Algebra for beginners

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Solu^ T li. rr ,<) Cf^ !^

\

HARVARD UNIVERSITY

LIBRARY OF THE

Department of Education

COLLECTION OF TEXT-BOOKS

Con1ribate<3 by the Publishers

TRANSFERRED

3 2044 097 012 397

yJ6

ALGEBRA FOR BEGINNERa

Mm

o

ALGEBRA

FOR BEGINNERS

BT

H. S. HALL AND S. R. KNIGHT,

AuTUORS OF ''Elementary Algebra for Schools/* "Higher
Algebra," " Ei^ementary Trigonometry/* Etc. Etc.

REVISED AND ADAPTED TO AMERICAN SCHOOLS

BY

FRANK L. SEVENOAK, A.M., M.D.,

Professor of Mathematics and Assistant Principal in

tub Stevens School, Academic Department or

tub Stevens Institute of Tbounology

Wefa gotfe
MACMILLAN & CO,

AND LONDON

1895

AU rights reserved

M.trEiiiHliMlJbrar)f

fMhVMD COLLEeE LM/MV

TRANSFERRED FROM THC

LIDHARY OF THE

6RADUATE SCHOOL OF U^UCAHON

COPTKIOHT, 1895,

By MACMILLAN & CO.

Norb)ooti $Tra0 :

J. S. Gushing & Co. — Berwick & Smith.

Norwood, Mass., U.S.A.

PREFACE.

The rearrangement of the Elementary Algebra of
Messrs. Hall and Knight was undertaken in the hope
of being able to give to our advanced secondary schools
a work that would fully meet their requirements in this
important study. Many changes were made and ad-
ditional subject-matter introduced. The Algebra for
Beginners^ however, so fully meets the needs of the
class of students for which it was written, that we have
made only such changes as seemed to bring out more
clearly important points, and better adapt it to American
schools.

With reference to the arrangement of topics, we quote
from Messrs. Hall and Knight's preface to a former
edition :

"Our order has been determined mainly by two con-
siderations : first, a desire to introduce as early as pos-
sible the practical side of the subject, and some of its
most interesting applications, such as easy equations and
problems; and secondly, the strong opinion that all
reference to compound expressions and their resolution
into factors should be postponed until the usual opera-
tions of Algebra have been exemplified in the case of
simple expressions. By this course the beginner soon

V

VI PREFACE.

becomes acquainted with the ordinary algebraical proc-
esses without encountering too many of their difficulties ;
and he is learning at the same time something of the
more attractive parts of the subject.

" As regards the early introduction of simple equations
and problems, the experience of teachers favors the
opinion that it is not wise to take a young learner
through all the somewhat mechanical rules of Factors,
Highest Common Factor, Lowest Common Multiple,
Involution, Evolution, and the various types of Frac-
tions, before making some effort to arouse his interest
and intelligence through the medium of easy equations
and problems. Moreover, this view has been amply sup-
ported by all the best text-books on Elementary Algebra
which have been recently published."

The work will be foimd to meet the wants of all who
do not require a knowledge of Algebra beyond Quadratic
Equations — that portion of the subject usually covered
in the examination for admission to the classical course
of American Colleges.

FRANK L. SEVENOAK.
JuNB, 1895.

CONTENTS.

CHAP. PAOB

I. Definitions. Substitutions 1

II. Negative Quantities. Addition op Lire Terms . 7

III. Simple Brackets. Addition 11

IV. Subtraction , , 16

Miscellaneous Examples I. 19

Y. Multiplication 21

VI. Division 31

VII. Removal and Insertion of Brackets . . • • 38

Miscellaneous Examples II. 42

VIII. Revision of Elementary Rules 44

IX. Simple Equations 52

X. Symbolical Expression 59

XI. Problems Leading to Simple Equations . . . C4
XII. Highest Common Factor, Lowest Common Multiple
OF Simple Expressions. Fractions Involving

Simple Expressions onlt 71

XIII. Simultaneous Equations 77

XIV. Problems Leading to Simultaneous Equations . 85
XV. Involution 89

XVI. Evolution 93

XVII. Resolution into Factors. Converse Use of Factors 101

Miscellaneous Examples III 114

• ••

viu

CONTENTS.

OHAF.

XVIII.
XIX.

XXL

XXII.

XXIII.

XXIV.

XXV.
XXV^I.

Highest Common Factor of Compound Expressions

Multiplication and Division op Fractions

Lowest Common Multiple of Compound Expressions

Addition and Subtraction of Fractions

Miscellaneous Fractions

Harder Equations .

Harder Problems .

Miscellaneous Examples IY

Quadratic Equations

Problems Leading to Quadratic Equations

Miscellaneous Examples V* . . .

PAGE

116
122
126
129
139
146
i53
158
162
173
177

ALGEBRA.

CHAPTER I.

Definitions. Substitutions.

1, Algebra treats of quantities as in Arithmetic, but with
greater generality ; for while the quantities used in arithmetical
processes are denoted by figures which have one single definite
value, algebraical quantities are denoted by symbols which may
have any value we choose to assign to them.

The symbols employed are letters, usually those of our own
alphabet ; and, though there is no restriction as to the numerical
values a symbol may represent, it is understood that in the same
piece of work it keeps the same value throughout. Thus, when
we say "let a = l," we do not mean that a must have the value
1 always, but only in the particular example we are considering.
Moreover, we may operate with symbols without assigning to
them any particular numerical value at all ; indeed it is with
such operations that Algebra is chiefly concerned.

We begin with the definitions of Algebra, premising that the
symbols +, -, x, -r-, will have the same meanings as in
Arithmetic.

2, An algebraical expression is a collection of symbols;
it may consist of one or more termS) which are separated from
each other by the signs -f and -. Thus 7a+56-3c-^-f2y is
an expression consisting of five terms.

Note. When no sign precedes a term the sign + is understood.

3, Expressions are either simple or compound. A simple
expression consists of one term, as ba. A compound expression
consists of two or more terms. Compound expressions may be

^ H.A. A

2 ALGEBRA. [chap.

farther distinguished. Thus an expression of two terms, as
3a — 26, is called a binomial expression ; one of three terms, as
2a-36+c, a trinomial; one of more than three terms a multi-
nomial.

4. When two or more quantities are multiplied together the
result is called the product. One important difference between
the notation of Arithmetic and Algebra should be here remarked.
In Arithmetic the product of 2 and 3 is written 2x3, whereas
in Algebra the product of a and h may be written in any of
the forms ax 6, a. 6, or ab. The form ah is the most usuaL
Thus, if a=2, 6=3, the product a6 = a x 6 = 2 x 3=6 ; but in
Arithmetic 23 means " twenty -three," or 2 x 10+3.

6. Each of the quantities multiplied together to form a pro-
duct is called a factor of the product. Thus 5, a, 6 are the
factors of the product 5a6.

6, When one of the factors of an expression is a numerical
quantity, it is called the coefficient of the remaining factors.
Thus in the expression 5a6, 5 is the coefficient. But the word
coefficient is also used in a wider sense, and it is sometimes
convenient to consider any factor, or factors, of a product as
the coefficient of the remaining factors. Thus in the product
6a6c, 6a may be appropriately called the coefficient of he, A
coefficient which is not merely numerical is sometimes called a
literal coefficient.

Note. When the coefficient is unity it is usually omitted. Thus
we do not write la, but simply a.

7. If a quantity be multiplied by itself any number of times,
the product is called a power of that quantity, and is expressed
by writing the number of factors to the right of the quantity
and above it. Thus

a X a is called the second power of a, and is written c? ;

axaxa third pov^er of a, a^;

and so on.

The number which expresses the power of any quantity is
called its index or exponent* Thus 2, 5, 7 are respectively
the indices of a^, a\ a^.

Note, a' is usually read **a squared"; a' is read **a cubed";
a* is read " a to the fourth " ; and so on.

When the index is unity it is omitted, and we do not write
a}, but simply a. Thus a, la, a^, la^ all have the same meaning.

I.] DEFINITIONS. SUBSTITUTIONS. 3

8. The beginner must be careful to distinguish between
coefficient and index.

Example 1. What is the difference iu meaning between 3a and a^ ?

By 3a we mean the product of the quantities 3 and a.

By a^ we mean the third power of a ; that is, the product of the
quantities a, a, a.

Thus, if a = 4,

3a = 3xa = 3x4 = 12;

a' = axaxa = 4x4x4=s64.

Example 2. If 6 = 5, distinguish between 46^ and 26*.
Here 462 = 4x6x6 = 4x6x5 = 100;

whereas 26* = 2x6x6x6x6 = 2x5x5x5x5 = 1250.

Example 3. If jc = 1, find the value of 5a:*.

Here 6ar* = 5xa;xa;xa;xa; = 5xlxlxlxl = 5.

Note. The beginner should obserye that every power of 1 is 1.

9. In arithmetical multiplication the order in which the
factors of a product are written is immaterial. For instance
3x4 means 4 sets of 3 units, and 4x3 means 3 sets of 4 units ;
in each case we have 12 units in all. Thus

3x4 = 4x3.
In a similar way,

3x4x5=4x3x5=4x5x3;

and it is easy to see that the same principle holds for the
product of any number of arithmetical quantities.

In like manner in Algebra ah and 6a each denote the product
of the two quantities represented by the letters a and 6, and
have therefore the same value. Again, the expressions ahc,
acb, hac, hca, cab, cba have the same value, each denoting the
product of the three quantities a, 6, c. It is immaterial in
what order the factors of a product are written ; it is usual,
however, to arrange them in alphabetical order.

Fi'actional coefficients which are greater than unity are
usually kept in the form of improper fractions.

13
Example 4. If a = 6, a; = 7, z = 5, find the value of — axz.

Here l?aa;z = ^x6x7x5 = 273.

4 ALGEBRA. [CHAP.

EXAMPLES I. a.

If a = 5, 6 = 4, c = 1, a: = 3, y = 12, z = 2, find the value of

1. 2a. 2. a-. 3. 3z. 4. z". 5. c*.

6. 4c. 7. 462. Q ^ 9 a,3. iQ^ 3a,

11. 7y2. 12. 8a3. 13. 62*. 14. 5s«. 15. 7c«.

If a = 6, /) = 4, g = 7, r = 5, a: = 1, find the value of

16. ap. 17. 3p^. 18. Sqx. 19. Syj^. 20. Saqx,

21. 7>gr. 22. 8a^n 23. Iqrx. 24. 2a/>a:. 25. ^x*.
26. 3;)^ 27. 8r*. 28. Qapqx. 29. Ga:^. 30. ar^^

If A = 5, ^ = 3, a; = 4, y = 1 , find the value of

31. Ji». 32. Ihc. 33. Jy'. 34. ^hkx. 35. j^.
36. Jar*. 37. ^t». 38. j|gA». 39. Jy". 40. i*fc»T^.

10. Wlien several difi'erent quantities are multiplied together
a notation similar to that of Art. 7 is adoj)ted. Thus aahhhhcddd
is written a%*cd\ And conversely 7a^cd^ has the same meaning
asl xaxaxaxcxdxd.

Example 1. If c = 3, d = 5, find the value of 16c*cP.

Here 16c*rf* = 16 x 3* x 5^ = (16 x 6^) x 3* = 2000 x 81 = 162000.

Note. The beginner should observe that by a suitable combination
of the factors some labour has been avoided.

Example 2, If 2? = 4, g' = 9, r = 6, « = 5, find the value of

S2qr^

81p»
•a 32g?'«_32x9x6^_ 32x9x6x6x6 _3

^^^ Sip'' 81x45 81x4x4x4x4x4 4'

11, If one factor of a product is equal to 0, the product must
be equal to 0, whatever values the other factors may have, A
factor is usually called a zero factor.

For instance, if x—^ then ali^xy^ contains a zero factor.
Therefore ah^xy^=0 when ^=0, whatever be the values of a, 6, ?/.

Again, if c = 0, then c^=0 ; therefore a6V=0, whatever values
a and h may have.

Note. Every power of is 0.

l] DEFINITIONS. SUBSTITUTIONS. 5

EXAMPLES I. b.

If a = 3, 6 = 2, 7? = 10, g = 1, a; = 0, z = 7, find the valae of

1, 36p. 2, Sax. 3. Spgz. 4, ^qz. 5. ^/>s«

6. 36Y 7, az2. 8. 9^. 9. gz*. 10. ^f^^px*.

11. a3/>4a;. 12, 8i?V. 13. «»'«» 14. i?a:V. 15. 8aV.

If A; = 1, ^ = 2, m = 0, JO = 3, g = 4, r = 5, find the value of

16. ^4- 17. ^- 18. ^. 19. ^'. 20. i^.
26. ^. 27. g. 28. 5. 29. '-^. 30. ^^

12. We now proceed to find the numerical value of expres-
sions which contain more than one term. In these each term
can be dealt with singly by the rules already given, and by
combining the terms the numerical value of the whole expres-
sion is obtained.

13. We have already, in Art. 8, drawn attention to the
importance of carefully distinguishing between coefficient and
index; confusion between these is such a fruitful source of
error with beginners that it may not be unnecessary once more
to dwell on the distinction.

Example. When c = by find the value of c* - 4c + 2c^ - 3c^.
Here c* = 5* = 5x5x5x5 = 625 ;

4c = 4 X 5 = 20 ;

2c' = 2 X 53 = 2 X 5 X 5 X 5 = 250 ;

3c2 = 3 X 52 = 3 X 5 X 5 = 75.
Hence the value of the expression

= 625-20 + 250-75 = 780.

14. The beginner must also note the distinction in meaning
between the sum and the product of two or more algebraical
quantities. For instance, ah is the product of the two quan-
tities a and 6, and its value is obtained by multiplying them
together. But a-f-6 is the sum of the two quantities a and 6,
and its value is obtained by adding them together.

6 ALGEBRA. . [CHAP. L

Thusif a=ll, 6 = 12,

the sttm of a and 6 is 11 + 12, that is, 23 ;
thQ product of a and 6 is 11 x 12, that is, 132.

15. By Art. 11 any term which contains a zero factor is
itself zero, and may be called a zero term.

Example, If a = 2, 6 = 0, a; = 5, y = 3, find the value of

The expression = (5 x 2^) - + (2 x 5« x 3) +0

= 40 + 150=190.
Note. The two zero terms do not affect the result.

16. In working examples the student should pay attention
to the following hints.

1. Too much importance cannot be attached to neatness of
style and arrangement. The beginner should remember that
neatness is in itself conducive to accuracy.

2. The sign = should never be used except to connect
quantities which are equal. Beginners should be particularly
careful not to employ the sign of equality in any vague and
inexact sense.

3. Unless the expressions are very short the signs of equality
in the steps of the work should be placed one under the other.

4. It should be clearly brought out how each step follows
from the one before it ; for this purpose it will sometimes be
advisable to add short verbal explanations ; the importance of
this will be seen later.

EXAMPLES I. c.

If a = 4, 6 = 1, c = 3, /= 5, </ = 7, A = 0, find the value of

1, 3/+6A-7&. 2. 7c-9A + 2a. 3, 4^ -5c -96.

4. 3gf-4/i + 7c. 5. 3/-2gf-6. 6. 96-3c + 4A.

7. 3a-96 + c. 8. 2/-3^ + 5a. 9. 3c-4a + 76.

10. 3/+5A-2c-46 + a. H. 6A - 76 - 5a - 7/+ 9(7.

12. 7c + 56-4a + SA+3sr. 13. 96+a-3(7 + 4/+7A.

14. fg + gh-ah. 15. gb-Shc+fh. 16. /h + hh-3hc.

17. /2-3a2 + 2c5. 18. 63_2A3+3a2. 19. 3b^-2b^-\-ih^-2h*,

CHAPTER 11.
Negative Quantities. Addition of Like Terms.

17. In his arithmetical work the student has been accus-
tomed to deal with numerical quantities connected by the signs
+ and — ; and in findinff the value of an expression such as
l|-f7§-3j + 6-4i he understands that the quantities to which
the sign + is prefixed are additive, and those to which the sign
— is prefixed are subtractive, while the first quantity, 1|, to
which no sign is prefixed, is counted among the additive terms.
The same notions prevail in Algebra ; thus in using the expres-
sion *7a+3b-4c-2d we imderstand the symbols 7a and 36 to
represent additive quantities, while 4c and 2d are subtractive.

18. In Arithmetic the sum of the additive terms is always
greater than the sum of the subtractive terms ; if the reverse
were the case the result would have no arithmetical meaning.
In Algebra, however, not only may the sum of the subtractive
terms exceed that of the additive, but a subtractive term may
stand alone, and yet have a meaning quite intelligible.

Hence all algebraical quantities may be divided into positive
quantities and negative quantities, according as they are
expressed with the sign + or the sign - ; and this is quite
irrespective of any actual process of addition and subtraction.

This idea may be made clearer by one or two simple illus-
trations.

(i) Suppose a man were to gain $100 and then lose $70, his
total gain would be $30. But if he first gains $70 and then
loses $100 the result of his trading is a loss of $30.

The corresponding algebraical statements would be

$100 -$70 =-|-$30,
$70 -$100= -$30,

8 ALGEBRA. [CHAP.

and the negative quantity in the second case is interpreted as a
debt, that is, a sum of money opposite in character to the positive
quantity, or gain, in the first case ; in fact it may be said to
possess a subtractive quality which would produce its effect
on other transactions, or perhaps wholly counterbalance a sum
gained.

(ii) Suppose a man starting from a given point were to walk
along a straight road 100 yards forwards and then 70 yards
backwards, his distance from the starting-point would be 30
yards. But if he first walks 70 yards forwards and then 100
yards backwards his distance from the starting-point would be
30 yards, but on the opposite side of it. As before we have

100 yards - 70 yards = + 30 yards,
70 yards — 100 yards = - 30 yards.

In each of these cases the man's ahsolvie distance from the
starting point is the same ; but by taking the positive and
negative signs into account, we see that — 30 is a distance from
the starting point eqival in magnittide but opposite in direction
to the distance represented by +30. Thus the negative sign
may here be taken as indicating a reversal of direction.

(iii) The freezing point of the Centigrade thermometer is
marked zero, and a temperature of 15° C. means 15** above the
freezing point, while a temperature 15" below the freezing point
is indicated by — 15° C.

19. Many other illustrations might be chosen ; but it will
be sufficient here to remind the student that a subtractive
quantity is always opposite in character to an additive quantity
of equal absohUe value. In other words subtraction is t/ie reverse
of addition.

20. Dkfinition. When terms do not differ, or when they
difier only in their numerical coefficients, they are called like,
otherwise they are called unlike. Thus 3a, 7a ; 5a'^6, 2a26 ;
3a^b^, — AaW are pairs of like terms ; and 4a, 36 ; 7a^, 9a^6 are
pairs of unlike terms.

Addition of Like Terms.

Rule I. The sum of a number of like terms is a like term.
Rule II. If all the terms are positive, add the coefficients.

n] NEGATIVE QUANTITIES. ADDITION OF LIKE TERMS. 9

Example, Find the value of 8a + 5a.

Here we have to increase 8 like things by 5 like things of the
same kind, and the aggregate is 13 of such things ;

for instance, 8 lbs. + 5 lbs. =13 lbs.

Hence also, 8a + 6a = 13a.

Similarly, 8a + 5a + a + 2a + 6a = 22a.

Rule m. If all the terms are negative, add the coejictenta
numerically and prefix the minus sign to the sum.

Example, To find the sum of - 3a:, - 5af, — 7ar, - x.

Here the word sum indicates the aggregate of 4 subtractive
quantities of like character. In other words, we have to take away
successively 3, 5, 7* 1 like things, and the result is the same as
taking away 3 + 5 + 7 + 1 such things in the aggregate.

Thus the sum of -3a:, —5a:, -7a:, -x is -16a:.

Rule IV. If the terms are not all of the same sign, add to-
gether separately the coefficients of all the positive term^ and the
coefficients of all the negative terms ; the difference of these two
results, preceded by the sign of the greater, will give the coeffi^dent
of the sum required.

Example 1. The sum of 17a; and —8a; is 9a;, for the difference of
17 and 8 is 9, and the greater is positive.

Example 2. To find the sum of 8a, - 9a, - a, 3a, 4a, - 11a, a.

The sum of the coefficients of the positive terms is 16.

negative 21.

The difference of these is 5, and the sign of the greater is nega-
tive ; hence the required sum is - 5a.

We need not however adhere strictly to this rule, for the
terms may be added or subtracted in the order we find most
convenient.

This process is called collecting terms.

21. When quantities are connected by the signs + and — ,
the resulting expression is called their algebraical 8Um.

Thus 11a — 27a + 13a = —3a states that the algebraical sum
of 11a, —27a, 13a is equal to —3a.

22, Tlie sum of two quantities numerically equal but with
opposite signs is zero. Thus the sum of 5a and — 5a is 0.

10 ALGEBRA. [CHAP. ll.

EXAMPLES n.

Find the sum of

1. 2a, 3a, 6a, a, 4a, 2. 4x, x, 6x, 6a;, 8a;.

3, 66, 116, 86, Oft, 56. 4. 6r, 7c, 3c, 16c, 18c, 101c.

5. 2p, Py 4/>, 1p, 6p, 12/?. 6. d, 9rf, 3d, Id, id, 6rf, lOd,

7. -2a;, -6a;, -lOx, -8a;. 8. -36, -136, -196, -56.

9. - y, - 4y, - 2y, - 6y, - 4y. 10. - 17c, - 34c, - 9c, - 6c.

11. -2\y, -5y, -3y, - 18y. 12, -4m, -13w, -17wi, -59m.

13. -48, 3«, 8, 2«, -2«, -s. 14. lly, -9y, -7y, 5y, 7y.

15, 3a;, -10a;, - 7a;, 12a;, 2a;. 16. 8a6, -6a6, 5a6, -4a6.

17. 2a;y, -4a;y, - 3a;y, ary, 7a;y. 18. 5pq, -Spq, Spq, -4pg.

19. a6c, - 3a6c, 2a6c, - 5a6c. 20. - a?yz, - 2a:yz, 7a;y2, - a;y2.

Find the value of

21, -9a2+lla2+3aa-4o«. 22. 368-26'+ 763-968.

23. -Ila3}-3a8-8a8-7a8 + 2a*. 24. 2x^-Sx^-63^-0a^.

25. a262 - 7a262 + 8a-62 + 9a262. 26. a'^x-Ua^x + Sa^x-2a^x.

27. 2/) V - 31;? V + 1 7/^ V- 28. 1rn*n - I6m*n + 3w*».

29 9a6cd-lla6cd-41a6cd. 30. 13/?ga? - 5a;pg - 19g/?a;.

CHAPTER III.
Simple Brackets. Addition.

23. When a number of arithmetical quantities are connected
together by the signs + and - , the value of the result is the
same in whatever order the terms are taken. This also holds
in the case of algebraical quantities.

Thus a- 6 + c is equivalent to a + c-by for in the first of the
two expressions b is taken from or, and c added to the result ; in
the second c is added to a, and b taken from the result. Similar
reasoning applies to all algebraical expressions. Hence we
may write the terms of an expression in any order we })lease.

Thus it appears that the expression a-b may be written in
the equivalent form -b + a.

To illustrate this we may suppose, as in Art. 18, that a rep-
resents a gain of a dollars, and —b a, loss of b dollars: it is
clearly immaterial whether the gain precedes the loss, or the
loss precedes the gain.

24. Brackets ( ) are used to indicate that the terms enclosed
within them are to be considered as one quantity. The full use
of brackets will be considered in Chap. vii. ; here we shall deal
only with the simpler cases.

8 +(13 + 5) means that 13 and 5 are to be added and their
sum added to 8. It is clear that 13 and 5 may be added
separately or together without altering the result.

Thus 8 + (13+5) = 8 4-13 + 5 = 26.

Similarly a+(b + c) means that the sum of b and c is to be
added to a.

Thus a+(6+c)=a + 6 + c.

8 + (13 — 6) means that to 8 we are to add the excess of 13 over
5 ; now if we add 13 to 8 we have added 5 too much, and must
therefore take 5 from the result.

Thus 8 + (13-5) = 8 + 13-5 = 16.

Similarly a + {b — c) means that to a we are to add 6, diminished
bye.
Thus a+(6-c)=a+6-c

12 ALGEBRA. [chap.

In like manner,

a + 6-c+(c?-e-/)=a+6-c + c?-c-/.

By considering these results we are led to the following rule :

Rule. When an expression within brackets is preceded by the
sign + , the brackets can be removed vdthout making any change in
the expression.

25. The expression a-(6 + c) means that from a we are to
take the sum of b and c. The result will be the same whether
b and c are subtracted separately or in one sum. Thus

a — ib'\-c)=a — b-c.

Again, a — ijb — c) means that from a we are to subtract the
excess of b over c. If from a we take b we get a-b; but by so
doing we shall have taken away c too much, and must therefore
add c to a - 6. Thus

a-(6 — c)=rt — 6+c.
In like manner,

a — b-'{c — d-e)=a-b — c-\-d-\-e.

Accordingly the following rule may be enunciated :

Rule. When an expression icithin brackets is preceded by the
sign — , the brackets mxiy be removed if the sign of every term within
the brackets be changed.

Addition of Unlike Tenns.

26. When two or more like terms are to be added together we
have seen that they may be collected and the result expressed
as a sinqle like term. If, however, the terms are unlike they
cannot oe collected; thus in finding the sum of two unlike
quantities a and 6, all that can be done is to connect them by
the sign of addition and leave the result in the form a + b.

27. We have now to consider the meaning of an expression
like a 4- ( — 6). Here we have to find the result of taking a
negative quantity —b together with a positive quantity a.
Now — b implies a decrease, and to add it to a is the same in
effect as to subtract 6 ; thus

a-\-{ — b) = a-b ;

that is, the algebraical sum of a and — 6 is expressed by a — 6.

28. It will be observed that in Algebra the word sum is used
in a wider sense than in Arithmetic. Thus, in the language of
Arithmetic, a — b signifies that b is to be subtracted from a.

III.]

ADDITION.

13

and bears that meaning only ; but in Algebra it is also taken
to mean the sum of the two quantities a and — h without any
regard to the relative magnitudes of a and 6.

Example 1. Find the sum of 3a - 56 + 2c, 2a + 36 - d, - 4a + 26.

Thesum = (3a-56 + 2c) + (2a + 36-fZ) + (-4a + 26)
= 3a-56 + 2c + 2a + 36-d-4a + 26
= 3a + 2a-4a-56 + 36 + 26 + 2c-d
= a + 2c - (^,
by collecting like terms.

The addition is however more conveniently effected by the
following rule :

Rule. Arrange the expressions in lines so that the like terms
inay he in the same vertical columns: then add each column
beginning with that on the left.

3a- 56 + 2c
2a + 36 "d

4a + 26

a

+ 2c-d

The algebraical sum of the terms in the
first column is a, that of the terms iii the
second column is zero. The single terms
in the third and fourth columns are
brought down without change.

Example 2. Add together - 5a6 + 66c - 7ac ; 8ab + 3ac - 2ad ;

- 2ab + 4ac + 5ad ; be - Sab + 4ad.

- Sab + 66c — 7ac

Here we first rearrange the ex-
pressions so that like terms are in
the same vertical columns, and tlien
add up each column separately.

8a6
-2a6
-3a6 +

+ 3ac - 2ad

+ 4ac + bad

be +4ad

- 2ab + 76c

+ 1ad

EXAMPLES m. a.

Find the sum of

1. 3a + 26-5c

2. 3a: + 2y + 6z

3. 4;; + 3(7 + 5r

4. 7a-56 + 3c

5. 8^-2m + 57i

-4a + 6-7c; 4a-36 + 6c.
x-Sy-Zz; 2x + y-Sz.
-2p + SqSr; p-q + r.
lla4-26-c; 16a + 56-2c.
- 6/ + Tm + in ; -I- 4m - 871.

6. 6a-76 + 3c-4d; 66-5c + 3d; 6 + 2c-f/.

7. 2a + 46-5a:; 26-5a;; -3a + 2y; -Qb + Sx + y.

8. 7a:-6y-7«; 4x + yi 6z; 6a?-3y + 2».

14 ALGEBRA. [cjhap.

9, a-26 + 7c + 3; 26-3c + 6; 3c + 2o; a-8-7c.

10. 5-x-y; T + 2x; Sy-2z; -4 + a;-2y.

11. 25a-156 + c; 4c- 106+ 13a; a-c + 206.

12. 2a-36-2c + 2a:; 6a: + 36-7c; 9c-6a:-2a.

13. 3a-6c + 26-2d; 6 + 2d-a; 5c + 3/+ 3e -2a - 36.

14. p-q-\-7r; Qq + r-p; q-Zp-r; Qq-7p.

15. nab-lSkl-dxy; Ixy; I2kl-5ab; Sxy-4kl-ab.

16. 2aa; - 36y - 2cs ; 2by-ax + 1cz; ax-4cz-\-7hy ; cz-Qby.

17. 3aa; + C2-46y; Ihy-Sax-cz; -36y + 9aa:.

18. 3 + 6cd; 2/[7-3«t; l-5cd; -4 + 2«<-/r7-

19. 5ca: + 3/y-2 + 28; -2/y + 6-9«; -3«-4 + 2ca:-/y.

20. -3a6 + 7crf-65^r; 2ry + 8qr-cd; 2cd-3qr-\-ab-2ry.

29. Different powers of the same letter are unlike terms ;
thus the result of adding together ^jc^ and 3jc^ cannot be ex-
pressed by a single term, but must be left in the form 2x^-\-3a^.

Similarly the algebraical sum of ba^b\ -.3a6^, and -6* is
5a-62 - 3a6^ - 6*. This expression is in its simplest form and
cannot be abridged.

Example. Find the sum of 6x^ - 5a;, 2ar^, 6x, - 2a:*, - 3a:^, 2.

The8um = 6ar'-5a:+2a:2^5a,_2a:S_3aJ! + 2

= 6ar^-2ar» + 2a:a-3ar»-5a: + 5a; + 2

= 4ar»-a;2 + 2.
This result is in descending powers of x.

30. In adding together several algebraical expressions con-
taining terms with different powers of the same letter, it will be
found convenient to arrange all expressions in descending or
ascending powers of that letter. This will be made clear by
the following example.

Example 1. Add together Zx^ + 1 + ^ - bx^ -, 2x^ - % ~ Qx;
4a:-2ar» + 3ar^; 3ar^ - 9a: - a;^ ; x-7?-7^Jr4.

3ar» - 5a:2 + 6a: + 7
2a:2 - 9a: - 8
-2a:8 + 3a;2 + 4a:
3a:3- a:2-9a;

- x^ - x^+ a: + 4

3a:3_2a:2-7a: + 3

In writing down the first expression
we put in the first term the highest
power of a:, in the second term the
next highest power, and so on till the
last term in which x does not appear.
The other expressions are arranged in
the same way, so that in each column
we have like powers of the same letter.

-263 +

3a6-^

+ a3

—

a6H

5a26

-3a3

5^3

+ 8a«

ab'^ +

9a^b

-2a3

HI.] ADDITION. 15

Example 2. Add together
3a62-263 + a3; Sa^ft - ai^ - 3a' ; SaHSfr'; 9a'6-2o' + a6a

Here each expression contains
powers of two letters, and is
arranged according to dencend-
iiKj powers of h, and ascending

363 + 3a6H14a26 + 4a3 powers of a.

EXAMPLES m. b.

Find the sum of the following expressions :

1. a;2 + 3a?y-3y-; -Say^ + xy + 2y^; 2ar»-3ay+y».

2. 2a:2-2a: + 3; -2x^ + 5x + 4; si^-2x-6.

3. 5ar»-a;5+a;-l ; 2x^-2x + 5; -53^-\-5x-4.

4. a^-a^h + 5ah^ + ¥; - a» - lOaft^ + ^3 . 2a2fe + Saft^ - fts.

6. 3a:3-9a:3-lla; + 7; 2x!^-5x^+2; 6x^ + l5x'^-7x; 8a?-9.

6. 07^-50:3 + 82;. "J^xr^ + ix^ + Sx; 8a:3-9a:; 2x^-'Jsi^-4x.

7. 4m3 + 2m2-5m + 7; 3m3 + 6m*-2; -5m2+3w; 2wi-6.

8. aa:3-46a^* + ca;; ^bx^-2cx-d; bx^ + 2d; 2ax!^ + d.

9. py^-9qy + lr; -2py^ + Sqy-Qr\ 1qy-4r; Spy\

10. 6y3 + 20y2 + 3y-l; -2y + 5-7y^ -3y2-4 + 2y»-y.

11. 2-a + 8a2-a3; 2a3-3a2 + 2a-2 ; -Sa + la^-Sa^

12. l+2y-3y2-6y8; -l+2y2_y; 5y3 + 3y2 + 4.

13. aV-3a3ar^ + a;; 5a? + 7a3ar«; 4a3a~» - aV - 5a;.

14. x^-4tX*y"5s^^ ; 3x!h/ + 2oi^i^-Ga:y* ; iijpy^+Qxy*-y'^,

15. a3-4a26 + 6a6c; a^b - lOabc + c^ ; b^ + Sa^b + abc.

16. ap^-6bjP + 7cp; 5-ecp + 6bp^; 3-2ajt)«; 2c/)-7.

17. c7-2(r'+llc«; - 2c7 - 3c« + 5c5 ; 4(fi-l0c^; 4^-c^

18. 4/i3-7 + 3A^-2A; 7/t-3A3 + 2-A*; 2A* + 2^3-5.

19. 3a:3 + 2y2_5a; + 2; 7a:3 _ 5^2 + 7^. _ 5 . 9a:3+ll -8a; + 4y2 ;

6a; - y2 - i8ar^ _ 7.

20. x^ + 2xy + 3y^; Sz^ + 2yz-{-y^ ; a^* + 3^* + 2a:z ; z^-Sxy-Syz;

a:y+a»+yz-6z2-4y2-2ar*.

CHAPTER IV.
Subtraction.

31. The simplest cases of Subtraction have already come
under the head of addition of like terms, of which some are
negative. [Art. 20.]

Thus 5a-3a= 2a,

3a - 7a = — 4a,

— 3a - 6a = - 9a.

Since subtraction is the reverse of addition,

+6-6=0;

.'. a=a + b—b.

Now subtract —b from the left-hand side and erase —6 on
the right ; we thus get

a — ( -b) = a+b.

This also follows directly from the rule for removing brackets.
[Art. 25.]

Thus 3a-(-5a)=3a + 5ti

= 8a,

and - 3a - ( - 5a) = - 3a+5a

= 2a.

Subtraction of Unlike Tenns.

32. We may proceed as in the following example.

Example, Subtract 8a -- 2& - c from 4a - 36 + 6c.
The difference

The expression to be subtracted is
first enclosed in brackets with a minus
sign prefixed, then on removal of the
brackets the like terms are combined
by the rules already exclaimed in
Art. 20.

=4a-36-|-5c-(3a-26-c)
=4a-36 + 5c-3a+26+c
=4a-3a-36+26+6c+c
—a—h-\-Qc,

CHAP. IV.] SUBTRACTION. 17

It is, however, more convenient to arrange the work as follows,
the signs of all the terms in the lower line being changed.

4a - 36 -f- 5c
-3a + 26 + c

by addition, a- h + Qc

The like terms are written in
the same vertical column, and each
column is treated separately.

Hule. Change the sign of every term in the expression to ho
subtracted^ and add to the other expression.

Note. It is not necessary that in the expression to be subtracted
the signs should be actually changed ; the operation of changing
signs ought to be performed mentally.

Example 1 . From 5x^ + xy take 2a;- + %xy - 7y-.

6qiP+ xy
2ar^ + 8 OT/-7y^
3ar^-7a:y + 7y2

In the first column we combine mentally 5a:^
and - 2a:-, the algebraic sum of which is 3j-^. In
the last column the sign of the term - 7//"^ has to
be changed before it is put down in the result.

Example 2. Subtract Zx^ - 2x from 1 - a:*.

Terms containing different powers of the same letter being urdike
must stand in different columns.

^7^ +1

3a:^-2x
~a.'3-3a;2 + 2a;+l

In the first and last columns, as there is
nothing to be subtracted, the terms are put
down without change of sign. In the second
and third columns each sign has to be changed.

The re arrangement of terras in the first line is not necessary^ but
it is convenient, because it gives the result of subtraction in descend-
ing powers of x.

EXAMPLES IV.

Subtract

1, a + 26-c from 2a + 36 + c.

2, 2a-b + c from Sa -5h-c.

3, Sx-i y-z from x-4y-\-3z.

4, a: + 8y + 8z from 10a;-72/-6z.

5, -m-Zn+p from -2m hw-3p.

6, Sp-2q + r from 4p - 7(7 + 3r.

7, a -76 -3c from -4a + 36 + 8c.

8, -a-6-9c from -a + 6-9c.

H.A. B

18 ALGEBliA. [CHAP.

Subtract

9. 3a:-5y-7z from 2a:+3y-42.

