2 Vi^. 6. 3'V^i28iz:3z/';2 —• ^x \^i6yz.
537. Find the Product of the following Madicals:
1. (« 4- ?/)i X (^ + 70^. 3- (^' + y)^ X (x H- 2/)i
2. 4 + 2^/2x2 — \/2. 4- 3^ V^^ + ^ X 4 a/^.
538. I>ividing one Radical by Another.
1. (fl^%)- -^ («a;)^. 4. (^ + yf -J- (^ + 2^)".
2. 24a; ^ay -^ 6 V^. 5. 4« a/«^ -i- 2 V«c-
3. "s/ i6a^ — \2aH -^ 2a. 6. yoyp-i-yViS.
539. Required the Factors which will Rationalize the fol-
lowing Radicals:
1. 2 ^/a + V7. 4. \/5 — \/5.
2. ic + Vy. 5- 4 V2S — 5 Vy.
3. Thedenom. of-^. 6. The d. of -— -^-=
2V3 V3+V2+I
540. Solve the following Madical Equations:
1. Given Vx + i = Vn + ^j to find x.
2. Given \/x + 18 — ^5 = V^ — 75 to find x,
3. Given Vx^ —11=5. 5. (13 + ^23 + y"^)^ = 5.
6 _
4- /--— =^ = V 3 + ^- 6. 2 V^ = V iz^ + 3«.
290 APPEKDIX.
541. Solve the following Quadratics,
2. —
0-^ -
^~3
"~~ '
.. J ^
i6
X
lOO —
4^2
gx ^
= 3.
^
a^
I
2
4
32
Vi^
L±_? _
4 —
V5
4 -(- Va; \^x
5. Find two numbers whose differenco is 12, and the suni
of their squares 1424.
6. Eequired two numbers whose sum is 6, and the sum
of their cubes 7 2.
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 diiference 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.
Required 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 jrrs. older, the product
of our ages would still be 120? How old is each ?
12. Vx^ + a/^ = 6 Vx.
13. X 4- '\/x + 6 = 2 + 3 a/^ + 6.
14. A man bought 80 lbs. of pepper and 100 lbs. of
ginger for £65, at such prices that he obtained 60 lbs. more
of ginger for £20 than he did of pepper for £10. Whai
did he pay per pound for each ?
COLLEGE EXAMINATION PROBLEMS.
542. I. Divide i^x^ — ^~ hjx
5<? c
2. Divide a* — ¥ hy {a — b),
3. Solve tne equation a; + = 12 — -^ — •
4. Multiply 3 V45 — 7 V7 ^y Vif + 2 Vpf
5. Divide a^b^ by «^5i 6. Divide xy~^ by a;%~^,
7. Given ^cd^ -[- 2X — 9 = 76, to find x.
8. Given ia;^ _ t^; + 7I = 8, to find x.
9. Find two numbers, the greater of which shall bw to
the less as their sum to 42, and as their difference to 6.
10. Find the value of i + | + ^f + ^f + ^^^* ^^ infinity.
11. Find the third term of {a + by^.
12. Expand to four terms (i + x^)~k
543. I. Divide . ^ + - by — ' — ^ ;
2. Find the product of a^, a\ a^ and a ^^.
2a
Va^ + ^
3. Solve the equation x + V«^ + ^^
4. Solve ^^ ^1 -^— = o.
a; X -\- I X -{■ 2
5. Solve V^^ I = a? — !•
6. Find the value of f + i + J +, etc., to infinity.
7. Given x-{-'\/x : x—Vx :: 3 ^^ + 6 : 2 V^, to find x,
8. Expand to four terms (a^ + x)'^.
9. Expand to four terms {x^ — y^)~^.
544. I. Find the g. c. d. and the I. c. m. of (243aioj5 _^ ^ j
and {^ia%^ — i) by factoring.
T^. ., 6 V^ 1 20c Vl^
2. Divide ^— r by ^— •
25 v«^ 2iaJ wd^
3. Solve the equations 2X — y — 21, 20^2 + ^/^ = 153.
292 APPENDIX.
4. A person buys cloth for $90. If lie 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^^ — ^y\
7. Multiply 3 Y ? by 2 W |-
8. Given x + 2y = y and 22:4-3^ = 12, to find z and y,
9. Eeduce a V4Sa^d and Vj|- to their simplest forms.
10. Given 4- - = 12 , to find x.
32 3 ^
X or — — rt'
645. I. From ^x ■{ — = subtract x —
2h c
2. Multiply together -^^ — , ~~~2> ^^^ ^ "^ ~ — *
I -J- y X -f- X I — X
3. Extract the square root of Sal)^ + a* — 4a^ -f- 4 J*.
4. From 2 V3I0 take 3 'V^4o.
5. Divide a"^^^ by a^b~^. 6. Solve a;* + 40^2 — 12.
7. Solve a;2 — X Vs = x — ^ \/^.
8. What two numbers are those whose sum is 2a, and the
sum of their squares is 2^* ?
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 120;^ — 192 by ^x — 6.
2. Divide a;* H 7 by 7 — m,
3. Solve the equation 21 + ^-^^ = 5^ + 2Zzi2?.
^ 16 82
4. Find the product of a^, «^, «^, and «~^.
5. Solve the equation ~ — g— = 2.
3/ 2X
6. Find 6 arithmetical means between i and 50.
COLLEGE PEOBLEMS. 293
7. How many different combinations may be formed of
eight letters taken four at a time ?
8. Expand {a — l)~^ to four terms.
9. Divide 150 into two such parts that the smaller may
be to the greater, as 7 to 8.
10. Given 5a; + 2?/ = 29 and 2^ — a; = — i, to find
X and y,
2 T
547. I. Solve the equation {- 4 = a
2. What is the relation between a, ap, and ar^ ?
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.
4. Give the first three and last three terms of (2« ) •
5. Find the r/. c. d, of a^ — W and a^ — lab + V^,
6. Find the I, c. in, of (p? — ^), and 4 (a — x\ and
7. Add -=^-^ f , — ^ =:r, and ^—, tto*
5 (a — ^ 5 (« — *) 5_(« — ^)^
8. Find the value of x in "^x + a = V^ + a.
648. I. Eeduce r 5 to its lowest terms.
c? — x^
2. Multiply ar^m by -^ ; and divide ar^W by -^
o 1 J.1 . • nx — 6 X — 5 <^
3. Solve the equation 7 ^^— = —
^ 35 6;r — loi s
4.Give.Z£±J_(._£^) = ,.
a^ X
5. Solve the equation - -^ [- 7I = 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":^ COLLEGE PROBLEMS.
7. Find tlie sum of n terms of the series i, 2, 3, 4,
5^ 6, etc.
8. Expand to five terms {cfi — ^3)-^.
9. Find the sum of the radicals A/300 and a/tJ.
10. Solve the quadratic - -| = 2^,
549. I. Find the sum and difference of \/i8«^and
2. Multiply 2 Vi - a/3^ by 4 V3 - 2 a/^.
3. Solve the equation """ - 4- ^^ ~~ ""^ z=z n ^ ^ -r ^^
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^ — ]/)^,
7. Simplify the radical (cfi — 20^!) -\- aW)^,
8. A and B together can do a piece of work in 3I days,
B and 0 in 4f days, and 0 and A in 6 days. Required the
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 - + -==2, - + - = 3, --{-- = 3; find
X, y, and z.
2. Given ^ =: , to find x.
S — X 3 12
3. Find the I. c. m. and g. c. d. oi x^ ■\' ^ — 21 and
x^ — X — $6,
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 ?/ + 3 f or a; in >* — a? -{■ 2X^ — t, \ simplify
and arrange the result.
ANSWERS.
Page 15.
I, 2. Given.
3. 2 cts. A, 6 cts. 0.
4. $8, h. ; I32, c.
5. 9 and 27.
6. 4^,0; 8^, B; 16^?, A
7. 121/, son ; z^y> father.
Pagre 16*
8. $20, B's ; I80, A's.
9. 15. 30. 45-
10. $7, calf; $56, cow.
II. $5.25, bridle;
$10.50, saddle;
$110.25, horse.
12. $3000, daughter;
I6000, son ;
$27000, wife.