10, -4a;-2y + llz from -x + 2y-13z.

11, '•2x-5y from a: + 3y-2z.

12, 3a;-y-83 from x + 2y,

13, Tn-2tt-p from ?» + 2n.

14, 2p-Sq-r from 2q - 4r,

15, ab~2cd-ac from -ab-^d+2ac. ^

16, Sab + 6cd-3ac-5ld from Sab + 5cd - iac - fibd,

17, -ary + ys-za; from 2a:y + za:.

18, -2pq-Sqr + 4r8 from gr-4r«.

19, - nin + 1 l»p - Spm from - 1 Inp.

20, ar^y - 2a:y2 + 3xyz fpQ^ 2a:V + 3a:ry2 - aryz.

From

21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40

a^-3x^-{x take -ar» + 3a~»-«.
-2a^»-ar2 take x^-x^-x.
a^ + I^-Sabc take l?-2abc.

- 8 + Gbc + 62c2 take 4 - 36c - 5b^cK
32^ - 2p^q + Ipq^ take j^q - 3pq^ + g^.
7 + a:-ar take 6 - a: + ar' + a:^.

- 4 + ar^y - a:yz take - 3 - 2a:-y + 1 laryz.
-8a2a:2 + 5a;2+15 take 9a^x^-Sx^-5,
j^ + r^-Spqr take r^-\-^ + 3pqr.
l-3ar' take ar^-3a;2+l.

2 + 3a;-7a:2 take 3ar»-3x-2.

ar*+lla:2 + 4 take 8a;2-6a;-3.

a3 + 5-2a2 take Sa' + Sa^-?.

a:* + 3ar^-a;2-8 take 2ar* + 3x^-a: + 2.

l-2a; + 3a:2 take 7ar»-4ar*+3a;+l.

ar^yz + y^za; take - 3y^a: - 2a:yz^ - a:^z.

40^2^- Sass^ + a^ take 3a^x^ + 7a^x^-a^

l-x-^x^-x*-x^ take ar* - 1 + a; - ar*.

-Smn^ + 15mhi+n^ take m^-n^ + Smn^-'jTJihi.

1 - p^ take 2p^ - 3pq^ - 2q\

33. The following exercise con tains miscellaneous examples
on the foregoing rules.

IV.] SUBTRACTION. 19

MISCELLANEOUS EXAMPLES L*

1. When a: = 2, y = 3, e = 4, find the value of the sum of 5x^,
- 3xy, z^. Also find the value of 3s* + 3x^.

2. Add together Sah + bc-ca, -oi + ca, ah-2bc + 5ca. From
the sum take Sea + bc- aft.

3. Subtract the sum of x-y+Zz and -2y~2z from the sum of
2x-5y-Sz and - 3a: + y + 4z.

4. Simplify (1) 36-2Z»2-(26-362).

(2) 3a-26-{26 + a)-(a-56).

5. Subtract Sc^ + 8c - 2 from c^ - 1.

6. When a: = 3, a = 2, y = 4, z = 0, find the value of

(1) 2x2-3ay+4a»». (2) ^.

oy

7. Add together 3a- -7a + 5 and 2a* + 5a -3, and diminish the
resultby 3a2 + 2,

8. Subtract 26^-2 from -26 + 6, and increase the result by
Sfe-7.

9. Find the sum of 3a:2-4a: + 8, 9a:-3-ar», and 2a:*^-2, and
subtract the result from 6a:^ + 3.

10. What expression must be added to 5a^-3a+12 to produce
9a2-7?

11, Find the sum of 2a?, -v?, 3a~», 2, -bx, -4, Zt?, -&x\ 8;
arrange the result in ascending powers of x.

. 12. From what expression must the sum of 5a2-2, 3a + a2, and
7 - 2a be subtracted to produce Za^ + a - 5 ?

13, When a: = 6, find the numerical value of the sum of 1 - a; + a:-,
2a:^-l, and x-x^,

14, Find the value of Gaa: + {2by -cz)- (2aa: - Sby + icz) - {cz + ax),
when a = 0, 6 = 1, c = 2, x = 8, y = 3, z = 4.

15, Subtract the sum of a;^ _ 3^,2^ 2x^ - 7a?, 8a? - 2, 6 - 3a^, 2x3 - 7
froma?» + «2+a?+l.

20 ALGEBRA. [cHAP. IV.

16, What expression must be taken from the buqi of p*-^p^f
2p + 8, 2;>^, 2j^ - 3p*i in order to produce 4p* - 3 ?

17.- What is the result when -3a:' + 2ar^-lla; + 5 is subtracted
from zero.

18. Ky J^ow much does b + c exceed b-cl

19. Find the algebraic sum of three times the square of a:, twice
the cube of a;, -a^-^x- 2ar, and ar^-x-x^ + l.

20. Take jr^-q^ from Spq-^q^y and add the remainder to the
sum of 4p5 -p^- 3g' and 2p- + 6^^^

21. Subtract 36^ + 26- -8 from zero, and add the result to
6^-263 + 36.

22. By how much does the sum of -ni* + 2m-l, m^-Sm,
2»i3 - 2m^ + 5, 3w^» + 4m^ + 5w + 3, fall sliort of 1 1 m-* - Sm^ + 3m ?

23. Find the sura of 8x^-4x^y\ 1x*y-xy*, .3ary + 2a:V + 5a:y*,
y^-4xy* + 3^y^, x^-y^ + Qi^y^ + xy*-xh^ + 3x*yy and arrange the
result in descending powers of x.

24. To what expression must Sx - 4a:* + 7a;- + 4 be added so as to
make zero ? Give the answer iu ascending powers of x,

25. Subtract Tx^ - 3a: - 6 frpm unity, and x - 5ar^ from zero, and
add the results.

26. When a = 4, 6 = 3, c = 2, rf = 0, find the value of

2/>2c

(1) 3a^-2bc-ad + Sb'^cd. (2)

\)a

27. Find the sum of a, -3a^ 4a, -6a, 7, -18a, 4a^, -6, and
arrange the result in descending powers of a.

28. Add tocTcther 4 + Sx'^ + x^, a:=' - a-2 - 1 1 , ar' - 2ar2 + 7, and sub-
tract 2a;^ + a:- - 7 from the result.

29. If a = 5x-oy-\-Zy 6= -2a: + y-3z, c-x- 5y + 6z, find the
value of a + 6-c.

30. If a: = 2a2-5a + 3, y= -3a^-{-a + 8, z = 5a2-6a-5, find the
value of a; - (y + z).

-4

CHAPTER V.
Multiplication.

34. Multiplication in its primary sense signifies repeated
addition.

Tims 3x5 = 3 taken 5 times

= 3 + 3 + 3 + 34-3.
Here the multiplier contains 5 imits, and the number of timea
we take 3 is the same as the number of units in 5.

Again axb-a taken b times

= a + a + a + a+..., the number of terms being h.
Also 3x5 = 5x3; and so long as a and b denote positive whole
numbers it is easy to shew tliat

axb = bxa.
Hence abc = axbxc=(axb)xc = bxaxc = bac

= bx{axc) = bxcxa = bca.

Similarly we can show that the product of three positive
integiTil quantities a, 6, c is the same in whatever order the
factors are written.

Examjile. 2ax36 = 2xax3x6 = 2x3xax6 = 6a6.

35. When the quantities to be multiplied together are not
positive whole numbers, the definition of multiplication has to
be modified. For examjile to multiply 3 by f , we perform on
3 that opei-ation which when performed on unity gives f ; that
is, we must divide 3 into 7 equal parts and take 4 of them.

By taking multiplication in this sense, the statement ab = ba
can be extended so as to include every case in which a and b
stand for positive quantities.

It follows as in the previous article that the produ(;t of a
number of positive factors is the same in whatever order the
factors are written.

36. Since, by definition, a^ — aaa, and a°=aaaaa ;

.*. a^xa^ — aaa x aaaxia = aaaaaaaa = a^ = a*^ ■*^ " ;

that is, the index of a in the product is the sum of the indices
of a in the factors of the product.

Again, 5a-=5aa, and la? = laaa ;

.*. ba^ X la? = 5 X 7 X aaaaa = 35a°.

22 ALGEBRA. [chap.

When the expressions to be multiplied together contain
powers of different letters, a similar method is used.

Example. 5a'6' x Sa-ftar* = ^acuibb x Saabxxx

- 40a«6»ar».

Note. The beginner must be careful to observe that in this pro-
cess of multiplication tJie indices of one letter cannot combine in any
icay loith those of another. Thus the expression ^a^li^T? admits of
no further simplification.

37. Rule. To midtiply two simple expressions together,
multiply the coefficients together and prefix their product to the
product of the different letters^ giving to each letter an index equal
to the sum of the indices that letter has in the separate factors.

The rule may be extended to cases where more than two
expressions are to be multiplied together.

Example 1. Find the product of a;^, a:*, and a^.
The product = a:^ x a:» x a:^ = a;2+3 x x-s = 3^2+3+8 = 3.13.

The product of three or more expressions is called the con-
tinued product.

Example 2. Find the continued product of bx^y^, 8yV, and 3ass*.
The product = bxh^ x 8yV x Sarz* = I'iQT^yh^

38. By definition, (a+ 6) m = m + m + w + ... taken a + 6 times

=(m+wi+wi + ... taken a times),
together with (m + wi + w + . . . taken h times)

= am + hm.

Also (a-6)m=m + m+w+ ... taken a - 6 times

= (m + wi+wi+... taken a times),

diminished by {m-{-m-{-m+... taken h times)

==am—hm.

Similarly {a-h+c)m—am~hm-\-cm.

Thus it appears that the product of a compound expression hy
a single factor is the algebraic sum of the partial products of each
term of the compound expression by that factor.

Examples. 3(2a + 36 - 4c) = 6a + 96 - 12c.

(4a;2 -ly- ^) x Zxy- = U^x^y"- - 2\xif - 2ixyh^

v.] MULTIPLICATION. 23

EXAMPLES V. a.

Find the value of

1. 5a: X 7. 2. 3x26. 3. ar'xa^s, 4^ 5a;x6a~^.

5. 6c3x7c*. 6. 9y3x5y». 7. Sm'xSm'. 8. 4a«x6a*.

9. 3ii:x4y. 10. 5ax6/>2. H, 4c^x5(P. 12. 3//*x5r7».

13. 6aa: x 5aa;. 14, Sryr x 4qr, 15, a& x ab, 16, 3ac x 5ad.

17. a^xxa*x^. 18. 3a:V^x42^. 19. a^b^xa^b*, 20. a^xSa^t^.

21. a2 X a^b X 5a6*. 22. pr* x C/rV x 7i?r*.

23. 6x^yxxyxdaih/^. 24. Ta^xSt^xSc*.

25. Qx7/^x1y^xxz\ 26. 3a&cd X 56ra2 X 4ca6d.

Multiply

27. ab-ac by a^c. 28. ar^y- A + 4ys« by ar^^-.

29. 5a2-362 by SaftV. 30. a26-6a6 + 6a by 3a'6.

31. a2-263 by 3a;2. 32. 2aa~^-6V + 3 by a-xy.

33. 7/?2g-2?g2 4.i by 2p\ 34. m^ + Smn-S?!^ by 4m-n.

35. a?^* - 3a;2z ^ 2 by 3y«. 36. a^ - Sa^a; by 2a'bx.

39. Since (a-b)m=am — bmy [Art. 38.]

by putting c — c? in the place of m, we have

(a - b){c -d)=a{c-d)-h{c-d)
= (c — d)a — (c— ci?)6
= (ac — f/o?) — (6c — bd)
= ac — ad-bc + bd.

If we consider each term on the right-hand si<le of this result,
and the way iu which it arises, we find that

( + a)x{ + c) = +ac.
(-b)x{-d) = -\-bd,
(-6)x(-fc) =-bc,
( + a)x{-d)= -ad.

These results enable us to state what is known as the Rule
of Signs in multiplication.

Bule of Signs. The product of two terms loith like signs is
positive ; the 'product of two terms with unlike signs is negative.

24

ALGEBRA.

[chap.

40. Tlic rule of sirrns, and especially the use of the negative
multiplier, will probably present some difficulty to the beginner.
PerhajM the following numerical instances may be useful in illus-
trating the interpretation that may be given to multiplication
by a negative quantity.

To multij>ly 3 by - 4 we must do to 3 what is done to unity
to obtain - 4. Now — 4 means tliat unity is taken 4 times and
the result made negative : therefore 3 x ( - 4) implies that 3 is
to be taken 4 times and the product made negative.

But 3 taken 4 times gives + 12 ;

.-. 3x(-4)=-12.

Similarly — 3x -4 indicates that —3 is to be taken 4 times,
and tlie sign changed ; the first operation gives — 12, and the
second +12.

Thus (-3)x(-4)=-M2.

Ilence, multiplication hy a negative quantity indicates that we are
to proceed just as if the multiplier were positive^ and then change
the sign of the product.

Example 1. Multiply 4a by —36.

By the rule of signs the product is negative ; also 4a x 36 = 12a6 ;

/. 4ax(-36)= -12a6.

Example 2. Multiply - 5a6'a: by - a6'a:.

Here the absolute value oE the product is Sa^/Za:^, and by the rule
of signs the product is positive ;

.-. ( - Soft^ar) X ( - oft'a:) = 5a26V.

Example 3. Find the continued product of 3a^6, - ^a^h^^ - ab*.

3a-h X ( - 2a^62) = - 6a«63 ;
( - 6a-'63) X ( - a6*) = +6a«67.

Thus the complete pro-
duct is 6a®6^.

This result, however, may be
written down at once : for

3a26 X 2aW x ab* = 6a%7^

and by the rule of signs the re-
quired product is positive.

Example 4. Multiply Ga^ - 5a^b - 4a63 by - Zah\

The ])roduct is the algebraical sum of the partial products formed
according to the rule enunciated in Art. 37 ;

thus (Ga3 - 5a26 - 4a62) x ( - 3«62) =. _ i sa-*^^ + \ba%'^ + 12a26*.

v.] MULTIPLICATION. 26

EXAMPLES V. li.

Multiply together

1. a, -2. 2. -3, 4ar. 3. -a^, -«*. 4. -6m, 3m«.

6. -4^,3g2. 6. -4y8, -4y». 7. -3m3, 3m8. 8. 4x*, -4a?*.

9. -3a:, ~4y. 10. -Sa^, 4a;. H. -3pa, -4g«. 12. 3a6, -4a5.
13. 3a2, - 62^ 2ah. 14. - a, - 6, - c^. 15, Sa^, - 26, - 4c3, - d

16, - Sah, - 4ac, She. 17. - 2a\ - Sa^b, - 6. 18. - 2p, - 3q, 4«, - 1.

Multiply
19. -o6+ac-6c by -a6. 20. - Sa^ - 4ax + Boi^ hy -oV.

21. a2c-ac3+c*by -a^c 22, -2a6 + cd-«/*by -3a:V«

41. To further illustrate the use of the rule of signs, we add
a few examples in substitution where some of the symbols,
denote negative quantities.

Example 1. If a = - 4, find the value of a'.

Here a^ = ( -4)8 = ( -4) x ( - 4) x ( -4) = -64.

By repeated applications of the rule of signs it may easily be
shown that any odd power of a negative quantity is negative^
and any even power of a negative quantity is positive.

Example 2. If a = - 1, 6 = 3, c = - 2, find the value of - 3a*6c'.

Here -3a*tc3= ~3x(-l)*x3x(-2)»

= -3x( + l)x3x(-8)
= 72.

We write down at
once, (-1)*= +1, and
(-2)8= -8.

EXAMPLES V. c

Ifa=-1, 6 = 0, c=-2, »=1, g=-3, find the value of

1. 3c. 2. -5a. 3, an. 4. (-a)*.

6. -3c2. 6. (-#. 7. -2a8. 8. -ac.

9. ah. 10, -acn. H. -3a<. 12. 4{-cf.

13. 2a6c2. 14. -c\ 15. -(a)*. 16. -3aY

17> -aV. 18. ac^. 19. -a»c>. 20. c»«».

26 ALGEBRA. [chap.

If a = - 3, c = 1, i- = 0, a: = 5, y = - I, fiud the value of

21, 3a-2y + 4it. 22. -4r-3a: + 2y. 23. -ia + 5y-x.

24. ac-3cy-2/A-. 25. 2ai/-kx + ik\ 26. a2-2c2 + 3y2.

27. -a^-ay-{-:'y^. 28. aar-yar-ry. 29. c2-y2-c'+y'.

30. a3-x2-2y. 31. cy-2oc2+ci[;2^ 32. acy-y* + 2a2.

Mnltiplication of Gomponnd Expressions.

42. To find the product of a-\-h and c+d.

From A rt. 38, (>( -f- h)m = am + ^tti ;
replacing m by c+rf, we Lave

{a + h){c + d) = a{c-\-d) + b{c-{-d)
= (c + d)a + (c-^d)b
= ac+ad-{-bc+bd.
Similarly it may be shown that

{a — b)(c-{-d)=ac-\-ad—bc — bd;
{a + bXc — d) = ac — ad + be — bd ;
(a - b)(c - d) = ac- ad — bc + bd.

43. When one or both of the expressions to be multiplied
together contain more than two terms a similar method may be
used. For instance

(a — b+c)m = ajn—bm-\-cm;

replacing mhy a; — 7/, we have

(a - 6 + c)(^ -y)— ^C^ —y)~ ^(-^ —y) + <^~y)

= {ax — (n/) -{bx — bi/)-\-{cx — ci/)
= ax — ay - bx -\-by-\-cx- cy.

44. The preceding results enable us to state the general rule
for multiplying together any two compound expressions.

Rule. Multiply each term of the fi?'st expression by each term
of the seco7id. When the terms inuHi plied together have like signs^
j))'(fix to the product the sign +, ichen uidike prefix — ; the
ahjelyraical sum of the partial prodiicts so formed gives the com-
plete product,

45. It should be noticed that the product oi a-\-b and x~y
is briefly expressed by {a-\-b){x—y)^ in which the brackets
indicate that tlie expression a-\-h taken as a whole is to be
multiplied by the exffnession x-y taken as a whole. By the

v.]

MULTIPLICATION.

27

above rule, the value of the product is the al^ehraical sum of
the partial products +ar, +ba^, —f^Ut —^y\ ^^^ ^\^n of each
product being determined by the rule of signs.

Example 1. Multiply a: + 8 by a: + 7.
The product = (a; + 8)(a; + 7)

= a;- + 8a; + 7a: + 5G

= a:^ +15x4-16.
The operation is more conveniently arranged as follows :

a; + 8

by addition,
Example 2.

a; + 7
x^+ 8a;

±_lx + 5(5_
a;- + 15a; + 513.

We begin on tlio left and work
to the right, placing the second
result one place to the right, so
that like terms may stand in the
same vertical column.

by addition,

Multiply 2a; - Sy by 4a; - ly.
2x -3y

Ax -^ty

8a;2-12;ry

-14a;y + 21 yg
8a;2-26a:y + 21y2.

Find the product of

1. a + 7, a + 5.

4. y-4, y + 4.

7, ^ + 5, ^-5.

10. a-14, a + 1.

13. a;-4, -a; + 4.

16. ar-lO, -a; + S.

19. 2a - 5, 3a + 2.

22. 3y-5, y + 7.

25. a; 3a, 2a; + 3a.

28. a - 2a;, 3a + 2a;.

31. ^x-i)yy Ax + y.

34, 2a; -3a, 2a; + 36.

EXAMPLES V. d.

2. a;-3, a; + 4. 3.

5. a; + 9, a;-8. 6.

8. m-9, m+12. 9,

11. P-IO, 7>flO. 12.

14. -y + '-h -y~S. 15.

17. -^-f-i, -^-7. 18.

20. a; -7, 2a; + 5. 21.

23. 5m - 4, 7m - 3. 24.

26. 3c -2?), 2a + 36. 27.

29. 76 + c, 76 -2c. 30.

32. 2y-3z,2y + Sz. 33.

35. 3a; -4//, 2a + 36. 36.

a - 6, a - 7.

c-8, c + 8.
a;-12, a;+ll.
rf + 7, cZ+7.
-a + 4, -a + 5.
-2/-7,-2/-7.
.Sa;-4, 2a; -f 3.
7;?-2, 2;> + 7.
5c 4- 4cZ, 5c - id.
2a - 5c, 2a + 5c.
xy + 26, xy - 26.
mn-2>i 2xy + Zz,

28

ALGEBRA.

[chap.

46. We shall now give a few examples of greater difficulty.

Example 1. Find the product of 3a;- -2a: -6 and 2x-5.

Each term of the first expression is
multiplied by 2a;, the first term of the
second expresaion ; then each term of the
first expression is multiplied by - 5 ; like
terms are placed in the same columns and
the results added.

3ar»- 2a; -6
2a? - 5

6a:*- 4x2 -10a;
-15ar^+10a; + 25

6ar»-19a:2

+ 25.

Example 2. Multiply a - fc + 3c by a + 26.

a - 6 + 3c
g + 26
a^- ah + ^CLC

2ab -2h' + 66r

a2+ ab + Sac-2h' + 6ljc,

47. If the expressions are not arranged according to powers,
ascending or descending, of some common letter, a reaiTange-
ment will be found convenient.

Example. Find the product of 2a^ + 46' - 3a6 and 3a6 - 5a^ + 46-.

2a« - 3a6 + 46'

- 5a'^+ 3a6 +46^

-10a"»+15a=^6 -20a262

+ 6a36- 9a262 + 12a63

Sa'^b^-\2alr^+\Gh*

- 10a* + 21a36 - 21a262 + 166*.

The re-arrangement is not
necessary, but convenient,
because it makes the collec-
tion of like terms more
easy.

EXAMPLES V. e.

Multiply together

1, x^-Zx-2, 2x-\.

3. 2y2-3y + l, 3y-l.

5. 2a2-3a-6, a-2.

7. 3ar^-2a; + 7, 2a;-7.

9, x^ + x-2, ar'-a; + 2.

11, 2a2-3a-6, a2-a + 2.

13, a+6-c, a-6 + c.

2, 4a2-a-2, 2a + 3.

4. 3a;2 + 4a; + 5, 4a; -6.

6. 562-26 + 3, -26-3.

8. 5c2-4c + 3, -2c + l.

10, a;2-2a; + 5, a;2-2a; + 5.

^^ ^iC" "~ oA/ "" 1, ijk/ -~ a; X.

14. a-26-3c, a-26+3c

v.] MULTIPLICATION. 29

15, ar^-ary+y2, ^^xy+'i^, 16. a^-2az+2a^, a»+2aa:+2x».

17. a2-63_3c3, _a2-62-3c2. 18. ar'-3rr»-ar, ar^-3a;+l.

19. a3-6a+5, a3 + 6a-5. 20. 2y^-4ya+l, 2^/4-4^2 _i,

21. 57?i2 + 3-4m, 5-4m+3w2. 22. Sa^ - 2a2 - 3a, Sa^ + 1 - 6a.

23. 2a: + 2a:3 _ 33.2^ 3a; + 2 + 2a:2. 24. a^ + ft^ - aS^, a^^a - a^ + 6^

25. a^ + s^+^aa^-i-^^x, a^+Sax^-x^-3ah:.

26. 5/?*-p3+4/>2-2/? + 3, p2-2p + 3.

27. m5-2m*+3»i*-4m3, 4w«-3»i5+2m*

28. a*+l+6a2-4a3-4a, a5-l + 3a-3a«.

29. a^ + b^+c^-\-ah+ac-bc, a-h-c.

30. ar*+6ar^2 + j^4_4ajy_4py3^ - a:* - y* - Gary - 4a:y8 - 4aJV.

48. Although the result of multiplying together two binomial
factors, such as ^+8 and ^ — 7, can always oe obtained by the
methods already explained, it is of the utmost importance that
the student should soon learn to write down the product rapidly
by inspection.

This is done by observing in what way the coefficients of the
terms in the product arise, and noticing that they result from
the combination of the numerical coefficients in the two bi-
nomials which are multiplied together ; thus

{x - S){x - 7) =^2 „ Sjf - 7x+ 56
=a?2_ 15^+56.

(a:+8)(:F-7)=a;2+8a?-74?-56

=^+x-56.
(:F-8)> + 7)=^-&F+7:r-66

=jf2_^ — 56.

In each of these results we notice that :

1. The product consists of three terms.

2. The first term is the product of the first terms of the two
binomial expressions.

3. Tlie third term is the product of the second terms of the
two binomial expressions.

4. The middle term has for its coefficient the sum of the
numerical quantities (taken with their proper signs) in the second
terms of the two binomial expressions.

30 ALGEBRA. [chap. v.

I'lio in termed iato Rtep in the work may be omitted, and the
products wiitten tlown at once, as in the following examples :

(./• + :>)(.i' + 3) = X' + bx + 6.
{x - 3)(.r + 4) = 0,- + ^- - 1 2.
(.r + 6)(.c - 9) = A'2 - 3a: - 54.
{x - 4^)(.r - 10//) = ^ - 1 4j:y + 40^^.
(-'^ - 6^)(-^' + 4i^) = 0:2 - 2.ry - 24^.

By an easy extension of these principles we may write down
the product of ani/ two binomials.

Thus {2x + 3y)(.r -y) = 2^ + 3jr?/ -^xy- Zf-

(ar - 4 y )(2.r + ?/) = 6u:2 - 8a:y + a^y - 4y2

(a: + 4)(.r-4)=ar + 4jr-4A'- 16

(2:r+5y)(2.r-5y)=.4a:- + 10.ry-10^-25y*

= 4r2-25/.

EXAMPLES V. f.

Write down the values of the following products :

1. (a + 3)(a-2). 2. (a-7)(a-6). 3. (a: - 4)(a: + 5).

4. (6-6)(6 + 4). 5. (y-7)(2/-l). 6. (a-l)(a-9).

7. (c-5)(c + 4). 8. (x-9)(a;-3). 9. (y-4)(y + 7).

10. (a-3)(a + 3). 11. (a; - 5)(a: - 8). 12. (a + 7)(a-7).

13. (/j-6)(^-6). 14. (a-5)(a + 5). 15. (c + 7)(c + 7).

16. (p + 9)(p-10). 17. (s + 5)(=-8). 18. (a:-9)(a: + 9).

19. (a;-3a)(a; + 2a). 20. (a - 2i;)(a + 26). 21. [x - Aij){x - ^y).

22. (a + 4c)(a + 4c). 23. (c - 5c/)(c - 5c?). 24. {p-2q){p-\-2q).

25. (2a:-3)(3a: + 2). 26. (3a;- l)(2a; + l). 27. (5a;-2)(5a: + 2).

28. (3a; + 2a)(3a?-2a). 29. (6a: + a)(6y-2a). 30. (7a; + 3y)(7a; - y).

CHAPTER VI.

Division.

49. The object of division is to find out the quantity, called
the quotient, by which the divisor must be multiplied so as to
produce the dividend.

Division is thus the inverse of multiplication.

The above statement may be briefly written

quotient x divisor = dividend,

or dividend -i- divisor = quotient.

It is sometimes better to express this last result as a frac-
tion ; thus

dividend .. ,

-——, = quotient.

divisor

Example 1. Since the product of 4 and x is 4x, it follows that
when 4a; is divided by x the quotient is 4,

or otherwise, 4a;-^a: = 4.

Example 2. Divide 27a' by 9a'.

We remove from the divisor
and dividend the factors com-
mon to both, just as in arith-
metic.

n»e quotient = ?1»- = ?I„???^

ya^ 9aaa

= .3aa = 3a»

Therefore 27a« -^ 9a^ = Sa^.

Example 3. Divide 35a^6'c' by "Jah^c^,

m i.' i. 35aaa . hb . ccc k__ ^ p. «>
The quotient = — „ — , , = oaa . c = oa-c.

la . hb , cc

In each of these cases it should be noticed that the index of any
letter in the quotient is the difference of the indices of that letter in
the dividend and divisor.

32 ALGBBRA. [chap.

GO. It is easy to prove that the rule of signs holds for
division.

Thus a6-a=''-*=''-ii*=6.

a a

a a

a6^(-a)=^J-°)><<-^)= -6.
— a -a

— a —a

Hence in division as well as multiplication

like signs prodiLce + ,
unlike signs produce — •

Bnle. To divide one simple expression by another :

The index of each letter in the quotient is obtained by subtracting
the index of that lettei* in the divisor from that in the dividend.

To the result so obtained prefix with its proper sign the quotient
of the coefficient of the dividend by thai of the divisor.

Example 1. Divide %^a^x^ by - 12a*a:.

The quotient = ( - 7) x a*-^-i
= -7aa;2

Or at once mentally,

84a»a:»-^(-12a*5i:)= -7aa;2.

Example 2. - 4oa«62a4^( « %a%x^) = ba^bm^.

Note. If we apply the rule to divide any power of a letter by the
same power of the letter we are led to a curious conclusion.

Thus, by the rule o'-f o' = a^-s = a® ;

a?
but also o' -r a' = 5 =1,

a'

.'. a»=l.

This result will appear somewhat stranee to the beginner, but
its full significance is explained in the Theory of Indices.
[See Elementary Algebra, Chap, xxxi.]

Rule* To divide a compound expression by a single factor,
divide each term separately by that factor, and take the algebraic
sum of the partial quotients so obtained.

This follows at once from Art. 38.

Examples, (9a;-12y + 3z)-f (-3) = -Sx + 4y-z.

(36a»62 ^ 2ia'b^ - 20a^62) -f ia'^b = 9ab - 6b* - 5a^b.

VI.] DIVISION. 33

EXAMPLES VI. a.

Divide

1. 2a:3 by ar». 2, 6a« by 3a. 3. 5a7 by a*.

4. 2167 by 76^ 5. JB'y^ by -xi/. 6. - 3a;y3 by 3y.

7. 4;^9^ ^y -2/)g. 8. ISm'w by -5m. 9. -Pm^ by -/m.

10. -48x» by -6ar». 11. 35z" by -Jz*. 12. -Ta^ft by -76.
13. - ^P^q by 28/)«. 14. -'Jafihy - t?. 15. 2ixy:^ by - Sz^.
16. -1262c8 by 66V. 17. -Oifc^^ by -B\ 18. 2W by -H.
19. - 45a*63c" by ^^l^c^K 20. - ar'y**" by - ar^//s«.

21. - 168a26»cx2 by - 7a6ar». 22. - 35a86V by - 7a26V.

23. ^^ - 2a: by a:. 24. 6a»6 - 7a63 by ab.

25. 48i>2g - 24pga by Spg. 26. -1 53^^+250:* by -Sar^.

27, a;2-jcy-a»by -a:. 28. 10a' - 5a26 f a by -a.

29. 4a;H36ax2-16a; by -4x. 30, 3a«-9aa6-6a62 by -3rt.

When the Divisor is a Compound Expression.

51. Rule. 1. Arrange divisor and dividend in ascending or
descending powers of some common letter,

2. Divide the term on the left of the dividend by the term
on the left of the divisor^ and put the result in the quotient,

3. Multiply the whole divisor by this quotient^ and put the
product under the dividend,

4. Subtract and bring down from, the dividend as many terms
as may he necessary.

Repeat these operations till all the terms from the dividend are
brought doxon.

Example 1. Divide ar' + llx + SO by a:+6.

Arrange the work thus :

a? + 6)a?3+lla:+30(

divide a:^, the first term of the dividend, by a;, the first term of the
divisor ; the quotient is x. Multiply the whole divisor by Xf and put
the product a^ + 6a; under the dividend. We then have

X + ^)v^+\\x + ZQ{x

7?+ %X

by subtraction, 5a; + 30

On repeating the process above explained, we find that the next
term in the quotient is + 5.

B.A.

34

ALGEBRA.

[chap.

The entire operation is more compactly written as follows :

r + 6)ar- + llx + 30(x + 5

5a; + 30
5a?-h30

Tlie reason for tlie nile is this : the dividend is separated
into Rs many ])arts as may be convenient, and the complete
cjiiotieut is found by taking the sum of all the mrtial quotients.
iJy the alK)ve process ^-2+11^' + 30 is 8e]>arate(i into two parts,
namely .r-' + Or, and 5.r + 30, and each of these is divided by :f4-6 ;
thus we obtain the partial quotients +j; and +5.

Example 2. Divide 2 ix'^ - G5xy + -21^= by Sx - Sy.

Sx-Sij) '2^X' - Goxt/ + 2 1 y- ( 3x - ly
2\x'- '^xy _

-r){Sxy-^2\y'

Divide 24a:- by 8a:; this
gives 3a:, the first term of the
quotient. Multiply the whole
divisor by 3a;, and place the
result under the dividend.
]>y subtraction we obtain

-5riJ7/ + 2Iy-. Divide the first term of this by 8a;, and so obtain

- ly, the second term of the quotient.

Example 3. Divide IGa- - ^Ga- + 39a - 9 by Sa - 3.

8a-3)16a3-46a2 + 39a-9(2a2-5a + 3
IQq^- Ga-

-40a2 + 39a
-40^2+ 15a

24a -9
24a-9

Thus the quotient is 2a- - 5a + 3.

EXAMPLES VI. b.

Divide

1. a2 + 2a + l by a + 1.

3. a;- + 4a; + 3 by x + \.

5, a;- + 5a:-G by a:- 1.

7. ?>'' + 3/;-40 by p + 8.

9. a- + 5a -50 by a +10.

11, a:2 + aa;-30a2 by a: + Ga.

2. h'^ + ^h + 2 by 6 + 2.

4. y2 + 5y + 6 by y + 3.

6. ar' + 2a;-8 by a:-2.

8. ^'--47-32 by ^ + 4.

10. «i- + 7m - 78 by m - 6.

12. a- + 9a6 - 3662 by a + 126.

VI.] DIVISION. 35

Divide

13. - a;2 + 18a; - 45 by a: - 15. 14. x^ 42a; + 441 by a; 21.

15, 2a;2 - 13a; - 24 by 2a; + 3. 16. 5ar^ + 16a; + 3 by a; i r>.

17. 6a;2 + 5a: - 2 1 by 2a; - 3. 18. ISa^ + aa; - 6a;2 by 3a - 2a;.

19, -5a;2+a;y + 6y2 by -a;-y. 20. 0«- ac-35c2 by 2a-6c.

21. Vlf - Upq + 12g2 by 2p - 12g.

22. 4wi2 - 497i2 by 2m -f- In.

23. 12a2-3ia6 + 20Z/2 by 4a -66.

24. -25a;2+ 492/2 by -5a; + 7y.

25. 21/i2 + ll;?/7-4052 )jy 3^ + 5^.

26. 8a;3 + 8a;2 + 4a,+ l l^y 2x-\^\,

27. -2ar^ + 13a:2-17a; + 10 by -a; + 5.

28. x^ + aa;2 - 3a2a; - Ga^ by a; - 2a.

29. Q^y - xhf - "ixy^ + 12y* by 2a; + 3y.

30. 8ar^ - 12a;2 - 14a; + 21 by 2a; - 3.

52. The process of Art. 51 is applicable to cases in which
the divisor consists of more than two terms.

Example, 1. Divide a^ - 2a? - loP- + 8a + 12 by a^ - a - 6.
a2-a- 6 )a*-2a3-7a2 + 8a + 12 (a2-a- 2

-g^-g"- + 8a
-rt3 + a2 + 6g

-2g2 + 2a+12
-2a2 + 2a + 12

Example2, Divide 4t7?-^x^ + Qx^-\^-x/^-x by 3 + 2a;2-a;.

First arrange each of the expressions in descending powers of x.

2a;2-a; + 3)6a;5- x^ + ^x^-^x^-x- 15 ( 3a;3^a;2_2a;-5
6a:^-3a;^ + 9a;»

2ar*-5ar^- 5a;2
2a;^- ar^+ 3a;2

-43;'"^- 8a;2- x
-4a;*+ 2^2 -6a;

-10a:2 4-5a;-15
-10a;2 + 5^-15

36 ALGEBRA. [chap.

Example 3. Divide 23r»-2ar*-4a:»4-12+aJ^-31a: by a:»-7x+5.
«»-7ar+6)a:»-2r*-4ar»+23ar»-31ar+12(jc2-2a:+3

-2ar*+3a:»+iar»-31a:
-2a?* +14a:»~10a:

3a:»+ 4ar*-21ar+12
3a:» -21ar + 15

4a:« - 3

Now 4a:' is not dlTisible by a:', so that the division cannot be
carried on any further ; thus the quotient is a;' -2a; + 3, and there is
a remainder 40^^—3.

In all cases where the division is not exact, the work should be
carried on until the highest power in the remainder is lower than
that in the divisor.

53. Occasionally it may be found convenient to arrange the
expressions in ascending powers of some common letter.

Example. Divide 2a5 + 10 - 16a - 390* + 15a* by 2 - 4o - Sa'.

2-4a-5a2)10-16a-39a«+ 2a«+15a*( 5 + 2a-3a«
10-20a-25a'

4a-14a2+ ^a^
4a- 8a2^10a3

- 6a2 + 12a3+15a*

- 6a2+12a3+15a*

EXAMPLES VL c.

Divide

1. a3-6a2+lla-6 by a2-4a+3.

2. arJ-4ar» + a:+6 by x^-x-%

3. 2r' + 2/'-9y + 12by y2_3y+3.

4. 21w3-m2 + m-l by 7m2 + 2m+l.

5. 6a3-5a2-9a-2 by 2a2-3a-l.

6. ei-s-F-HJC' + a by 3F+4Jt-l.

7. 6a:3+iiaJ5_39a,_65 by 3ar' + 13a:+13.

8. 12ar»-8aa~'-27a2a;+18a3 by Qx^-\^ax+^\

VI.] DIVISION. 37

Divide

9. m2^+ 14arV - 129a:y2 _ 15^3 by 8a:' + 27a:y + 3y«.

10. 21c3 - 5M - Sc(P - 2il^ by Tc^ + 3cd + dK

11. 3a:*-10a:3+i2a:2_iia. + c by 3x2-a: + 3.

12. 30a*+lla3-82a2-12a+48 by 3a2+2a-4.

13. a^-a^-Sx-lS by ar»+3a:+3.

14. a+3a3+6-10a2 by a2-4a+3.

15. 2l77i3 - 27m - 26m2 + 20 by 3m + Tw^ - 4.

16. ISa^ + 24a3 - 40a^x - 9aar^ by Oa;^ + Ta* - 18aa:.

17. 3y*-4y8+lV + 3y-2 by 2r*-y2 + 3y + 2.

18. 6a5 + l + 10a*-4aa by 5a3-2a+l.

19. l2x*-{-5a^-oSx^-Sx+l6 by 4a;2-a:-r).

20. 7>*-6/>3+13i)2_10/) + 7 by ;?2-3/? + 2.

21. 28a;* + 69a: + 2-71ar'-35a;2 by 4ar» + 6-13a:.

22. 5a5-7a4-9a3-lla2-38a + 40 by -Sa'+Ha-lO.

23. a:"-8a3 by x^ + 2ax + ia\ 24, y* + V + 8I by y2_3y + 9.
25. ar* + 4y4 by ar» + 2a:y + 2i/2. 26. 9a*-4a2 + 4 by 3a2-4a + 2.

27. a8 + 64 by a*-4a3 + 8.

28. 16a:4 + 36a^J + 81 by 4a;2 + 6a; + 9.

29. 4m'' - 29m - 36 + 8m2-7m3 + 6m* by mS-2m2 + 3m-4.

30. 15a?* + 22 - 32a:3 ^ 30^: + SOar' by 3 - 4ar + 5x\

31. 3a2 + 8a6+462 + 10ac + 86c + 3c2 by a+26 + 3c.

32. 9ar»-4y2 + 4yz-z2 by -3a: + 2i/-z.

33. 4c2-12c-cP + 9 by 2c + d-3.

34. 9/)2- 16^2 + 30;? + 25 by -3p-4g-5.

35. a^-ic^ + x^y^-a^ + x^-'i/^ by a^-x-y.

36. iK^ + xh/-as^y^ + 9t^-2xy^ by x^ + xy-y\

37. a3&3 + a6-9-6* + 3&3 + 36-a*-3a»-3a by 3-6 + a\

38. ic^+l by a:3 + a;2 + a.+ i. 39^ 2a6 + 2 by a^ + 2a^ + 2a + l.
40. a:»-6a:*-8«8-l by jr3-2a;-l.

CHAPTER VIL

Removal and Insertion of Brackets.

54. Quantities are sometimes enclosed within brackets to
indicate that they must all he operated upon in the same way.
Thus in the expression 2a — 36 - (4a — 2b) the brackets indicate
that the expression 4a — 2b treated as a icJiole has to be sub-
tracted from 2a — 36.

It will be convenient here to quote the rules for removing
brackets which have already been given in Arts. 24 and 25.

When an expression within brackxts is preceded by the sign +,
the brackets can be removed without making any change in tlie
expression.

When an expression within brackets is preceded by the sign — ,
the brackets may be removed if the sign of every term within the
brackets be changed.

Example. Simplify, by removing brackets, the expression

(2a - 36) - (3a + 46) - (6 - 2a).

The expression = 2a - .16 - 3a - 46 - 6 + 2a

-a- 86, by collecting like terms.

55. Sometimes it is convenient to enclose within brackets
part of an expression already end used within brackets. For
this i)urpose it is usual to em])l()y brackets of different forms.
The brackets in common use are (),{},[]•

56. When there are two or more ])airs of brackets to be
removed, it is generally best to begin with the innermost pair.
In dealing with each pair in succession we apply the rules quoted
above.

Examjde. Simplify, by removing brackets, the expression

a - 26 - [4a - G6 - {3a - c + (2a - 46 + c)}].
Removing the brackets one by one,

the expression = a - 26 - [4a - 66 - {3a - c + 2a - 46 + c}]
= a - 26 - [4a - 66 - 3a + - 2a + 46 - c]
= a - 26 - 4a + 66 + 3a - c + 2a - 46 + c
= 2a, by collecting like terms.

Note. At first the beginner w ill find it best not to collect terms
until all the brackets have been removed.

CHAP. VII.]

REMOVAL OF BRACKETS.

39

EXAMPLES VII. a.

Simplify by removing brackets and collecting like terms :

1

8
5
7
9

10
11
12
13
14
15
16
17
18
19
21
22
23
24

2. a + 26 -(2a -36).

4. a - 2 - (4 - 3a).

6. a + 26-3c-(6-a-4r).

8. 4x - {2y + 2x) - (3a; - 5y).

a +26 + (2a -36).

2a -36 -(2a + 26).

(x - 32/) + {2x - 4y) - (a? - 8y).

(a: - 3y + 2z) - (z - 4y + 2a;).

2a+ (6 - 3a) - (4a - 86) - (66 - 5a).

771 - (n - p) - (2m - 2/> + 3to) - (» - m + 2p).

a - 6 + c - (a + c - 6) - (a + 6 + c) - (6 + c - a).

5x - [ly + 3a;) - (2y + 7a;) - (3a; + 8y).

(p-g)-(g-2p) + (2p-g)-(/?-2g).

2ar^ - (3y2 - ar^) _ (a^J - 4y2).

(m2 - 2?i2) - (2n2 - 3m2) - (3wi2 - 47i2).

(a; - 2a) - (a; - 26) - {2a - a; - (26 + a;)}.

(a + 36)-(6-3a)-{a + 26-(2a-6)}.

p2 _ 2<^ - (y2 + 2p2) - {pa + 3gr2 - (2/>2 _ g2)}.

a;_[y + {a;-(y-a;)}], 20. (a-6)-{a-6-(a + 6)-(a-6)}.

p-[p-(g+p)-{p-(2p-g')}].
3a;-y-[a;-(2y-z)-{2a;-(y-z)}].
3a2 - [6a2 - {862 _ (9^2 _ 2^2)}].

[3a - {2a - (a - 6)}] - [4a - {3a - (2a - 6)}].

57. A coefficient placed before any bracket indicates tliat
every term of the expression within the bracket is to be multi-
plied by that coefficient ; but when there are two or more
brackets to be considered, a prefixed coefficient must be used aa
a multiplier only when its own bracket is being removed.

Examples 1. 2a; + 3(a;-4) = 2a; + 3a;- 12 = 5x-12.
2. 7a;-2(a;-4) = 7a;-2ar + 8 = 5a; + 8.

Example 3. SimpUfy 5a - 4[10a + 3{a; - a - 2(a + a;)}].

The expression
= 5a - 4[10a + 3{a; - a - 2a - 2a;}]
= 5a - 4[10a + 3{ - a; - 3a}]
= 5a-4[10a-3a;-9a]
= 5a — 4[a - 3a;]
= 5a — 4a + 12a:
= a + 12a;.

On lemoving the innermost
bracket each term is multiplied
by - 2. Then before multiply-
ing by 3, the expression within
its bracket is simplified. The
other steps will be easily seen.

40 ALGEBRA. [chap.

58* Sometimes a line called a vincnlfini is drawn over the
symbols to be connected ; thus a — b-\-c is used with the same
meaning as a - (6 + c), and hencea-6 + c=a-6-c.

Note. The line between the numerator and denominator of a

fraction is a kuid of vinculmn. Thus — -— is equivalent to ^[x - 5).

o

Example 4. Find the value of

84 - 7[ - 1 la; - 4{ - 17a:+3(8 - 9 - 5x)}].
The expression = 84 - 7[ - 1 la; - 4{ - 17a: + 3(8 - 9 + 5ar)}]

= 84-7[-lla;-4{-17a; + 3(5a;-l)}]
= 84-7[-lla;-4{-17a; + 15x-3}]
= 84-7[-lla;-4{-2a;--3}]
= 84-7[-lla; + 8a;+12]
= 84-7[-3a;+12]
= 84 + 21a;-84
= 21a;.

When the beginner has had a little practice the number of steps
may be considerably diminished.

Insertion of Brackets.

59. The rules for insertion of brackets are the converse
of tliose given on jmge 12, and may be easily deduced from them.

For the following equivalents have been established in
Arts. 24 and 25 :

a+h-c=a-{-{b-c),

a — b — c=a — {b+c),

a — b + c=a- (b — c).

From these results the rules follow.

Bnle. 1. Ani/ part of an expression may be enclosed within
brackets and the sign -f prefixed, the sign of every term within the
brackets remaining unaltered.

Examples. a-b+c-d-e = a-b + {c-d-e).

7? - a;x + bx - ab - {x^ - ax) + {Jbx - ab).

Rule. 2. Any part of an expression may be enclosed within
brackets and the sign — prefixed, provided the sign of every term
within the brackets be changed.

Examples, a-b + c-d-e = a-{b-c)-{d + e),
xy-ax-by+ah=^ {xy - by) - {ax - ah).

VII.] REMOVAL AND INSERTION OF BRACKETS. 41

60. The terms of an expression can be bracketed in various
ways.

Example. The expression ax - bx + ex - ay +by - cy
may be written (aa: - 6a:) + {ex - ay) + [by - cy),

or (ax - bx + ex) - {ay -by + cy),

or {ax - ay) - {bx - by) + {ex - ey),

61. When every term of an expression is divisible by a com-
mon factor, the expression may be simplified by dividing each
term by this factor, and enclosing the quotient within brackets,
the common factor being placed outside as a coefficient.

Thus ar-21 = 3(^-7);

and a^ — 2ax + 4a^ =a^- 2a{a: - 2a)t

EXAMPLES Vn. b.

Simplify by removing brackets :

1. 3{x-2y)-2{x-iy). 2. x - 3{y - x) - 4{x - 2y),

16-3(2a;-3)-(2a; + 3). 4. 4(a: + 3) - 2(7 + a:) + 2.

8(a; - 3) - (6 - ar) - 2(a; + 2) + 5(5 - a;).
2a; - 5(3a:-7 + y) + 4(2a: + 3y - 8) - 7y.
2a:-5{3a:-7(4a;-9)}.

x^ + Z{x^y + xy^)-\-y^'X^-:]{x-y-xy^)-~y^,
4x-^x-{l-y) + 2{\-x)}.
x-{y-z)-[x-y-z-2{y + z)].
a?-[a^-{x'-{z^-x^-i/^)-2y^} + y'],
5x + 4(y - 22) - 4{a; + 2(y - z)}.

3
5
6
7
8

9
10

11

12
13
14
15
16
17
18
19
20
21
22

a + {-2b + 3{c-d-e)}.

{a2 - (62 - c2)} - [2a2 - {a^ - (&2 - c^)} - 2(^2 - c^)].

Sp-{oq-[Gq + 2{l0q-p)]}.

Sx-2[2x-{2{x-y)-y}-yl

3(5 - 6a;) - 5[a:-5{l - 3(a; - 5)}].

12 - [6a - (7 - a- 5) - {5a + (3 - 2^)}].

62- {a2 ^ab- {a^ + 62)} _ [a^ _ {Sab - (62 - a2)} ].

2[4a;- {2y + (2a:-y) - (a: + y)}] -2( -a;-27^).

20(2 -x) + 3{x - 7) - 2[a; + 9 - 3{9 - 4(2 - a:)}].

- 4(a + y) + 24(6- a;) -2[a: + y + a-3{y + a- 4(6+0?)}],

42 ALGEBRA. [chap.

23. Multiply

2a: - 3y - 4(a; - 2y) + 5{3a: - 2(a: - y)}
by 4x-{y-x)- .3{2y - 3(a: + y)}.

Jq each of the following expressions bracket the powers of a; so
that the signs before the brackets may be (1) positive,
(2) negative.

24. aa?*+2a:»-ca^»+2a^»-6a:»-a:*.

25. ax^+aV-hafl-&u^-ca^,

MISCELLANEOUS EXAMPLES H

L Find the sum of a- 26 + c, 36-(a-c), 3a-6+3c.

2. Subti'act 1 - ar" from 1, and add the resnit to 2y - a:^.

3. Simplify a + 26 - 3c f- (6 - 3a + 2c) - (36 - 2a - 2c).

4. Find the continued product of 3a:^, 2a:y^, - 7x}^j - 5ary.

5. What quantity must be added to p + q to make 2q^ And
what must be added to ;:>' - 3pq to TDa,kep^+2pq-\-q^^

6. Divide 1 -6a:*+6a:» by 1 -a: + 3a:».

7. Multiply 362 + 2a» - 5a6 by 2a + 36.

8. When a; = 2, find the value of 1 - a: + a;"* - ^

1 + a;

9. Find the algebraic sum of 3aa:, - 2a:z, 9aa?, - Iscz, 4ax, - 4xz,

10. Simplify 9a - (26 -c) + 2d -(5a + 36) + 4c -2d, and find its
value when a = 7, 6 = - 3, c = - 4.

11. Subtract cux^ - 4 from nothing, and add the difiference to the
sum of 2a:^ - 5x and unity.

12. Multiply *6xh/-4xfz+2sch/h^ by -6ar»yV and divide the
result by SaryV.

13. Simplify by removing brackets 5{x - 4{a: - 3(2a; - 3a: + 2) }].

14. Simplify 2x^ - {2xy - Zy^) + 4y2 + (5a:y - 2ar2) + a;^ - (2a:y + Q»f).

15. Find the product of 2a: -7y and 3a: + 8y, and multiply the
result by a; + 2y.

16. Find the sum of 3a+26, -5c-2d, 3e+5/, 6-a+2d,
-2a-36 + 6c-2/.

17. Divide a:*-4a:8- 18a;2-lla? + 2 by a;2-7a:+l.

18. If a=-l, 6 = 2, c = 0, d = l, find the value of

od + ac - a^ - crf + c^ -a + 2c + o26 + 2a*.

vn.] MISCELLANEOUS EXAMPLES H. 43

19. Simplify 3[1 - 2{ 1 - 4(1 - 3a;)}], and find what quantity must
be added to it to produce 3 - 82;.

20. Divide the sum of I0x^-7x{l-btx^) and S{x*+9^+2) by
3(x^ + l)-{x-hl).

21. Simplify 5ar* - Sa:^ - (2a:a _ 7) _ (^^4+ 5) + (3a:3 _ ^.j^ ^nd subtract
the result from 4ar* - a: +2.

22. M a = 0, 6 = 1, c = 3, d=-2, c = 2, find the value of

(l)3c*-d«; (2) (c + a)(c-a) + 6'-»; (3) c +a».

23. Find the product of 7a?' - y(ar - 2y) and a;(7a: + y) - 2//'.

24. Subtract (a' + 4) + (a^ - 2) from (a« + 4)(o« - 2).

25. Express by means of symbols

(1) V% excess over c is greater than a by 7.

(2) Three times the sum of a and 26 is less by 5 than the

product of h and c.

26. SimpUfy

3a2_(4a-62)-{2a2-(36-o«)-26-3a}-{56-7a-(ca-62)}.

27. Find the continued product of

ar' + a^y + y*, 7^-xy+y^, a?*-a:y+y*.

28. Divide Aa?- Oft' - 4ac + c' by 2a - 36 - c.

29. If a = 3, 6 = -2, c = 0, d = 2, find the value of

(1) c(a+6) + 6(a + c) + a(c-6); (2) a^+d*.

30. From a rod a+6 inches long 6-c inches are cut off; how
much remains ?

31. A boy buys a marbles, wins 6, and loses c ; how many has
he then ?

32. Simplify 2a - { 5a - [8a - (26 + a)] }, and find the value of

(a-6)[a*+6(a + 6)] when a=l, 6 = 2.

33. Divide l-6a:*+4a:5 by ar"-2a:+l.

34. Multiply the sum of 3a:^-6a:y and 2a:y-y' by the excess of
3ar*+y^ over 2y*+3a:y.

35i Express in algebraical symbols

(1) Three times x diminished by the sum of y and twice z.

(2) Seven times a taken from three times 6 is equal to five

times the product of c and d.

(3) The sum of m and n multiplied by their difference is

equal to the difiference of the squares of m and n.

36. If a = 2, 6 = 1, c = 0, d = -l, find the value of
{d - 6)(c - 6) + (ac - hdf + (c» - d)(2c - 6).

CHAPTER VIIL
Eevision of Elementary Eules.

[If preferred, this chapter may be postponed until the chapters
on Simple Ecjuatious and Problems have been read.]

Substitntions.

62. Definition. The squaxe root of any proposed expression
is that quantity whose square, or second power, is equal to the
given expression. Thus the square root of 81 is 9, because 9- =81.

The square root of a is denoted by Jja, or more simply ^fa.

Similarly the cube, fonrth, fifth, &c., root of any expression
is that quantity wliose third, fourth, fifth, &c., power is equal to
the given expression.

The roots are denoted by the symbols J/, ij, Ij, &c

Examples. %/27 = 3 ; because 3' = 27.
\/S2 = 2 ; because 2^ = 32.

The root symbol ^ is also called the radical sign.

Example 1. Find the value of 5 ^/(Ga^Mc), when a = 3, 6=1, c = 8.
5 ^((ki^b*c) = 5 X V(6x33xl4x8)
= 5 X V(6 X 27 X 8)

= 5 X V(3 X 27) X (2 X 8)

=5x9x4

= 180.

Note. An expression of the form ^{Mb^c) is often written oJod'^h^c ,
the line above being used as a vinculum indicating the square root
of the expression taken as a whole,

Example2. If a=-4, 6= -3, c= -l,/=0, a: = 4, find the value of

7 V{a"cx) - 3 s/¥? + 6 V(/^iP)-
The expression = 7 ?/("-4)-PT)4 - 3 V( -3)*(- 1)^+0
= 7l/(-G4)-3V81
= 7x(-4)-3x9
= -56.

CITAP. VIII.] REVISION OF ELEMENTARY RULES. 45

EXAMPLES Vm. a.

If a = 4, 6 = 1, c = 6, d = 0, find the value of

1. V&*. 2. \^9ab\ 3. V6^. 4. s/d^\

5. n/46^2. 6. iJOaMfc. 7. a\/9ac. 8. 36^/3aV.

9. sfam-sl^, 10. 3N/a3cd3_cj3/2^L^,+^6^^

If a = -3, 6 = 2, <j = -l, a: = -4, y = 0, find the value of

11. VcT^. 12. V3^. 13. v^6^. 14. 5v^.

15. V3^6V^. 16. N/a=*c2. 17^ 3^/5^^ 13^ i/3^^.

19. 'J^c-Jcx + 'J^ac, 20. 'Jd^+>j2c^h-J^.

21. If a;=^100, y = 81, z=16, find the value of

V:

J- \ly+ \^4z.

4

22. If a = -6, 6 = 2, c = -l, a; = -4, y = 0, find the value of

2x/S^ - 2^^254^+ ^8^.

Fractional Coefficients and Indices.

63. Fractional Coefficients. The rules which have been
already explained in the case of integral coefficients are still
applicable when the coefiicients are fractional.

Example 1. Find the sum of \x'^ + \xy-\y'^, - a;* - |a:y + 2y',
W-xy-iy\

ix'^-hixy-ly^

- x^-lxy + ^y^

%a^- xy~\y^

Ix^-^xy + y^

Here each colunm is added up separately,
and the fractional coefficients combined
by the rules of arithmetic.

Example 2, Divide iar^ + yV^^+xVy' by ^x-\-\y,

\x^^) \o^ + ^xy^ + ^^r^ ( ^^ - ^xi/ + iy^

46 ALQEBIUL [CHAP.

64. Fractional Indices. In all the examples hitherto ex-
plained the indices have been integers, but expressions involving

fractional and negative indices such as a^, ^"*, 3^+0''^ — 2,
a~^ — 4a-\v — S.%^ may be dealt with by the same rules. For
a complete discussion of the theory of Indices the student is
referred to the Elementary Algebra^ Chap. xxxi. It will be
sufficient here to point out that the rules for combination of
iudices in multiplication and division given in Chapters v. and vi.
are universally true.

ExampUl. a;^xa:^ = a:' * = a:".

Example 2. a"'* x a* = «"*+* = a® = 1. [See Note, Art. 50.]

ExampltZ. 2a^6'^x3a"M = 6a^"^6"^"*"i = 6o06'^ = 66"i

Example 4. 3a; y^ -r oj'y = Zx ' 'y * ' ' = 3a: " ^y * .

Exampleh. a~%^ ^ah'^ =a^'%^'^^ = a'%^.

It will be seen from these illustrations that the rules for
combining indices in multiplication and division may be con-
cisely expressed by the two statements,

(1) a'"xa"=a"*+", (2) a"*^a"=a**-*;

where m and n may have any values positive or negative,
integral or fractional.

65. We shall now give some examples involving compound
expressions.

Example!. Multiply a:* -3a:* +4 by 2a:* -1.

a:*-3a:^+ 4

2a:^-l

2a: -6a:*+ 8x^
-- x^+ 3a;^-4

2a: -7a:^ + na;^-4

Example^, Multiply c' + 2c-'-1 by 5-3c-*+2c*.

c« -7 + 2c-*
2c« +6-3C-*

2c2«-14c«+ 4

+ 5c*-35 + 10c-«

- 3 + 21c-*-6c-^

2c*"- 9c»-84+81o-«-6c-2*

Here the expressions have
been arranged in descending
powers of c, and it should be
noticed that in this arrange-
ment the numerical terms - 7
and +5 stand between the
terms involving c* and o~*.

\

VIII.] REVISION OF ELEMENTARY RULES. 47

Example 3. Divide

24a;^- 16a;' *+a;^- 16a:"* -6a;* by 8a;'*-2a;*+a;*-4a;*.
ArraDge divisor aud dividend in descending powers of a;.

a;*-2a:*-4a;*+8a;"*)a:*-6a;*+24a;*-16a;'*-16a;~*(a;*-3-2a;'*

1 ft 8 1

x*-2x^- 4a;^+ 8a;*

» » 1 _i

-3a;'+ 4a;^+16a;^-16a;

-3a;^+ 6a;* + 12a;^-24a; *

- 2a;^+ 4a;*+ 8a;"* -16a;

- 2a;*+ 4a;* + 8a;" i- 16a;

EXAMPLES Vm. b.

1. Find the sum of -^m- 1 n, -|m+|n, -2m-«.

2. Add together \a-\h-\-\c, ^-^6, ia+^g^+Jc, -Ja+Aft-ic

3. From a + \h-\c take ^-6 + Jc.

4. Subtract \a^ + iab-\ly^ from ia^ - Jafr + \l^.

5. Multiply iar^ + i,y^ by ia; - iy.

6. Find the product of \x^-ix-\-\ and \x+\,

7. Divide |ar^-|y3 by ia;-|y.

8. Divide a? - 2a% + ^^^ - Jfe^ by a^ - fa6 + Jft^.

9. Simplify i(2a; - 3y ) - i(3a; + 2y) + A(7a; - 6y).

10. Find the sum of

11. Find the product of

ia;-Jy+i(z-iy) and i(a;-z)- J(y-ia;).

12. Simplify by removing brackets

8(|-3.5{2.-3(a-|)}.

13. Divide ia;8 _ j{a^ + Ja: - i by f a; - i.

14. Subtract ^^x - 9y) from i(a; - 3y) - \{y - 2a;).

15. Add together (a; - \y){\x + y) and (2a: - \y){\x - y).

16. Multiply \d? - \(3?x -^-yx^y^y \a.- 2x,

17. Divide 36a2 + 1 J2 ^ ^ _ 4^6 - 6a + i6 by 6a - ^6 - J.
la Simplify 6{a; - «y - i)} {i(2a; - y) + 2(y - 1)}.

48 ALGEBRA. [chap.

19. Multiply ^^-^ah-\-l^ by a'+|a6-|&2, and verify the resalt

when a = 1, 6 = 2.

20. Multiply X - x^i/^ + y by a:^ - y^.

4 2 1 t ' ' i

21. Divide a; ' + x^y^ + y by a:' - x'y * I- y * .

22. Find the product of x^y+y^ and x'-y'.

23. Multiply a^-a:^ by ar^ + xK

24. Divide C-3-8C-1-3 by c-i-3.

25. Divide

4a?^y-2-12a;V-i + 25-24a;"*y + 16a;"V ^7 2a;^y-i-3 + 4a:"V

26. Find the value of (ax~^ + a~^x){ax~^-Sa~^x).

1 _i

27. Find the square of a^ -l-a ^.

28. Find the continued product of Sa'^h'^Xj ax^ - 6 , and ax^ + b.

SI 1 r 1 It

29. Divide a:-y by a:''y^ + a:^y"^^ + a?*^y^

30. Multiply a2 + 2a-2-7 by 5 + a^-2a-^,

31. Find the value of (3ar"y " « - a; " ''y^){x^y - x " «y " ^).

Important Cases in Division.

66. The following example in division is worthy of notice.

Example. Divide d^' + b^ -{■ c^ — Sahc by a + b + c.

a + b + c)a^-^bc+ ¥+c^{a^-eib-ac-\-V^-bc + c^
a^+ a^b + orc

- a?b - CL'C - Sabc

— a%-afr- abc

-a'c + iib''-2abc

- a^c - abc - ac^

aU^- abc-hac^ + l^
ab- +lP + ¥c

- abc + a,c'^- b^c

- abc - b^c - bc^

ac:-j-b(?+ c3
ac' + bc^+ c^

Here the work is arranged in descending powers of a, and the
other letters are taken alphabetically ; thus in the first remainder
a^b precedes a-Cy and d^c precedes 3aic. A similar arrangement will
be observed throughout the work.

VIIT.]

REVISION OF ELEMENTARY RULES.

49

67. The following examples in division may be easily veri-
fied ; they are of great importance and should be carefully
noticed.

^=^+y,

x-y

I. i

x-y

x-y

\ X —

and so on ; the divisor being x-y^ the terms in the quotient all
positive, and the index in the dividend either odd or even.

'L =x^-xy +y\

II. \

x-\-y
*^=oi^ — x^y + ary'^ — xy^ + ?y*,

^ ^ =ofi- a^y + ^y2 - .v^y^ + x^y^ - xy^ + y^,

and so on ; the divisor being x+y^ the terms in the quotient
alternately positive and negative, and the index in the dividend
always odd.

I X

III. I

V
x+y ^'

x+y

x^-aFy+xy^-y^,

•^-y^ ^rs-

a^ — x^y+x^y^ — xhJ^-^-ocy^ — 'if*,

\ x-\ry

and so on ; the divisor being x-\-y, tlie terms in the quotient
alternately positive and negative, and the index in the dividend
always even.

TV. The expressions a^+y^, ^+^> xfi+y^, ... (where the
index is even, and the terms both positive) are 7ievei'' exactly
divisible by x+y or by x—y.

All these diflferent cases may be more concisely stated as
follows :

(1) af*—y"ia divisible by a?- y if w be a7iy whole number.

(2) x^+y" is divisible by x-\-y if n be any odd whole number.

!3) x" —y" is divisible by x+y if n be any even whole number.
4) x'^+y'* is never divisible by x+y or by x — y, when n is
an even whole number.

H.A. D

60 ALGEBRA. [CHAP.

Dimension and Degree.

^ 68. Each of the letters composiDg a term is called a dimen-
sion of the term, and the number of letters involved is called
the degree of the term. Thus the product cdw is said to be of
three dimennonsy or of the third degree ; and or* is said to be o/^
iive dimensions^ or of the fifth degree.

A numerical coefficient is not counted. Thus Bd?i^ and aV^
are each of seven dimensions.

69. The degree of an expression is the degree of the term
of highest dimensions contained in it; thus a* — Sa^+Sa — 5 is
an expression of the fourth degree, and a^x — Wa^ is an expression
of the fifth degree. But it is sometimes useful to speak of the
aimensions o^ an expression with regard to some one of the
letters it involves. For instance the expression ax^ — ftx^+cx— c?
is said to be of three dimensions in x.

70. A compound expression is said to be homogeneous when
all its terms are of the same degree. Thus 8a*— a*62+9aft^ is a
homogeneous expression of the sixth degree.

It is useful to notice that the product of two homogeneous
expressions is also homogeneous.

Thus by Art. 47,
(2a2 -Zab + 46^) ( - 5a2 + 3a6 + 462) = - 10a* + Sla^ft _ 2la^b^ + 166*.

Here the product of two homogeneous expressions each of
two dimensions is a homogeneous expression of four dimensions.

Also the quotient of one homogeneous expression by another
homogeneous expression is itself homogeneous.

For instance in the example of Art. 66 it may be noticed
that the divisor is homogeneous of one dimension, the dividend
is homogeneous of three dimensions, and the quotient is homo-
geneous of two dimensions.

EXAMPLES Vm. c.

1. Divide a^ + 30a6 - 12568 + 8 by a - 66 + 2.

2. Divide a^-\-y^-z^+Sacyz by x+y-z,

3. Divide a3-63+l+3a6 by o-6 + l.

4. Divide 18c<i+l + 27c5-8d3 by l+3c-2<i.

x-l'

6.

a^ + b^
a + b'

7.

x*-a*
x-a

8.

a:* -a*

a; -fa *

1+a

10.

2 + 6'

11.

a^ + W
a + b'

12.

a-6'

a:3 + 27y8
x+Sy '

14.

a^-oifi

a + x

15.

c' + l
c + l'

16.

VIII.] REVISION OF ELEMENTARY RULES. 51

Without actual division write down the qaotients in the
following cases :

5.

9.

13.

17. In the expression

2a362 + 3a64 + ^a^^x -7^ + 2Qa^l^ - 11a* + 7a»62,
which terms are likey and which are liomogeneoufi ?

18. In each term of the expression

7a36c2 - ab^c + IW^C^ - b\

introduce some power of a which will make the whole expression
homogeneous of the eighth degree.

19. By considering the dimensions of the product, correct the
following statement

(Sar^ - bxy + y^){%x^ -2xy- Sy^) = 24a?* - 46x^y + 9xY + ^^^/ " 3/,

it being known that there is no mistake in the coejficienta,

20. Write down the square of 3a^-2a6-6^ having given that
the coefficients of the terms taken in descending powers of a
are 9, -12, -2, 4, 1.

21. Write down the value of the product of 3a-6 + 6a^-a6^ and
db^ + 5a^ - Sa^bf having given that the coefficients of the terms when
arranged in ascending powers of b are 25, 0, - 9, 6, - 1.

22. The quotient of a;^ - y^ - 1 - Sxy by a; - y - 1 is

x^^ + xy + x + y^-y+l.

Introduce the letter z into dividend, divisor, and quotient so as to
make them respectively homogeneous expressions of three, one, and
two dimensions.

CHAPTER IX.

Simple Equations.

71. An eqnation asserts that two expressions are equal, but
we do not usually emj)loy the word equation in so wide a sense.

Thus the statement :r4-3+^ + 4= 2^7+7, which is always
true whatever value a: may have, is called an identical equation,
or briefly an identity.

The parts of an equation to the right and left of the sign
of equality are called members or sides of the equation, and
are distinguished as the ri^ht side and left side,

72. Certain equations are only true for particular values of
the symbols employed. Thus 3a; =6 is only true when a; =2,
and is called an equation of condition, or more usually an
equation. Consequently an identity is an equation which is
always true whatever be the values of the symbols involved;
whereas an equation (in the ordinary use of the word) is only
true for particular values of the symbols. In the above example
3^=6, the value 2 is said to satisfy the equation. The object
of the present chapter is to explain how to treat an equation of
the simplest kind in order to discover the value which satisfies it.

73. The letter whose value it is required to find is called
the unknown quantity. The process of finding its value is
called solving the equation. The value so found is called the
root or the solution of the equation.

74. An equation which involves the unknown quantity in
the first degree is called a simple equation* It is usual to
denote the unknown quantity by the letter a:.

The process of solving a simple equation depends only
upon the following axioms :

1. If to equals we add equals the sums are equal.

2. If from equals we take equals the remainders are equal.

3. If equals are multiplied by equals the products are equal.

4. If' equals are divided by equals the quotients are equal.

CHAP. IX.] SIMPLE EQUATIONS. 53

76. Consider the equation 7a: =14. ^

It is required to find what numerical value x must have to
satisfy this equation.

Dividing both sides by 7 we get

a: =2, [Axiom 4].

Similarly, if - = — 6,

multiplying both sides by 2, we get

a: = — 12, [Axiom 3] .

Again, in the equation 7a; — 2a:— a; =23 + 15 — 10, by collecting
terms, we have 4a: =28.

Transposition of Terms.

76. To solve 3a: - 8 = a: + 12.

This case differs from the preceding in that the unknown
quantity occurs on both sides of the equation. We can, how-
ever, transpose any term from one side to the other by simply
changing its sign. This we proceed to show.

Subtract x from both sides of the equation, and we get

3a:-a:-8 = 12, [Axiom 2].

Adding 8 to both sides, we have

3a: -a; = 12 + 8, [Axiom 1].

Thus we see that -\-x has been removed from one side, and
appears as — a: on the other ; and — 8 has been removed from
one side and appears as -f 8 on the other.

Hence we may enunciate the following rule :

Rule. Any term may be transposed from one side of the equation
to the other by changing its sign.

It appears from this that we may change the sign of every term
in an equation ; for this is equivalent to transposing all the terms,
and tnen making the right and left hand members change
places.

Example, Take the equation —3x— 12= a;— 24.

Transposing, ^ a; +24= 3a; +12,

or 3a;+12=-a:+24,

which is the original equation with the sign of every term changed.

54 ALGEBRA. [CHAP.

77. To solve f-3=f+f.

2 4 5

Here it will be conveoient to begin by clearing the equation
of fractional coefiicient 3. This can al ways be done by multiplying
both sides of the equation by the least common multiple of the
denominators. [Axiom 3.]

Thus, multiplying by 20,

10a:-60=5j7+4jr;

transposing, \^x-hx- 4.v = 60 ;

/. ^=60.

78. We can now give a general rule for solving any simple
equation with one unknown quantity.

Bnle. First, if necessary, clear of fractions ; then transpose
all the terms containing the unknoion quantity to one side of tlie
equation, and the known quantities to the other. Collect the terms
on each sidej divide both sides by the coefficient of the unknown
qtbantity, and the value required is obtained.

Example 1. Solve 6(a: - 3) - 7(6 - ar) + 3 = 24 - 3(8 - a;).

Removing brackets, 5a; - 15 - 42 + 7a: + 3 = 24 - 24 + 3a; ;
transposing, 5a; + 7a; - 3a; = 24 - 24+ 15 + 42 - 3 ;

.-. 9a; = 54;
a; = 6.

Example^. Solve (a; + l)(2a; - 1) - 6a; = (2a; -3)(a?- 6) +47.
Forming the products, we have

2a~» + ar-l-5a; = 2ar«-13a: + 15 + 47.
Erasing the term 2a;^ on each side, and transposing,

a;-5a;+13a;=15 + 47 + l;
.-. 9a; = 63 ;

a; = 7.

79. It is extremely useful for the beginner to acquire the
habit of verifying, that is, proving the truth of his results ;
the habit of applying such tests tends to make the btudent
self-reliant and confident in his own accuracy.

In the case of simple equations we have only to show that
when we substitute the value of x in the two sides of the equation
we obtain the same result

DC.] SIMPLE EQUATIONS. 55

Examjple. To show that x = l scUiafiea the equation

(a: + l)(2a; - 1 ) - 6a: = (2a: - 3)(a; - 5) + 47.

When a: = 7, the left side (a: + 1 )(2a: - 1 ) - 5a:

= (7 + l)(14-l)-35 = (8xl3)-35 = 69.

The right side (2a: - 3)(a: - 5) + 47

= (14 - 3)(7 - 6) + 47 = (11 X 2) + 47 = 69.

Thus, since these two results are the same, a; = 7 satisfies the
equation.

EXAMPLES IX. a.

Write down the solutions of the following equations :

1. 7x = 21. 2. 3a:=15. 3. 9a:=18. 4. 5a; = 5.

5. 12a: =132. 6. 33=1 la:. 7. 4a: = -12. 8. -10 = -5a;.