13- 234, 702, 936.
Page 19.
1-3. Given.
4. 98^.
5. 18.
INTRODUCTION.
6. 10.
7- 34-
8. 17.
Page 21c
1. 60.
2. 40.
3. ac + 85.
4. 55 — 2t7.
5- 35-
6. 24.
7. 3^ + 21/ + «5.
8. 6b — jcx + sa,
9. Z>a;^ + ca:?/.
Page 22,
10. -^-^ + a.
2Z
b — a ,
II. f- 2;^.
12. 3^ + i?^«/ + %2;.
^3- d
' 14. 92.
15. 120.
ax — Qty — Zia; + by
29G
S U B T K A C T I 0 If .
ADDITION.
Page 24,
l^wgre 26.
1, 2. Given.
I. 240^ + 25—36?.
3. 2iah.
2. iGmn—xy-^-hc
4. 11 xy.
3. i6hc -\- xy -—mn
5. isa\
4. 4fl!5— 3w/^ + 2^
6. — 22,hcd.
5. i52:?/-i-«5 + 5.
7. — i6:r3|/2.
6. Given.
8. 45a^2.
7. 21 {a + h).
9. -39«^^y.
8. igc{x—y).
10. 2gWd7n\
9. 7«V^y.
II. Given.
10. 6^/^?.
12. 4.
II. loVa; — y.
13. 5.
Page 27.
12. Given.
J*asre ;^5.
13. a{T~6h + ^d
14, 15- Griven.
—3>m).
16. 82;.
14. 2/(^^ + 3 — 2C
17. «^c.
-5m).
18. — 12J.
15. m{g + ab--jc
19. — 12?/.
+ 3^).
20. — 2m.
16. i?;(l3«r_3^_|_^
21. I.
-3^+?^).
22. 75-
17. a;?/(a4-5-c).
23. Given.
I. Given.
4.
5-
6.
7.
8.
9-
10.
II.
12.
13.
14.
15-
16.
17.
18.
19.
20.
. 16 cts., b ;
30 cts., k.
Page 28,
26 peaches ;
49 pears.
15 and 70.
15 K 25 g.
I.
7.
7.
356, A; 94, B.
36 and 141.
8.
9 cts., top;
23 cts., ball.
$9? b; I31, s.
26 cts., A. M.;
74 cts., p. M.
Given.
14.
7.
12.
60.
20.
I'cffire 31,
I, 2. Given.
3. i4^yz.
4. — 62«J.
5. I gab.
i^UBTRACTION.
6. 2 7ip?/.
7. 43(ic.
8. 37fl;:?:l
9. 5i«2J.
10. — 44.7:^^2.
Page 32,
11. z^a^b.
12. o.
1 3- —lim^x.
14. 53^^«/-
MULTIPLICATION
297
15-
$150.
28.
9(«— ^-f-a;).
38.
xy{%—ab-^c
i6.
25°.
29.
5(« + J).
-d).
17.
$420.
30.
-7ix'-yy
39-
c(2a + bm + d)
18.
4xy—6a.
31.
$600, A's.
19.
iS^+i6am.
32.
60°.
20.
i8a;2 + ?/2-f6«.
rage 34,
21.
13^^ + ^— iC
I-
■3. Given.
—5^ + 3^^-
JP«rflre 33,
4-
b—c + d—m.
22.
gcd—ab—2m
33.
Given.
5.
Sx-\-y—ab
+ 3^ + 4^.
34.
{2h — C-i-d)3^.
+ 4^.
23.
iSm— 23.
35-
(ab—c—d
6.
2a—b—c-{-x
24.
12a;' — 13a;.
+x)y'
+y+d.
25.
i6«Z'+ i^c+d.
36^
a^d-b + c).
7.
a—b-{-c—a
26.
a—b + c.
37.
x(ab—2iC—d
•\-c-^c—a-\-b.
27.
6(a + b).
-m).
MULTIPLICATION.
rage 36,
1-4. Given.
5. 42abc.
6. ^^abcxy.
7. 2>dmxy.
8. dibcdxyz.
9. ^6abxy.
10. 42acdx.
11. $4bcdm.
12. d^adfxyz.
Page 37.
13. Given.
14. — 45rt5a:^.
15. 42abcd.
16. i^2abcxy.
17. — 41 4abcxy.
18. g4^bcdxy.
Page 38.
19-21. Given.
22. i$x^y^.
23. 24aW.
24. a^x^y^.
25. «2j'«+«.
26. dx^yH.
27. i8«5^V.
28. 18.
29. 240.
30. 6a^^.
31. — i8a'*J%.
32. 4^^1/2^
33. 2iaW.
34. 4oc^xy.
35. — 28a4^3.
36. —6x^y^.
37. 2jaWc^,
38. — 28a3c*.
39. a:2^V.
I, 2. Given.
3. 6fljca;2 + 8c2d
4. i^aWx—6acdx
+ 3«^-
5. —8^25^
-\-6ab^d—2bdm>
6. — 15«%
+ 2oa2^cH-ioaV
7. 8. Given.
I. 6aa; + 3Ja;
+ 2ay + Jy.
298
DIVISION-.
Page 40,
24. 24a^x^—6a^yK
16. a^b^+2adcd
2. 2,a^+4ay
25. 6/2_|_^^^_j_^^^
+ c^d^.
—lbx—\ly.
+ ^ca;2.
17. 9a^—4y^*
3. \2M—2>cd
26. Given.
18. rz;4__^2.
—4ad + ac,
27. ^2_^2.
19. x^—2xy'^-\-y^.
4. 6ixt/—2ab
28. «^ + a3_|.^_|.j^
20. 4«^— ic2.
'\-6cxy—2ac,
29. x^-\-^x^y
5. z^x-\-4'bx—cx
+3^y'+f'
Pwgre 45.
—Z(iy—Aby-\-cy.
30. a2'^ + 2a'^Z»«
6. ^ax-^^ay + az
4-52-.
I, 2. Given.
•\-ibx-\-zhy-\-lz.
31. a;2 4-2:ry + 2a;2?
3. 36.
7. i^cdmx—Galm
+ ^2+2y;2; + ^2.
4. 24 chickens.
— 2icdnx-{-gah7i.
5. Given.
8. 24abcx-\-i2mx
6. 48.
—Z2ahcy~\(imy,
JPagre 45.
7. i960.
II. labd^xyz"^.
I. a^-{-2a+i.
8. I960.
12. \iaUH'''^\
2. 4a^ + 4a+i,
9- 25.
13. aV+^
3. 4«2— 4c^^ + ^l
10. i6w~
14. c:z;(«4-^)^.
4. 2:2+2a;y + ?/2.
II. 56.
15. 5^(«_^)5.
5. x^—2xy\-y\
12. 77.
16. abc(x-{-yY'^''.
6. I— ;rl
13. 36 apples.
1^, ~3^(« + ^)^.
7. 49^^-14^^+^^
14. 70 sheep ;
8. i6m^—gn^.
100, both.
9. ic4_^2.
15. 16 and 12
P«9re 41,
10. I — 4gx^.
16. 24 plums.
20, a^-{-h^.
II. i6:z;2— 80:4-1.
17. 42.
21. a^^^s^H^**.
12. 25^2^105+1.
18. 144.
22. a:^ + a;2-j-i.
13. I — 2a; + 2:1
19- 2if; i4f
23- 3^ + 4^^i/
14. i+4^ + 4''i'^-
20. i4yV bu., one;
— I3z2— 4:z;2^2
15. 64b^—48ab
6^ bu., other,
+ 222:^-30.
+ 9«^.
DIVISION.
Page 47.
4. 5^^.
7. sab'
I, 2. Given.
5. 5-
8. z^c.
3. 2^5.
6. 2/7i
9. 4mn,
DIVISION"
299
10, II. Given.
12. — 3C.
£3. 7J.
14. 5^-
15. -65.
t6, ydf.
17. —gag.
Page 48,
r8. Given.
20. a:®,
^i. c^.
22. 2:^.
23. 4^»
24. — •
y
25. Given.
26. 8a^»c2.
27. — 6^2f.
28. 5a5.
29. 72;2?/.
30. «Jc.
31. 2a5c.
32. Sx^y^z,
S3. Sa^c.
34. i2d^x^y.
I2iC2
a
36. 12X^z\
37. iim^n.
35-
Pagre 49.
1-3. Given.
4. I^+(^-\-d*.
5- 3^+5-
6. 35^—14-45.
8. — 2a;+y.
9. y^+z—i,
10. — 5a— 45 + 6.
11. sab—sa.
12. — 4a;2 — scP
■{■ax,
13- a^— 5^ + 25.
14. i+5«— 9«^.
15. 2a—4h—sc.
16. 2(^ + 5)2
+ 3^(«4-2')2.
17. 9i?r-9^.
18. a;(5— c)
19. 3^2 — 2a.
20. a— a^+tt^.
I, 2. Given.
3. i»+y.
4. «— 5.
5. «2_2a5 + 52,
6. c-^-d.
7. a;— d
8. 20; + 32/.
9. «— &.
10. ic+y.
11. a^+ah-^-l^.
12. 3<Z + 26.
13. a4-2._
14. c^—2ax + a?.
15. 2a.-3+4a:2 + 8a;
+ 16.
16. rc + 5.
17. a;— 2.
18. c — X.
19. a + J.
20. 2 (a — Z>).
1. 10 yrs., son ;
46 yrs., father.
2. 15, F.'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.
Page 52,
6. 20 cows ;
180 sheep.
7. 13I, less;
43 J, greater.
8. 9.
9. 5 hours.
10. 8.
11. 7.
12. 5 of each.
13. 4 hours.
14. 32>%'
15. 10.
16. 7 m., A's No. ;
14 m., B's ;
21 m., C's.
300
FACTOEIiq^G
17. ^2x, A'sm.;
^x, B's m. ;
$Sx, C's m. ;
$14^, all.
^^' 5> 15^ and 20.
19- 10, A's;
20, B's;
30, C's.
20. 8, 16, 24
21. 24.
FACTORING,
T'af/e 54,
I, 2. Given.
2, 3j 3j cf^^h,
2, 2, $hxxxyy.
5, 7» aaabicc.
3, 7. ^?/«/;2;;a;;?.
iixxyyyz.
S* 5^ 5^ abhcxxx,
9' 1, uaahccd.
io» 5> ^Ziifnmnnnx,
Page 55,
1-3. Given.
4. ^ (y + <? + 3^).
5. 2a (a; + ?/ - 2^;).
2,hc{x — 2x—'a).
Mm {n — 3).
-ja {sm + 2:r).
2']cl{bx ~ 2my).
^axy (3.T + 5).
J2. 5(5 +3^^ — 4^W
13. x{i -\-x + x^).
14- 3 (^ + 2 - 3?/).
15. i9«5(a;— i).
I, 2. Given.
3- (« + J) (« + J).
6.
7.
8.
9-
10.
II.
4. {x^y)(x-y).
5. (m 4- 27^) (m + 2n)
6. (4a + i) (405 + j).
7- (7 + 5) (7 + 5).
8. (2« — 3Z») (2« — 3J)
9- (y + i) (2/ + i).
10. (l — c2) (l — c2).
11. C:?;"* + ^») (a;»^ 4- 3^«).
12. (2a«-- i) (2a»— i).
13. {a^ -h h') (a^ + i^),
14. («^ + y) («2;8 -f- y).
1. Given.
2. (« + re) (« _- x).
3. (3^ + 4«/)(3^-4y)
4- (^+2)(y-.2).
5. (3 + ^O (3 -- x).
6. («5 + i) (« - i).
7. (i + ^) (i - Q.
(5« + 4^) (5«5 - 4^).
{2X + y){2X^y),
(i +4«)(i — 4«).
(5 4- I) (5 -I).
{X^ i- tf) (x^ - y2),
(ax + %) (ax — %).
(«'« 4- b") (a^ — 5«).
8
9
10,
II.
T2.
13.
14.
MULTIPLES,
301
12. (a -f- i)(fl' — 0^ + a^
Page 59.
-a^
4. a3 — a^-\. a — i).
I. Given.
13. Given.
2. (x^i)(a^ + x+ i).
3- (^ - 2^) (^ + ^y + ^y
Page 60.
4- a;y + a;^ + y^).
14. (ar + «/) (a:* - a:3y + a;y
4. (:?;— i)(a;+ i).
- ^if + ^).
5. (I -6^) (I +6^).
15- (a+i)(a2_-a4. i).
6. Given.
16. {a+ i)(a« — a8 + a2_.^
7. (b^x)(b + x).
+ 1).
8. (^ + ;^)(^_^;2 + 6Z22-;23).
17. (I +^)(i-y + ^^)-
9. (« 4- b) (a^ - a*b + a^^
18. (i +a)(i _« + a2_«?
--aW + «54 _ j5).
+ A
lO. (.T + i) {t^ — X^ -\-X— l).
19. (i + J)(i_J + J2_63
II. (i + a)\i ^a-\-a^ — a^
4.54_^5_|_ J6).
4- a* - a^).
20-28. Given.
DIVISORS.
Page 61.
Page 63.
7. a + J.
I, 2. Given.
I, 2. Given.
8. J 4- 2.
3. ^'
3. 3«^-
9. iP + 3-
4. J.
4. 2«a:y.
10. a — 2.
5. «<7.
5. 4a^T^z\
II. « + 3.
6. 20;.
6. 6aa:2;2;2.
12. a; 4- I.
7. 7^i-
13. a — h.
8. 6al).
Paflre «(>.
14. a — 5.
1-5. Given.
15. a;3 4. 3a;2 4. 32:
6. a; — 2/.
+ 1.
MULTIPLES.
Page 68.
7. si5x^f2^.
13. Q^ + a?'— a;— I.
I, 2. Given.
8. Z^mhiY*
14. 6a^ 4- iia*
3. 56fl4^,2cs^.
Page 69.
-3«-2.
4. 2>oxY^'
5. 9oa3^%'5.
6. 42oa^J*.
9-11. G
12. a* -f
_ 7,;
iven.
— 2.
302
BEDUCTIOK OF FBACTION^S.
REDUCTION
Page 74,
1-3. Given.
I
d'
I
a — h
x — y
zy — 3^,
2X — 2Z
I
X
I
OF FRACTIONS.
X'
'-\-f
X
a
+ x
I
a
— I
I
x + y
3- h
Page 75.
1. Given.
2. a — x.
a
4. h ^ c.
5.5 + . + ^--.
6. a — h.
7. 5 + 7
b — a
8. a + a; +
9. 3a; + I -
«
fl^ — X
I. 2. Given.
^ A^y — h
, X^ — X
o. •
a; 4- I
i2«c — a + 5
'• — 3^ —
5^
J'asre 76.
I. Given.
i2ywa;
2.
6m
24a%x
iSac^ 4- 24b(^
6^
a^ — y^
66?V?/ — 45a;2y
3«2 — 2b
Page 77*
I, 2. Given.
ab
2I«2
49«
^ — y^
^' Q? — 2a;?/ + y^
6.
32Q;3(a:4-y)
REDUCTION or FRACTIONS.
303
Page 78,
I, 2. Given.
3
2CX 2ld CpX
2dx' 2dx^ 2dx
6.
lO.
II.
12,
ac^y 2hxy 2(^x^
zc^xy' 2(?xy^ 2c^xy
2c? 4- 2db 2>^x
x^ — 2rry + y'^
a::^ -[- 2xy + ^2
«2 -{- «5 15^ — 3
3« ' 3«^ *
2Wc—2T)^d :^ac—T,ad
2it^c^l6^d
zWc—^d
2'bxy hz 4az
2bz ' 2bz' 2bz
ax + ay 6x -\- 6y
2X + 2y 2X + 2y
2X^
^y
2X + 2y
c? — 2ax 4- a^
a^ + 2ax -\- a^
a^ — Q?
2.
Page 79.
I. Given.
2acx 4i^c^ hxy
4dcx' 4bcx' 4bcx
2,(?d 2'bcx 2>bxy
^ay 4by zcy
5-
\2X
12?/' 12?/' 12?/' \2y
4abc^ Sod bx^y
i6a^ jSa^c 24a;
6. —- ^ , —
24a^c' 24a^c' 240^0'
24a^c
ac 2cd 2xy
2bc' 2bc' 2bc
8.