9. 4j;=18. 10. 12a: = 42. 11. 30 = -6a:. 12. 4a: = 0.

13. 6a: = 26. 14. 0=lla:. 15. l = lla:. 16. 3a: = -27.

17. = -2a:. 18. 6a: = 3. 19. 5 = 15a:. 20. -24 = -8a:.

Solve the following equations :

21. 6a:+3 = 15. 22. 6a:-7 = 28. 23. 13 = 7 + 2ar.

24. 15 = 37 -11a:. 25. 4a:-7 = ll. 26. 7a: =18 -2a:.

27. 3a:-18 = 7-2a:. 28. 4a: =13 -2a:- 10.

29. 3a: = 7-2a: + 8. 30. = 11 -2a: + 7- 10a:.

31. 8a;-3-5a:-5 = 7a:. 32. 7a: -13 = 12-5a:-5.

33. 5a;-17+3a:-5 = 6a:-7-8a?+115.

34. 7a:-21-4a:+13+2a: = 41-5a:-7 + 6a:.

35. 15-7a:-9a:-28+14a:-17 = 21-3a: + 13-9a:+8a:.

36. 5a;-6a:+30-7a: = 2a;+10-7a: + 5a:-20.

37. 5(a:-3) = 4(a:-2). 38. 11 (5 -4a;) = 7(5 -6a:).
39. 3-7(a;-l) = 6-4a;. 40. 5-4(a:-3) = a:-2(a:- 1).

41. 8(a:-3)-2(3-a;) = 2(a?+2)-6(5-a;),

42. 4(5-a;)-2(a:-3) = x-4-3(a;+2).

43. lx+lx = .-z. 44. 'f-y='f+l.

45. -f-|=3+| 46. .^-|-ii=^-+2.

47. (a; + 3)(2a:-3)-6a;=(a;-4)(2a: + 4) + 12.

48. (a; + 2)(a; + 3) + (a;-3)(a:-2)-2a:(a:+l) = 0.

49. (2a:+l)(2a: + 6)-7(x-2) = 4(a;+l)(a:-l)-9x.

50. (3a; + 1)2 + 6 + lS(a;+ 1)2 = 0a;(3a:-2) + 65.

56 ALGEBRA. [CHAP.

51. Show that a; = 5 satisfies the equation

6a:-6(a;-4) = 2(a: + 5) + 5(a;-4)-6.

52. Show that a; = 15 is the solution of the equation

7(25 - a;) - 2a; - 15 = 2(3a: - 25) - jc.

53. Verify that a: = 3 satisfies the equation

2(a;+l)(a; + 3) + 8 = (2a;+l)(ar+5).

54. Show that a; = 4 satisfies the equation

(3a: + 1 )(2a; - 7) = 6(a; - 3)* + 7.

80. We shall now give some equations of greater difficulty.

Example 1. Solve 5a: - (4a; - 7)(3a; - 5) = 6 - 3(4a: - 9)(a; - 1).

Simplifying, we have

5x-(12ar»-41a: + 35) = 6-3(4ar»-13a:+9);
and by removing brackets

5a: - 12ar^ + 41a: - 35 = 6 - 12a~5 + 39a: - 27.
Erase the term - 12a:^ on each side and transpose ;
thus 5a; + 41a:-39a; = 6-27 + 35;

.-. 7a: = 14 ;
.-. a: = 2.

Note. Since the - sign before a bracket affects every term within
it, in the first line of work we do not remove the brackets until we
have formed the products.

Example 2. Solve 4 1— = — _ - .

Multiply by 88, the least common multiple of the denominators ;

352-ll(a:-9) = 4a:-44;

removing brackets, 352 - 1 1 a: + 99 = 4a: - 44 ;

transposing, -lla:-4a: = -44- 352 - 99 ;

collecting terms and changing signs, 15a: = 495 ;

. • a: — t>t5.

a: — 9
Note. In this equation - — -— is regarded as a single term with

8
the minus sign before it. In fact it is equivalent to - -(a: - 9), the

o

vinculmn or line between the numerator and denominator having the
same effect as a bracket. [Art. 58.]

In ceitain cases it will be found more convenient not to multi-
ply throughout by the l.c.m. of the denominator, but to clear
of fractions in two or more steps.

rx.] SIMPLE EQUATIONS. 67

Example^. Solve ^zi+2^z3 = 5^z32_£±9^

^ 3 35 9 28

Multiplying throughout by 9, we have

transposing, ^-^^^^ + — ^^ = 2a! - 20.

Now clear of fractions by multiplying by 5 x 7 x 4 or 140 ;
.-. 72a:- 108 + 45a: + 406 = 2800? -2800 ;

/. 2800 -108 +405 = 280a: -72a? -46a:;

.-. 3097 = 163a:;

.-. a: = 19.

81 , To solve equations whose coefficients are decimals, we m ay
express the decimals as common fractions, and proceed as before ;
but it is often found more simple to work entirely in decimals.

Example, Solve -3753: - 1 -875 = 12a: + 1 185.

Transposing, '3753: - 12a: = 1 -186 + 1 -875 ;

collecting terms, ( '375 - -12)3: = 3 '06 ;
that is; •256a? = 3 06;

. ^_306

= 12.

EXAMPLES IX. b.

Solve the equations :

1. (a: + 15)(a:-3)-(a:-3)2 = 30-16(a:-l).

2. 15-3a: = (2a:+l)(2a:-l)-(2a:-l)(2a;+3).
8. 21-a:(2a;+l) + 2(a:-4)(a; + 2) = 0.

4. 3(a:+5)-.3(2a;-l) = 32-4(a:-5)2 + 4a:2.

5. 3a:2..7a;_(a.+2)(a;-2) = (a:+l)(a:-l) + (a:-3)(a:+3).

6. (a:-6){2a:-9)-(ll-2a;)(7-a:) = 5a:-4-7(a:-2).

17 a:-l,a:-9_o o a:,a:-8 i.a:-6

^- -5'+~2~""^- ^- 6^"ir"^'*^"3-

Q a: + 8 o_La?-6 ia 6a;-2^3a: + 5 1

8. -3-"^'^"r* ^^' "9~'^"l8"=3-

58 ALGEBRA. [CHAP. ix.

Solve the equations :

11, 12* + ]-l = 5x-2. 12. x+3 + ^-? = 7 + 2a:.

5 o

IK a? + 5 a:-f-l _ a; + 3 -,« ll-6a;_9-7a?_ 5(x- 1)

^^- T" ~9 4"' •^^* ~5 2 6

,„ 47-6a: . ^.4(2: -7) ,« 4-5a: \-1x 13

17. -^-(x-6) = _^^. 18. -g— 3-=-2-

,Q .3x-l x-l_2a:-3I „^ l-2a: 2-3x_,l x
^^- "lO 4 3~- '^"- ~7~ 8 ^2"^r

21. |(«-l)-|(^-4) = ^(^-6)+J.

22. ?(x-4)-i(2x-9) = l(x-l)-2.
5 o 4

23. ^(a: + 4) -^(x-3) = I(3x - 5)-?(x- G)-l(x-2).

24. ?(3-8x)-^(7-2x) + ^^ = 2-x-l(l-6x).

25. l(x + 4)-3(20-x) = l(5x-l)-l(5x-13) + 8.

„Q x+1 6x + 9 X+6.R x-12

26. -2— -^8-=^-*-^--3-
„„ -_10x + l_x_13x + 4 5(x-4)
^'* ~27~ 8"~18 T"'

28. •^-^-(a:-3)-l(x + 10)+^±5=.0.
29- ^'-|(l--) = 2-|(6-5x)-l(x.-4).

^^' ri'*"44''"2(n"3^3)"66"^3(^"^}

31. ^x -3*36 = 6-4 -3-2x. 32. -5x4- "25+ •1 + 1-25= •4x.

33. 3 ^Sx - 'Ihx = 9 + 1 -Sx. 34. •2x - -Olx + -OOSx =117.

35. -Sx - 'hx = ^Sx - 11. 36. •4x - -SSx = -7 - -3.

37. Find the value of x which makes the two expressions

(3x-l)(4x-ll) and 6{2x-l)(x-3) equal.

38. What value of x will make the expression 77x - 3(2x - l)(4x - 2)

equal to 337 - 8(3x - l)(x + 1) ?

CHAPTEE X.

Symbolical Expression.

82. In solving algebraical problems the chief difficulty of
the beginner is to express the conditions of the question by
means of symbols. A question proposed in algebraical symbols
will frequently be found puzzling, when a similar arithmetical
question would present no difficulty. Thus, the answer to the
question " find a number greater than a: by a " may not be self-
evident to the beginner, who would of course readily answer an
analogous arithmetical question, " find a number greater than 50
by 6." The process of addition which gives the answer in the
second case supplies the necessary hint ; and, just as the number
which is greater than 50 by 6 is 50+6, so the number which ia
greater than jchy aia x'+a,

83. The following examples will perhaps be the best intro-
duction to the subject of this chapter. After the first we leave
to the student the choice of arithmetical instances, should he
find them necessary.

Example 1. By how much does x exceed 17 ?

Take a numerical instance ; " by how much does 27 exceed 17 ? "
The answer obviously is 10, which is equal to 27 - 17.
Hence the excess of x over 17 is a; - 17.
Similarly the defect of x from 17 is 17 - x.

Example 2. If x is one part of 45 the other part is 45 - x.

A ft

Example 3. If jc is one factor of 45 the other factor is — .

X

Example 4. How far can a man walk in a hours at the rate of
4 miles an hour ?

In 1 hour he walks 4 miles.

In a hours he walks a times as far, that is, 4a miles.

60 ALGEBRA. [CHAP.

Example 6. If ^20 is divided equally among y persons, the share

20
of each is the total sum divided by the number of persons, or ^ —

y

Example 6. Out of a purse containing $x and y half-dollars a
man spends z quarters ; express in cents the sum left.

$x=4x quarters,
and y half-dollars =2^ quarters;

.*. the sum left=(4x-\-2y—z) quarters,
= 26 (4x -I- 2^—2?) cents.

EXAMPLES X. a.

1. By how much does x exceed 5 ?

2. By how much is y less than 15 ?

3. What must be added to a to make 7 ?

4. What must be added to 6 to make h ?

5. By what must 5 be multiplied to make a ?

6. What is the quotient when 3 is divided by a ?

7. By what must 6a: be divided to get 2 ?

8. By how much does 6a? exceed 2aj ?

9. The sum of two numbers is x and one of the numbers is 10;
what is the other ?

10. The sum of three numbers is 100 ; if one of them is 25 and
another is a;, what is the third ?

11. The product of two factors is 4a; ; if one of the factors is 4,
what is the other ?

12. The product of two numbers is p, and one of them is m ;
what is the other ?

13. How many times is x contained in 2y ?

14. The difference of two numbers is 8, and the greater of them
is a ; what is the other ?

15. The difference of two numbers is x, and the less of them is 6 ;
what is the other ?

16. What number is less than 30 by y ?

17. The sum of 12 equal numbers is 48a; ; what is the value of
each number ?

18. How many numbers each equal to y must be taken to make
15a;y ?

19. If there are x numbers each equal to 2a, what is their sum ?

20. If there are 5 numbers each equal to x, what is their product ?

X.] SYMBOLICAL EXPRESSION. 61

21 . If there are x numbers each equal top, what is their product ?

22. If there are n books each worth y dollars, what is the total
cost?

23. If ^ books of equal value cost x dollars, what does each
cost?

24. How many books each worth two dollars can be bought
for y dollars ?

25. If apples are sold at x for a dime, what will be the cost in
cents of y apples ?

26. What is the price in cents of n oranges at six cents a score ?

27. If I spend n dimes out of a sum of $5, how many dimes
have I left ?

28. What is the daily wage in dimes of a man who earns $12
in p weeks, working 6 days a week ?

29. How many days must a man work in order to earn ^6 at
the rate of y dimes a day ?

30. If ^ persons combine to pay a bill of ^y, what is the share
of each in dimes ?

31. How many dimes must a man pay out of a sum of ^p so as
to have left 30x cents ?

32. How many persons must contribute equally to a fund con-
sisting of $x, so that the subscription of each may equal y quarters ?

33. How many hours will it take to travel x miles at 10 miles
an hour ?

34. How far can I walk in p hours at the rate of q miles an hour ?

35. If I can walk m miles in n days, what is my rate per day ?

36. How many days will it take to travel y miles at x miles
a day?

84. We subjoin a few harder examples worked out in full.

Example 1. What is (1) the sum, (2) the product of three con-
secutive numbers of which the least is n ?

The two numbers consecutive to n are n + 1 and w + 2 ;
/. the sum = w+(n + l) + (w + 2)
= 371 + 3.
And the product = 7i{n + 1 )(n + 2).

Example 2. A boy is x years old, and five years hence his age
will be half that of his father : how old is the father now?

In five years the boy will be a: + 5 years old ; therefore his father
will then be 2{x + 5)i or 22:+ 10 years old; his present age must
therefore be 2a; + 10 - 5 or 2a; + 5 years.

62 ALGEBRA. [CHAP.

Example 3. A and B are playing for money ; A begins with $p
and B with q dimes. B wins $x ; express by an equation the fact
that A has now 3 times as much as B.

What B has won A has lost ;

. '. A has p—x dollars, that is 10(jp— x) dimes,

B has q dimes +x dollars, that is g+lOx dimes.

Thus the required equation is 10(p— a;)=3(g+10x).

Example 4. A man travels a miles by coach and b miles by train ;
if the coach goes at the rate of 7 miles an hour, and the train at
the rate of 25 miles per hour, how long does the journey take ?

The coach travels 7 miles in 1 hour ;
.* 1 ihour;

that is, a ^ hours.

Similarly the train travels h miles in — hours.

;. the whole time occupied is - + -— hours.

'^ 7 25

Example 5. How many men will be required to do in /> hours
what q men do in np hours ?

np hours is the time occupied by q men ;

.*. 1 hour qxnp men;

that is, p hours iL^--P men.

P
Therefore the required number of men is qn.

EXAMPLES X. b.

1, Write down three consecutive numbers of which a is the least.

2, Write down four consecutive numbers of which b is the
greatest.

3, Write down five consecutive numbers of which c is the
middle one.

4, What is the next odd number after 2n - 1 ?

5, What is the even number next before 2n ?

6, Write down the product of three odd numbers of which the
middle one ia2x-\-l.

7, How old is a man who will be x years old in 15 years ?

8, How old was a man x years ago if his present age is n years ?

9, In ^ years a man will be y years old, what is his present age ?

X.] SYMBOLICAL EXPRESSION. 63

10. How old is a man who in x years will be twice as old as
his son now aged 20 years ?

11. In 6 years a boy will be x years old ; what is the present
age of his father if he is twice as old as his son ?

12. A has $m and B has n dimes ; after A has won 3 dimes
from B, each has the same amount. Express this in idgebraical
symbols.

13. A has 25 dollars and B has 13 dollars ; after B has won
X dollars he then has four times as much as A. Express this in
algebraical symbols.

14. How many miles can a man walk in 30 minutes if he walks
1 mile in x minutes ?

15. How many miles can a man walk in 50 minutes if he walks
X miles in y minutes ?

16. How long will it take a man to walk p miles if he walks
15 miles in q hours ?

17. How far can a pigeon fly in x hours at the rate of 2 miles in
7 minutes t

18. A man travels x miles by boat and y miles by train, how
long will the journey take if the train goes 30 miles and the boat
10 miles an hour ?

19. If X men do a work in 5a; hours, bow many men will be
required to do the same work in y hours ?

20. How long will it take p men to mow q acres of com, if each
man mows r acres a day ?

21. Write down a number which, when divided by a, gives a
quotient h and remainder c.

22. What is the remainder if x divided by y gives a quotient z ?

23. What is the quotient if when m is divided by n there is a
remainder r ?

24. H a bill is shared equally among n persons, and each pays
75 cents, how many dollars does the bill amount to ?

25. A man has $x in his purse, he pays away 25 dimes, and
receives y cents ; express in dimes the sum he has left.

26. How many dollars does a man save in a year, if he earns
$x a week and spends y quarters a calendar month ?

27. What is the total cost of 6x nuts and 4a; plums, when x
plums cost a dime and plums are three times as expensive as
nuts?

28. If on an average there are x words in a line, and y lines in
a page, how many pages will be required for a book which contains
z words ?

CHAPTEE XL
Problems leading to Simple Equations.

85. The principles of the last chapter may now be employed
to solve various problems.

The method of procedure is as follows :

Bepresent the unknown quantityby a symbol, as x, and express
in symbolical language the conditions of the question ; we thus
obtain a simple equation which can be solved by the methods
already given in Chapter IX.

Example I. Find two numbers whose sum is 28, and whose
dififereDce is 4.

Let X be the smaller number, then a; + 4 is the greater.
Their sum is a: + (a: + 4), which is to be equal to 28.
Hence a: + ar+4 = 28;

2a; = 24 ;
.-. X = 12,
and a;+4= 16,

80 that the numbers are 12 and 16.

The beginner is advised to test his solution by finding
whether it satisfies the conditions of the question or not.

Example II. Divide $47 between A, B, C, so that A may have
9 10 more than B, and B |8 more than C.

Let X represent the number of dollars that C has ; then B has
05+8 dollars, and A has x -1-8 + 10 dollars.

Hence a;+ (x+8) + (a;+8+10) =47 ;

a;+x+8+a;+8+10=47,

3a;=21;

BO that C has $7, B $15, A $25.

CHAP. XI.] PROBLEMS LEADING TO SIMPLE EQUATIONS. 65

EXAMPLES XI. a.

1. Six times a number increased by 11 is equal to 65 ; find it.

2. Find a number which when multiplied by 11 and then
diminished by IS is equal to 15.

3. If 3 be added to a number, and the sum multiplied by 12, the
result is 84 ; find the number.

4. One number exceeds another by 3, and their sum is 27 ; find
them.

5. Find two numbers whose sum is 30, and such that one of them
is greater than the other by 8.

^ 6. Find two numbers which differ by 10, so that one is three
times the other.

7. Find two numbers whose sum is 19, such that one shall exceed
twice the other by 1.

8. Find two numbers whose sum shall be 26 and their differ-
ence 8.

9. Divide $100 between A and B so that B may have f 30 more
than A,

10. Divide 1^ between A, B, and C so that B may have
more than A^ and C ^14 more than B.

11. A^ jB, and C have #72 among them ; G has twice as much
as J?, and B has $4 less than A ; find the share of each.

12. How must a sum of 73 dollars be divided among A^ B^
and C, so that B may have 8 dollars less than A and 4 dollars
more than C ?

Example III, Divide 60 into two parts, so that three times the
greater may exceed 100 by as much as 8 times the less falls short
of 200.

Let X be the greater part, then 60 - a; is the less.

Three times the greater part is 3^, and its excess over 100 is

3a: - 100.
£ight times the less is 8(60 - x), and its defect from 200 is

200 - 8(60 - x).
Whence the symbolical statement of the question is

3ar - 100 = 200 - 8(60 - a:) ;
3a? -100 = 200 -480 + 8a?,
480 -100 -200 = 8a: -3a:,
5a; = 180 ;
.*. X = 36, the greater part,
and 60 - a; = 24, the less.

H.A. B

66 ALGEBRA. [chap.

Example IV. ^ is 4 years older than B, and half ^'s age
exceeds one-sixth of ^'s age by 8 years ; find their ages.

Let a; be the number of years in B^a age, then A 's age is a; -f 4 years.

One-half of A*b age is represented by i(a;+4) years, and one-sixth
of B'b age by ^2; years.

Hence J(a; + 4)-|a; = 8 ;

multiplying by 6 3a: + 12 - a: = 48 ;

.-. 2a; = 36 ;

.-. X - 18.

Thus B*B age is 18 years, and A* a age is 22 years.

13. Divide 75 into two parts, so that three times one part may
be double of the other.

14. Divide 122 into two parts, such that one may be as much
above 72 as twice the other is below 60.

15. A certain number is doubled and then increased by 5, and
the result is less by 1 than three times the number ; find it.

16. How much must be added to 28 so that the resulting number
may be 8 times the added part ?

17. Find the number whose double exceeds its half by 9.

18. What is the number whose seventh part exceeds its eighth
part by 1 ?

19. Divide 48 into two parts, so that one part may be three-fifths
of the other.

20. If ^, B, and C have $76 between them, and A*a money is
double of ^'s and (7's one-sixth of ^'s, what is the share of each ?

21. Divide $511 between Ay B^ and G, so that B^a share shall be
one-third of ^'s, and CPa share three-fourths of -4's and -B's together.

22. ^ is 16 years younger than A^ and one-half ^'s age is equal
to one-third of -4's ; how old are they ?

23. -4 is 8 years younger than 5, and 24 years older than (7;
one-sixth of ^'s age, one-half of ^'s, and one-third of (Ta together
amount to 38 years ; find their ages.

24. Find two consecutive numbers whose product exceeds the
square of the smaller by 7. [See Art. 84, Ex. 1.]

25. The difference between the squares of two consecutive
numbers is 31 ; find the numbers.

86. We shall now give examples of somewhat greater
difficulty.

Example I. A has $6, aiMl B has six dimes ; after B has won
from A a certain sum, A has then fiTe-sixths of what B has ; how
much did B win ?

XI.] PROBLEMS LEADING TO SIMPLE EQUATIONS. G7

Suppose that JB wins x dimes, A has then 60 -x dimes, and B
has 6 -fa; dimes.

Hence 60 - a)= f (6 -f x) ;

36O--6a;=30+5a5,

11»=330 ;

a;=30.

Therefore ^ wins 30 dimes, or $3.

Example II. A is twice as old as B, ten years ago he was four
times as old ; what are their present ages ?

Let B^a age be x years, then -<4's age is 2x years.

Ten years ago their ages were respectively as— 10 and 2x--10
years ; thus we have

2a;-10=4(a;-10);

2aj-10=4a5-40,

2x=30 ;

.'. x=16,

so that B is 15 years old, A 30 years.

EXAMPLES XI. b.

1, A has $12 and B has $S ; after B has lost a certain sum to A
his money is only three-sevenths of ^'s ; how much did A win ?

2, A and B begin to play each with $16 ; if they play till J5's
money is four-elevenths of -4's, what does B lose ?

3, A and B have $28 between them ; A gives $3 to B and then
finds he has six times as much money as B ; how much had each
at first ?

4, A had three times as much money as B ; after giving $3 to
B he had only twice as much ; what had each at first ?

5, A father is four times as old as his son ; in 16 years he will
only be twice as old ; find their ages.

6, -4 is 20 years older than 5, and 5 years ago A was twice as
old as B ; find their ages.

7, How old is a man whose age 10 years ago was three-eighths
of what it will be in 16 years ?

8, Ai& twice as old as J? ; 5 years ago he was three times as
old ; what are their present ages ?

9, A father is 24 years older than his son ; in 7 years the son's
age will be two-fifths of his father's age ; what are their present
ages?

68 AL6EBBA. [CHAP.

Example III. A person spent $56.40 in baying geese and
ducks ; if each goose cost 7 dimes, and each duck 3 dimes, and if
the total number of birds bought was 108, how many of each did
he buy ?

In questions of this kind it is of essential importance to have all
quantities expressed in the same denomination; in the present
instance it will be convenient to express the money in dimes.

Let X be the number of geese, then 108— x is the number of ducks.

Since each goose costs 7 dimes, x geese cost 1x dimes.

And since each duck costs 3 dimes, 108— x ducks cost 3(108— x)
dimes.

Therefore the amount spent is

7x+3(108-x) dimes.

But the question states that the amount is also $56.40, that is 564
dimes.

Hence 7x4-3(108-x) =564 ;

7x+324-3x=564,

4x=240,

.-. x=60, the number of geese,

and 108— x=48, the number of ducks.

Note. In all these examples it should be noticed that the un-
known quantity x represents a number of dollars, ducks, years, etc. ;
and the student must be careful to avoid beginning a solution with
a supposition of the kind, **let x=A^a share" or "let x=the
ducks," or any statement so vague and inexact.

It will sometintes be found easier not to put x equal to the
quantity directly required, but to some other quantity involved
in the question ; by this means the equation is often simplified.

Example IV. A woman spends $1 in buying eggs, and finds
that 9 of them cost as much over 25 cents as 16 cost under 75 cents ;
how many eggs did she buy ?

Let X be the price of an egg in cents ; then 9 eggs cost 9x cents,
and 16 eggs cost 16x cents ;

.-. 9x-25=76-16x,
25x=100;

Thus the price of an egg is 4 cents, and the number of eggs
= 100-f-4=25.

10, A sum of $30 is divided between 50 men and women, the
men each receiving 75 cents, and the women 50 cents; find the
number of each sex.

XI.] PROBLEMS LEADING TO SIMPLE EQUATIONS. 69

11. The price of 13 yards of cloth is as much less than $10 as
the price of 27 yards exceeds $20 ; find the price per yard.

12. A hundredweight of tea, worth $68, is made up of two
sorts, part worth 80 cents a pound and the rest worth 50 cents a
pound ; how much is there of each sort ?

13. A man is hired for 60 days on condition that for each day
he works he shall receive $2, but for each day that he is idle he
shall pay $1 for his board : at the end he received $90 ; how many
days had he worked ?

14. A sum of $6 is made up of 30 coins, which are either quar-
ters or dimes ; how many are there of each ?

15. A sum of $11.46 was paid in half-dollars, quarters, and
dimes ; the number of half-dollars used was four times the number
of quarters and ten tunes the number of dimes ; how many were
there of each ?

16. A person buys coffee and tea at 40 cents and 80 cents a
pound respectively ; he spends $16.10, and in all gets 24 lbs. ; how
much of each did he buy ?

17. A man sold a horse for a sum of money which was greater
by $08 than half the price he paid for it, and gained thereby $18;
what did he pay for the horse ?

18. Two boys have 240 marbles between them ; one arranges
his in heaps of 6 each, the other in heaps of 9 each. There are 36
heaps altogether ; how many marbles has each ?

19. A man's age is four times the combined ages of his two sons,
one of whom is three times as old as the other ; in 24 years their
combined ages will be 12 years less than their father's age ; find
their respective ages.

20. A sum of money is divided between three persons, A^ B, and
C in such a way that A and B have $42 between them, B and C
have $46, and C and A have $53 ; what is the share of each ?

21. A person bought a number of oranges for $3, and finds that
12 of them cost as much over 24 cents as 16 of them cost under 60
cents ; how many oranges were bought ?

22. By buying eggs at 16 for a quarter and selling them at a
dozen for 16 cents a man lost $1.50 ; find the number of eggs.

23. I bought a certain number of apples at four for a cent, and
three-fifths of that number at three for a cent ; by selling them at
sixteen for five cents I gained 4 cents ; how many apples did I buy ?

24. If 8 lbs. of tea and 24 lbs. of sugar cost $7.20, and if 3 lbs.
of tea cost as much as 46 lbs. of sugar, find the price of each per
pound.

70 ALGEBRA. [CHAP. XI.

25. FoTir dozen of port and tliree dozen of sherry cost $89 ; if
a bottle of port costs 25 cents more than a bottle of sherry, find the
price of each per dozen.

26. A man sells 50 acres more than the fourth part of his farm
and has remaining 10 acres less than the third ; find the number
of acres in the farm.

27. Find a number such that if we divide it by 10 and then
divide 10 by the number and add the quotients, we obtain a result
which is equal to the quotient of the number increased by 20 when
divided by 10.

28. A sum of money is divided between three persons, -4, B,
and C, in such a way that A receives $10 more than one-half of
the entire amount, B receives $10 more than one-third, and C the
remainder, which is $10 ; find the amounts received by A and B.

29. The difference between two nimibers is 15, and the quotient
arising from dividing the greater by the less is 4 ; find the numbers.

30. A person in buying silk found that if he should pay $3.50
per yard he would lack $15 of having money enough to pay for it ;
he therefore purchased an inferior quality at $2.50 per yard and
had $25 left ; how many yards did he buy ?

31. Find two numbers which are to each other as 2 to 3, and
whose sum is 100.

32. A man's age is twice the combined ages of his three sons,
the eldest of whom is 3 times as old as the youngest and f times as
old as the second son ; in 10 years their combined ages will be 4
years less than their father's age ; find their respective ages.

33. The sum of $34.50 was given to some men, women, and
children, each man receiving $2, each woman $1, and each child
50 cents. The number of men was 4 less than twice the number of
women, and the number of children was 1 more than twice the
number of women ; find the total number of persons.

34. A man bought a number of apples at the rate of 5 for
3 cents. He sold four-fifths of them at 4 for 3 cents and the
remainder at 2 for a cent, gaining 10 cents ; how many did he
buy?

35. A farm of 350 acres was owned by four men, A^ B, C, and
2). B owns five-sixths as much &s A, C four-fifths as much as B,
and D one-sixth as much as A, B, and C together ; find the num-
ber of acres owned by each.

CHAPTER XII.

Elementary Fractions.

Highest Oommon Factor of Simple Expressions.

88. Definition. The highest common factor of two or
more algebraical expressions is the expression of highest dimen-
sions [Art. 68] which divides each of them without remainder.

Tlie abbreviation H.C.F. is sometimes used instead of the
words highest common factor,

89. In the case of simple expressions the highest common
factor can be written down by inspection.

Example 1. The highest common factor of a"*, a^, a^, a^ is a^.

Example 2. The highest common factor of a^6*, a^ft-'c^, a^b^c is
a^b* ; for a^ is the highest power of a that will divide a^, a^, a"* ;
h* is the highest power of b that will divide b*, ¥, W ; and c is not a
common factor.

90. If the expressions have numerical coefficients, find by
Arithmetic their greatest common measure, and prefix it as a
coeflicient to the algebraical highest common factor.

Example. The highest common factor of 21a^ar'y, 35a^a;*y, 2Sa^xy
is la^xy ; for it consists of the product of

(1) the greatest common measure of the numerical coefficients ;

(2) the highest power of each letter which divides every one of

the given expressions.

EXAMPLES Xn. a.

Find the highest common factor of

1. 3a62, 2a68. 2. a^y^ 4a:y. 3. 6c«, Sb^c. 4. 4ar', 2xy'^,

5. a^l^c,a%c\ 6. 3a26, 9a6c. 7. 6a;yz, 2a:y. 8. ISy^ Sary^zS.

9. I2a^bc\ 18a6V, 10. Tar^i^^z*, 2\xhp^,

11. %ax, 6a2y, Kkitf^^, 12. aVy, ¥xy^, cxSf\

13. 146c2, 636a^ 5662c. 14. loa:^, GOaryz^ 2bx^z^.

15. iTa^z, SlaryzS Uxhfz. 16. lla%^c^, ^^Wc\ ab^c\

72 ALGEBRA. [CHAP.

Lowest Common Multiple of Simple Expressions.

91. Definition. The lowest common multiple of two or
more algebraical expressions is the expression of lowest dimen-
sions which is divisible by each of them without remainder.

The abbreviation L.C.M. is sometimes used instead of the
words lowest common mvltiple,

92. In the case of simple expressions the lowest common
multiple can be written down by inspection.

Example 1. The lowest common multiple of a^, a?^ a?^ a* is a^.

Example 2. The lowest common multiple of a^h\ a6*, am^ is
aW ; for a* is the lowest power of a that is divisible by each of the
quantities a*, a, a^ ; and ¥ is the lowest power of h that is divisible
by each of the quantities 6*, 6^ V.

93. If the expressions have numerical coefficients, find by
Arithmetic their least common multiple, and prefix it as a co-
efficient to the algebraical lowest common multiple.

Example, The lowest common multiple of 2\a^Qi^, SSa^ar^y,
28a^a:y is 420a*ar*y ; for it consists of the product of

(1) the least common multiple of the numerical coefficients ;

(2) the lowest power of each letter which is divisible by every

power of that letter occurring in the given expressions.

EXAMPLES XII. b.

Find the lowest common multiple of

1. xyz, 3i/3. 2. a26*, ahc. 3. 27?y, Zxyh,

4. 4a2, Zaha^, 5. 4a*6c3^ 5aj2, ^ 2ab, 4xy,

7. mn, nlf Im. 8. ^l/^t ^z\ 2z3tP. 9. 2xy, Syz, 42a:.

10. P^qr, ^p^, 7pq^ 11. loxh/, 25xy7^. 12. 9ah^ 2la^c.

13. 27a^ 8163, lSa%\ 14. 5asifi, 6cy, la^x^iH,

15. ISa^fts, 20aar^y, SOx^. 16. 72p2^r*, 108p«gV.

Find both the highest common factor and the lowest common
multiple of

17. 2a62, 3a263, 4a^6. 18. 15a:V, Sar^z^ 19. 2a*, Sa^feV.

20. 57aar2/, 76a:yV. 21. ^^a'^Vc, ^MW.

22. 51m>S jon, S4mnp*. 23. 49a*, 566*c, aioc*.

24. 66a263ca;S 56ab'^z^, I2l7?yz\

3ni.]

ELEMENTARY FRACTIONS.

73

Elementary Fractions.

94. Definition. If a quantity x be divided into h equal
parts, and a of these parts be taken, the result is called the

fraction =- of x. If x be the unit, the fraction y- of j? is called

simply " the fraction =- " ; so that the fraction j- represents a

equal parts, h of which make up the unit,

95. In this chapter we propose to deal only with the easier
kinds of fractions, where the numerator and denominator are
simple expressions.

Their reduction and simplification will be performed by the
usual arithmetical rules. For the proofs of these rules the
reader is referred to the Elementary Algebra for Schools,
Chapter xv.

Bnle. To reduce a fraction to its lowest terms: divide
numerator and de nominator by every factor which is common to
them bothy that is by their highest common factor.

Dividing numerator and denominator of a fraction by a com-
mon factor is called cancelling that factor.

Examples, (1) ^a^c 2a

(2)

(3)

9ac2 3c

Ixhfz 1
2%7^yz^ \xz

S5a^bh_5a*b_f. ..
lab'^c 1

EXAMPLES XII.

Reduce to lowest terms :
, 2a o 3a2 « 26

4ab'

"• ISabc^ '
,n I5l^m*

o 3aa

^- 9^-

o I2mn

^- TElm'

,. 21a*b^aP

n 26c2
^* Wc

rj I4xy^
'• 2l^'

11. 1^2/".
24a;^y*

4.
8.

2ahc

90^6
12a6V

•1 Q Sm^n^p^
I5mn^p'

,^ 42^y2^

74 ALGEBRA. [chap.

Multiplication and Division of Fractions.

96. Bnle. To mvltiply algebraical fractions : as in Arith-
meticy miUtiply together all the numerators for a new numerator ,
and all the deiiominators for a new denominator,

^36 2a^b 2x Zbx2a^x2x 2a'
by cancelling like factors in numerator and denominator.

^^ 6c^ 3a» 76-^ '
all the factors cancelling each other.

97. Bule. To divide one fraction hy another: invert the
divisor and proceed as in mvliiplication.

^-'»^«- 4^><S-

28aV
1562a:jr*

• _ 7a8 ^6c»a:^

1562a^

9c

28a2c3

all the other factors cancelling each other.

EXAMPLES Xn. d.

Simplify the following expressions^:

1. '^x"''^. 2.

ab xy^

a6 ^4c*rf
2cd^ ab''

3.

4a^x

^ lal^'\2ax' °-

3a62 1562c«
563c ' 9a26 '

6.

56c2

25c2
146c'

„ ahn ^2cd:^ ^9my
'• 6V 3a6 4m=^'

9xy

8a'b''

^ 2aV»^ 1062.6V
^* 6ax* 4a;2 ' 3a:«*

la '^x

zar

I7y.

a: V •

34y3

X^y'

„ 8a V ^ 9aar» ^ oSyS
^^- 26y 6a22 26V

12. r >

40c

^4d',

' a6c

,Sld^
' 27c2-

Reduction to a Common Denominator.

98. In order to find the sum or difference of any fractions,
we must, as in Arithmetic, first reduce them to a common
denominator ; and it is most convenient to take the lowest com-
mon multiple of the denominators of the given fractions.

xn.] ELEMENTARY FRACTIONS. 75

Example. Express with lowest common denominator the fractions

a h c

Zxy* %xyz 2yz

The lowest common multiple of the denominators is Qxyz, Multi-
plying the numerator of each fraction by the factor which is required
to make its denominator Gxyz, we have the equivalent fractions

2az h Sex
Qxyz %xyz Qxyz

Note. The same result would clearly be obtained by dividing the
lowest common denominator by each of the denominators in turn,
and multiplying the corresponding numerators by the respective
quotients.

EXAMPLES XII. e.

Express as equivalent fractions with common denominator :

•. X 2x n y ^t/ n Sa 4a a ^ n.

2a a So 2c oc 2y

^ m Sm n X 2a „ a 2a q x Sx

4n 5n Sy x h o^ Sy y^

9» r> -> !• 10» r> -» 2a. H, 3, ^^^ - — .

by ha 26 2a

tqIII ^n a h c i^wwjo

OC ca ab 2xy Syz zx n m q

Addition and Subtraction of Fractions.