10.
II.
13.
(a-^by (a-bf a^^W-
a^ — b^' a^ — W' a2_j2
4xy{x^y) 6a{x-hy)
6xy(x-\-y)' 6xy{x + y)'
abxy
6xy{x + y)'
ad bx
aW' aW
b'^cdx acdm
aWchV aWcH' aWM
'^^ ayz + byz dy^
x^z
xy\
xyH
' xyH
4cmx^ -\- 4cnx^
6acm — 6acn :^a^m^x
i2a^cx^ ' i2a^cx^
304
ADDITION OF FRACTIONS.
ADDITION
Page 80.
I, 2. Given.
2xy
Sgdxz
Sabc
X
ya + 2b
7. Given.
rage 81,
8. Given.
10.
II.
12.
13.
4$a + 12:?: + 20^
60
Sax -]- 6 -\- <)ay
12a
ah — ac + hx -\- ex
z^ — y
2xy
2a + 2ax 4- 3
ay
OF FRACTIONS.
am — dy
ax — ay + ahx + «5^
5g<^ + 3Q^^ + zbda?
^' iSdx
^ah — 2dn — d^
16.
17.
18.
my
nx — mx
my — 7iy
19. — 6.
4adx + 6hcx — Mm
20.
I. Given.
bdx
2. a + c +
hx -f 2d
2X
3' x +
am — ay — dx -\-bd
bm — by
4. $d +a -\- b — c
xy + z
Sdh
5- 5^ +
2flj — by
2b~^'
Page 82.
6. Given.
^bd 4- 2«
7. ^-^— .
8 ^ + ^ — 4gy.
c
X -\- y ^ a^
a
3^2 _ 2a;y — y^ 4- a — b
x-^y
X — y — a^ -i- Sab — 5^
II. ^ 7 "^ —
a — b
2x^-\-2xy— 2x—2y4-a-\-h
X— I
10
12.
MULTIPLICATIOK Of FRACTIOi^S.
305
SUBTRACTION OF FRACTIONS.
rage 84.
rage 83
I, 2. Given.
gahc
3-
4-
d
5, 6. Given.
ay — dm ■\- hm
^' " my
8.
hy — f/y -f- J?;i
12
/i?/ 4- 7im 4- dm
10. -^— ^= ~
my
h — my
II. ^.
y
, M + c7t
12. a A 9 — •
cd
6« + 55 — 2^/
13-
14.
«<? +
5<? +
ex
15. flf
bd—dx-^ly — xy
2X + 3<Z«/
2V
, a^ — y^ — loa + lob
16. -^^ -^
10:?; -\- loy
MULTIPLICATION OF FRACTIONS.
rage 85.
1-4. Given.
5. h + ^d.
. ab
6. — •
4
6ca; — 9c?/ + 4dx — 6dy
ISC + 4d
8. 2abc,
a + b
10.
4 + 5^
2«2 _f- 2^25
Z> + I
11. 8a;2 + 122;.
12. 2ax — 2bx + 3« — 35.
13. abc,
a + b
5
5
9c — 3^
•
4
17. 3^y + 3^-
18 ^'
rz; — j5
Page 87.
I, 2. Given.
3. 6xy.
306
MULTIPLICATIOK OF FRACTIOKS
4.
5-
6.
7.
8.
9>
II.
12.
13-
14.
15-
16.
I.
2.
3-
4.
5-
6.
'ay
0? — y^
yh + «/;2;2*
3
8« -1- 24a;
^ (o? + ^)
10. Given.
y — 3^
2xy
h — 2a
2,ab
ly
xy -\- 2X -\- y'^ -\- 2y
xy
xS — y^
~^
2& + 4.
Page 88.
Given.
abdx
dbd 4- acd
xy
mx + nx
— — — •
4
4«c + 4cA
sr— I
8. 7a; — 7«:r.
aco; — acy
10.
12.
13
14
Zac
2aa^ 4- 2«:p
3^ — 3
Sx^y
a-\-b
6am
X ■{- I
2dbxy 4- y^xy
4« 4- 6
15. I — w.
^cx — 3^a?
1, •
4
2. 3a;(?/4- i).
xy -\- 2X -\- y^ -\- 2^
^' ■" ^^^
4. 9a:.
S-6-
6. X,
x — z
8. 6ay.
9. a?-5^3
2i?7 4- y
10. -^-
5
X/^ — IP
12. 254-4.
13. 2a (c 4- d).
y — 3^
2xy
15. y"'
14
DlVISiei^ OF FRACTIONS.
307
I
5-
6.
7-
8.
9-
lo.
II.
12.
-4. Given.
2«
Wo'
a;
Vb
DIVISION OF FRACTION
4. Given.
7.^.
21/
8.
a; — I
a^ — a
10.
II.
12.
h -\-c
Given.
3 times.
5-
9. Given,
a^ + 20? + I
a?' — 20^+1
oc^ — y'^
^ — xy^ + ^y — ?/3
2x^y
3
11. 6.
12. -
2a + 2^
^ z^j^zy^
abcxy
X — a
15.
3<2:2; + 3^^
a — h
13-
16. -4^
3«2J + ilz
18. — .
' 4^y°
I. Given.
abdmy
ex
mx -f wa;
ax + a;2
253^2 — ^xy
y
6 50^^ + 5«
rr + I
7. 32; — 3fl5:r.
jPasre 9fl.
5^
243^J/^'
24^^
x-\-y
22,XZ
308 SIMPLE EQUATIONS.
U (x — y)
6.
c
7. Given.
a
8.
£0.
« 4- a;
c + 2
3« (c2 __ ^) '
2C
11
a2 _ «c 4- c2
12 4 (q^^ — zaa; 4 a^)
IX
13-
(«^ 4 hf
X
a^ -\- 2X •\- I
c
X
SIMPLE EQUATIONS,
Page 97.
5. 24f
12. 24.
I, 2. Given.
6. Given.
13. 14.
3. a — J 4 c — ^.
4ad
14. -i^.
4. a 4 ^ —ab + c.
15. Given.
8. '^^-
2h — c
]
t6
Page 98,
9' 7i*
5. Given.
i6c
Page 102,
6. 2 — a 4 5.
10.
15
16. 12.
7. ^>-|-c — « — 3.
17. 60.
8. «6Z — ^c 4 2m
18. 4.
— 8.
Page .
101.
19. 20.
9. 17 — sai — d.
I. 8.
20. aJ 4 ac.
10. 4^6? 4 6? — 3M
2. 50.
dn
21. — •
— I.
3. 30-
a
II. 32 — c 4 6?.
4. 9.
6c
12. II.
5- 7.
22. T'
3^4 2b
6.9.
ad — be
23. •
7. 72.
^ ac
Pagre 100.
8. 3a
2ab
I, 2. Given.
9. 5-
24.
ac— 2C
3. 15.
10. 28.
5« 4 13-^
4. 12.
II* 12.
^* 24
ONE UKKKOWK QUANTITY.
309
24^ + I.
50c — 2005
52
b^
26.
0 ^r
3« — 6
45
a— I
2
27.
4
^'- 2
28.
I
32. 7*-
2a — I
a^(c — a ■}- en-
-ac)
3Z' 34.
29.
C2
rage 103,
Page 105.
Page 106.
34.
20.
I. Given.
15.
13-
35.
24.
2. $8, vest ;
16.
30. days.
S^'
I.
$32, coat.
17.
240 m., one;
37-
If
3. $1500, A;
180 m., other.
38.
i6f
$3000, B ;
$4500, C.
18.
12 in., one;
39-
lA-
4. 40 men;
16 in., other.
40.
36.
80 boys;
19.
$25, H. ;
41.
II.
880 women.
$175, 0.
42.
1200.
5. 40 miles ;
20.
8h. 24m. A. M.
43-
i4f
80 miles.
21.
9 A days.
44.
5-
6. 1 33 J barrels.
22.
32 of each.
45-
4f.
7. 12 p., ist;
1
23-
30, 75, and 45.
46.
!(.-
^).
24 p., 2d ;
60 p., 3d.
24.
25 cts., ch. ;
47.