99. Rule. To add or subtract fractions: compress all the
fractions with their lowest common denominator ; form the algebrai-
cal sum of the numerators^ and retain the common denominator.

Example I. Simplify -t+'.^~'/.'

3 4 o

The least common denominator is 12.

20x+9x-Ux 15a? 5x

The expression =

12 12 4

Example 2. Simplify ^"g-^-

rr,, • Q(ib-5ab-ah ^
The expresBion = ^ = ito = *'

76 ALGBBRA. [CHAP. Xil.

Example 3. Simplify ^„ - -^.

a^cr 3ca'

The expression = ~ ^^, and admits of no farther simplification.
Note. The beginner must be careful to distinguish between

erasing equal terms with different signs, as in Example 2, and

cancelung equal factors in the course of multiplication, or in
reducing fractions to lowest terms. Moreover, in simplifying frac-
tions he must remember that a factor can only be removed from
numerator and denominator when it divides each taken 08 a whole.

Thus in °if 75^ , c cannot be cancelled because it only divides cy

and not the whole numerator. Similarly a cannot be cancelled
because it only divides Goa; and not the whole numerator. The
fraction is therefore in its simplest form.

When no denominator is expressed the denominator 1 may
be understood.

Example^. 3x - ^' = ?f _ .^M^^

4y 1 4y 4y

If a f racoon is not in its lowest terms it should be simplified
before combining it with other fractions.

n I K aa xhi ax x ^ax-2x

Example 6. -^^ - ^ = -^ - « = n — •

^ 2 3a:y 2 3 6

EXAMPLES Xn. f.

Simplify the following expressions :

1.

2 + 3-

2.

3^4-

O X X

3" 4-5-

A ^y.y

^ 3^C*

5.

a b

6.

m 2n
"8 20'

'' r21-

5a 6
^' 12 "4-

9.

? + l
X y

10.

X a

y-b'

,, 2a 4a

,„ ah x^y
■^- 3 Gary

13.

a a a
4 8^12'

14. ^"^
^^ 3

x,9x

15.

a 2a 3a

xy yz zx

la

xy 2y Ay
Ix 3 8*

17. 2 +

a h^
b ah'

18.

a h c
p2 pf^ q2'

19.

.t3

20.

2 m2'

21- ^-3-

22. w

28.

d?x „dx
dy" ~dy

'2y»-

OA «' «' , ac

CHAPTER XIII.

Simultaneous Equations.

100. Consider the equation 2,2>+5y=23, which contains two
unknown quantities.

By transposition we get

5y=23-2a?;

that is, y=?5^ (1).

From this it appears that 'for every value we choose to give
to a; there will be one corresponding value of ^. Thus we shall
be able to find as many pairs of values as we please which
satisfy the given equation.

For instance, if ^==1, then from (1) we obtain y=--.

5

27
Again, if ^= —2, then v=—- ; and so on.

5

But if also we have a second equation containing the same un-
known quantities, such as 3aj+4y=24,

24 — 3:f
we have from this y= — - — (2).

If now we seek values of sp and y which satisfy both equa-
tions, the values of y in (1) and (2) must be identicsd.

Therefore 23-2^^24-3^

5 4

Multiplying across 92 - &f= 120 - 15.?? ;

.-. 7a?=28j

Substituting this value in the first equation, we have

8 + 5^^=23;
.•. 5y=15;

and

Thus, if both equations are to be satisfied by the same /alues
of ^ and y, there is only one solution possible.

78 ALGEBRA. [chap.

101. Definition. When two or more equations are satisfied
by the same values of the unknown quantities they are called
simnltaneons equations.

We proceed to explain the different methods for solving simul-
taneous equations. In the present chapter we shall contine our
attention to the simpler cases in which the unknown quantities
are involved in the first degree.

102. In the example already worked we have used the
method of solution which best illustrates the meaning of the
term simtdtaneous eqtuition ; but in practice it will be found that
this is rarely the readiest mode of solution. It must be borne
in mind that since the two equations are simultaneously true,
any equation formed by combining them will be satisfied by the
values of x and y which satisfy the original equations. Our
object will always be to obtain an equation which involves one
omy of the unknown quantities.

103. The process by which we cause either of the un-
known quantities to disappear is called elimination. We shall
consider two methods.

Elimination by Addition or Subtraction.

Example 1. Solve 3a;+7y=27 (1),

5a;+2y=16 (2).

To eliminate x we multiply (1) by 5 and (2) by 3, so as to make
the coefficients of a; in both equations equal. This gives

15a:+35y=135,

I5a:+ 6y = 48;

suhtradiTig, 29y = 87 ;

.-. y = 3.

To find Xf substitute this value of y in either of the given
equations.

Thus from (1), 3a:+21 = 27 ;

and

Note. When one of the unknowns has been found, it is immaterial
which of the equations we use to complete the solution. Thus, in
the present example, if we substitute 3 for y in (2), we have

6aj + 6 = 16 J

.*• a; = 2, aa before.

xm.] SIMULTANEOUS EQUATIONS. 79

Example2, Solve 7a; + 2y=:47 (1),

5iB-4y= 1 (2).

Here it will be more convenient to eliminate y.
Multiplying ( 1 ) by 2, 14a: + 4y = 94,
and from (2) 6a; - 4y = 1 ;

adding^ 19a; ? 95 ;

a; = 5.
Substitute this value in (1),

.-. 36+2y = 47;
.-. y = 6,

Note. Add when the coefficients of one unknown are equal and
unlike in sign ; subtract when the coefficients are equal and like in
sign.

Elimination by Substitution.

Example 3. Solve 2x=6y-\- 1 (1),

24-7a;=3y (2).

Here we can eliminate x by substituting in (2) its value obtained
from (1). Thus

24-j5y+l) = 3y;

.-. 48-35y-7 = 6y;
.-. 41=41y;

and from (1)

a; = 3.)

104, Any one of the methods given above will be found
sufficient ; but there are certain arithmetical artifices which
will sometimes shorten the work.

Example, Solve 28a;-23y = 22 (1),

63a;-55y = 17 (2).

Noticing that 28 and 63 contain a common factor 7, we shall make
the coefficients of x in the two equations equal to the least common
multiple of 28 and 63 if we multiply (1) by 9 and (2) by 4.

Thus 252a; - 207y = 198,

252a:-22()ya 68;

subtracting, 13y = 130 ;

that is, y = 10,

and therefore from (1), a; = 9.

80 ALGEBRA. [chap.

EXAMPLES Xm. a.

Solve the equations :

1. a:+y=19, 2. x+y = 23, 3. x + y = n,

x-y:= *J. x-y= 5. x-y = -9.

4, a;+y = 24, 5. a;-y = 6, 6. a:-y = 25,

x-y= 0. x+y = 0. x+y = 13,

7. 3x+6y = 50, 8. a;+5y=18, 9. 4ar+y=10,

4a:+3y = 41. 3:^+2^ = 41. 6a; + 7y = 47.

10. 7a:-6y = 25, 11. 6a:+4y=:7, 12. 3a:-7y=l,

52;+4y = 51. 4a:+6y = 2. 4x+ y = 53.

13. 7a:+5y = 45, 14. 4a; + 5y = 4, 15. lla;-7y = 43,

2a;-3y = 4. 5a;-3y = 7a 2a:-3y = 13.

16. 4a:-3y = 0, 17. 2a:+3y = 22, 18. 7a; + 3y = 65,

7a:-4y = 36. 6a?+2y = 0. 7a;-8y = 32.

19. 13a:-y=14, 20. 9a;-8y=14, 21. 14a:+13y = 35,
2a;-7y = 9. 15a:-14y = 20. 21a; + 19y = 56.

22. 5a? = 7y-21, 23. 55a: = 33y, 24. 6x-1y = ll,

21a: - 9y = 75. 10a: = 7y - 15. 18a: = 12y.

25. 13a:-9y = 46, 26. 6a:-5y=ll, 27. lly-lla; = 66,
lla:-12y=17. 28a; + 21y = 7. 7a:+8y = 3.

28. 6y-5a: = ll, 29. 3a:+10=5y, 30. 4y = 47 + 3a:,
4a: = 7y-22. 7y = 4a;+13. 6a: = 30-16y.

31. lla:+13y = 7, 32. 13a:-17y = ll, 33. 19a: + 17y = 7,
13a:+lly=17. 29a;-39y = 17. 41a: + 37y = 17.

105. We add a few cases in which, before proceeding to
solve, it will be necessary to simplify the equations.

Example, Solve 5{a:+2y)-(3a:+lly) = 14 (1),

7a:-9y-3(a;-4y) = 38 (2).

From (1), &x + lOy-Sx - lly = 14 ;

.-. 2a:-y=14 (3).

From (2), 7a:- 9y - 3a: + 12y = 38 ;

/. 4a: + 3y = 38 (4).

From (3), 6a:-3y = 42,

By addition, 10a; = 80 ; whence a; = 8. From (3) we obtain y = 2.

Xm.] SIMULTANEOUS EQUATIONS. 81

106. Sometimes the value of the second unknown is more
easily found by elimination than by substituting the value of
the unknown already found.

Example, Solve 3a;-?^ = i^^ (1),

?y+±A{2x-6) = y (2).

Clear of fractions. Thus
from (1), 42a: - 2y + 10 = 28a: - 21 ;

.-. 14a:-2y = -31 (3).

From (2), 9y + 12 - 10a; + 25 = 15// ;

.-. 10a: + 6y = 37 (4).

Eliminating y from (3) and (4), we lind that

^--13-
Eliminating x from (3) and (4), we find that

^ 26

107. Simultaneous equations may often be conveniently

solved by considering - and - as the unknown quantities.

X y

ExampU. Solve ?_? = ! (1),

X y

-L"+? = 7 (2).

« y

Multiply (1) by 2 and (2) by 3 ; thus

1618^,^
X y "" '

X y
adding, ^ = 23 ;

X

multiplying up, 46 = 23a; ;

;. a; = 2 ;
and by substituting in f 1), ^ = 3.

B.A. F

82 ALGEBRA. [chap.

EXAMPLES Xm. b.

8olve the eqiiations :

1. 2a:-y = 4, 2. 4a:-y = ], 3. x+2y:=lS,

_ + _»5. 2^y*- IT'S"*'

4. 3^+y = i. 5. ^'f = 0, 6. ^-y = 7,

a;+3y = 2. 4a:-3y=l. 4x+5y = 0.

7. 5a: = 4y, 8. a;-y = 0, 9. a; + y=-2,

3 "5-^- 3^ 2^""^*- 4^6""-

10. ^(a:+3) = 0, 11. ?a;-2y = 20, 12. |a:-ly = 0,

ia:-y = 4i. ~(y+8) = 2. 3x = 2y.

13. 3(x-y)+2(a: + y) = 16, 3(x + y)+2(a;--y) = 25.

14. 3(a: + y-5) = 2(y-a:), S{x-y-1) + 2{x+y-2) = 0,

15. 4(2a; - y - 6) = 3(3a: - 2y - 6), 2(a:-y+l) + 4a; = 3y + 4.

16. 7(2a;-y) + 5(3y-4x) + 30 = 0, 6(y - a; + 3) = 6(y - 2a?).

a:-12_y-4-18_2a; + 3y
""3 2~^

X y

?+?=-3i
a: y

108. In order to solve simultaneous equations which contain
two unknown quantities we have seen that we must have two
equations. Similarly we find that in order to solve simul-
taneous equations which contain three imknown quantities we
must have three equations.

Rule. Eliminate one of the unknowns from any pair of the
equations, and then eliminate the same unknown from another pair.
Two equations involving two unknowns are thus obtained^ which may
he solved by the rules already gipen. The remaining unknoum is
then found by substituting in any one of the given equations.

17 a: + 4_y-4
^'- 6 7

= 2a!+y+4. 1&

x~\2
4

X y

20. 1+5 = 37.

21,

5-1 = 2i.
X y

X y

Xin.J SIMULTANEOUS EQUATIONS. 83

Example, Solve 7a:+6y-7z=-8 (1),

4a:+2y-3z = (2),

6a:-4y+42 = 35 (3).

Choose y as the unknown to be eliminated.
Multiply (2) by 5, 20a: + lOy - 15s = ;
Multiply (1) by 2, 14a: + IQy - 142 = - 16 ;
by subtraction, 6a:-z = 16 • (4).

Multiply (2) by 2, 8a: + 4y - 6z = ;

from (3), " 5a: - 4y + 4z = 36 ;

by addition, 13a; - 2z = 35.

Multiply (4) by 2, 12a: - 2z = 32 ;

by subtraction, a; = 3.

From (4) we find 2 = 2,

and from (2), y = - 3.

109. Some modification of the foregoing rule may often be
used with advantage.

Example, Solve ^-l=|+l = ? + 2,

2 6 7

From the equation ? - 1 == ^ + 1, •
we have 3a:-y = 12 (1).

Also from the equation ^-1=1+2,

2 7

we have 7a:-2z = 42 (2).

And from the equation ^+|= 13,

we have 2y+3z = 78 (3).

Eliminating z from (2) and (3), we have

21a: + 4y = 282;
and from (1) 12a; - 4y = 48 ;

whence a; = 10, y s 18. Also by substitution in (2) we obtain z » 14.

84 ALGEBRA. [CHAP. xin.

EXAMPLES XIII. c.

Solve the equations :

1, 3a;-2y + s = 4, 2. 3a:+4y-6z = 16,

2x + 3y-2 = 3, • 4r+ y- z = 2A,

a:+ y + z = 8. a;-3y-2z= 1.

3. a: + 2y + 3z = 32, 4. x- y+ z= 6,

4a?-5y + 6z = 27, 6a; + 3y + 2z = 84,

7a; + 8y-9z=14. 3a:+4y-52= 13.

5. 7a:-4y-3z= 0, 6. 4a: + 3y- e= 9,

6x-3y+2z=12, 9a:- y+5z=16,

2x+2y-5z= 0. a;+4y-3z= 2.

7. 3y-6z-5a: = 4, 8. 3y + 2z + 5a: = 21,

2z-3a:- y = 8, 8a:-3z+ y= 3,

a:-2y+2z+2 = 0. 22 +2a:-3y = 39.

9. l^+ y+l. = l. 10. i-iy-5-i..

x+ y-92=l. 2y + 1 = hz'X),

4

11. |a:+l(y + z) = l§, 4a: + l(z-y) = ll, ^(z-4a:) = y.

12. 2a:-i(z-2y) = 2, |(a:+y) = l(3-z), a: = 4y+3».

13. 7+|Z^ = y-a: = a:-z = z-3.

14. |-|=y + | = a: + y+2+2 = 0.

,_ 2a:-y-z 2y-z-a; 2z-a:-w ^

15. 2^ — =-—3 = ^-^-=a:-y-«-6.

16. |+y = l, g-« = 3, z + 2y + 3a: + 8 = 0.

CHAPTEE XIV.

Problems leading to Simultaneous Equations.

110. In the Examples discussed in the last chapter we have
seen that it is essential to have as many equations as there are
unknown quantities to determine. Consequently the statement
of problems which give rise to simultaneous equations must
contain as many independent conditions, or different relations
between the unknown quantities, as there are quantities to be
determined.

Example 1. Find two numbers whose difference is 11, and one-
fifth of whose sum is 9.

Let X be the greater number, y the less ;
then aj— y=ll (1).

Also ^±i^=9,

6

or x-\-y—^b (2).

By addition 2x=56 ; and by subtraction 2y=34.

The numbers are therefore 28 and 17.

Example 2. If 15 lbs. of tea and 10 lbs. of coffee together cost
$15.50, and 25 lbs. of tea and 13 lbs. of coffee together cost $24.55 ,*
find the price of each per pound.

Suppose a pound of tea to cost x cents,
and coffee y

Then from the question we have

15a; -f 10y= 1550 (1),

25a4-13y=2456 (2).

Multiplying (1) by 5 and (2) by 3, we have

75a; +50y= 7750,
75x+39y=7365.
Subtracting, . lly=385,

y=35.
And from (1), 15a;+350=1550 ;

whence 1 5x = 1200 ;

.-. x=80.

. •. the cost of a pound of tea is 80 cents,
and the cost of a pound of coffee is 35 cents.

86 ALGEBRA. [chap.

Example 3. In a bag oontainin^ black and white balls, half the
number of white ia equal to a third of the number of black ; and
twice the whole number of balls exceeds three times the number of
black balls by four. How many balls did the bag contain ?

Let X be the number of white balls, and y the number of black
balls ; then the bag contains x+y balls.

We have the following equations :

? = y (1)

2(a:+y) = 3y+4 (2).

Substituting from (1) in 2, we obtain

^ + 2y = 3y+4;

whence y = 12 ;

and from (1), a; = 8.

Thus there are 8 white and 12 black balls.

111. In a problem involving the digits of a number the
student should carefully notice the way in which the value of a
number is algebraically expressed in terms of its digits.

Consider a number of three digits such as 435 ; its value is
4 X 100 + 3 X 10 + 5. Similarly a number whose digits beginning
from the left are x, y, z

—X hundreds +y tens+s units

= lOO^+lOg^+0.

Example, A certain number of two digits is three times the sum
of its digits, and if 45 be added to it the digits will be reversed ;
lind the number.

Let x be the digit in the tens' place, y the digit in the units' place ;
then the number will be represented by 10a; + y, and the number
formed by reversing the digits will be represented by lOy + x.

Hence we have the two equations

10a; + y = 3(a: + y) (1),

and 10a;+y + 45 = lOy+o: (2).

From (1), 7a; = 2y ;

from (2), y-x = 5.

From these equations we obtain x = 2, y = 7«

Thus the number is 27.

XIV.] PROBLEMS LEADING TO SIMULTANEOUS EQUATIONS. 87

EXAMPLES XI7.

1, Find two numbers whose sum is 54, and whose difference is
12.

2, The sum of two numbers is 97 and their difference is 51 ; find
the numbers.

3, One-fifth of the difference of two numbers is 3, and one-third
of their sum is 17 ; find the numbers.

4, One-sixth of the sum of two numbers is 14, and half their
difference is 13 ; find the numbers.

6, Four sheep and seven cows are worth $131, while three cows
and five sheep are worth t^QQ. What is the value of each animal ?

6, A farmer bought 7 horses and 9 cows for $330. He could
have bought 10 horses and 5 cows for the same money ; find the
price of each animal.

7, Twice -4's age exceeds three times B's age by 2 years j if
the sum of their ages is 61 years, how old ar« they ?

8, Half of -4's age exceeds a quarter of B's age by 1 year, and
three-quarters of B^s age exceeds ^'s by 11 years ; find tjie age of
each?

9, In eight hours C ]jalks 3 miles more than D does in six hours,
and in seven hours D walks 9 miles more than C does in six hours ;
how many miles daea each walk per hour ?

10, In 9 hours a coach travels one mile more than a train does
in 2 hours, but in 3 hours the train travels 2 miles more than
the coach does in 13 hours ; find the rate of each per hour.

11, A bill of $16 is paid with half-dollars and quarters, and
three times the number of half-dollars exceeds twice the number of
quarters by 6 ; how many of each are used ?

12, A bill of $8.70 is paid with quarters and dimes, and five
times the number of dimes exceeds seven.times the number of quar-
ters by 6 ; how many of each are used ?

13, Forty-six tons of goods are to be carried in carts and wag-
ons, and it is found that this will require 10 wagons and 14 carts,
or else 13 wagons and 9 carts ; how many tons can each wagon and
each cart carry ?

14, A sum of $14.50 is given to 17 boys and 15 girls ; the same
amount could have been given to 13 boys and 20 girls ; find how
much each boy and each girl receives.

15, A certain number of two digits is seven times the sum of
the digits, and if 36 be taken from the number the digits will be
reversed ; find the number.

88 ALGEBRA. [CHAP. XIV.

16. A certain number of two digits is four times the sum of the
digits, and if 27 be added to the number the digits will be reversed ;
find the number.

17. A certain number between 10 and 100 is six times the sum
of the digits, and the number exceeds the number formed by re-
versing the digits by 9 ; find the number.

18. The digits of a nimiber between 10 and 100 are equal to
each other, and the number exceeds 6 times the sum of the digits
by 8 ; find the number.

19. A man has $380 in silver dollars, half-dollars, and quarters ;
the number of the coins is 852, and their weight is 235 ounces. If
a dollar weighs ^ oz., a half-dollar ^ oz., and a quarter ^ oz., find
how many of each kind of the coins he has.

20. A man has $22 worth of silver in half-dollars, quarters, and
dimes. He has in all 70 coins. If he changed the half-dollars for
dimes and the quarters for half-dollars, he would then have 180
coins. How many of each had he at first ?

21. Divide $100 between 3 men, 6 women, 4 boys, and 3 girls,
so that each man shall have as much as a woman and a girl, each
woman as much as a boy and a girl, and each boy half as much as
a man and a girl.

22. If 17 lbs. of sugar and 5 lbs. of coffee cost $2.50, and 10
lbs. of sugar and 10 lbs. of coffee cost $3.80, find the cost per lb. of
sugar and coffee.

23. The value of a number of coins consisting of dollars and
half-dollars amounts to $22.50 ; the number of dollars exceeds five
times the number of half-dollars by 6. Find the number of each.

24. A sum of $23.80 is divided among 11 men and 16 women;
the same sum could have been divided among 19 men and 6 women.
Find how much each man and woman receives.

26. Two articles A and B are sold for 20 cents and 30 cents
per lb. respectively ; a person spends $6.50 in buying such articles.
If he had bought half as much again of A and one-third as much
again of 5, he would have spent $9.00. What weight of each did
he buy ?

CHAPTER XV.

Involution.

112. Definition. Involution is the general name for multi
plying an expression by itself so as to find its second, third
fourth, or any other power.

Involution may always be effected by actual multiplicatioa
Here, however, we shall give some rules for writing down at
once

(1) any power of a simple expression ;

(2) the square and cube of any binomial ;

(3) the square of any multinomial.

113. It is evident from the Rule of Signs that

(1) no even power of an^ quantity can be negative;

(2) any odd power of a quantity will have the same sign as
the quantity itself.

Note. It is especially worthy of remark that the sqiuire of every
expression, whether positive or negative, is positive,

114. From definition we have, by the rules of multiplication.

(-3a3)*=(-3)V)* = 81a".

Hence we obtain a rule for raising a simple expression to any
proposed power.

Bule. (1) Raise the coefficient to the required power by Arith-
metiCy arid prefix the proper signfouTid hy the JRule of Signs.

(2) Multiply the index of every factor of the expression hy the
exponent of the power required.

90 ALGEBRA. [cHAP.

Examplea. {-2x^f =-32a:i^

( - 3a63)6 = 729a«6i8.

\3xrh/) 8lar8y4'

It will be seen that in the last case the numerator and the denomi-
nator are operated upon separately.

EXAMPLES XV. a.

Write down the square of each of the following expressions :
1. a^b, 2. 3ac3. 3. 5xy\ 4. 66^.

5. 4a26c3. 6. -SxY- 7. -2a''lPc. 8. -Sdx*.

Q a^c ,^ 2x^ n 3a3 .„ 5

13.' -?^. 14. ^'. 15. -_J_. 16. -?Z^-

3 9a:y* ' ^t^t? ' 5a^a:

Write down the cube of each of the following expressions ;
17. 2a;. 18. 3tt62. 19. 4ar^. 20. -3a26.

21. -4a;3y2. 22. -Vcd?. 23. -ey*. 24. -^i^cf.

^Ot ~4~^' 26. — — ,-o-q» 27. ~ 1 — q- 28. ~ o^*

/rr* ah^<? 4yz^ 3

Write down the value of each of the following expressions :

29. (^^'Y. 30. i-^y)'. 31. (-2m2n3)6. 32. {-xh/y,

^ & «• (-©)' '^ (-«'■ '«• (-H"

To Square a Binomial.

115. By multiplication we have

(a-hby={a-hb)(a-hb)

i=a2+2a6 + ft2 (1).

(n-by=(a-b)(a^b)

= a2_2a6+ft2 (2).

XV.] INVOLUTION. 91

These formulae may be enunciated verbally as follows :

Rule 1. The square of the sum of two quantities is equal to
the sum of their squares increased hy troice their product.

Bale 2. The square of the difference of two quantities is equal
to the sum of their squares diminished hy twice their product.

Example 1. {x + 2yf = 7^ + 2 . x . 2y + {2yf

= ar' + iary+iy^.
Example 2. (2o8_362)2= (2a8)2-2 . 2a« . 362+ (362)2

=4a5-12a862_|.964.

To Square a Multinomial.

116. By the preceding article

(a + 6 + c)2={(a + 6) + c}2

=:(a + 5)2+2(a + 6)c + c2
= a2 + 62 ^_ c2 _|_ 2a6 + 2ac + 26c.
In the same way we may prove
(a-6 + c)2 = a3+62 + c2-2a6 + 2ac-26c.
(a+6 + c+o02 = aH62+c2+ci?2 + 2a6+2ac+2ac;+26c+26G?+2co?.

In each of these instances we observe that the square con-
sists of

(1) the sum of the squares of the several terms of the given
expression ;

(2) twice the sum of the products two and two of the several
terms, taken with their proper signs ; that is, in each product
the sign is + or - according as the quantities composing it
have like or unlike signs.

Note. The sqiiare terms are always positive.

The same laws hold whatever be the number of terms in the
expression to be squared.

Rule. To find the square of any multinomial: to the sum of
the sqitares of the several terms add twice the product {with tlie
proper sign) of each term into each of the terms that follow it.

Ex. 1. (a:-2y-as)2 = a;2 + 4y2 + 922-2.a:.2y-2.a;.3z + 2.2y.3z

= a;2 + 4y2 + 92* - 4a:y - Qxz + 12yz.

Ex. 2. (l+2a:-3a~5)2=l+4ar^+9ar*+2.1.2a:-2.1.3a;2-2.2x.3a:«

= 1 + 4a~^ + 9a:* + 4a; - 6a;= - 12ar*
= l+4a;-2a:2-12a:3 + 9a^^

by collecting like terms and rearranging.

92 ALGEBRA. [OHAP. rv.

EXAMPLES XV. b.

Write down the square of each of the following expressions :

1. x + 2y, 2. x-2y. 3. a+36. 4. 2a-36.

5. 3a + 6. 6. a;-5y. 7. 2m + 7n. 8. 9-x.

9. 2-db, 10. aftc + l. 11. ab-cd. 12. 2ah-{-xy.

13. l-a:*. 14. 3 + 2i?g. 15. ar^.s^.. ^g 2a+a6.

17. a + 6-c. 18, a-6-c. 19, 2a + 6 + c.

20. 2a:-y-z. 21. a: + 3y-2z. 22. o^ + ar+l.

23. 3a; + 2/>-g. 24. l-2a;-3a:2. 25. 2-3a: + a:2.

26. x+y+a~6. 27. m-n+p-^. 28. 2a+36+x-2y.

To Cube a Binomial.

117. By actual multiplication, we have

(o+6)3=(a + 6)(a+6)(o+ft)
= a3+3a26 + 3a62+6'.
Also (a - 6)3 = a' - 3a26 + 3a^>2 - 63,

By observing the law of formation of the terras in these
results we can write down the cube of any binomial.

Example 1. {2x + yf = (2a;)» + 3(2a:)V+3(2a;)y« + y»

= 8a^-{-l2a^y + 6xy^ + ^.

Example 2. (3x - 2a^f = {Zxf - 3(3a;)2(2a2) + 3(3a;)(2a2)2 - (2a*)3

= 27ar» - 543i^a^ + 36a»* - 8a».

EXAMPLES XV. c.

Write down the cube of each of the following expressions ;

1. P + ^' 2. m-n, 3. a-26. 4. 2c + d.

5. x + 3y. 6. x + yz, 7, 2xy-l. 8. 5a+2.

9. x^'l' 10. 2a;«+V. 11^ 2a»-36». 12. 4y2-3

CHAPTER XVL
Evolution.

118. Definition. The root of any proposed expression is
that quantity which being multiplied by itself the requisite
number of times produces the given expression.

The operation of finding the root is called Evolution: it is
the reverse of Involution.

119. By the Rule of Signs we see that

(1) any eve9i root of a positive quantity may be either positive
or negative ;

(2) no negative quantity can have an even root ;

(3) every odd root of a quantity has the same sign as the
quantity itself.

Note. It is especially worthy of remark that every positive
quantity has two square roots equal in magnitude, but opposite
in sign.

Example, /JdcM = ±3ax^.

In the present chapter, however, we shall confine our attention to
the positive root.

Examples, ija^b* = a%^, because {a^b^f = a^h*,
\l -7? = -7^y because ( - ^)^ = - ic®.

Vc^ = c*, because (c*)* = c^.
VSI^ = ^x^, because (Sa:*)* = Ut?^,

120. From the foregoing examples we may deduce a general
rule for extracting any proposed root of a simple expression :

Rule. (1) Find the root of the coefficient hy Arithmetic, and
prefix the proper sign.

(2) Divide the exponent of every factor of the expression hy the
index of the proposed root.

Examples. V— 64x* = -4a:*.

^

Vl6aS=2a3.

§I^o_9aH^
25c* 6?*

94 ALGEBRA. [CHAP.

EXAMPLES XVI. a.

Write down the square root of each of the following expressions :
1. 9x*y^ 2. 25o«6*. 3, 49c2d«. 4. a«6V«

5. 36x«y3«. 6. 16a:8. 7. a^V^^. 8. 9A^'.

^- I6^* ■^"- 36* ^^' ■26"* ■*-'^' '^'

Write down the cube root of each of the following expressions :
13. A'. 14. 'a%\ 15. Sx^. 16. -27a:».

17 -^^ IQ S«'^^^ 10 125a»x2i „^ 64a2763

1/. 2^. 10. -IB . iW. 27c« ' ^^' ~~1^'

Write down the value of each of the following expressions ;

21. V^. 22. W^^ 23. V - ar^y^.

24. V(Ua4^ 25. V^^^6^ 26. V/>«g^.

27. V - x^^y^. 28. VsIiV^. 29. V32a«6iV«.

121. By the formulae in Art. 115 we are able to write down
the square of any binomial.

Thus (207 + 3y )2 =4a;^+l2xy-\-9f.

Conversely, by observing the form of the terms of an expres-
sion, it may sometimes be recognised as a complete square, and
its square root written down at once.

Example 1. Find the square root of 253:^ - 40xy + 16y^.

The expression = (5a:)* - 2 . 20xy + (4y)2

= (5a:)2-2{5a:)(4y) + (4y)2

= (5a; - 4y)*.
Thus the required square root is 5a; - 4y.

Example 2. Find the square root of --r^ +4 + *-—»

The expression = (|f)'' + (2)'' + 2(-'^)

* ft/*

Thus the required square root is - - +2.

XVT.] EVOLUTION. 95

122. When the square root cannot be easily determined by
inspection we must have recourse to the rule explained in the
next article, which is quite general, and applicable to all cases.
But the stvdent is advised, here and elsewherCy to employ methods
of inspection in preference to rules.

To Find the Square Root of a Compound Expression.

123. Since the square of a + 6 is a^-\-2ab-\-h^, we have to dis-
cover a process by which a and h, the terms of the root, can be
found when a'^-\-2ah-\-h'^ is given.

The first term, a, is the square root of a^.

Arrange the terms according to powers of one letter a.
The first term is a^, and its square root is a. Set this down as
the first term of the required root. Subtract a^ from the given
expression and the remainder is 2ab + b^ or (2a -\-h)xh.

Now the first term 2a6 of the remainder is the product of
2a and h. Thus to obtain h we divide the first term of the
remainder by the double of the term already found ; if we add
this new term to 2a we obtain the complete divisor 2a + 6.

The work may be arranged as follows :

a^-\-2ah + h^{a + h
2a+h

a2

2ab + b^
2ab + b^

Example Find the square root of Oar^ - i2xy + 49y^

9ar»- 42a;y + 492/2 {Sx-7y
9ar»

6ar-7y

-i2xy^4:9y^
-42xt/ + 49y^

Explanation. The square root of 9oi^ is 3x, and this is the first
term of the root.

By doubling this we obtain 6a;, which is the first term of the
divisor. Divide - 422^, the first term of the remainder, by 6a; and
we get - 7y, the new term in the root, which has to be annexed both
to the root and divisor. Next multiply the complete divisor by - 7y
and subtract the result from the first remainder. There is now no
remainder and the root has been found.

96 ALGEBRA. [chap.

124, The rule can be extended so as to find the square root of
any multinomial. The first two terms of the root will be
obtained as before. When we have brought down the second
remaiThder^ the first part of the new divisor is obtained by
doubling the terms of the root already found. We then divide
the first term of the remainder by the first term of the new
divisor, and set down the result as the next term in the root
and in the divisor. We next multiply the complete divisor by
the last term of the root and subtract the product from the
last remainder. If there is now no remainder the root has
been found ; if there is a remainder we continue the process.

Example, Find the square root of

25a:2a3 _ yixa? + 16a;* + 4a* - 2i3^a.

Rearrange in descending powers of x.

16a:* - 24a:3a + 250:2^2 _ i2ara« + 4a* ( 4jb2 - 3aro + 2a2
16ar*

8a:^-3a»
8ar*-6a:a + 2a*

- 2^7^a + 25ar^a2
-24ar3a+ Oar^a^

16ar'a2-12a:a8+4a*
16ar*a2-12a;a8 + 4a*

ExplancUion, When we have obtained two terms in the root,
^a^ - 3a;a, we have a remainder

16ar'a2-12a:a3 + 4a*.

Double the terms of the root already found and place the result,
Sa:^ - 6a;a, as the first part of the divisor. Divide 16a;^a^, the first
term of the remainder, Dy SiC^, the first term of the divisor ; we get
+ 2a* which we annex both to the root and divisor. Now multiply
the complete divisor by 2a^ and subtract. There is no remainder
and the root is found.

125. Sometimes the following method may be used.

Example. Find by inspection the square root of

4aa + 62+c2 + 4a&-4ac-26c.

Arrange the terms in descending powers of a, and let the other
letters be arranged alphabetically ; then

the expression = \a^ + ^kib - ^ac + h^- 2bc + c-

= 4a2 + 4a(6-c) + (6-c)2

= (2a)2 + 2. 2a(6-c) + (6-c)2;
whence the square root is 2a + (6 - c). [Art. 121. ]

ana.] evolution. 97

EXAMPLES ZVI. b.

By inspection or others? ise, find the square root of each of the
following expressions :

1

3
5
7

9

11

13
14
15
16
17
18
19
20

21
22

23

24
25
26

27
28

a2-8a+16. 2. x^+Ux + ^9,

64+48a:+9a;2. 4. 25-30m + 9m2.

36»4-84w2 + 49. 6. 8l + 144y3 + 64y8.

afi - 6aV55< + 93^8. 8. 4a264 - 12a6V + 9c^<>.

l^-S^^9y^. 10. '^^'-^-^'-^

16a:*-32ar» + 24ar^-8a: + l.

25 - 30a + 29a2 - 12a3 + 4a4.

9a8-12a«-2a*+4a2+l.

25/)* - 30p» + 121 - 101i>a + 66p.

8a:3 + l+4a:*-4x.

201a2 - 108a3 + 100 + 36a* - I8O0.

a2 + 62+c2 + 2a6-2ac-26c.

yV + z2a^» + a: V - 2a^Vz + 2a?y^ - 2a:yz?.

a4-2a»+2L'-?P + l.
2 2 16

a* 4a2? X* A a cue^ « •

4 16 4 2
9a:* + 144a;2 + 12aa:2 + 4^2 _ ^g^^s _ ^^^x^

a2+962 + c2-6a6 + 66c-2ac.

m* ow" . -»*^ >iWi ■ A
— - 2 — + o — — 4 — + 4.

n* n* n^ »

9a2 -^62 6a_^26
F"^^a2"T+a-

98 ALGEBBA. [CHAP.

[ If preferred, the remainder of this chapter may be postponed
and taken at a later stage.]

To Find the Cube Root of a Compound Expression.

126. Since the cube of a-\-b is a^+Sa^ft + SaJ^-^js^ ^fe have
to discover a process by which a and b, the terms of the root,
can be found when a^-\-daV)-\-3ab^-\-b^ is given.

The first term a is the cube root of a\

Arrange the terms according to powers of one letter a ; then
the first term is a\ and its cube root a. Set this down as the
first term of the required root. Subtract a' from the given
expression and the remainder is

3a^b + 3a62 + ftJ or {3a^ + Sab + b^) x b.

Now the first term of the remainder is the product of Sa^
and b. Thip to obtain b we divide the first term of the re-
mainder by three times the square of the term already found.

Having found b we can complete the divisor, which consists
of the following three terms :

1. Three times the square of a, the term of the root already
found.

2. Tliree times the product of this first term a, and the new
term b.

3. The square of b.

The work mey be arranged as follows :

a^ + 3a^b + 3ab- + b^a-{-b

a3

3(a)2 =3o2
3xax6= +Zab
(6)2 = +62

3a2 + 3a6 + 62

3a-6 + 3a62+63

Zarb-^3ab^+P

Example 1. Find the cube root of Sxr^ - S6xh/ + 54xy^ - 27j^.