2a — 25
4-c
8. 28f feet.
75 cts., goose;
26
9. $120.
$1.50, turkey.
48.
3a — 6
10. $50, B's sh. ;
25.
8 ft. 8 in.
4
$100, A's sli.;
26.
40 and 60.
66?
$150, C's sh.
ad ,
49.
5^*
II. 18 yrs., w. ;
27.
— . — -7? less.
c + d'
50.
I — 8«
I + 8a
7,6 yrs., m.
12. $3000.
ac ,
c + d' ^''''^'■
a
13. 16 and 41.
28.
$1128.
^i-
2.
4
14. $6000,
310
SIMPLE EQUATIONS,
Page 107.
29. Given.
30. 60 lbs., b.;
1 20 lbs., m.
31. isyrs., B's;
30 '' A's.
32. 32|yrs., C's;
37l " B's;
40J " A's.
33. 1 1 75 votes d.;
1325 " s.
34. 164 artillery;
472 cavalry;
564 infantry.
35- ^533i B's;
$633^, A's;
Ussh O's.
36. $56.25, one;
^93-75> other.
rage 108.
37- h3^> V^- one.
I280, " other
SS, i4yrs.,y'ngest;
16 " next;
18 " eldest.
39. i6|days.
40. 550.
41. 30 and 18.
42.
13
43. 225 acres, A;
315 " B.
44. 2f hrs.
45. 5, istpart;
8, 2d "
2, 3d "
24, 4th '*
46. 9.
47. 47 sheep.
48. $120.
rage 109,
49. 60 min.
d
50.
m — n
51. 300 leaps.
52.
a — c
he
one.
, other.
53. 72 lbs.
54. 36 hours;
312 miles.
55. 2oyrs., s.;
40 yrs., f.
56. 280.
57. $324, ist;
I108, 2d ;
$144, 3d.
58. Given.
Page 110,
59. 8, ist part;
12, 2d ''
16, 3d **
60. 9 in. and 1 2 in
61. $75.
62. 27 days.
63- ^J575> one;
$2625, other.
64. 12 days.
65. $720.
dd, $384, sum ;
I162, A'ssh.;
$118, B's *'
$104, C's "
67. Given.
Pasre m.
(i%, 6 and 8.
69- 3456, one;
2304, other.
70. 3 m. an hour,
71. 400 in.,
or 33i ft.
72. 8 k. of one name
6 k. of another;
3 k. « "
2 k. " "
73. 7 and 8.
74. 240 leaps of d.
75. 100 days;
30000 m., ist;
24000 m., 2d ;
SIMPLE EQUATIOifS.
311
Page 114:.
1. Given.
2. x^% y=4.
3. ir=i2, ^=6.
4. a;=i8, «/=2.
5. ic=i, 2^=3.
6. rr=i6, «/=35-
7. tr=3, ?/=2.
8. Given.
9. .T=4, 2/=5-
10. x—df y=i2.
11. a;=5, y=6.
12. ic=io, ?/=3.
13. a:=ii, 2^=9.
14. a;=3, 2^=2.
Page 116,
15. 16. Given.
17- ^-3. 2/=5-
18. a:=:4, y=T,
19. a;=7, «/=2.
20. x=i6, 2/=35-
21. a:=3, ^=2.
1. a:=4, «/=5.
2. ir=8, y=2,
3. a;=5, ?/=3.
4- ^=Zy y=4'
5. ir=3, ?/=4.
6. 0^=12, y=3.
7. ^=3. y=5'
8. t?;=4j 5^=3-
TWO UNKNOWN QUANTITIES.
Page 117.
9- x=S4> y=4^'
10. a;=4, ^=2.
11. x=i6, ^=7.
12. a;=8, y=i»
13. a;=6o, 2/=36.
14. x=io, y=z20.
15- ^=5? y=2.
16. x=2, y=4.
17. a.=8, y=6,
18. a;=4, ^=9.
19. ic = 6,
«/=I2.
20. ir=i8, y=i4.
1. a;=43, ^=27.
2. 4 cts., lemons ;
6 cts., oranges.
3- 233 v.; 142 v.
4. 21 and 54.
5. $48, cow;
$96, horse.
6. 40 1.; 50 g.
7. 3 and 2.
8. mil m., one;
9999 m., other.
9. 56.
10. $320, B's;
$250, A's.
12. I900, A's;
$2400, B's.
13. 31 and 17.
14. $6000 h. ;
$2500 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.
23. 10 1. ; 30 g.
24. 3 oxen ;
21 colts.
25. 53-
26. $5000, B's cap.;
$4800, A's "
27. $21 or 63 g.
an
28. x =
y =
n+i*
a
n-\-i
3VZ
GEKEBALIZATIOK.
THREE OR MORE UNKNOWN QUANTITIES,
rage 121,
6. ir=24, y
= 6, ^=23.
I. Given.
7. ^=7i
y
= 10, z=^.
2. x=7, y=s,
Z=4*
8. ic=24, y
= 60, ;z=i20.
3. x=2, y=s,
2=5-
9-1 1. Given.
4. x=S, y=4,
Z = 2.
12. W=2
, X-.
=3» y=A, ^=5
5. x=4, «/=3,
z=S'
13. X=2
> y--
=3, ^=4.
4. 630 men, ist;
7.
18 = ist;
rage 123.
675 " 2d;
22 = 2d;
I. 12 yrs., ist;
600 « 3d.
io==3d;
15 " 2d;
17 " 3d.
2. I5, s. ; $4, 1- ;
$20, c.
5. 50 cts., ist;
60 " 2d;
80 " 3d.
8.
40 = 4th.
46 m., A's;
9 " B's;
7 '' C's.
6. 105 min.. A;
9-
$64, A's;
3. 5, 8, and 11.
210 " B;
I72. B's;
420 " C.
I84. C's.
G
ENERALIZATIO
N.
rage 125.
12. 22|-hrs.
24.
I5625.
I. 3 chickens.
13. I67.32.
25.
1842I A.
2. 30 rods.
Page 129.
26.
$55.80.
3. 12.
27.
$190,32.
4- S^TTrFS-
14. 9077.
28.
I1253.
5. 7 ft
6. 34.
15. 1x036.12^.
29.
$418.60.
16. 587.19 bu.
30-
$5250.
7. I205, $187.
17. i2f per cent.
1 8. 40 per cent.
31-
32.
$3865.86.
$1339.29.
Pagres 127, 128.
8. $961, A's;
$614, B's.
19. 60 per cent.
20. $3000.
Pages 130-133.
33-
34.
35.
36.
$222.22+.
2i yrs.
Given.
3h. i6T^m.p.M.
9. 1248, g.;
21. 137500-
37-
6h.32y8ym. P.M.
902, 1.
10. 4f days.
22. I31250.
23. $2700, B's;
38.
9h.49-^m.P.M.
II. 5|lirs.
$230C
. C'8.
INV0LUTI02«"
313
Page 137.
2. aW(^'
3. aWc^,
4. .T^^/V.
5. a^b^c^,
6. i6x^y*.
7. 2\6a^h^.
8. 62sa>W(^.
9. 64«i2^,6^i2.
10. aW(^(P.
II. a;y5f.
INVOLUTION.
rage 138
12. (a+^y^.
13. (« + ^')^
14. (a: — 2^)'««.
15. (^ + ^r-
16. («3 4-J3)2.
17. aWii2.
2'ja^^
19
20.
8^3
21. -
22.
23
49^
2»»
rtvnn'lMy.n
26. ic3_j_6/p2^ _j_ 62^3
4-120:^2^242:^
+ i2X-{-2>y^
+ 243/2 + 24^
+ 8.
1. «4 + 4fl3J + 6aW + 4a5s + J*.
2. i«5 _ 5^4§ 4. loaW — IOa2^,3 _{. 50,^4 _ J5.
3. c^ + 1(^d + 2ic56?2 + 356^^ + z$(^d*^\-2lc^d^-{■^cd^+d^,
4. ic^ -f 6a:5y + i5.T^?/2 ^ 20:2:^2/^ 4- isa:^?/^ + 6xy^ 4- ?/^.
5. x"^ — 7a:^y 4- 2 la^y^ — zs^y^ 4- 35^^^ — 2 1 a:^^^ 4- ^xy^- -y'^.
6. ^^® 4- loyh 4- 45«/^;2;2 4- \2oy*7? 4- 2io?/V 4- 2<^2y^^
4- 2\oif^ 4- 120?/%^ 4- 45?/22« 4. io?/;2;3 4- 2;io.
7. a^—ga% 4- 36a'^^2 _ 34^6^ 4- i26«5J4 _ 126^4^,5 ^84^8^
8. m^^ 4- iim}^n 4- $$m^n^ 4- 165WW 4- 33om'';i* 4- 46 2772^^2^
4- 462771^7^^ 4-33om%''4-I65m3/^8 4 55^27294-11^7^*0 4- w'^.
9. a:'^ — i2x^^y 4- 66a:io^2 _ 22ox^y^ 4- 4952:^7/* — 7920:'^^
4- 924a::6/ — 'jg2X^y'^ -\- 4gsc(^y^ — 22o:x?y^ 4- 66:z^yo
— 122:7/1* -f 7/*2.
10. a*4-wa"~^J-f- w
— I
I „_o 70 , n—i n
- fl^** 2 52 _j_ ^ ^ _
(?"-3 5s
>»-
n — 2 7i —
X X
3 4
#4- etc.
314
EVOLUTION,
13. 0^ -{. ^x^ -\- ^x + I,
.14. h^ — 4b^ + 6b^ — 45 + 1.
15- I — 505 + loa^ — io«3 -I- 5^4
16. I -j- ^« 4-
n — I
a^ + n
71—1 n — 2
X
a^ 4- etc.
2 23
J*a</es 143, 144.
17. a:34_3^2^4.3^22;4.3^^2_|_6^^2; + 3:6-;2;2_f_^3_^2^2^_j_3^^2^^3
18, 19. Given.
2o. x^ + 2X (y -\- z) -\- y^ + 2yz + z\
ii. a^ — 2a\b — c) -\- If^ — 2bc + c\
22. a^ ^ 2a{x -{-J/ -{- z) + x^ -\- 2x(y + z) + y^ + 2yz + 2;2
23. Given.
ga^ + 12a + 4
24
25.
4^2 — 4ac + 6^
26.
27.
36 — iSSabc + i()6aWc^
49
^(2 — 6hmxy -f gm^x^y^
m^
28, 2g. Given.
MULTIPLICATION AND DIVISION OF POWERS.
Pages 145,
I, 2. Given.
146,
10. a^y-h'^.
11. Given.
12. all.
20-
24.
23. Given,
a
3. ««.
4. ars.
13. a;-ii.
14. &2.
25.
ay-^
b
5. b-\
6. «"*+«.
15. C-2.
16. x'''y~^zK
26.
a
7. «-7^.
8. a^c^cR
17. 4«&c~*.
18. 32^2/.
27.
bx-""
a
9. ^6^.
19. I2a^C^.
EVOLUTION.
Pages 148,
149.
15. #.
16. «-25.
19.
20.
b'\
x^\
13. ai
17. «-2.
21.
^.75;
14. tci
j8. a-6.
22-
■25. Given.
RADICAL QUANTITIES
315
t*a(/es 151, 152,
I, 2. Given.
3- a^'
4. a^.
5. 4*a;%i
6. 20^1^.
8. 2a'".
9. 35a52;8,
10. 6^2^.
11. 2 3 a; 3^ 3,
12. 8«t^>3,
13- (i3)^^V-
14. ya^y.
15. 30^3^7.
16. 1^.
1. Given.
2. ic 4- 2.
3. a— I.
4. I -[-X,
5. ^ + 1.
6. « — J.
7. ^ + -•
8. Given.
9. a; + «y 4- 2;.
10. a — 2(3> -f I.
11. a^ -\- 2b — 2.
12. I — 2^2 4- ir.
13. 20^ — 4a 4- 2.
14. a
X y
15. ~*
y »
Page 156
1-5. Given.
6. aVb-
7. 2a v^.
8. 6\/xy.
, 3 /-
9. 6v3.
10. i5V^5.
11. ^SaV^b.
12. 3a\^2^.
13. 2m\/i — 3^>
14. 4^V^2^-
15. sa\^b.
16. 6a'v/i3C.
17. I2a'\/ll.
rage 157»
I. Given.
RADICAL QUANTITIES.
4. («3^(i296)i
5. v^i5625,
v/87, ^8.
6. ^^4^8^ '^T^.
7. v^«~^ v^i^".
8. ::;7^, "^i-.
10. v^c^+Ty,
II. ^{x — y)\
2. ^/8^.
3. ^/(zfl^ 4- yf-
4. V(« — 2^)2.
5. a/9«^^-
6. ^8^7^.
7. yi6x^y^z.
8. l/"^.
27
9. ^27 (« — J)'
10.
11. V^rti2^.
12. \/(« — bf,
13. a/«^.
J><«</es J5«, 159,
1. Given.
2. (^3)i^ (^^)i.
3. 9^, (i25)i
V^l^ 4- yf.
I, 2. Given.
3- (3*)S (4^)i
4. (a'»)i (J")i
5. (a*)J, (J^)i.
6. (ai)i, (JtI)I.
7. (asr, (^^3",
316
DIVISIOK OF RADICALS.
ADDITION OF RADICALS.
Page 160.
1-3. Given.
4. 5V3.
5- 2^/5+4^/3.
6. (2^ + 3a) V^.
7. (a2 4- 3c)V'3«^.
8. (90; 4- 8a) a/2«.
9. 25^/2.
10. 118V3.
11. 30a v^.
12. (5^a;-|-6ic2) ^/c^
13. 40!^^/?/
SUBTRACTION OF RADICALS,
Page 161.
1. Given.
2. 81/7.
3. 4V30 — i2a/7-
4. 52^/5- _
5. {2ix—io)^/ax
6. 2'V^« + ^.
7. 7^Z>.
8. 95\/2&a;.
9. a~2.
10. ilVs-
MULTIPLICATION OF RADICALS.
Page 162.
1-3. Given.
4. 90A/10.
5. aJic.
6. a/«^
^.
7. \acxy.
' 8. a\/«^.
9. Given.
10. v«^ «
11. 42A/2.
12. 12a.
13. 6.
14. ^ax.
15. Given.
16. Vs-
17. 2^/5-
18. (w + w)'V^//z4-yi
19
V"
4. Vs^^^ or
«\/3«^-
5. 3\/^.
6. (a^ta:)i
DIVISION OF RADICALS
7. i2(a«/)i
8. 35\/^.
9. -^V&.
10. 2a
V^f
11. (« + 5)«.
12. i5a:V^.
13. 'v/a; — y.
14. l6\/2.
15. 32-
KADICAL EQUATIONS.
317
INVOLUTION OF RADICALS.
Page 164,
t, 2. Given.
3. at
4. 18a;.
5. Sa.
xi
6. — V2X.
4
7. Saa^Vct.
8. 9&2.
9. a2_j_2ay^-}-^.
EVOLUTION OF RADICALS.
rage 165,
1. Given.
2. 3V^a.
3. 2V^3:r.
4. ^9^«/-
5. 'V/SR^
6. \/^bc,
'• ^r
8. aci
9. 2«^V
10. fl^^J^*^.
11. Vza.
12. fl«^6w^«.
1-3. Given.
4. ak
1 a
5. a^c^,
6. (« + h)l
7. Vttc.
8. \^x-{-y,
9. 'V^« + J.
10. a/« + 5 + c.
Pai/e i67.
I, 2. Given.
3. a; — 4^/9.
4. 3-
5. A/7 — v«-
6. 31.
7- Vs^ + ^3^.
8: a — 5.
9. 3'v/a — Vs.
10. 4^2^ + sVb.
Page 168,
1-3. Given.
xys/ci -\- Vc)
A/3 — I
A/3 4-
RADICAL EQUATIONS
Pages 169, 170,
1-3. Given.
4. (d — a— cy.
5- 25.
6. 4f|.
7. 256.
8. 1 100. ' I — fl
9- 3^-
10, 21.
11. 252.
12.
2«
13. Given.
I
14.
IS- «a/}.