8ar» - 36a:2y + 54iry2 _ 27^* ( 2ar - 3y

3(2a;)3 = I2a^

3x2a:x(-3y)= -I8xy

(-3y)2= +9y2

l2x^-lSxy + 9y^

-36x^ + 6ixy^-27y^

-36a2y + 54ay*-273r*

XVI.]

EVOLUTION.

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100

ALGEBRA.

[chap. XVL

EXAMPLES ZVI c.

Find the cube root of each of the following expressions :

1. aS+12a2 4-48a + 64. 2. 8ar» + 12a:a + 6a; + l.

3. 64a:3-144ar»4-108a;-27. 4. 8p«-36/>* + 54p2_27.

5. m'-18m2 + 108«i-216. 6. a:«+6icV+12xV+8y«.

7. l-3c + Gc2-7c»+6c<-3c«+c«.

8. 8 + SQm + 66m2 + GSm^ + SSm* + 9ot« + 7w«.

9. 216-108ifc+342F~109ifc»4-171ifc*-27P+27it«.

10. 48y«4-108y + 6V-90y8-27 + 8y«-80y«.

11. 64+192X;+240ifc3+160i[:«+60it*+12ifc« + l^.

12. 7^ -6a^y -SA + I2xy^+I2xyz+ 3ocz^ -Sy^ -I2i/hi -6yz^- 7?,

[For additional examples see Elementary Algebra."]

127. The ordinary rules for extracting square and cube roots
in Arithmetic are based upon the algebraical methods explained
in the present chapter. The following example is given to
illustrate the arithmetical process.

Example, Find the cube root of 614125.

Since 614125 lies between 512000 and 729000, that is between
(80)^ and (90)^, its cube root lies between 80 and 90 and therefore
consists of two figures.

a+h

614125(80+5 = 85

512000

3a» = 3 X (80)2 _ 19200

3xax6=3x80x5= 1200

62 = 5x5= 25

20425

102125

102125

In Arithmetic the ciphers are usually omitted, and there are
other modifications of the algebraical rules.

CHAPTER XVIL

Resolution into Factors.

128. Definition. When an algebraical expression is the
product of two or more expressions each of these latter quanti-
ties is called a factor of it, and the determination of these
quantities is called the resolution of the expression into its
factors.

In this chapter we shall explain the principal rules by which
the resolution of expressions into their component factors may
be effected.

Expressions in which Each Term is divisible by a Common

Factor.

129. Such expressions may be simplified by dividing each
term separately by this factor, and enclosing the quotient within
brackets ; the common factor being placed outside as a coefficient.

Example 1. The terms of the expression 3a^ - 6a5 have a common
factor 3a ;

.-. 3a»-6a6 = 3a(a-26).
Example 2. 6a%iKr^ - Uahx^ - 2Qh^x^ = bht^a^x - 3a - 4^),

EXAMPLES XVII. a.

Resolve into factors :

1. ar» + aa;. 2. 2a2-3a. 3. a^-aK 4. a^-a^h,

5. 3m2-6mn. 6. T^+2j^q, 7. a:fi-5ar». 8. y^ + xy.

9. 5a«-26a26. IQ. 12a; + 48a;Y 11. 10c3-25c<d

12. 27-162a:. 13. T^yh^ + Zxy, 14. 17a;2-51a:.

15. 2a8-a2+a. 16. 3ar» + ea^ar' - 3a»a;.

17. Ip'-ljfi+Up^ 18. 4^/5 + 6a2&3- 262.

19. si^f'-x^^+2xy. 20. 26a365+39a*6a.

102

ALGEBRA.

[chap.

Expressions in which the Terms can be so grouped as to
contain a Compound Factor that is Common.

130. The method is shown in the following examples.
Example 1. Resolve into factors x'^—ax-\-bx—ab.

Since the first two terms contain a common factor x, and the
last two terms a common factor b, we have

a^'-aX'\-bx-dh = {x^ - ax) \- {bx - ab)

= x{x - a) + b{x - a)

= (x-a) taken x times plus {x - a) taken b times
= {x-a) taken (a; + b) times
= {x-a){x + b).

Example 2. Resolve into factors 6x^ - 9ax + ibx - 6a6.

ea^-9ax + 4bx-Qab = {6x^-dax) + {4bx-6ab)

= 3a:(2a: - 3a) + 26(2a; - 3a)
=:(2a:-3a)(3a: + 26).

Example 3. Resolve into factors 12a'* + bx^ - 4a5 - ^xK

12a2 + b7^- 4a6 - 3aar* = ( 1 2a2 - 4a6 ) - (3aar» - bx^)

= 4a(3a -b)- a:2(3a - b)
= (3a-6)(4a-a;2).

EXAMPLES XVn. b.

Resolve into factors :

1, x^-\-Qcy + xz^yz,

3, a2 + 2a + a6 + 26.

5, 2a + 2a; + aa; + a~*.

7, am -bm-an + bn.

9. PQ "•' Q''^ ~ 1^ ~ ^•

11, ax-2ay-bx-\-2by,

13. ac^ + b-{-bc^ + a.

15. a^-a^ + a-l.

17. a^x-aby-\-2ax-2by.

2, tx^-ocz + xy-yz.

4. a^ + ac + 4a + 4c.

6. Sq-Sp+pq-j)"".

8, ab-by-ay + y^.

10, 2mx +nx + 2my + ny.

12. 2a2 + 3a6-2ac-3/'c.

14. ac2-2a-6c2 + 26.

16. 2ar' + 3 + 2a; + 3a;2.

18. axy-\-bcxy—az—bcz.

Trinomial Expressions.

131. In Chap. V. Art. 48 attention has been drawn to the
way in which, in forming the product of two binomials, the
coefficients of the different terms combine so as to give a trino-
mial result.

XVII.] RESOLUTION INTO FACTORS. 103

Thus (^+5)(^+3)=jr2 + 8^+15 (1),

(^-5X^-3)=^-8.?7+15 (2),

(^+5X.'P-3) = :r2 4. 2^-15 (3),

(:r-5X^+3)=^-2^-15 (4).

We now propose to consider the converse problem : namely,
the resolution of a trinomial expression, similar to those which
occur on the right-hand side of the above identities, into its
component binomial factors.

By examining the above results, we notice that :

1. The first term of both the factors is x,

2. The product of the second terms of the two factors is
equal to the third term of the trinomial ; e.g. in (2) above we
see that 15 is the product of —6 and —3 ; and in (3) we see
that - 15 is the product of +5 and —3.

3. The algebraic sum of the second terms of the two factors
is equal to the coefficient of x in the trinomial ; e.g. in (4) the
sum of — 5 and + 3 gives - 2, the coefficient of x in the tri-
nomial.

The application of these laws will be easily understood from
the following examples.

Example, 1 . Resolve into factors a;^ + 1 1 a: + 24.

The second terms of the factors must be such that their product
is +24, and their sum +11. It is clear that they must be +8
and +3.

.-. a:2+lla; + 24 = (a: + 8)(a: + 3).

Example 2. Resolve into factors a?^- 10a; + 24.

The second terms of the factors must be such that their product
is +24, and their sum - 10. Hence they must both be negative^ and
it is easy to see that they must be - 6 and - 4.

.-. a:2 - 10a; + 24 = (a; - 6)(a; - 4).

Example 3. ar* - 18a; + 81 = (a; - 9)(.a; - 9)

= {x - 9)2.

Example 4. a;* + lOa;^ + 25 = (ar^ + 5)(ar^ + 5)

= (a;2 + 5)2.

Example 6. Resolve into factors x^ - 11 aa? + lOa^.

The second terms of the factors must be such that their product
is + lOa^, and their sum - 11a. Hence they must be - 10a and - a.

:. ar»-llaa; + 10a2 = (a;-10a)(a;-a).

Note. In examples of this kind the student should always verify
his results, by forming the product {mentally, as explained in
Chap. V. ) of the factors he has chosen.

104 ALGEBRA. [chap.

EXATiffPLES XVn. c

Resolve into factors :

1. a:*+3ar+2. 2. y'+Sy+e. 3. y»+7y+12.

4. a2-3a+2. 6. o*-6a+8. 6. &»-56+6l

7, 6'+136 + 42. 8. 6»- 136+40. 9. z«-13b+36.

10. a:*-15a: + 56. 11. a:»-15x+54. 12. 2*+lfe+44.

13. 6*- 126+36. 14. o»+15a + 56. 15. o»-12o+27.

16. x^+9x+20. 17. a:*-10r+9. 18. a:*-16r+64.

19. y*-23y+102. 20. y*-24y+95. 21. y*+54y+729.

22. a2 + 10a6 + 216». 23. a2+12a6 + 116«. 24. o«-23a6+ 13262.
26. m*+8»i«+7. 26. m*+9iiiV + 14»*. 27. 6-5x+a:*.

28. 54-15a + a«. 29. 13 + 14y + y». 30. 216-35a + a».

132. Next consider a case where the third term of the tri-
nomial is negative.

Example 1. Resolve into factors 3^ + 2x-35.

The second terms of the factors mnst he such that their product
is -35, and their algebraical sum +2. Hence they must have
opposite signs, and the greater of them mast he positive in order
to give its sign to their snm.

The required terms are therefore + 7 and - 5.

.-. a:2+2x-35 = (x + 7)(a;-6).

Example 2. Resolve into factors a^-Zx-5^,

The second terms of the factors must he such that their product
is -54, and their algebraical sum -3. Hence they must have
opposite signs, and the greater of them must he negative in order
to give its sign to their sum.

The required terms are therefore - 9 and + 6.

.-. a:2-3ar-54 = (a:-9)(a:+6).

Remembering that in these cases the numerical quantities
must have opposite signs, if preferred, the following method may
be adopted.

Example 3. Resolve into factors xh^ + 23a:y - 420.

Find two numbers whose product is 420, and whose difference is

23. These are 35 and 12; hence inserting the signs so that the
positive may predominate, we have

xY + 23a:y - 420 = {xy + 35)(a:y - 12).

XVII.] RESOLUTION INTO FACTORS. 105

EXAMPLES XVn. d.

Resolve into factors :

1, iK^+x-2. 2, a;2-a:-6. 3. a:*-a:-20.

4. jr*+4y-12. 5. y^+4:y-2l. 6. y'-5y-36.

7. a^+Sa-S3. 8. a^-lSa-SO. 9. a2+a-132.

10. 6^-126-45. 11, 6^ + 146-51. 12. 6^ + 106-39.

13. m2-m-56. 14. m2-5m-84. 15, m^+m-56.

16. p2-82>-65. 17. p^ + Sp-lOS. 18. p2+l?-110.

19. a:2+2a:--48. 20. ar»-7a;-120. 21. ar»-a:-132.

22. ?/<+13y2-48. 23. y«+4a:y-96a:2. 24. y*+7a:y-98ar».

25. a* + 0^6^-726*. 26. a3+ aft -24062. 27. U-6a-aK

28. 35-26-63. 29. 96-46-62. 30, 72 + 6-62.

133. We proceed now to the resolution into factors of tri-
nomial expressions when the coefficient of the highest power is
not unity.

Again, referring to Chap. v. Art. 48, we may write down the
following results :

(3a?+2)(^+4)=3a;2+14^+8 (1),

(3a7-2X^-4)=ar2_i4r+8 (2),

(307 +2)(:r- 4) =3^72-1007 -8 (3),

(3a7-2X^+4)=ar'4.iOa7-8 (4>

The converse problem presents more difficulty than the cases
we have yet considered.

Before endeavouring to give a general method of procedure
it will be worth while to examine in detail two of the identities
given above.

Consider the result 3o72-14o7 + 8 = (3o7- 2)(a: - 4).

The first term 3^*2 is the product of 30? and x.

The third terra +8 -2 and -4.

The middle term — 14^ is the result of adding together the
two products 307 X - 4 and o? x - 2.

Again, consider the result Zx^ - lOo?— 8 = (3o7 + 2Xo7 ~ 4).

The first term 307^ is the product of 3o7 and 07.

The third term -8 + 2 and -4

The middle term - lOo; is the result of adding together the
two products 3o7 X — 4 and 07 x 2 ; and its sign is negative because
the greater of these two products is negative.

106 ALGEBRA. [CHAP.

134. The beginner will frequently find that it is not easy
to select the proper factors at the first trial. Practice alone will
enable him to detect at a glance whether any pair he has chosen
will combine so as to give the correct coefiicients of the expres-
sion to be resolved.

Example, Resolve into factors It^ - 19a; - 6.

Write down (7a; 3)(a: 2) for a first trial, noticing that 3 and 2
must have opposite signs. These factors give *Jx^ and -6 for the
first and third terms. But since 7x2-3x1 = 11, the combination
fails to give the correct coefficient of the middle term.

Next try (7a; 2)(a; 3).

Since 7x3-2x1 = 1 9, these factors will be correct if we insert
the signs so that the negative shall predominate.

Thus 7a;2 - 19x - 6 = (7a; + 2)(a; - 3).

[Verify by mental multiplication.]

135. In actual work it will not be necessary to put down
all these steps at length. The student will soon find that the
different cases may be rapidly reviewed, and the unsuitable
combinations rejected at once.

It is especially important to pay attention to the two follow-
ing hints :

1. If the third term of the trinomial is positive, then the
second terms of its factors have both the same sign, and this
sign is the same as that of the middle term of the trinomial.

2. If the third term of the trinomial is negative, then the
second terms of its factors have opposite signs.

Example 1. Resolve into factors 14a;' + 29a;- 15 (1),

14a:2-29a;-Io (2).

In each case we may write down (7a; 3)(2a: 5) as a first trial,
noticing that 3 and 5 must have opposite signs.

And since 7 x 5 - 3 x 2 = 29, we have only now to insert the proper
signs in each factor.

In (1) the positive sign must predominate,

in (2) the negative

Therefore 14a;2+29a; - 15 = (7a; - 3)(2a; + 5).

Ut? - 29a; - 15 = (7a; + 3)(2a; - 5).

xvn.j

RESOLUTION INTO FACTORS.

107

Example 2. Resolve into factors 5a:* + 17a:+6 (I),

5ar»-17a: + 6 (2).

In (1) we notice that the factors which give 6 are both positive.

In (2) negative.

And therefore for (1) we may write (5a: + ){x+ ).

(2) (5a:- )[x- ).

And, since 5x3 + 1x2 = 17, we see that

5a;2 + 17a: + 6 = (5a: + 2)(a; + 3).

6x^ - 17a; + 6 = (5a; - 2)(a; - 3).

Note. In each expression the third term 6 also admits of factors
6 and 1 ; but this is one of the cases referred to above which the
student would reject at once as unsuitable.

EXAMPLES XVn. e.

Resolve into factors :

1

4
7
10
13
16
19
22
25
28
31

2a2 + 3a + l. 2.

2a2 + 5a + 2. 6.

6a2 + 7a+2. 8.

2a:2 + 9a; + 4. H.

3y2 + y-2. 14.

2fe2 - 56 - 3. 17.

4m2 + 5w-6. 20.

4a:2-8a;y-5y2. 23.

12a2-17a6 + 662. 26.

2 - 3y - 2y2. 29.

4 + 17a;-15a:2. 32.

3a2+4a+l. 3.

3o2 + 10a + 3. 6.

2a2 + 9a+10. 9

2a;* + 6a; -3. 12

3y2_7y_6. 15

663 + 76-3. 18

4m2-4m-3. 21

^7^-*Jxy-\-2y\ 24

M - bob - 662. 27

3 + 2.3y-8y2. 30

6-13a+6a2. 33

4a»+6a+l.

2a' + 7a + 3.
2a2 + 7a+6.
3a;2+5a;-2.
2y2 + 9y-5.
262 + 6-15.
6m2-7m-3.
Qx^-\Zxy+2y\
6aa+35o6-66a.
8 + 18y-5y3.
28-316-562.

When an Expression is the Difference of Two Squares.

136. By multiplying a + 6 by a — 6 we obtain the identity

(a + 6)(a-6)=a2-62,
a result which may be verbally expressed as follows :

The product of the sum and the difference of any two quantities
is equal to the difference of their squares.

Conversely, the difference of the squares of any two quantities
is equal to the product of the sum and the difference of the two
quantities.

Thus any expression which is the difEerence of two squares
may at once be resolved into factors.

108 ALGEBRA. [CUAP,

Example, Resolve into factors 25a:' - 16y^.

25ar»-16y2 = (5a:)2-(4y)2.
Therefore the first factor is the sum of bx and 4y,
and the second factor is the difference of bx and 4y.

.-. 25ar» - 16y* = (5a: + 4y)(5a: - 4y).

The intermediate steps may usually be omitted.
Example. 1 - 49c« = (1 + 7c»)(l - Vc^).

The difference of the squares of two numerical quantities is
sometimes conveniently found by the aid of the formula

Example, (329)2- (171)' = (329 + 171)(329- 171)

= 500x158
= 79000.

EXAMPLES XVn. f.

Resolve into factors :

1. a^-9. 2. a«-49. 3. a^-S\. 4. a* -100.

5. a:2-25. 6. a:2-144. 7. 64 -a:^. g^ 81-4ar2.

9. V-1- 10. y--9a«. 11. 4y2-25. 12. 9y2-49ar.

13. 4m2-81. 14. 36a2-l. 15. P-64^2.

16. 9a2-2663. 17, 121-16y». 18. 121 -36a^*.

19. 25 -c*. 20. a^b^-x'y^ 21. 4da*-l00h\

22. 64ar»-4922. 23. 4/>V-81. 24. a*b*c^-Q,

25. a;«-4a*. 26. a:* -252*. 27. a^^-p'q*.

28. 16ai»-96». 29. 25x^-4. 30. a^b^c^-Qx'.

Find by factors the value of

31. (39)2 -(31)2. 32. (51)2 -(49)2. 33. (1001)2-1.

34. (82)2 -(18)2. 35^ (275)2 -(225)2. 36. (936)2 - (64)2.

When an Expression is the Sum or Difference of Two Cubes.

137. If we divide a^-\-b^hy a-\-b the quotient is a^—ab-\-b^;
and if we divide a^ — b^hy a — b the quotient is a^-{-ab-\-b^.

We have therefore the following identities :

a8+68=(a+6)(a2-a&+62) ;
a8-68=(a-ft)(a24-a6 + &2).

These results enable us to resolve into factors any expression
which can be written as the sum or the difference of two cubes.

XVTI.] RESOLUTION INTO FACTORS. 109

Example 1. Bar* - 27y3 = {2xf - {Zyf

= {2x - 3y)(4a;2 + Qxy + 9if).

Note. The middle term 6xy is the product of 2x and 3y.

Example 2. 64a3 + 1 = (4a)3 + (1)3

= (4a+l)(16a2-4a+l).

We may usually omit the intermediate step and write down
the factors at once.

Examples. 343a« - 27a:3 = (7a2 - 3a:)(49a* + 2la^x + 9x^),
8x8 + 729 = (2a:3 + 9)(4^_ i3jj^ + 81).

EXAMPLES XVII. g.

Resolve into factors :

1.

a3 - 63,

2.

a^ + b^ 3.

l+a:3.

4. l-j^.

5.

Sx^ + l.

6.

a:3-8z3. 7,

a3 + 2763.

8. a^-l.

9.

1 - 8a3.

10.

63-8. 11.

27 + x^.

12. 64-;>3.

13.

125a3 + l.

14. 216-63.

15.

a:3y3 + 343.

16.

1000a:8+l.

17. 512a3-l.

18.

a363c3 - 27.

19.

8ar» - 343.

20. ar» + 216y3.

21.

a;0 - 272^.

22.

m3 - 1000n«.

23. a«- 72963.

24.

125a«+ 51263.

138. We shall now give some harder applications of the
foregoing rules, followed, by a miscellaneous exercise in which
all the processes of this chapter will be illustrated.

Example 1. Resolve into factors (a + 26)^ - I6x^,
The sum of a + 26 and 4a? is a + 26+4a:,
and their difference is a + 26 - 4a:.

.-. (a + 26)2-16a;2 = (a + 26 + 4a:)(a + 26-4a;).

If the factors contain like terms they should be collected so
as to give the result in its simplest form.

Example 2. (3a; + 7y)- - (2a: - 3y)^

= {(3a: + 1y) + (2a: - 3y)} {(3a: + 7y) - (2a: - 3y )}
= (3a: + 7y + 2a: - 3y)(3a: + 72/ - 2a: + 3y)
= (5a:+4y)(a: + 10y).

110 ALGEBRA. [chap.

139. By suitably grouping together the terms, compound
expressions can often be expressed as the difference or two
squares, and so be resolved into factors.

Example 1. Resolve into factors 9a^ - c^ + 4ca; - ia?,

9o2-<r> + 4ca;-4a~» = 8a2-(c2-4ca:+4a:2)

= (3a)2-(c-2a;)a
= (3a + c - 2a:)(3a - c + 2a:).

Example 2. Resolve into factors 2M -a*-c^ + 6^+£P+ 2ac.

Here the terms 2bd and 2ac suggest the proper preliminary
arrangement of the expression. Thus

2W-a2-c2+62 + cP + 2ac = 62+26d + da-aa + 2ac-c«

= (6 + d)2-(a-c)a

= (6 + d + a-c)(6+d-a+c).

140. The following case is important

Example, Resolve into factors x^ + jb^^ + y**

ar* + a:2y2 + y4^ (ar* + 2arV' + y*)-ary

= {x^ +y^ + xy)(xi^ -hy^ - xy)
= («» + a:y + y2)(ar* - a:y + y^).

141. Sometimes an expression may be resolved into more
than two factors.

Example 1. Resolve into fstctors 16a* -816*.

16a* - 816* = (4a2 + 962)(4a2 - 962)

= (4a2 + 962)(2a + 36)(2a - 36).

Example 2. Resolve into factors afi-i/^,

sfi -i^ = {aP + y^){a^ - y^)

= (x + y){x^-xy + y^)(x-y){aP+xy + y^),

Note. When an expression can be arranged either as the dif-
ference of two squares, or as the difference of two cubes, each of the
methods explained in Arts. 136, 137 will be applicable. It will,
however, be found simplest to first use the rule for resolving into
factors the difference ox two squares.

XVII.] RESOLUTION INTO FACTORS. HI

142. In all cases where an expression to be resolved contains
a simple factor common to each of its terms, this should be first
taken outside a bracket as explained in Art. 129.

Example, Resolve into factors 28a;*y + ^7?y - 60a:^.

2&7^+^^y - 60a;2y _ Aa^y{jx^ + 16a: - 15) i

EXAMPLES XVn. h.

Resolve into two or more factors :

1
4
7
10
12
14
16
18
20
22
24
27
30
33
36
38
40
42
44
46
47
49
51
53

[x+yf-z^. 2. {x-yf-z\ 3. {a+'ibf-cK

(a + 3c)2-l. 5. {2x-\f-a\ 6. a^'{h+cf.

4a2_(6-i)2. 8. 9-(a + a;)2. 9, (2a-36)2-c2.

(I8a;+y)a-(17a:-y)3. H. (6a + 3)2 - (5a - 4)2.

4a2 - (2a - .36)2. 1^3^ ar» - (26 - 3c)2.

{x+yf-{m-nf, 15. (3a; + 2y )« - (2a; - 3y )2.

a2-2aa; + a;2_42,2^ yi^ ar' + a* + 2aa; - z^

l-aa-2a6-62. 19. 12a;y + 25-4a;2-92^.

c2-a2-62 + 2a6. 21. a;^ - 2a: + 1 - m^ - 4mn - 4w2.

a:*+y*-z*-a*+2a^y-2a2z2. 23. {m-\-n-\-pf-(m-n+pf.

a* + a2+l. 25. a*6*-16. 26. 256a;* -Sly*.

16a*62-6«. 28. Um^-miA 29. v^-^\

a^h^-^larh. 31. 400a2a:-a:3. 32^ l-729y».

2166« + a365. 34, 250^3 + 2. 36. 1029 -3a;».

oa:* - aa:^ _ 240aa:. 37. o^^ + ^^ - <^x - ^*

m* + imhiV + 4nV- 39. 8a:2y3 _ ^a^

%xY-\-\bx^y^-^Qicf, 41. 2m V - 7mV - 4»8,

98ar* - 72r2y2 _ y4. 43^ a'^h'^-a^-h^+l,

a:3-2a:2_a. + 2. 45^ (^ + 6)8+1.

a2a:3 _ g^y _ 4^23^3 + 3262^-.,

2p-3^ + 4jy2-9g2. 48^ 119+10m-m2.

24a262 - 30a63 - 366*. 50. 240ar' + a:«y* - a:i V-

a:*+4ar»+16. 52. a:* + y* - 7a^'y2.

a*-18a262+64. 54, 7^-\-7^-vl.

[EoT additional examples see Elementai'y Algebra»'\

112 ALGEBRA. [CHAP.

Oonverse Use of Factors.

143. The actual processes of multiplication and division can
often be partially or wholly avoided by a skilful use of factors.

It should be observed that the formulae which the student
has seen exemplified in this chapter are just as useful in their
converse as in their direct application. Thus the formula for
resolving into factors the difference of two squares is equally
useful as enabling us to write down at once the product of the
sum and the difference of two quantities.

Example 1 . Multiply 2a + 36 - c by 2a - 36 + c.

These expressions may be arranged thus :

2a + (36-c) and 2a-(36-c).

Hence the product = {2a + (36 - c)} {2a - (36 - c)}

= (2a)* - (36 - cf
= 4a»--(962-66c + c2)
= 4a2-962+66c-c3.

Example 2. Find the product of

a:+2, x-2, a;2-2a: + 4, a:2+2a:+4.

Taking the first factor with the third, and the second with the
fourth,

the product = {(a; + 2)(a:2 -2x + 4)} {{x - 2){7^ + 2x+ 4)}

= (ar^ + 8)(a:3-8)

= a:«-64.

Example^, Divide the product of 2a?2 + a;-6 and 6a:^-5a:+l
by 3a^J+6a:-2.

Denoting the division by means of a fraction,
the required quotient = (^+''-JW^-^''+'^)

_ (2a?-3)(a?+2)(3a?-l) ( 2a?-l )
(3a;-l)(a: + 2)

= (2a;^3)(2a;-l),

by cancelling factors which are common to numerator and denomin-
ator.

XVII.] CONVERSE USE OF FACTORS. 113

Example 4. Prove the identity

17(6a: + Sa)* - 2(40a: + 27a)(5a: + 3a) = 25ar» - ^a?.

Since each term of the first expression contains the factor 5a; + 3a,
the first side = (5a; + 3a){17(5a; + 3a) -2(40a:+27a)}
= (5a; + 3a)(85a; + 51a - 80a; - 54a)
= (5a;+3a)(5a;-3a)
= 25ar» - ^a\

EXAMPLES XVII. k.

Employ factors to obtain the product of

1, a-h-\-Cy a-b-c. 2, 2x-y + z, 2x + y + z.

3. l+2a;-ar», l-2a;~a;2. 4, c2 + 3c + 2, c2-3c-2.

5. a + b-c + d, a + 6 + c-d. 6, p-Q + x-y, p-q-x+y.

7. a3-4a26 + 8a62-863, a^+4a^b + Sab^-\-8b^.

Find the continued product of

8. (a-W (a + 6)^ (a2+62)a.

9. (l-a:)8, (l+a;)3, (l+a;2)3.

10. a2-4a + 3, a3-a-2, a2 + 5a + 6.

11. 3-y, 3 + y, 9-3y + y2, 9 + 3y + y2.

12. l + c + c^, l-c + c2, l-c2 + c*.

13. Divide a\a + 2)(a2 - a - 66) by a^ + 7a.

14. Divide the product of ar^ + a;-2 and oi^ + 4x+3 by a;2-f5a; + 6.

15. Divide ^{x + 4){x^ - 9) by ar« + a; - 12.

16. Divide the product of 2ar» + lla-21 and 3a2-20a-7 by

a2 - 49.

17. Divide (2a2 - a - 3)(3a2 - a - 2) by 6a2 - 5a - 6.

18. Divide a;« - 7a;^ - 8 by (a; + 1 )(ar^ + 2a? + 4).

Prove the following identities :

19. ia + b)^-{a-bf(a+b) = 4ab{a + b),

20. c*-d4_(c-d)3(c + d) = 2cd(c2-cP).

21. {in- n){m + n)^ -m*+n^ = 2mn{m^ - n^).

22. {X + yf - 3a;y(a: + yf = (a; + y){a^ + y^).

23. 3a6(a - bf + (a - h)^ = (a - b){a^ - b^).

H.A. H

114 ALGEBRA. [CHAP.

MISCELLANEOUS EXAMPLES m.

1. Find the product of lOa:^ ._ 12 - 3a: and 2a; - 4 + 3«",

2. If a = 1, 6 =- 1, c = 2. d = 0, find the value of

a?-h^ b^-cd c^-b^
a2 + 62+2^Hcrf 3a6c'

3. Simplify 2[4a; - {2y + (2x-y)-(x + y)}].

4. Solve the equations :

(l) ig-3 _2-a;_l-2a;. .ov 3a;-4y = 25,

^ ' 5 3 15 ' ^ ' 5a;+2y = 7.

5. Write down the square of 2ar' - a; + 5.

6. Find the H.C.F. and L.C.M. of 3a^'b^c, 12a*6V, Iba^if^c

7, Divide 0^+46* by a2-2a6+262.

8, Find in dollars the price of 6k articles at 8a cents eaeh.

9, Find the square root of re* - 8a:' + 240^^ - 32a: +16.

10. If a = 5, & = 3, c = 1, find the value of

(a-&)a^(&-c)2^(a-c)a

a+b b+c a+c

11. Solve|(7a: + 6)-7f = 13-|(a:-|).

12. A is twice as old as ^; twenty years ago he was three
times as old. Find their ages.

13. Simplify (1 - 2a;) - {3 - (4 - 6a:)} + {6 - (7 - 8a:)}.

14. The product of two expressions is

6a:* + 5x^y + %a^^ + bx}^+^,
and one of them is 23:^ + 3a;y + 2y^ ; find the other.

15. How old is a boy who 2a; years ago was half as old as his
father now aged 40 ?

16. Find the lowest common multiple of 2a^, 3a&, 5a^&c, 6a6^c, la?h,

17. Find the factors of

(1) a;2 - a;!/ - 72y2. (2) 6a:2-13a;+6.

13. Find two numbers which differ by 11, and such that one-
third of the greater exceeds one-fourth of the less by 7*

xvn.] MISCELLANEOUS EXAMPLES III. 115

19. If a = 1, 6 =- 1, c = 2, d = 0, find the value of

a + b.c + dad-hc_c^-(l^
a-b c-d hd + ac a^-hb^'

20. Simplify |^-y-{2^-5y-7-g-4) + (2-Ja;)}

21. Solve the equations :

(1) (3a;-8)(3a: + 2)-(4a;-ll)(2a:+l) = (a:-3)(a;+/);

(2) ^+^ = ^, a:-fy-5 = Jy.,).

22. A train which travels at the rate of p miles an hour takes
q hours between two stations ; what will be the rate of a train
which takes r hours ?

23. Fuid the sum of

-a— a?, 1 — -, ^ — {2a- 1, -x—a,
4 3' 2' 3 V 2/' 3 4

24. Resolve into factors

(1) I2si^+ax-20a^; (2) a^- 16-6aar+9^.

25. Solve

(1) a;+l+2(a; + 3) = 4(a:+6);

(2) 4a: + 9y = 12, 6a:-3y = 7.

26. Findthev^lueof :^L^|=2^, whena:=-^^

27. Find the quotient when the product of fe^ + c^ and ft^-c*
is divided by 6* - 2lr^c + 26c* - c^.

28. -^f ^) ^^^ ^ have $168 between them ; ^'s share is greater
than ^'s by f^^, and C'a share is three-fourths of A^a, Find the
share of each.

29. Find the square root of 9a:« - r2a:* + 22ar» + a:* ^ |2a: + 4.

30. Simplify by removing brackets a* - [(6 - c)* - {c* - (a - 6)2}].

CHAPTER XVIII.

Highest Common Factor.

144. Definition. The highest common factor of two or
more algebraical expressions is the expression of highest dimen-
sions which divides each of them without remainder.

Note. The term greatest common measure is Bometimes nsed
instead of highest common factor ; but this usage is incorrect, for in
Algebra our object is to find the factor of highest dimensions which
is common to two or more expressions, and we are not concerned
with the numerical values of the expressions or their divisors. The
term greatest common measure ought to be confined solely to arith-
metical quantities, for it can easily be shown by trial that the
algebraical highest common factor is not always the greatest
common measure.

145. We have already explained how to write down by
inspection the highest common factor of two or more sim,ple
expressions. [See Chap, xii.] An analogous method will enaole
us readily to find the highest common factor of compound ex-
pressions which are given as the product of factors, or which
can be easily resolved into factors.

Example 1. Find the highest common factor of

4car' and 2ca:' + 4c^.

It will be easy to pick out the common factors if the expressions
are arranged as follows :

Ac2^ = 4ca:',

2car3 + 4c2ar2 = 2^k^[x + 2c) ;

therefore the H.C.F. is 2car'.

Example 2. Find the highest common factor of

Resolving each expression into its factors, we have

3a2+9a6 = 3a(a + 36),
a' - 9a62 = a{a + 3fe)(a - 3/>),
a3 + 6a26 + Oaft^ = a[a + 36)(a + 3B) ;

therefore the H.C.F. is a(a + 3/>).

CHAP. XVIII.] HIGHEST COMMON FACTOR. 117

146. When there are two or more expressions containing
different powers of the same compound factor, the student should
be careful to notice that the highest common factor must contain
the highest power of the compound factor which is common to
all the given expressions.

Example 1. The highest common factor of

x(a-x)^y a(a-a;)', and 2ax{a-xf is (a -a:)'.

Example 2. Find the highest common factor of

a7? + 2a^x + a\ 2aai^ - ia^x - ^\ ^{ax + a^)\

Resolving the expressions into factors, we have

aar'+2a2a; + a8 = a{x^ + 2ax+a^)

r^a{x-vaf (1),

2aa3 _ 4a2a; - Ga' = 2a(ar» - 2aa? - 3a2)

= 2a(a; + a)(a;-3a) (2),

3(cw; + a2)a = 3{a(x + a)}2 = 3a2(a; + a)a (3).

Therefore from (1), (2), (3), by inspection, the highest common
factor isa(2t;+a).

EXAMPLES XVm. a.

Find the highest common factor of

1

3

5

7

9

11

13

15

17

19

21

22,

x'^-y\ x^-xy.

3a3-2a26, 3a2-2o6.

ah^ia - xf, 2aV(a - a;)^.
ax+x, a*x + aM,

2. 3(a-6)3, a2-2a6 + &2.

4. 9a^-ih\ 6a2 + 4a6.

6. ix^-x*y\ x^y'^ + xy,

8. 2a:2-8a; + 8, (a:- 2)3.

10. iKV-2^, ay + y^, xy-y\

7?+x^, x^+xy, a^+xy\ 12. ^y^-f, y^ixy-jf^f,

(a2-aa:)», (aa;-a:2)3, 14. {ahc-U^f, {a^c-ac^)\

ar* - a:2 _ 42a;, a?<-49ar». 16. {^-^x'^)\ ar'-8a?*+15a:3.

a3-36a, o8 + 2a2-48a. 18. 3a2 + 7a-6, 2a2 + 7a + 3.

2ar»-9a:+4, 3ar»-7a:-20. 20. 3c*+5c»-12c2, 6c» + 7c*-20c».
4m*-9mS 6m'-5m»-6m, 6m*+5m3-6m2.

3aV - 8aSa:8 + 4aV, Za^x^ - Wah? + Ga'ar^,
3aV+16a3a:3-12aV.

118

ALGEBRA.

[chap.

147. The highest common factor should always be determined
by inspection when possible, but it will sometimes happen that
expressions cannot oe readily resolved into factors. To find
the highest common factor in such cases, we adopt a method
analogous to that used in Arithmetic for finding the greatest
common measure of two or more numbers.

148. We shall now work out examples illustrative of the
algebraical process of finding the highest common factor ; for
the proof of the rules the reader may consult the Elementary
Algebra, Arts. 102, 103* We may here conveniently enunciate
two principles, which the student should bear in mind in reading
the examples which follow.

I. // an expression contain a certain factor, any multiple of
the expression is divisible by that factor,

II. If two expressions have a common factor, it will divide
their sum and their difference; and also the sum and the difference
of any multiples of them.

Example, Find the highest common factor of

4a:8-3ar*-24a;-9 and %7^-23?-^Zx-Z9.