16. Given.
17. 4.
18. ^.
I
19,
1 —a
318
AFfECTED QUADRATICS.
PURE QUADRATICS,
14.
^= ± 2.
2. a? = ± 8.
Page 173.
15.
X=:±S,
3. 40 rods.
I. Given.
16.
X=: ± 1.
4. 30 rods.
2. X=±^.
17*
^=± ' .
5. 12, one;
3- ^ = ± 3-
4. ar = ± 4.
18.
a— I
Given.
30, other.
6. $6.
5- a: =±5.
19.
x=±s.
7. 80.
6. x= ±4,
20.
a; = ± 2«.
8. 15, less.
7. a;==±6.
21.
x = ±Vc^ + (P
60, greater.
8. a; = ± 4.
22.
x=±\.
9. 27 yds.;
9. ic = i V6,
23-
X=i±^/w^^l^
I1.50, price
10. x= ±: 7.
24.
x= ±26.
10. x=z ±: 16.
II. a; = ± 2.
II. 77 ft.
12. 0;= Jt «•
Page 174=,
12. x= ± 16.
13. a;=±i.
I.
X^±Z6,
AFFE
CTED QUADRA
TICS.
rages 178, 179,
7.
a
Page 181.
1-5. Given.
'^Tb
I, 2. Given.
6. a; = 6 or 2.
^/ah-{-d-{- f,.
3. 3 or - 4*.
7. a; = 9 or — I.
V 4^^
4. 5 or — 6.
8. a: = 3a
8.
— 205
5. J or - 2.
9. Given.
6. 2 or — J.
7. 4 or — 4|.
±Vb'i- Aa\
10. 15 or —4.
9-
7 or — 5.
8. 4 or — I.
II. 20 or — 7.
10.
3 ± 2V— I.
9. 3 or — 4-|.
12. Given.
II.
6 or — 3.
10. 9 or 6,
I. Given.
II. 4 or - 3f
12.
2 or — 3.
2. 10 or — 7.
0
3. 2 or — 5.
13-
^
Page 182*
4. 3 or i|.
±
2(;
I. 3 or I.
5. 4 or - I J.
/ W
\ hd—ch-\ — z.
2. 4 or I.
6. a; = 2.
V 4^
3- 3ori.
AFFECTED QUADRATICS.
319
4. 2 or — 12.
5- liorf.
6. 1 1 or 3.
7. I J or — li-
8. - J or - iM.
9. 4 or 2^.
10. i±V—a^+i'
IT — m
12. I or — if.
13. I or — 28.
14. 10 or — ^.
15- -ior-i
16. 4 or — I.
17. 4 or — if.
18. 5 or — 4f.
19. I J or — |.
20. 4 or — I.
21. if or — f.
22. 4ior 4.
23. 3 or — if
24. 4 or — i|.
25- |orf
26. f or — I.
27. n ±m.
28. 3^ or 3a— 3 J.
1-3. Given.
± 2 or ± ^2.
± A/3 or
J or — J.
8. I or -
9. 4iorJ.
10. 4 or —
2lf
Pagres 184, 185.
1. 8 or 4, one ;
4 or 8, other.
2. $60 or $40.
3. 6 or 4, one ;
4 or 6, other.
4. 1 6s., $5 each.
5. 5 or — 6|.
6. 16 scholars.
7. $30 or $20;
$20 or $30.
8. 60 or 40, one ;
40 or 60, other.
9. 36 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;
b " 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. ;
15 ft., hind-w.
28. 2 or —18, one;
18 or —2, oth.
29. I and i,
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^ = 3 or 4.
3. a; = 7 or 5 ;
^ = 5 or 7.
4. x = S, y =z 6.
5. a:=io or — 12;
y=i2 or — 10.
6. tz; = 10;
y=i2^or7.
>j3iO
ARITHMETICAL PROGRESSION?'.
7. iC = 9;
y = 4or Iff.
9. a; = 5 or 4;
^ = 4 or 5.
ro. a; = 5 or —3;
«^ = 3 or —5.
ti. a; = 3; 2^=2.
^2. x=±s;
15. 2?= 15 or 12;
y= 12 or 15.
3. J.
4. |.
5. f
6. 2fl^.
7. 3^-
8. 10.
9- i-
Piagre 204,
2. 4.
3. 6400.
4. 12.
16. a; = 21 or — 7,
?/ = 7 or — 21.
17. xz=62s;
y=i6.
18. a;= 2 or i;
2/ = I or 2.
1. 8 or -4, gr.;
4 or —8, less.
2. 30 yrs., wife ;
3 1 " man.
3. 0^ = 9^2, gr.;
^=±^2, less.
RATIO.
10. J.
a
11. —
2
12. X — y.
14. f
2
15. —
•^ 3a:
PROPORTION
5. 32 and 24.
6. 10 and 3.
7. 16 and 12.
8. 6 and 4.
9. 48 and of.
\. 40 rows ;
25 trees in each
;. 40 yds, length;
24 " breadth.
9 and 3.
1 1 and 7.
31 rds., length;
19 " width.
± 7 and i: 4.
25 m. and 23 m.
1 2 and 4.
3 or — 2, one ;
2 or —3, other.
19. Equality.
20. Equality.
21. Gr. inequality.
22. Less inequal'y.
23. f I > ¥•
25. 7.
26. 98.
10.
TI.
12.
13-
430 r., length ;
320 r., breadth,
20 r. ; 30 r.
9 and 15.
20 and 16.
2. 9.
3. (>^'
ARITHMETICAL
4. — 5-
5. If
6. .91.
8. 43-
PROGRESSION
Page 207*
9-
15-
10.
44i.
II.
49^.
12.
3^''
a.
GEOMETKICAL PROGRESSION".
321
Page 208,
2. 762J.
3. 216.
4. 1400.
5- 25i
6. 610.
7. 175-
8. 810.
rage 209.
1. 58.
2. 278.
3. II.
4. —43-
5. 2}.
6. -fj.
7. 1024.
8. 192.
Page 210.
1. 175-
2. 1 130.
3. 6.
4. 6.
5. 259.
6. 13.
7. —II.
8. o.
9- 255-
10. 62.
II 61.
1. I, 7, 13. 19.25,
31-
2. 3. 7 J, 12, i6|,
21,25^,30,34^,
39> 43|. 48.
1. 47.
2. —6.
3. 102.
4. 2, 1 1 1, 2I§, 31,
4of , 5oJ» 60.
5. i683f
6. 98A.
7. 5776-
8. foioo.
J'aflre 213.
9. 5-
10. 6, 13I 2o|, 28,
35i 421, 50.
Slh 64I, 72.
11. 12, 21.6, 31.2,
40.8, 50.4, 60,
69.6, 79.2, 88.8.
98.4, 108.
12. 975-
14. 3, 5, and 7.
15. loioo yards, 01
5|: mi., nearly.
Page 2 14.
16. 156.
17. $62.50.
18. $667.95.
19. 30c,
20. $1.20, int.;
$2.20, ami
2f. 20, 40, and 60.
22. 16.61 + days.
23. 30. 40, 50. 60.
24. 3 days.
25. 140.
26. $i78,]astpay't;
I5370, debt.
GEOMETRICAL PROGRESSION
Page 216.
1. 160.
2. 4374.
3. 4i.
4. 320.
5. 112.
6. —31250.
P^
iges 217-221.
3- 2.
2.
1 17 18.
4. 3-
3-
9999.
5. 6.
4-
27305-
6. 5.
5-
3885.
I. 242.
6.
8525.
2. 2.
I.
30000.
3- 500
2.
15625.
4. 5.
322
BUSINESS FORMULAS.
5- 215-
6. 567.
2. i, 2, 8, 32, 128.
I. 4371.
3- 7174453-
4. 2imu.
5- 9565938.
6. 43046721.
9. $196.83, 1. c. ;
I295.24, wh. c.
10. $10.23.
11. 2, 6, 18.
12. I4294967.295.
13. 10, 30, 90, 270.
Page 223,
14. $120, $60, $30
15- 3. 15. 75. 375^
1875.
16. $108, I144,
$192, I256.
17. 1^,01 I.I.
18. 8, 4, 2, I.
INFINITE SERIES.
Page 231.
1. li.
2. J.
3. I.
4. if
Pagre 257.
2. 1548.86.
3- I-973-
4. Ill
6. 78.
7. .0375-
8. 14.38.
9. 2.723.
10. 2906.3.
5. 2.
6. 9.
7. 10.
8. }.
LOGARITHMS.
11. .1814.
12. — 4.619.
Page 239.
14. .0003321.
15- 33-335-
16. 191.77.
18. 5.23.
19. 1.0836.
9- I-
10. •
a — I
ir. 50 rods.
20. 2.504.
21. 2.124.
Page 240*
23- -342 + .
24. .546+.
25. .324-f.
26. Given.
BUSINESS
Pages 245-260
2. $349.60.
4. i6| per cent.
6. $12600.
FORMULAS,
8. $840.
IP. $600.
1 2. 1 6f years.
13. 10 years.
^5* 2| per cent.
17. $2010.14.
19. $5414.28.
21. $2769.23, pr.w.;
P-77:
disc.
22. $6000, pr. w. ;
$1800, disc.
24. $1718.75-
26. $2125.
28. $2.33j.
29. $8.83+.
31- 5MI^ per ceiki.
32. 12^ per cent
33' 9ts per cent.
34. 0. 8 per cents.
^6. $24630.54, in.;
$369.46, com.
38. $1332.
3?- $6290.15.
TEST EXAMPLES,
323
40. $905.80.
41. $3278.69.
42. $2278.48.
44. $6130.67.
45. $2767.60.
47. $2336.25.
49. $14166.67.
51. $14775-
53. $249.77.
Pages 265-268,
-6.
4. 6v — I.
5. V—xy,
7. I-
V?
y
a/2.
10. 5
c
a
2. Impossible.
3. Impossible.
I.
2.
3-
4-
5-
6.
7-
8.
9-
10.
II.
12.
13-
14.
15-
16.
Pagre 274,
75«.
57^+ 7-
3ff2; + 3fl5j-f2C(^
7^6? + 3Cf?
— io:c?/+ 5rym.
31.
c (3^ _ 6Z»2c
— ccl).
3^ (2/ — 3^
2x2 (2« — i).
X (a— l).
20, II.
17. 24 shots.
EST EXAMPLES.
rage 275,
18. ±^/ab
±
« + I
4
19.
20
a — h
x{zxy + 2z),
22. (3^_c)(3^ — c).
23- •
24. 640 rodSc
25. 29 J miles.
b-i
26.
27.
28.
29.
30-
^ I — «^
32. 120.
33. $50.
b+ I
I.
a — b
3^2
«^
34.
35-
36.
37.
38.
39-
40.
41.
42.
43-
44.
45-
46.
47-
48.
49.
50-
51-
Page 276.
6 hours.
x = s; y
2.
85 f miles.
8ig.;5il.
x = 4', y = 6\
x=2) y = ^',
X r= $40 A,
y = $60 B,
z = $80 0.
3 meters f. w.
6 " h. w.
8 ft. one; 10 ft
49 ana 77.
48 meters.
$2.46 a meter.
$100 horse,
$200 carriage.
324
TEST EXAMPLES.
52. .
53. 65 hectares y.
100 " elder.
54. 26.
55. I5 1.; $6s.
56. 577 v.; 848 V.
57. 9oyrs.A;45y.
B; yy. C.
58- 9 Vsj
59. 2/ a/i + «.
Page 278,
60. {:c^)^ (f)i.
61. V27 (a — h)\
62. 90 cts.
63. $10; $18.
64. 15 men.
65. 28 persons.
66. I4 b. %(k w.
67. 9.
68. «.
69.-!-.
^ I — «
70. « -f J.
71. 240 liters.
72. .
73. 12 and 8.
74. ^180.
75- A-
76. $90.
77. ^321.
78. 24 s. $5 pr.
79. 81.
80. 45 m. ; 105 m.
81. 80 A; 70 B.
82. Vx — ^7.
83. V3^ + vsy-
84. ^2 ^ 6t? — 2^2^/
+ 9 — 6^^+^^.
85. 1 1 2000 Mayor;
$1200 Clerk.
S6. 216 g. ; 3of 1,
l^agre 280.
87. 8 and 6.
^^. 8 cts.
89. 10 days.
90. I1407 B;
$469 A.
91. $50 cow;
I200 h.
92. 6 miles.
93. 100 ft. x6oft.
94. lh 12^ 17,
2 if, 26f
95- 637!.^
96. 50 pair.
97. 18 and 14.
98. 32 h.; 75 1.
99. 10 y. B;
30 y. M.
100. $105.
Page 281,
1 01. 8:43^7yo'cik
102. 250 pp. A;
320 " B.
103. 20 days.
104. i4f yrs.
T05. I30.
106. $9000 whole i
$4800 A;
$4200 B.
107. 184 V. and
185 V.
108. 24 d.; 48 d.
109. 45.
no. 6, 18, 54, 162,
111. 16 hats.
Page 282,
112. 97656250.^^
113. 10 ft.; 30 ft.;
50 ft.
114. 95; 42; 82.
115. 2, 4, 8.
116. 12 weeks.
117. |i8o; $120.
118. |!30.
119. 45 A.; 55 A
120. 14348906.
121. 125 p. 175 b
122. 375 men.
" OF THE
DIVERSITY
^D 21-100^-8/34
Is and
*rooklyn
been to
e lower
itiaustive
ic which
but that
enabling
e author
n posses-
ience into
products,
lies are fol-
bff is made
to take the
' Sentential
d through to
lar and High
.-. This work
ae of analyz-
e Btudent to
that is being
unfolded, to
'. G. 6. Albfe,
ool, Oshkosh,
York.
AText-Book on English Literature,
Witli copious extracts from the leading authors, English and Ameri-
can. With full Instructions as to the Method in which these are
to be studied. Adapted for use in Colleges, High Schools,
Academies, etc. By Brainerd Kellogg, A.M., Professor of
the English Language and Literature in the Brooklyn Collegiate
and Polytechnic Institute, Author of a " Text-Book on Rhet-
oric," and one of the Authors of Reed & Kellogg's " Graded
Lessons in English," and ** Higher J^sons in English."
Handsomely printed. 12mo, 478 p]
The Book is divided into the following Te\
Period I. — Before the Norman Con
Prom the Conquest to Chaucer's dej
From Chaucer's death to ElizabetT
beth's reign, 1558-1603. Period
Restoration, 1603-3660. Period
death, 1660-1745. Period VII
Revolution, 1745-1789. Pe:
1789, onwards.
Each Period is preced<
great historical events tl
ing the literature of th
The author aims ' '
help himself to.
their relations to
the authors' ti
helping- to
asshonlrl'
allow^
8el(
t
. Period II.—
0. Period III.—
eriod IV. — Eliza-
th's death to the
storationto Swift's
th to the French
rench Revolution,
brief resum^ of the
. shaping or m color*
that which he cannot
places in the line and
pil; it throws light upon
great influences at work,
it points out such of these
limits of a text-hook would
writers of each Period. Such are
c traits of their authors, both in
of these extracts have ever seen the
them have been worn threadbare by
iiie pupil's familiarity with them in the^
selections are to be studied, soliciting and
exactmg tii^ jm.«.i.^..u a.h ^y^.cj step of the way which leads from the
author's diction up through his style and thought to the author himself,
and in many other ways it places the pupil on the best possible footing with
the authors whose acquaintance it is his business, as well as his pleasure, to
make.
Short estimates of the leading authors, made by the best English and
American critics, have been inserted, most of them contemporary with us.
The author has endeavored to make a practical, common-sense text-
book : one that would so educate the student that he wot;dd know and
enjoy good literature.
'• I find the book in its treatment of English literature superior to any other I
have examined. Its main feature, which should be the leading one of all similar
books, is that it is a means to an end, simply a guide-book to the study of Enj;li!?h
literature. Too many studants in the past 'have studied, not the literature of the
English language, but some author's opinion of that literature. I know from ex-
perience that your method of treatment will prove an eminently successful one." —
Jarms H. Shults, Prin. qf the West High School, Cleveland^ 0.
Effingham maynard & Co., Publishers, New York.
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