2x

4ar» - 3a;2 - 24a: - 9

8a:3_2a:2-5.3a:-39

4a:3 „ 5a:2 _ 21a:

8a:»-6ar«-48a;-18

2^^- 3a;-9

4a:2- 5a;_21

27^- 6a;

4ar5- 6a:- 18

3a:-9

X- 3

.3a;-9

Therefore the H.C.F. is a: - 3.

Explanaiion, First arrange the given expressions according to
descending or ascending powers of x. The expressions so arranged
having their first terms of the same order, we take for divisor that
whose highest power has the smaller coefficient. Arrange the work
in parallel colunms as above. When the first remainder lar' - 5a: - 21
is made the divisor we put the quotient x to the left of the dividend.
Again, when the second remainder 2a:^ - 3a: - 9 is in turn made the
divisor, the quotient 2 is placed to the right; and so on. As in
Arithmetic, the last divisor a; -3 is the highest common factor
required.

149. This method is only useful to determine the compound
factor of the highest common factor. Simple factors of the
given expressions must be first removed from them, and the
highest common factor of these, if any, must be observed and
multiplied into the compound factor given by the rule.

xvin.]

HIGHEST COMMON FACTOR.

119

Example. Find the highest common factor of

24a;4 - 2a:3 _ 60a^ _ 32a; and 18a;* - ear^ - 39ar» - 18a;.
We have 24a;* - 2a^» - 60ar» - 32a; = 2a;(12a;3 _ 3.2 _ ^Qx - 16),
and 18a;* - 6a;S - 39ar« - 18a; = 3a;(6a;' - 2ar» - 13a; - 6).

Also 2a; and 3a; have the common factor x. Removing the simple
factors 2a; and 3a;, and reserving their common factor a;, we continue
as in Art. 148.

2a;

6aJ» - 2ar» - 13a; - 6

6ar»-8arg- 8a;

6a;2- 6a;-6

6a;^- 8a;-8

3a:+2

\2tx?- a;2-30a;-16
12a;3_4a.3_26a;-12

3a;2-
3a;2 +

4a;-
2a;

4

—

6a;-
6a;-

4
4

X

-2

Therefore the H. C. F. is a;(3a;+2).

150. So far the process of Arithmetic has been found exactly
applicable to the algebraical expressions we have considered.
But ill many cases certain modifications of the arithmetical
method will be found necessary. These will be more clearly
understood if it is remembered that, at every stage of the work,
the remainder must contain as a factor of itself the highest
common factor we are seeking. [See Art. 148, I. & II.]

Example 1. Find the highest common factor of

3a;3_i3a:2 + 23a;-21 and 6ar' + a;2-44a; + 21.

3a;«-13ar»+23a;-21

6x^+ a;2-44x + 21
6a;^-26a;» + 46a;-42

27ar^--90a; + 63

Here on making 27a;* -90a; +63 a divisor, we find that it is not
contained in 3a;'-13a;* + 23a;-21 with an integral quotient. But
noticing that 21 x^ - 90a; + 63 may be written in the f onny(3a;-- 10a; + 7 ),
and also bearing in mind that every remainder in the course of the
work contains the H. C. F., we conclude that the H. C.F. we are
seeking is contained in 9{^x^ - 10a; + 7). But the two original expres-
sions have no simple factors, therefore their H. C. F. can have none.
We may therefore reject the factor 9 and go on with divisor
da^-lOx+7.

120

ALGEBRA.

[chap.

Resuming the work, we have

X

-1

3a:»-13ar» + 23ar-21
Z3fi-I0a^+ Ix

- 3x2+ 16a: -21

- 3ar» + 10a;- 7

2)6a:-14

3ir3-iar + 7
3a:«- 7a:

- 3x + 7

- 3x+7

X

-I

3x- 7

Therefore the highest common factor is 3a: - 7.

The factor 2 has been remored on the same gronnds as the factor 9
above.

151. Sometimes the process is more convenient when the
expressions are arranged in ascending powers.

Example^ Find the highest common factor of

3-4a-16a2-9a» (1),

and 4-7a-19a2-8a5 (2).

As the expressions stand we cannot begin to divide one by the
other without using a fractional quotient. The difficulty may be
obviated by inirodticing a suitable factor, just as in the last case we
found it useful to remove a factor when we could no longer proceed
with the division in the ordinary way. The given expressions have
no common simple factor, hence their H.C.F. cannot be affected if
we multiply either of them by any simple factor.

Multiply (1) by 4 and use (2) as a divisor :

la

4- 7a- 19a2-. 8a»
5

20-35a- 95a2- 40a«

20 -28a- 48a2

- 7a- 47a2- 40a3
-5

35a+235a2 + 200a»

35a- 49a2- 84a3

284a2 284a2 + 284a3

1+a
Therefore the H.C.F. is l + o.

12-16a-64a2-36a8
12-21a-57a2-24a8

a

5a- 7a2-12a«

5- 7a -12a2
5+ 5a

-12a -12a2
-12a -12a2

-12a

XVIII.] HIGHEST COMMON FACTOR. 121

After the first division the factor a is removed as explained in
Art. 150 ; then the factor 5 is introduced because the first tenn of
4 - 7a - \9a^ - 8a* is not divisible by the first term of 6 - 7a - I2a^.
At the next stage a factor - 5 is introduced, and finally the factor
284a* is removed.

152, From the last two examples it appears that we may
multiply or divide either of the given expressions, or any of the
remainders which occur in the course of the work, by any factor
which does not divide both of the given expressions.

EXAMPLES XVni. b.

Find the highest common factor of

1. 2a^ + Sx^ + x + 6, 2x^ + x^ + 2x + S,

2. 2y3-9y2.f9y-7, y3-53r' + 5y-4.

3. 2a:3 + 8a~«-5a:-20, exi^-4a^-l5x+l0.

4. a3 + 3a2-16a+12, a^ + a^-lOa+S.

5. 6ar»-ar»-7a;-2, 2x^-7a^+x+6,

6. g'-3g+2, g3_5g2 + 7g_3.

7. a* + a'-2a2 + a-3, 5aS+3a2-17a+6.

8. 3y< - 3y3 _ 15y2 - 9y, 4y« - 16y* - 44y8 - 24y2.

9. I5x*-I53i^+I0x^-I0x, 30x^ + l2Ox* + 20a^+SOa^.

10. 2m* + 7m' + lOm* + 35m, 4m* + Hm' - 4m2 - 6m + 28.

11. 3a:*-9a:3+12a;2-12a;, Qsi^~6x^-l6x + 6,

12. 2a5-4a4-6a, a'^ + a* - 3a^ - SaK

13. x^+4x^-2x-l5y ar»-21a;-3e.

14. 9a^ + 2a2ar^ + a:*, 3a* - Sa'a; + 5a V - 2aa:».

15. 2-3a + 5a2-2a8, 2 - 6a + Sa^ ^ 3a3.

16. Sx^-5x^-l5x*-^x^, 6a:-7ar^-29a:8-12ar<.

[For additional examples see Elementary Algebra,'\

CHAPTER XIX.
Fractions.

153. The principles explained in Chapter xviii. may now
be applied to the reduction and simplification of fractions.

Reduction to Lowest Terms.

154. Rule. The value of a fraction is not altered if toe mtUtir
ply or divide the numerator and denominator by the same quantity.

An algebraical fraction may therefore be reduced to an equi-
valent fraction by dividing numerator and denominator by any
common factor ; if this factor be the highest common factor,
the resulting f inaction is said to be in its lowest terms.

Example 1. Reduce to lowest terms

The expression = — -

ISa^xr" - I2a^x^
24a^c^x^

Qa:^2^(Sa - 2x)
Act/?

3a-2a;'

Example 2. Reduce to lowest terms y-.

^ ^xy-\2y^

Tvv. • 2x(3x-4y) 2x

The expression = ^ ^ — -^i = - -.

^ 'Sy{3x-4y) 3y

Note. The beginner should be careful not to begin cancelling
until he has expressed both numerator and denomifiator in the most
convenient form, by resolution into factors where necessary.

EXAMPLES XIX. a.

Reduce to lowest terms :

CHAP. XIX.] FRACTIONS. 123

Reduce to lowest terms :

*• ba^y-hlOxh,' °' Qx'y-2xy^' "' Ss^^'6x'

„ 7? + Ax J. larx-lc^c _ q a;2 + a:-30

'• x^ + x-Vi: °* bcQt^-lOc^x^-hc^' 5:c- + 30:c"

,0 g^-4a: -21 ,- g2-2a?-15 ,p. 2a:3 + a;-3^

^' 3:c2-fiOx + 3' • "3ar'-12a;-15' ■*-°' 2x' + nx+\2

Ifi 3a3-24 ,„ 4ar^-25a:y» ,j, 18a3 + 6a ^a; -f 2aa;3

■*-^* 4a2 + 4a-24' '* 2a;2 + a:y- 15y^* * 27a»-ar*

155. When the factors of the numerator and denominator
cannot be determined by inspection, the fraction may be reduced
to its lowest terms by dividing both numerator and denomi-
nator by the highest common factor, which may be found by
the rules given in Chap, xviii.

Example. Reduce to lowest terms 3^ " b3x ^+23a;j-21

The H.C.F. of numerator and denominator is 3a;- 7.

Dividing numerator and denominator by 3a; - 7» we obtain as
respective quotients ar^ - 2a; + 3 and Sa;^ - a; - 3.

Thus 3a;3_i3^+^^_ 21 _ (3a;-7)(a^^-2a; + 3) _ x^ -2a; + 3

15a;3 _ 38a:2 _ 2ic + 21 ~ (3a; - 7)(5ar^ - a; - 3) " 5a;2 - a; - 3*

156. If either numerator or denominator can readily be
resolved into factors we may use the following method.

«j3 r 32;2 _ 42?

Example. Reduce to lowest terms -—^ — — •

^ 7ar^-18a;^ + 6a; + 5

The numerator = a;(ar^ + 3a; - 4) = a;(a; + 4)(a; - 1).

Of these factors the only one which can be a common divisor is

a; - 1. Hence, arranging the denominator so as to shew a; - 1 as a

factor,

., r x« a;(a; + 4)(a; - 1)

the fraction ==-57 . / . , , — tt—t, ,t

7ar^(a; - 1 ) - 1 la:(a; - 1 ) - 5(a; - 1)

a;(a; + 4)(a;-l) a'(a; + 4)

(a;-l)(7a;--lla;-5)"7a;2-lla;-6

124 ALGEBRA. [chap.

EXAMPLES XIX. b.

«

Reduce to lowest terms :

3ar» + 7ar5 + 2 ' a»-4a^ + 5a-6

n y »-2y^-2y-3 ^ w^- 7 /t^-2m

3y» + **y'^ + 4y + l' • m^-m^-m-2

g. a^-2aIy^^2W ^ 9x^-a^x-2a^

a3-4a'^6-21a62 "' 3a:3 _ lOaa:^ - Ta^a; - 4«3

,, 5a:»-4a:-l ^ c3 + 2c2d-12c6P-9d3

'• 2a:3-3a:»+l' °' &-{-~6c^d-2Sc(P-24(P

p a: *~21a:+8 ,^ 2^+63^ + 23^-92^

"• 8a:*-21a:3+l' •*■"• y*+7y3 + 3y2_ Hy

■^' 2-a:+9a:3' -»-^* 4 + 4a; + 9a;2 + 4a:3 _ 5a;4'

[For additional examples see Elementary Algebra.']

Mnltiplication and Division of Fractions.

157. Rule. To multiple/ together two or more fractions:
multiply the numerators for a new numerator^ and the denomt-
natorsfor a new denominator.

rrn a c ac

c,* M 1 ace ace

Similarly, -x^x^=j-^^;

and so for any number of fractions.

In practice the application of this rule is modified by re-
moving in the course of the work factors which are conimou
to numerator and denominator.

rr J a' T* 2a^ + 3a _ 4a2 - 6a

Example, Simplify ——. — x - — .

' '^ 4a^ 12a +18

mv . a(2a + 3) 2a(2a-3)

The expression = \ ^ - x — i — - /
*^ 4a3 6(2a + 3)

_ 2a-3

12a *

by cancelling those factors which are common to both numerator and
denominator.

158. Rule. To divide one fraction by another: invert the
divisor t and proceed as in multiplication.

XIX.] FRACTIONS. 125

Thus ^- c_a d_ad

a c be

Example. Simplify ^^ - °^ " ^' x -?— °L- -. J^iL.

The expression = —^::^- x 9^3^ x __^.^-

_ (3a:-2a)(2a? + a) ^ x-a ^ c(3^*v 2a)

a(x-a) (3a; + 2a)(3a:-2a) 2a; + a

= 1,
since all the factors cancel each other.

EXAMPLES ZIX. c.

Simplify

1 _^^ 2x^±^ o _ab + 2_ M--Sa^

• 2:2 + 3^ 3^ + x ' 4a2-12a6 a-6-^-4 '

o 2c^ + 3cd ^ c + (£ A 5y-10y3^1 -2y

^' 4c2 - 9cP ' 2crf - 3^2* *• 12y2 + 6y3'2y + 2/'^'

^ a;g-4 ^a;-2 ^ 6^-56 96^ -16a'

°' ar»+4a: + 4'a: + 2' ^* 36-4a 6^-25 *

„ a:' + 9a; + 20 ^a: ' + 7a?+10 o y--y-12 y'-2y-2 4

'• ar» + 5a; + 4 * a;"^ + 3a; + 2* °* y'.jg yj + gy + g-

P o» + 27 ^ a'-4a-21 ,^ 2a2-3a- 2 3fl'-8a-3

"• a2 + 9a+14' a2-49 ' ' d^-a~6 3a^-5a-2'

n y+125 256^-1 ,„ 3m'-w-2 ^ 4m2 + m-5

^2 2p2 + 4jp ^j?2_5^^e^ ^2_2p.-15
2*2 _ 9 p^-5p 2)2 _ 4

,- 64a268-l ar^-49 ^ a?-7
■^*' a:2_a._56 Sa^ft - a' ' a'a: - 8a2*

,p. 43^^4-4^-15 a; + 8 ^ 2a^^ + 5a;

°' ar« + 2x-48 2ar»-15a:+18 * '(a:-6)2'
,^ o2 + 8a6-96 ' ^ o2-7a6 + 1262 ^ a8^a26 ha b'^
■*-^* a2 + 6a6-2762 a»-6* a2-3a6-462"

,„ oar>-16a^ a:2^,aa; -20a 2 _^ a:^ - 8aa: + 16a'

a:2 - aa; - 30a2 aa;2 + 9a''^a; + 20a3 ' a;2 + 8aa: + Um^'

■*-°* a2_a6 + ac (a-c)--62 (a + 6 + c)2'
[For additional examples see Elementary Algebra*']

CHAPTER XX.

Lowest Common Multiple.

159. Definition. The lowest common mnltiple of two
or more algebraical expressions is the expression of loivest di-
mensions wliicli is divisible by each of them without remainder.

The lowest common multij)le of compound expressions which
are given as the product of factors, or which can be easily
resolved into factoi-s, can be readily found by inspection.

Example 1. The lowest common multiple of 6ar*(a - xf, 8a'(a — a;)^,
and \2ax{a - xf is ^i4a^7^{a - x)\

J or it consists of the product of

(1) the L.C.M. of the numerical coefficients ;

(2) the lowest power of each factor which is divisible by every
power of that factor occurring in the given expressions.

Example 2. Find the lowest common multiple of
3a2 + 9a&, 2a3-18a62, a^ + ^^ + ^aJr.

Resolving each expression into its factors, we have

3a2 + 9a6 = 3a(a + 3i),
2a3 - I8a62 = 2a(a + 36)(a - 3&),
a3 + 6a26 + Ooi^ = a{a + 36)(a + 36)
= a(a + 36)2.

Therefore the L.C.M. is 6a(a + 36)2(a - 36).

Example 3. Find the lowest common multiple of
{yz^-xyzfj y\xz^-7?), z^ + 2xz^+7^\

itesolving each expression into its factors, we have
{yz^ - xyzf = {yz{z -x)Y = yh?{z - a;)^,
y'^{xz^ - x^) = y'^x{z^ - x^) = Qcy\z - x){z + rr),
z* + 2xz^ + xH'^ = z2(s2 + 2xz-}-x^) = z\z + x)K

Therefore the L.C.M. is xj^\z+x)\z-x)K

CHAP. XX.] LOWEST COMMON MULTIPLE. 127;

EXAMPLES XX. a.

Find the lowest common multiple of

1. a^, a^-a\ 2. x^, x^-Za^. 3. 4m2, 6m»-8m2

4. ^y a:< + 3ar». 5. ^^ + K &^-&. 6. «^-4, ar'+S.

7. 9a26-6, 6a'» + 2a. 8. Xi^-ib+l, ifc^-1.

9, w2-5m + 6, m»+5w-14. 10. y^ + Sy^, y^-^.

11. 7?-^x + U, (x?-\-\x-\2. 12. a:* + 27y3, a:2^a^_6y2,

13. 62 + 964.20, 62 + 5_20. 14. c^-Sca;- l&r^, c'-8ca;+12a^.

15. a2-4a-5, a^-Sa+lS, a»-2a2-3a.

16. 2a:2_4ay_i6y2^ a-' - 6a:y + SyS 3ar'-12y2.

17. 3ar»-12a2a:, 4ar'+16(M:+16a2. 18. a«c-aV, (a^c + ac^)^.
19. (a^a: - 2aa?2)a, (2aa: - 4ar2)a. 20. {^a-a^f, ^a^-4a^+a\

21. 2a~»-ar-3, (2a;-3)2, ^-^.

22. 2ar«-7a:-4, 6a^»-7x-5, a:^ _ ga:^ + 16a:.

23. \Ox^y\^-y^), 16y*(a;-y)3, 12a:8y(a. _ y)(a;2 _ yS).

24. 2ar»+a;-6, 7ar«+lla:-6, (7ar»-3a;)2.

25. M-lah:-Zax^, lOa^a:- llaar«-6a:3^ lOo^ - 21aa; - lOa:^^

160. When the given expressions are such that their factors
cannot be determined by inspection, they must be resolved by
finding the highest common factor.

Example, Find the lowest common multiple of

2a:*+a:3-20a^»-7a;+24and2a:* + 3a;3_]3a;2_7a. + l5

The highest common factor is a:^ + 2a; - 3.

By division, we obtain

2a:* + a:^ _ 20ar^ _ 7a. + 24 = (a:2 + 2a: - 3)(2a:a - 3a: - 8).
2a:* + 33:* - 13a:2 - 7a: + 15 = (a:^ + 2a: - 3)(2a:2 _^_ 5),

Therefore the L.C. M. is (ar» + 2a: - 3)(2a:2 - 3a: - 8)(2a~' - a; - 5).

128 ALGEBRA. ICHAP. XX.

EXAMPLES XX. b.

find the lowest common multiple of

1. a»-ea:»-13a:-10 and x^ - x^.- lOx - S.

2. y»+3y2-3y-9 and y» + 3y2-8y-24.

3. m»+3m2-w-3 and m' + Gm^+llm + e.

4. 2a:*-2a?5 + ar»+Sa:-6 and 4x*-2x^ + Sx-9.

5. Find the highest common factor and the lowest common
multiple of (a: - x^)\ {a^ - 7i?)\ 7? -7^,

6. Find the lowest common multiple of ^a^-v^-j^^y (a^ + aa;)^,
(aa; - v?'^,

7. Find the highest common factor and lowest common multiple
of 6iC^ + 5a:-6 and 6x2 + a: -12; ^mj show that the product of the
H.C.F. and L.C.M. is equal to the product of the two given expres-
sions.

8 Find the highest common factor and the lowest common
multiple of a^ + bah + W, a^ - 4b\ a^ - Sah^ + 268.

9. Find the lowest common multiple of l-a:^-ar*+a:* and

l+2a; + a;2-a:*-a:6^

10. Find the highest common factor of {a^-4ab^)\ {a^+2a%)\
{ah:+2abx)^.

11. Find the highest common factor and the lowest common
multiple of (3a2 - 2ax)^, 2a^x{9a^ - 4ar^), 6a^x - ISa^x^ + 6ax^.

12. Find the lowest common mi'ltiple of a:'+a:^ + a:y^, oi^-y*^
ar"y + a:8y'+a:y*.

[For additional examples see ElemerUary Algebra. ]

CHAPTEK XXL
Addition and Subtraction of Fractions.

161. To find the algebraical sum of a number of fractions
we must, as in Arithmetic, first reduce them to a common
denominator. For this purpose it is usually most convenient to
take the lowest common denominator.

Rule. To reduce fractions to their lowest common denomin-
ator: find the L.C.M. of the given denominators^ and take it
for the common denominator ; divide it hy the denominator of the
first fra^tion^ and multiply the numerator of this fraction hy tlie
quotient so obtained; and do the same with all the other given
fractions,

Examxile, Express with lowest common denominator

^^ and ^^

2a{x-a) 3a;(a;2-a2)'

The lowest common denominator is Qaa:{x-a){x-\-a),
We must therefore multiply the numerators by *Sx{x+a) and 2a
respectively.

Hence the equivalent fractions are

l^'^jx + a) ^^^ 8a'

6aa:(ar - a){x + a) Qaa:{x - a){x + a)

162. We may now enunciate the rule for the addition or
subtraction of fractions.

Rule. To add or subtract fraction's: reduce them to the
lowest common denomin/xtor ; find the algebraical sum of the
numerators, and retain the common denominator,

rwy^„„ a , c ad-\-bc

■t a c_ad—bc

h^d W

H.A.

130 ALGEBRA. [chap.

163. We begin with examples in further illustration of
those already discussed in Chapter xii.

Example 1. Find the value of 2g + ? . 5a^'-4aa:^

The lowest common denominator is 9a^.
Therefore the expression = ^^^i^l^^^nf^

6aa: + 3a' + 5a;^-4aa: 3a^ + 2aa: + 5ar»
" 9a» " 9a2

Example 2. Find the value of ^-^ + ^^"^ ^ 3£-2a

xy ay ax

The lowest common denominator is axy.

Thus the expression = «(^-2y)+»K3y-a)-y(3x-2«)

aa:y
ax - 2ay + 3ary -ax- Zosy + 2ay
~ axy

= 0,
since the terms in the numerator destroy each other.

Note. To ensure accuracy the becinner is recommended to use
brackets as in the first line of work above.

EXAMPLES XXI. a.

Find the value of

, a-2 a-1 + 5 « 3 a; -1 x + S 2a?- 1

■^* 3 ■'^ 2 ^ 6 ' ^' "IT'^ 6 "^ 3 '

o 26-1 6-376 + 3 - 2m-5 m + 3 m-5

5 ^ 2 " 10 • 9 " 6 ■*■ 12 ■

R ^^-1 3a:- 1 x-2 « 2a;-5 a?- 4 ar»-4ar

^* ~ 6 + 7 35 * "• a: ~ a; " 3ar» ■

rr y-z z-x x-y o a + x a + 2x x-5a

I. ^^ — + + — ^- o. — -— +—

yz zx xy 2a 3a 6a

9 2a^-5a a^ + Sa^ 9a^-a* ,/x x-y x + y %xy-4tx^

a a^ a-^ y x Sxy

n a^-263-c3 3a»-3 c» ,„ o6-6c a^J^z^

•*"*■• 663 - 15^ • J-^* 26c "3c~ 2a6 '

^ q 2ay-a:y + 4a: _ i _ a -i 4^ a^- a6 6-c 2c^ - ac

•'•^* 2a:i/ 2^'* ■'■^* "^26 ~"6c ^^'

XXI.] ADDITION AND SUBTRACTION OP FRACTIONS. 131

164. We shall now consider the addition and subtraction
of fractions whose denominators are compound expressions.
The lowest common mvZtiple of the denominators should always
he written dovm by inspection when possible,

Wxamplel. Simplify 2a?-3a_2a;-a

X -2a x-a

The lowest common denominator is {x - 2a){x - a).

Hence, multiplying the numerators by a: -a and x-2a respec-
tively, we have

the expression = i?^z3^)(^--«)r(2^j:^(^_T^

^ {x-2a){x-a)

2aP -5ax-{-3a^ - {2aP - 5ax-h2a^)

" (x- 2a){x - a)

2a^-5ax-hSa^-2x- + 5ax-2a^

{x - 2a){x - a)

a^

(x-2a){x-a)

Note. In finding the value of such an expression as

- {2x - a){x - 2a),

the beginner should first express the product in brackets, and then
remove the brackets, as we have done. After a little practice he
will be able to take both steps together.

Example 2, Find the value of ^^+ ^"^

ar^-16 {x + 4f
The lowest common denominator is (x-'4){x + 4)K

Hence the expression = (?^±2)(^) + i^^^^^

_ 3a:^ + 14a r + 8 + ar^- 9a:+20

(a;-4)(a: + 4)2

^ 4a?^ + 5a: + 28
'^{x-4)(x-\-4f

165. If a fraction is not in its lowest terms, it should be
simplified before it is combined with other fractions.

132 ALGEBRA. [OHAP.

E.a^pU. Simplify ^^5^ J^%.

The «p^io„ = ^7^-^,J|^

_ 7^ + 5xy-4y^ _ y

_ a:^ + 6a:y - 4y® - y(a: - 4y)

W- 1 6y»
_ a* + 5a?y - 4^ - ary + 4^^

ar^-iey'-* (a: + 4y )(a: - 4y) ar-4y

EXAMPLES XXI. b.

Find the value of

1 _J_._J_ 2 _J_ _L. 3 _J L

■^* a-2^a--6 ^* ar-4"'a:-2* ^' 6-2"6 + 2'

M a h c a-a; a + a: /% a+3 a-3

4, — ,• Ot 1 • D. ^ — ^»

sc-a a;-o a + x a-x o-3 a + 3

7 ^ a?^ p 3a 1 q ar^ ■ a;-2y

'• ar-l'ar^-l* °* a-^-4"a + 2* ^" ^^:4P'^a: + 2y

.^11 ., 3a 2a

a(a-6) a(a + 6) ' 2aj(a;-o) 3ar(a; + a)

IQ _6_ _4a: ,„ 1 ^3( y+2)

"' a;-2"(i-2)(«+l)' ^^' y^-2,y-Z^^-y-%'

1A _i_ tt IK __3 2a; ,^ 36 2

■*•*• l-a"*"!!^^' ■^^* a? + y"(a:+y)2* ^^' {Jb-\-\f~h + \

17 2^ + y 2^y ^p 6 + c 6 -2c

•^'« a~^-y2'"(a:+y)2' •^°' 6a-26c + c2 ^^^'

19. _^_^ 20. -^^-4t-

xy-y^ ar-ary a^+a-2 a + 1

21. ^?!-^%. 22. ^ '

2a6-62 2a + t 3:^+1 x+\

^* 6»+8"^6 + 2" ^*' ar» + a:y + y2'*' arV-y**

XXI.] ADDITION AND SUBTRACTION OF FRACTIONS. 133
Find the value of

25. -^i—^- . 3 .^ . 26.

27. r+2-^. 28. 4 + ^-2a. 29. A+ ^ ^

a;-l * 2 + a ' ' x^ «+! a;*

30 ^ -?-4. ^ 31 3 1 _2

166. The following examples furnish additional practice in
the simplification of fractions.

E.a..^.. SimpUfy ^^- ^^^^^.

The expression = 3^ -^j + 2±f

_ 20(6-l)-6(&+l) + 764-5
16(6=»-1)

^ 216-21 _ 21(6-1) _ 7

15(62-1) 15(6^-1) 6(6+1)

167. Sometimes the work will be simplified by combining
two of the fractions together, instead of finding the lowest
common multiple of all the denominators at once.

Example, Simplify a- x

8(a-ar) 8(a + a;) ^oC^ + x^)
Taking the first two fractions together,

*!,« <^^^^<.o«;»» 3(a + x)-{a-x) a-2x
tne expression = -i———' — i- — ^ _ __ — __

a+2x a-2x

^a'-x^) 4{a^ + x^)

_ (o + 2a;)(a2 + s^)-{a- 2x)(a^-tx^)
4t(a*-x*)

_ a^ + 2a^x + ax^ + 2x^-{a^-2a^x-ax^+23^)

4(0^ -X*)

_ 4a^x + 2ax^ _ ax{2a + x)
" "ilia* ^^) "2(a* - X*) '

134 ALGEBRA. [CHAP.

EXAMPLES XXI. c.

Find the value of
1 6 32- 3a: „ 1 1 ^ 6c

2x-l 2x+l 4a;''-l 2a + 3c 2rt-3c 4a?-9c^

Q l_+2a_3a2 + 2a , . 2a; 5 _ 4ar'-9a;

3-3a 2-2a2 * 9-6a; 6 + 4a: 27 - 12ar«*

^ 1 ^ 2a ^ «« « 2 a-3 ^ 2

0. +7 ...+ , -V.,- 6. -. — ^„- , . ,.. +

x-a {x-a)^ (x-af "* (a+lp (a+l)* (a-hlf

rj a? a ab oll.l

• • 7 r».i~ f ~ r~ — TT^' O. — —^ — +

(a-bf a-h (a-b)'^ "* a; + 3 x a: + l

9 i - 1 10 5 _ _2_

2y2_y_3 2y2 + y-l* 4 + 3x-arJ 3+4a; + ar«'

n _i- + 1 - 2_ 19 _^ - ^ + 1

■*"^' z(z-l) ::(s+l) 22-1" ^^' (a;-2)2 ar» + 4 x-2'

13 2 ■ 3 1^_

''• 3 -a (2 + a)(3-a)(l-f2a) (3-a)(l+2a)*

14 y-2 _ 2(.v--3) y-4

(y-3)(y-4) (y-2)(y-4) "(y -2)(y-3)"

IR 1 _ 24^a; 2+3a: + 3a^»

1-a; (l-a;)(2--a:) (r-a;)(2-a;)(3 + a:)'

16.

2 3 ^ 1_

a;- - 6a:y + Gy^ ot^-xy- 6y^ x^ - 4y^'

17 ^^ a + 3 ?+l_

'^'* 6(a2~lj 2(a2~:ir2^^3) 30^-00 + 3'

^P x-5 2x _ 3a; - 6

a;2-4a:-5 x^ + 2x x^ + x-Q)

^^* a-b a'-^ab + h"^ a?-¥'

90 3(6 - a; ) a;-3 _ J_

^"' a;3 + 27 a;2_3a. + 9 a. + 3-

21. ^ 1 4a;2,

{x-yf ar» + 2a;y + y2 ^^^y^^y*'

99 a? _ °^ _ _^^ _

(a;-a)2 7?— or {x-af'

23. A: + .rk-.-^- 24. 77^-.-.,-,— „.+

2^ 2 + a; i+ar"' 4(l+x) 4(l-a;) 2(l + a:»)

XXI.] ADDITION AND SUBTRACTION OF FRACTIONS. 135

9R 3 3 2 no 2a-6 _ 2a-3

'^"' 2»i-4 27/1 + 4 3w2+i2* "^"^ a2-6a + 9 a^-a-6'

97 « _ ^ _ ^^ Qo a;-3 _a:-l 1

^'* a-6 a + 6 a^-\-b^' ^^' a;-4 x-2 (a;-2y-^*

QQ x—S_x—6 x-3^ X
^^' ^^ x^=3 "^ x^*

30 J^ 1 . 1 _ 1

a-6 3(a-2) 3(a+2) a+6*

168. To find a meaning for the fraction ■^, we define it as

—

the quotient resulting from the division of — a by —b; and this
is obtained by dividing a by 6, and, by the rule of signs, pre-
fixing + .

Therefore -11?= +?=? (1).

— 6

Again, -11— is the quotient resulting from the division of — a

by b ; and this is obtained by dividing a by b, and, by the rule
of signs, prefixing — .

Therefore ^=-? (2).

6 o

Likewise -^ is the quotient resulting from the division of

a by - 6 ; and this is obtained by dividing a by 6, and, by the
rule of signs, prefixing — .

Therefore -^=-? (3).

— o

These results may be enunciated as follows :

(1) If the signs of both numerator and denominator of a
fraction be changed, the sign of the whole fraction mil he un-
changed.

(2) If the sign of the numerator alone be changed, the sign of
the whole fraction will be changed.

(3) If the sign of the denominator alone be changed^ the sign
of the whole fraction vjill be changed.

JSxamplel. h_-a ^ - {h - a) -hj^ ^a-b^
y-x -iy-x) -y + x x-y

Example 2. — - — = rr = — - —

2y 2y 2y

136 ALGEBRA. [CHAP.

3a; 3a: 3a:

Example 3.

4-ar^" "■ -4 + ar'" a?»-4'

Example 4. Simplify -^ + ^ + ^^.

Here it is evident that the lowest common denominator of the first
two fractions is ar - a-, therefore it will be convenient to alter the
sign of the denominator in the third fraction.

rm- XI ■ o 2a: a(3a:-a)
Thus the expression = + _ -1- -?

x-\-a x-a ar-a*
_ a(x - a) + 2a:(a: + a) - a(3a: - a)
"" x^-d?

aa: — a^ + 2a:^ + 2aa: - 3aa: + a'

2x»

7? -a?"

Example. 5. Simplify =- + -r — - - — + ■

(a-6)(a-c) (6-c)(6-a) {c-a){c-h)

Here in finding the L.C.M. of the denominators it must be
observed that there are not h\x different compound factors to be
considered ; for three of them differ from the other three only in
sign.

Thus [a'-c)--{c—a\

(6 — a)=-(a-6),
(c-6) = -(6-c).

Hence, replacing the second factor in each denominator by its equi-
valent, we may write the expression in the form

111

(a-6)(c-a) (6-c)(a-6) (c-a)(6~c)'

NowtheL.C.M. is {^-c){c-a){a—h)\

and the expression = " ^\- c)-{c-a)-{a- b)
^ {b - c)(c - a){a - b)

_ -b+c-c+a-a+b

(b-c){c- a){a-b)

= 0.

Note. In examples of this kind it will be found convenient to
arrange the expressions cyclically, that is, so that a is followed by b,
b by c, and c by a.

XXL] ADDITION AND SUBTRACTION OF FRACTIONS. 137

169. If the sign of each of two factors in a product is changed,
the sign of the product is unaltered ; thus

(a- ;f)(6 -^)={ -(^- a)}{ - (^ - 6)}=(a?- a)(^- 6).

Similarly, (a - xf = (a? - of.

In other words, in the simplification of fractions we may
change the sign of each of tico factors in a denominator without
altering the sign of the fraction ; thus

11

(6-a)(c-6)'~(a-6)(6-cy

170. The arrangement adopted in the following example is
worthy of notice.

ExampU. Simplify -}-- -L^- J^- *^

a — x a + x a^-\-xr a* + JC*

Here it should he evident that the first two denominators give
L.C.M. a^-x^, which readily combines with a^ + ar* to give L.C.M.
a*-ar*, which again combines with a* -far* to give L.C.M. a^-sfi.
Hence it will be convenient to proceed as follows :

mv • a-\-x-(a-x)

The expression = , ^ „ — '-

a^-x*

2x 2x

d^ - x^ d^ + oc^
a*-ar* a*+a*

EXAMPLES XXL d.

Find the value of

1 5 _ 3ar 4- 13a: „ 10 _ 2 1

l+2a: l-2a: 4a;2-l* ^* 9-a« 3 + a a-3'

q 5a 1 1 4 2y _ 5 12y-f8

^' 6(a2- 1)^2(1 -a) 3(a+l)' 2y-3 6y + 9 27-1^

p. x + a_x-a 4ax « 3-2c 2c + 3 12

°* x^~a x+a c^^^' "• 3 + 2o 2c^ 4?-9*

138

ALGEBRA.

[chap. XXL

Find the value of

7.

a

a-b a+b b-a

a

a & a

a^-b^^ a^ + lf^ b^-a*'

9. --^+ ^

a

11.

12.

ar'-a? ar-a:* a:^-l'

(y-2)(y-a) (y-l)(3-y) (y-l)(y-2)"

g 6 , X

{x-a){a-b) (x-b){a-b) (a - x)(b - x)'

1 ^M I a

10 2 3

15.

a + c

b+e

{a - 6)(a; - a) (6 - a){b - x)

14.

. 16.

ah

a-b (a-b)^ {b-af
X - z y-z

{x •- y){a -x) (y - x)(y - o)

17 cH-6 2a a?-a?h

b a+b biP-d')

1ft g' - oft _ tt^ - ^' ^ g

19.
20.
21.
22.
23.
24.
25.

1

1

1

g + a: a-2x x-a 2x+a
3 _, l_+_3_+ 1

g + a? 3a; + g a;-g g-3a:
X y

+

+

(x-y)(a:-z) (y-2)(y-a;) (z-a;)(z-y)
g . 6 . c

-T +

+

(6-c)(ft-a) (c-a)(c-6) (a-6)(a-c)
y-z ^ z-x ^ a?-y

(a:-y)(a:-z) (y-s)(y-a;) (z-ar)(z-y)

1+r

1+P _+ 1+g

(p-g)(2?-r) (5'-r)(7-/)) (r-2?)(r-g)

1 _ 1 a? a:»

4(a; + a) 4(a-a:) 2{QC^-a^) a* -a:*'

„« 1 1 . 1_ ._2a*

^^' 2a*(a + a;)''2a3(a;-o) a2(a2-+a;2) ^TT^

g b . a' + 6' . g^

27.

28.

b g8 + 63

+77

+

a;-2"^(2 + a;)2^(2-x)^ a; + 2

CHAPTEE XXII.
Miscellaneous Fractions.

171. Definition. A fraction whose numerator and deno-
minator are whole numbers is called a Simple Fraction.

A fraction of which the numerator or denominator is itself a
fraction is called a Complex Fraction.

a a

rr h h

Thus -, - - are Complex Fractions.

c d

In the last of these types the outside quantities, a and dy
are sometimes referred to as the extremes, wliile the two middle
quantities, b and c, are called the means.

Instead of using the horizontal line to separate numerator
and denominator, it is sometimes convenient to write complex
fractions in the forms

V7

b a I ale

b r' bid

Simplification of Complex Fractions.

172. It is proved in the Elementary Algebra, Art. 141, that

a

b _a , c _a d _ad
c b ' d b c be
d

The student should notice the following particular cases, and
should be able to write down the results readily.

1 ^ a , b b
-=l-rT = lx-=-.
a a a

a I , ,

-=a-r \ — axo—ao,
1 o

140 ALGEBRA. [chap.

173. The following examples illustrate the simplification of
complex fractions.

Example 1. Simplify ^.

a? — —

The expre88ion = (:.+«-)-(«-?;) =^-?^

X a:* - a* 7? -a?'

.Eeampie 2. Simplify " ^ .

a 1 3

Here the reduction may be simply effected by multiplying the
fractions above and below by 6a, which is the L.G.M. of the
denominators.

Thus the expression = ^V^'-|-^
^ a2 + 3a-18

_ 2(og-6a + 9) 2(a-3)
"(a + 6)(a-3)" a+6

a2+&»_a2-6a

Example 3. Sunplify a^-^' ^t' + y

a+o a—

a-h a+h
The numerator = (^&!)M^!z^ 4^^ .

4a&

similarly the denominator =
Hence the fractions

(a + 6)(a-6)'
4a26a 4a6

(aa + 62)(a2-62) ' {a + h){a-h)
4a^h^ {a + b){a-h)

a-^ + 62-

Note. To ensure accuracy and neatness, when the numerator and
denominator are somewhat complicated, the beginner is advised to
simplify each separately as in the above example.

XXII.] MISCELLANEOUS FRACTIONS. 141

174. In the case of Continued Fractions we begin from
the lowest fraction, and simplify step by step.

Example, Find the value of — .

4 ?

2-f ^

The expressions

1-a;
1

4,3(1-- a:)

2 - 2a; + a ? 2-a?

1-a;
1

8-4a:-3-f3a; " 5-a;
2-a; 2-a;

_ 2-a;
b- X

EXAMPLES XXn. a.

Find the value of
1 ^ o tt Q 1-a M h

x+y. h-% 4-1

z d a^

5. f4 6. SI£^ 7. Hi

X a x~y h~d

a+- -5 y-3 + 2^

l + -3^5 y-i+3 7i-i^

a^ a y n

a;-2+_5_ ^_g_ 6 J?L_A

10 a? + 3 TQ 6 + 3 ,. 62 ^2

a;+3 6 + 3 a2 + a6 + 62

c + d c-d ^_ a-b x+S x + B

IR c-c? c-Hog 1/5 l-a6 -.„ 7 ~a; + 4

■'^* c + rf c-d ■*-°' ^_ a(a-6) ' ■*•'• a?-3 a;-3'
c-d'^c+d l-ab 4 "*"a;-l

18. 1+-^. 19. x+J~^ 20. 2 — L-.

l+± a;-i 4-1

a a; d

1-

a

P

- r

8.

JL

1

r

^g

-p^

1

3

4

n

«2-

«3

142 ALGEBRA [chap.

Find the valuo of
21. —2 — 22. \ 23.

24. ' * — 26. f 26.

X

2- *

?t*- ^^-»

1+ *.

,+ 1 - 1_ 1

«+? 1-1

a

27. — * w 1 28.

-y 6--^

3a-2c

3a-2c- ^

j_3(a-c)
a-1 3a-2c

175. Sometimes it is convenient to express a single fraction
as a group of fractions.

Example, S^ynl^^y'+l^ = J^ - 1^>?2^ + JV_

= 1-1+3^
2y a: 2ar»*

176. Since a fraction represents the quotient of the nume-
rator by the denominator, we may often express a fraction in an
equivalent form, partly integral and partly fractional

Example 1. ?-±| = (?1±3)+A = 1 + 5
■^ x + 2 x + 2 x + 2

ExampU 2. 5^ = 3"' + 5) -15 -2 ^ 3(£+ 5)-17 ^ 3 _ _17_,
a; + 5 x + 6 x + b a?4-5

Example 3. Shew that ?^ 7^1 ^^x-l- — .

x-S a:-3

By actual division, a; - 3 ) ^bi^ -*Jx-\{2x-l

27^-Qx

- a:-l

- a:+3

-4
Thus the quotient is 2a; - 1, and the remainder ~4)i

Therefore ^^ "' '^^" ^ = 2a; - 1 ~ -i-.

«-8 aj-3

xxn.] MISCELLANEOUS FRACTIONS. 143

177. If the numerator be of lower dimensions than the
denominator, we may still perform the division, and express the
result in a form which is partly integral and partly fractional.

Example. Prove that ^^ =2x-6x^+lSs^- ^^f ,

Bydiviflion l + 3a;2)2a: {2x~6si^+l^

2a:+6a:^

18a;g + 54ig7

-54a;7 ^
whence the result follows.

Here the division may be carried on to any number of terms in
the quotient, and we can stop at any term we please by taking for
our remainder the fraction whose numerator is the remainder last
found, and whose denominator is the divisor.

Thus, if we carried on the quotient to four terms, we should have
^ = 2a:-6a:3+i8aj6_54a,74. 162a:9

l + Sx^ l + 3ar»'

The terms in the quotient may be fractional ; thus if x^

is divided by a^-a^, the first four terms of the quotient are

1 a^ dP ofi a}-^

-A — - + _+__, and the remainder is -^,.

X a/^ x^ x^^ x^^

178. The following exercise contains miscellaneous examples
which illustrate most of the processes connected with fractions.

EXAMPLES xxn. b.

Simplify the following fractions :

1 l-ar^ 2 12a?^ + ig-l ^l+ 6a?+ftea

• l+2a; + 2arJ + a:3* 1 - Sa- + Itia:-^ ' 160:8-1 '

3a + 6 , 4a6 m a + 6 2a

a-6 62--a2* '* a^-ah-W' a'-W

144 ALGEBRA. [CHAP.

Simplify the following fractions :

ax^ + xy-ax-y ' * xVa-a: a-hSxJ a + 3a;'

9.

ar'-6a?+8 a8-llx+28 ix^-dx+U

3a-JL

1 rc3+a'

3

-a —

in 3o,, a -,-, X X

•^"' 3a+l 3^^ ^^' a3+a:3 +2.+^'

12. — ^ r^. 18. ^-'

1--^ _«__1 *^ «^^ 3(^-27^+54)

a;-l x+1 ^^ l-a;

14 c(f(a« + 62) + a5(ca + (P) ,j. / x l+a?\_l

16.

^

1 1

a«+4

-o

if4a2-l^
a^V 9/ 1

^r7 2 a' — 4 iQ %* \ Of X

a

10 2a:» + ar'-3a; ^5a:g-8a?-21 . 2a:S_3a._9

35a;2 + 24a:-35 t^+It^-^x' 1x^ + b\x-^0

20 g + r-p ^ r+p-g ^ p + g-r
* (I'-gKl?-^) (g-^Kg-p) {r-p){r-q)

ot^ + xh/+xy^+^

21 //a? + y + a;-y\^/a:+.y_a?-y\\_a:» + a;^

' \\a;-y x + y/ Vx-y x + y/f 23ir^ + 2xy^

on a^-{h-c)\h'-{c-a)\c^-{a-b)^

23. (^+?^)(^)--^+_^-4-.

\y a;/\y*-arv ar + xy xy-y* x+y

04 «'-J_ _^r a^-4a + 3 ^/ a^-Q ^ ag-a-2 \1

^^^ a2 + a-Q'La2-4a + 4* la* + a2 + l • a»+l /J*

QK /I . 1 \/l _ 1 \ a:^ - 4aa;

*"• V2a 2a-«A3a 3a -a:/ 6a^(a; - 2a)(a; - 3a)'

xzn.]

MISCELLANEOUS FRACTIONS.

145

26.
27.
28.

29.

Simplify the following fractions :
1 1.1 2aa

80.

32.

83.

84.

6a-6 6a+6 Sa^ + S 3a*+3

4ah^ 1 _ g 1

2a* +326* 8a +166 ia^ + lfife^ 8(26 -o)'

36« + 6 . 26-7 26^-36 3
662-1-6 1-26 462-86 + 3 *

('-a*(s*i) ■

31.

m^ ??r+m

1 -t»i8

m

2_

m*

m

m

\a6-62 ac-6c/Ij_6 j_c i_a)*

Y a 6 cl

a-c , ^ ^ a-c
+ c • a —

1 +ac

1 -etc

I _ c{a - c) , _ a(a - c)
1 + ac 1 -ac ,

1

3 1

• \c a/

(3a + a;)2
a

a; -3a 3{a: + 3a) (a;-3a)-

a

H.A.

CHAPTER XXm.
Harder Equations.

179. Some of the equations in this chapter will serve as a
useful exercise for revision of the methods already explained ;
but we also add others presenting more difficulty, the solution
of which will often be facilitated by some special artifice.

The following examples worked in full will sufficiently illus-
trate the most useful methods.

Example 1. Solve ^^'^ ^^ " ^

Clearing of fractions, we have

(6a; - 3)(a: + 6) = (3a: - 2)(2a: + 7),
6ar»+27a; - 15 = 6ar^+ 17x- 14 ;
/. 10a:=l;

" '"lO*

Note. By a simple reduction many equations can be brought to
the form in which the above equation is given. When this is the
case, the necessary simplification is readily completed by multiply-
ing across or ^* midtiplying up,*' as it is sometimes called.

Example 2, Solve ??+--t^=?^^-l.
^ 20 3a; + 4 5

Multiplying by 20, we have

8a;+23-?^^^±^ = 8a;+12-20.
3a: +4

By transposition, 31 = 20(5a:+2)^

3a: +4

Multiplying across, 93a: + 124 = 20(5a; + 2),

84 = 7x5

.*. a: = 12.

CHAP. XXIII.] HARDER EQUATIONS. 147

180. When two or more fractions have the same denomin-
ator, they should be taken together and simplified.

Example 1. Solve 2iz^+8^9 = J?- - 13.

ar — 2 4 — a? a? — 2

By transposition, we have

8ar-49_^,«_28-(24-5a?).

"4::^^^^ 1^^ '

. 3-5a: _ 4 + 5a?

4-a: aj-2

Multiplying across, we have

3a;-6ar^-6 + 10a;=16-4a;+20a:-6a:8.

that is, -3a; = 22;

.. X- _.

Example 2. Solve £ll?- + ^ = ^ + ^III.
^ oj-lO^x-e a;-7 ar-9

This equation might be solved by at once clearing of fractions, but
the work would be laborious. The solution will be much simplified
by proceeding as follows.

The equation may be written in the form

(ar-10 ) + 2 (a;-6) + 2 (a;-7) + 2 (a?-9) + 2 ,
a;-IO "^ x-% " x-1 "^ a;-9 '
whence we have

l + -^ + l+— ^ = 1+-A= + 1 +

a:-10 a:-6 x-1 a;-9'

u- I. • 1 111
which gives —. + = + -

a;-10 a; -6 a;-7 a;-9

Transposing. _L___L_ = JL_ __L_,

^ ^* a;-10 a;-7 aj-9 x-Q

(a;-10){a;-7)""(a;-9)(a:-6)

Hence, since the numerators are equal, the denominators must be
equal ;

that is, (a;-10)(a;-7) = (a;-9)(a?-6),

jB2-17a; + 70 = ar^-15a:+54;
.-. 16 = 2aj;

148 ALGEBRA. [CHAP.

EXAMPLES XXTTT. a.

Solve the following equations :

1. 3 l_ 2. 7 - ^^

14.
16.

17.

20.
21.

6x-9 4a;-10 "" 6a:-17 4a;-13

o 7, 3-4a? . 1 ■ 4 _Q

^' 9 r^l? *• 6-5a; 17x + 3 *

^ 5a?-8 _ 5g+14 ^ 8a?~l_ 4x-3

"• x-4~a:+7' 6a:+2 3a:-l'

„ 22a:-12_rt . 3a:+7 « 9a;-22 3a;-5 «

'• "8J:T""^'^4^+8- "• "2^:^ 2^^7"

Q 8a?-19_l_ 3rg-4 ,^ 7ar + 2 _1 6ar-l

**• 4a; -10 2 2a:+r ■*■"• 3(x-lj 3 3x+T

n a;~5 2a:-l _5a?--l ,2

•""*■• 2 3a; + 2"' 10 5'

IQ 5 a;-l7 . 2a:-ll _23_ 3a?-7

■*^'*' 13 -4a; 14 42" 21 *

1^ 3,_^-3_ l-9a; _4a; + 3_ 1 ^

2^

'S _

a;+l a: + 2 a; + 3 3a;+6

3t _ 18 _ 7 _ 4
a;-4 3a;-18 4a;-16 x-6*

16. 1 + 1 - « 1

a;+6 3a;+12 2a;+10 6(a;+4)

g-l _ a?-5 _ a;-3 _ a;-7
x-2 x-6 x-4 x-8'

18. 1 +^=^+^ 1

a;-9 a;-17 a;-ll a:-16

19. 1 + 1 = 1 + 1

2a:-l 2a;-7 2a;-3 2a;-6

x-l X _a;--4_a?-3
a:-2 a;-l a;-5 a:- 4*

5a?'-64 _ 4a;-55 _2a7-ll a;-6
a;-13"«-14"" a;-6 .a;-7"

XXIII.] HARDER EQUATIONS. 149

Solve the following equations :

no 5x + Sl 2a: + 91.a:-6 . 2a:-13
x + Q x + 5 x-5 x-6

„o 1 2a?+l 5 _ ll + 12a?
^^' 3a;- 1 1-9x3 l+3a; *

OA. 5x^ _a? + 3_^ ar-3
"^ ^^ ^^"^ x + S'

[For additional examples see Elementa/ry Algebra.li

Literal Equations.

181. In the equations we have discussed hitherto the co-
efficients have been numerical quantities. When equations
involve literal coefficients, these are supposed to be known, and
will appear in the solution.

Example 1 . Solve {x + a){x + 6) - c{a +c)^{x- c){x +c)'\-ah.
Multiplying out, we have

T^ + cue+hx+ah-ac-c^ =^x^-c^ + aJb ;

whence aa:+hx = aCf

(a + h)x = ac ;
. ^_ ow;
a + b

a b a-b

Example 2. Solve .

x-a x-b x-c

Simplifying the left side, we have

a{x -b)- b{x -a) _a-b
{x - a){x -b) x-c
{a-b)x __a-b ,
(x - a){x -b) x-c*
. X _ I

{x- a){x -b) x-c

Multiplying across, x^-cx = 0!^-ax-bxi-ahf

ax + bx-cx = ahy
(a + 6 - c)a: = a6 ;

a + 6-c

150 ALGEBRA. [chap.

Example 3. Solve the simultaneons equations :

ax-hy = c (1),

px + qy-r (2).

To eliminate y, multiply (1) by </ and (2) by h ;
thus aqx - hqy = cq,

hpx+hqy- hr.

By addition, {aq + hp)x -cq-\-hr',

•• •«• — - — .

aq + hp

We might obtain y by substituting this value of x in either of the
equations (1) or (2) ; but y is more conveniently found by eliminat-
ing X, as follows.

Multiplyiug (I) by p and (2) by a, we have

apx - bpy = cp,
apx + aqy = ar.

By subtraction y {aq + hp)y ^ar-cp;

. ,,_ ar-cp
.. y — —T—

(pj-^bp

EXAMPLES XXm. b.

Solve the following equations :

1. ax-\-h^ = a^-bx. 2. a:" - a' = (2a - «)«.

3. a^{a-x)-habxr=b^{x-b). 4. {b + l){x + a) = {b-l){x-a)

5. a(x + b)-b^ = a^-b(a-x). 6. c^a: - rfs = d^a: + c^.

7. a{x-a) + b{x-b) + c{x-c) = 2{ah + bc + ca),

8. ^'-6 = ^'+a. 9. ^ = ^+^-1,.

XX 2a b b^ 4a^

10. a: + (a:-a)(a:-6) + a3+62 = 6+3:8 -.a(6-l).

,, 2x-a _Sx-b_^a^-8b^ ^„ g-a; _ 6-a? _tt' + 6^

6 a ab ' * a-b a + b a^-b^'

^n ax-b bx-c _a-cx ,. x + a-b _x + b~c

■Iw* ' "^ E — • J-^ — TT": ; — TT*

c a a; + 6 + c a: + a + 6

15. p{p-x)-P-{x-q)^-p{p-q)+pq(^^-iy^O.

XXIII.] HARDER EQUATIONS. 151

Solve the following simultaneous equations :

16. x-y = a + b, 17. cx-di/ = c^ + (P, 18. ax = hy,

aa;+6y = 0. x + y = 2c, x-y = c.

19. ^+| = a+6, 20. ^-^- = 0, 21. "^+2/ = ^,

^^' 2 3 X y x-y b

ah ^ a+

22. ?-y="+^. 23. ^-^ = 0.

h a a P 9

a{a+x) = 6(6-y). ^ + ^+^ = ^3+^8.

„. 2a?-6 , 2y-fa , 3a;-fy qc aa; + 6y _l_ a ^-h^

^^ ~Z T a+26* 6a:+ay"'2 6a; + ay'

Irrational or Surd Equations.

182. Definition. If the root of a quantity cannot be ex-
actly obtained the indicated root is called a surd.

Thus V2, X/b, Xla\ ^Ja^ + b^ are surds.

A sard is sometimes called an irrational quantity; and
quantities which are not surds are, for the sake of distinction,

termed rational quantities.

183. Sometimes equations are proposed in which the un-
known quantity appears under the radical sign. For a fuller
discussion of surd equations the student may consult the Ele-
mentary Algebra, Chap, xxxii. Here we shall only consider a
few simple cases, which can generally be solved by the follow-
ing method. Bring to one side of the equation a single radical
term by itself : on squaring both sides this radical will disappear.
By repeating this process any remaining radicals can in turn
be removed.

Example 1. Solve 2fjx->sj4x-\l = 1.

Transposing, 2 ,Jx - 1 = Via; -11.

Square both sides ; then 4x - 4 ^x + 1 = 4^; — 1 1 ,

4^/a: = 12,

.-. a: = 9.

152 ALGEBRA. [chap. xxm.

Example 2. Solve 2 + Va: - 5 = 13.

Transposing, Ijx -6 = 11.

Here we must cvhe both sides ; thus a; - 5 = 1331 $
whence x = 1336.

Example Z, Solve hl'^j;L^^hJl±l,

Multiplying across, we have

{%^x-l\){^x + Q)^Zjx{2Jx+l)i
that is, %x-U^x + msJx-Qii = Qx+ZJx,
-llVa?+36V«-3Va? = 66,

22Va: = 66,
Va: = 3;

> . X *— «/.

EXAMPLES xxm. c.

Solve the equations :

1. >Jx~2=\, 2. \/5^^ = 7. 3. Vi^=2.

4. 2x/^Tl=3. 5. 3\^r^2^=-l. 6. i = l/2i.

7. N/l-6a: = 3Vl-a;. 8. 2 n/sS^ - 7 V^^ = 0.

9. >/4a:2-lla;-7 = 2a;-3. 10. 3 n/1 - 7a: + 4ar« = 5 - 6a;.

11. 1 + >/^-3a;a + 7a5-ll = a;. 12. slx^^ = ^x-1,

13. A/4a; + 13 + 2Va: = 13. 14. 3+ N/l2a:-33 = 2\/3^

IR \/ag-l _ /v/a?-3 ^/» Jx + 4, ^x + 5

^°* Va; + 3 ~^;7^ ^°- 3Va:-8"3V^^*

j,^^ 2V^-3^f2^ IB. 2^^£:i7^2+ ^^

19. \/l + 4a; + 2Va: = -7-. 20. ^/a: + Va: - 3 = -7==.

21. V4a; + 7- V;»+T= sfx - 3. 22. fslix^- sIx^Z = V«^^

CHAPTER XXIV.

Harder Problems.

184. In previous chapters we have given collections of
problems which lead to simple equations. We add here a few
examples of somewhat greater difficulty.

Example 1. If the nnmerator of a fraction is increased by 2 and
the denuniinator by 1, it becomes equal to | ; and if the numerator
and denominator are each diminished by 1, it becomes equal to 7:
find the fraction.

Let X be the numerator of the fraction, y the denominator ; then
Mie fraction is — .

y

From the first supposition,

a?-f2 _5

y+1 8
x-l I

from the second,

(1).

(2).

y-1 2

From the first equation, 8a: - 6y = - 11,
and from the second, 2x— y = 1 ;

whence a; = 8, y = 15.

g
Thus the fraction is — .

15

Example 2. At what time between 4 and 5 o'clock will the
minute-hand of a watch be 13 minutes in advance of the hour-hand?

Let X denote the required number of minutes after 4 o'clock ;
then, as the minute-hand travels twelve times as fast as the hour-

hand, the hour-hand will move over --- minute-divisions in x minutes.

At 4 o'clock the minute-hand is 20 divisions behind the hour-hand,
and finally the minute-hand is 13 divisions in advance ; therefore the
minute-hand moves over 20-1-13, or 33 divisions more than the hour*
hand.

154 ALGEBRA. [CH^-

Hence «=^+33,

li« = 33;

;. a; = 36.
Thus the time is 36 minutes past 4.

If the qnestion be asked as follows: "At what Himb between
4 and 5 o'clock will there be 13 minutes between the two hands ? "
we must also take into consideration the case when the minute-hand
is 13 divisions behind the hour-hand. In this case the minute-hand
gains 20 - 13, or 7 divisions.

Hence *~^'*"'^'

7
which gives x = 7^^.

7'
Therefore the times are 7=^ past 4, and 36' past 4.

Example 3. A grocer buys 15 lbs. of figs and 28 lbs. of currants
for $2. GO ; by selling the figs at a loss of 10 per cent., and the cur-
rants at a gain of 30 per cent., he clears 30 cents on his outlay ;
how much per pound did he pay for each ?

Let X, y denote the number of cents in the price of a pound of
figs and currants respectively ; then the outlay is

16x+28y cents.
Therefore 16a+28y=260 (1).

Tlie loss upon the figs is — x 16x cents, and the gain upon the

currants is — x 28y cents ; therefore the total gain is

^.?^ cents;
6 2 '

.-. ^-?? = 30;
6 2 '

that is, 28y-5a;=100 (2).

From (1) and (2) we find that a;=8, and y=5 ; that is, the figs
cost 8 cents a pound, and the currants cost 5 cents a pound.

Example 4. Two persons A and B start simultaneously from
two places, c miles apart, and walk in the same direction. A travels
at the rate of p miles an hour, and B at the rate of q miles ; how far
will A have walked before he overtakes B ?

XXIV.] HARDER PROBLEMS. 155

Suppose A has walked x miles, then B has walked x-c miles.

A walking at the rate of p miles an hour will travel x miles in

- hours; and B will travel x-c miles in hours; these two

p q

times being equal, we have

X x-c
p q

qx=px-pc;

whence a; = J^.

p-q

Therefore A has travelled ~^--- miles.

p-q

Example 5. A train travelled a certain distance at a uniform
rate. Had the speed been 6 miles an hour more, the journey would
have occupied 4 hours less ; and had the speed been 6 miles an hour
less, the journey would have occupied 6 hours more. Find the
distance.

Let the speed of the train be x miles per hour, and let the time
occupied be y hours ; then the distance traversed will be represented
by xy miles.

On the first supposition the speed per hour is a; + 6 miles, and the
time taken ia y-i hours. In this case the distance traversed will
be represented by (a? + 6)(y- 4) miles.

On the second supposition the distance traversed will be repre-
sented by {x - C){y + 6) miles.

All these expressions for the distance must be equal ;

.-. xy = {x + 6)(y - 4) = (a; - 6)(y + 6).

From t}iese equations we have

a;y = ary + 6y - 4a: - 24,

or 6y-4a; = 24 (1);

and xy = xy-Qy + 6x-36,

or 6a:-6y = 36 (2).

From (1) and (2) we obtain a; = 30, y = 24.
Hence the distance is 720 miles.

EXAMPLES XXIV.

1, If the numerator of a fraction is increased by 5 it reduces to §,
and if the denominator is increased by 9 it reduces to ^ : find the
fraction.

156 ALQKBRA. [chap.

2. Find a fraction such that it reduces to { if 7 be snbtracied from
its denominator, and reduces to § on subtracting 3 from its numeratur.

3. If unity i*) taken from the denominator of a fraction it reduces
to i ; if 3 is added to the numerator it reduces to f : required the
fraction.

4. Find a fraction which becomes | on adding 5 to the numerator
and subtracting 1 from the denominator, and reduces to ^ on sub-
tracting 4 from the numerator and adding 7 to the denominator.

5. If 9 is added to the numerator a certain fraction will be
increased by ^ ; if 6 is taken from the denominator the fraction
reduces to § : required the fraction.

6. At what time between 9 and 10 o'clock are the hands of a
watch together ?

7. When are the hands of a clock 8 minutes apart between the
hours of 5 and 6 ?

8. At what time between 10 and 11 o'clock is the hour-hand six
minutes ahead of the minute-hand ?

9. At what timo between 1 and 2 o'clock are the hands of a
watch in the same straight line ?

10. When are the hands of a clock at right angles between the
hours of 5 and 6 ?

11. At what times between 12 and 1 o'clock are the hands of a
watch at right angles 7

12. A person buys 20 yards of cloth and 25 yards of canvas for
$35. By selling the cloth at a gain of 15 per cent, and the canvas
at a gain of 20 per cent, he clears $5.75 ; find the price of each per
yard.

13. A dealer spends $1445 in buying horses at $75 each and
cows at $20 each ; through disease he loses 20 per cent, of the horses
and 25 per cent, of the cows. By selling the animals at the price
he gave for them he receives $1140 ; find how many of each kind
he bought.

14. The population of a certain district is 33000, of whom 835
can neither read nor write. These consist of 2 per cent, of all the
males and 3 per cent, of all the females : find the number of males
and females.

15. Two persons C and D start simultaneously from two places
a miles apart, and walk to meet each other ; if G walks ]? miles per
hour, and D one mile per hour faster than G, how far will D have
walked when tliey meet?

16. A can walk a miles per hour faster than B; supposing that
he gives B a start of c miles, and that B walks n miles per hour,
how far will A have walked when he overtakes B ?

XXIV.] HABDEB PBOBLEMS. 157

17. -4, B, C start from the same place at the rates of a, a-\-b,
a-|-2& miles an hour respectively. B starts n hours after A, how
long after B must C start in order that they may overtake A at the
same instant, and how far will they then have walked ?

18, Find the distance between two towns when by increasing
the speed 7 miles per hour, a train can perform the journey in 1
hour less, and by reducing the speed 5 miles per hour can perform
the journey in 1 hour more.

19. A person buys a certain quantity of land. If he had bought
7 acres more each acre would have cost ^4 less, and if each acre
had cost $18 more he would have obtained 16 acres less : how much
did he pay for the land ?

20. ^ can walk half a mile per hour faster than B, and three-
quarters of a mile per hour faster than O. To walk a certain dis-
tance C takes three-quarters of an hour more than Bj and two
hours more than A ; find their rates of walking per hour.

21. A man pays $90 for coal ; if each ton had cost 50 cents
more he would have received 2 tons less, but if each ton had cost
75 cents less he would have received 4 tons more ; how many tons
did he buy ?

22. -4 and B are playing for money ; in the first game A loses
one-half of his money, but in the second he wins one quarter of
what B then has. When they cease playing, A has won $10, and
B has still $25 more than A ; with what amounts did they begin ?

23. The area of three fields is 516 acres, and the area of the
largest and smallest fields exceeds by 30 acres twice the area of
the middle field. If the smallest field had been twice as large, and
the other two fields half their actual size, the total area would
have been 42 acres less than it is ; find area of each of the fields.

24. -4, B, C each spend the same amoimt in buying different
qualities of cloth. B pays three-eighths of a dollar per yard less
tiian A and obtains three-fourths of a yard more ; C pays five-
eighths of a dollar per yard more than A and obtains one yard less ;
how much does each spend ?

26. -S pays $28 more rent for a field than A ; he has three-
fourths of an acre more and pays $1.75 per acre more. C pays
$72.50 more than A ; he has six and one-fourtii acres more, but pays
25 cents per acre less ; find the size of the fields.

158 ALGEBRA. [chap.

MISCELLANEOUS EXAMPLES IV.

L When o=-3, 6 = 6, c=-l, d = 0, find the valne of

26c n/o* ^-d + 5bc^dc + cP".

2. Solve the equations :

(1) lx-^y = S-2x, iy-3a: = 3-y;

(2) l=y+z = 2(z + a;) = 3(a;+y).

3. Simplify

.|v a -a: 4a:^ a-3x ,

,9. 6«-3& 6^ + &-3 ^ y- 36^ -106
^■'^ 6^-26 + 4 6^ + 36-18' ¥+S '

4. Find the square root of

O u v oi

6. In a base-ball match the errors in the first four innings
are one-fourth of the runs, and in the last five innings the errors
are one-third of tlie runs. The score is 16, and the errors num-
ber 5 ; find the score in the first four innings.

1

6. Find the value of °'"'^ x ^.

1-A+l " + *
a^ ax a^

7. Find the value of

8. Resolve into factors

(1) 3a2-20a-7; (2) a*h^"-h*a^.

9. Reduce to lowest terms 4^^ + 7^2 - a: + 2

4ar + 5x--7a:-2

XXIV.] MISCELLANEOUS EXAMPLES IV. 159

10. Solve the equations :

(1) « -3-^=^-^.4 + ^-^;

(2) x+y-z = 0, x-y+z = 4, 5a: + y + z = 20;
#«» ttx + b dx + e 1

c /

n <?iTnT*lifv a?+3 a; + 2 ^ 4

11. oimpmy a^_5a; + 6"ar»-9a:+14'*"a:'»-10a; + 2r

12. A purse of sovereigns is divided amongst three persons, the
first receiving half of them and one more, the second half of the
remainder and one more, and the third six. Find the number of
sovereigns the purse contained.

13. If A =- 1, ife = 2, / = 0, m = 1, w = - 3, find the value of

hHm-D- 'Jihn + h k

14. Find the L.C.M. of

15. Find the square root of

(2) 1 - 6a + 5a2 + 12a3 + 4a*,

16. SimpUfy 20^±27r + 20^N:27x + 9

17. Solve the equations :

^jj a(a?-6) ^6(a?-o)_^ .
a-6 b-a *

(2) -A. + _JL=_^-I- 8
^ ' a;-4 a;-8 a:-9 a;-3'

18. A sum of money is to be divided among a number of per-
sons; if ^8 is given to each there will be i|3 short, and if $7.50 is
given to each there will be $2 over : find the number of persons.

160 ALGEBRA. £chap.

19. Hesolve into factors :

(1) 2ar» - 3a6 + (a - 66)a: ; (2) ^'^ - 4xy - I5y^.

20. 1*1 the expression t^ - 2x^ + 3a: - 4, substitute a -2 for x, and
arrange the result according to the descending powers of a.

21.

Simplify

(1) —V'

fl 4-

^ x'
a + '^

a

22. Find the H.C.F. of

3a:»-lla;2 + a;+15 and 5x*-1x^-20x^-nx-S.

23. Express in the simplest form

x y

(1) y"^ > (2) /'^zl+^_+;Vf-L+— \

g + y _ y + ^ ' \x-\ x+\ I \x-\ x+\J

y X

24. A person possesses $5000 stock, some at 3 per cent., four
times as much at 3^ per cent., and tlie rest at 4 per cent. : find the
amount of each kind of stock when his income is $176.

25. Simplify the expression

-3[(a + 6)-{(2a-36)-(5a + 76-16c)-(-13a+26-3c-5rf)}],
and find its value when a = 1, 6 = 2, c = 3, d = 4.

26. Solve the following equations :

(1) lly-a;=10, llar-lOly = 110 ;

(2) a; + y-2; = 3, a; + z-y = 5, y + z-x-1

27. Express the foUowiug fractions in their simplest form :

^^^ 12ar^^a:*T4x2 ' ^^' r~ 1^ '

2-ar

28. What value of a will make the product of 3 -8a and 3a + 4
equal to the product of 6a + 1 1 and 3 - 4a ?

XXIV.] MISCELLANEOUS EXAMPLES IV. 161

29. Find the L.C.M. of sbS - ar^ - 3a: - 9 and a:' - 2ar» - 5a: - 12.

30. A certain number of two digits is equal to seven times the
sum of its digits : if the digit in the units' place be decreased by
two and that in the tens' place by one, and if the number thus
formed be divided by the sum of its digits, the quotient is 10. Find
the number.

31. Find the value of

Qx^-6xy-Qy^ ^ Sa ^-xy-^y^ ^ 9ar^ - 6a:y - 8 y^
2x^+xy-y^ 2ar^ - 5xy + Sy"^ ' 2x^-Sxy + y^'

32. Resolve each of the following expressions into four factors :

(1) 4a*-17a262 + 464; (2) x^-25Qy^,

33. Find the expression of highest dimensions which will divide
24a'^h-2a^b^-9ab* and ISa^ + a*ly^ - 6a%^ without remainder.

34. Find the square root of

(1) a;(a; + l)(a: + 2)(a:+3) + l;

(2) (2a2 + 13a+15)(a2 + 4a-5)(2a3 + a-3).

35. Simplify

2a;~6 ^ a;^ + 3a;-4
a,-2-6a; + 9"' a:2 + a,_i2*

36. A quantity of land, partly pasture and partly arable, is sold
at the rate of $60 per acre for the pasture and |40 per acre for the
arable, and the whole sum obtained is J^IOOOO. If the average price
per acre were |I50, the sum obtained would be 10 per cent, higher :
tind how much of the land is pasture, and bow much arable.

H.A.

CHAPTER XXV.
Quadratic Equations.

185i Definition. An equation which contains the square
of the unknown quantity, but no higher power, is called a quad-
ratic equation, or an equation of the second degree.

If the equation contains both the square and the first power
of the unknown it is called an affected quadratic ; if it contains
only the square of the unknown it is said to he & pure quadratic.

Thus 2j;'^— 5x=3 is an affected quadratic,

and 5x^=20 is a pure quadratic.

Pure Quadratic Equations.

186i A pure quadratic may be considered as a simple equa-
tion in which the square of the unknown quantity is to be found.

9 25
Example, Solve = — .

^ a;-^-27 x'^-ll

Multiplying across, 9x2-99=25x2-675 ;

.-. 16x2=576;
.-. x2=36;
and taking the square root of these equals, we have

x=±6.

[In regard to the double sign see Art. 119.]

187. In extracting the square root of the two sides of the
equation x2=36, it might seem that we ought to prefix the
double sign to the quantities on both sides, and write ±x=±Q.
But an examination of the various cases shows this to be un-
necessary. For ±x= ±6 gives the four cases :

-|-x=-f6, -fx=-6, -x=+6, -x=-6,

and these are all included in the two already given, namely
x= +0, x= — 6. Hence, when we extract the square root of the
two sides of an equation, it is sufficient to put the double sign
before the square root of one side.

CHAP. XXV.] QUADRATIC EQUATIONS. 163

Affected Quadratic Equations.

188. The equation x2=36 is an instance of the simplest form
of quadratic equations. The equation (x— 3)"^ = 25 may be
solved in a similar way; for taking the square root of both
sides, we have two simple equations,

a:- 3= ±5.
Taking the upper sign, a? — 3 = + 5, whence x=S ;
taking the lower sign, ;r- 3= - 6, whence ^= - 2.
.*. the solution is a? =8, or -2.

Nuw the given equation (07 — 3)^=25

may be written a:2 — 6.r+ (3)^=26,

or 07^-6^7=16.

Hence, by retracing our steps, we learn that the equation

.r2-6^=16

can be solved by first adding (3)^ or 9 to each side, and then
extracting the square root ; and the reason why we add 9 to
each side is that this quantity added to the left side makes it a
perfect square.

Now whatever the quantity a may be,

^ + 2aj7 + a- = (or + a)',

and ^-2aj7+a^ = (^-a)^;

so that if a trinomial is a perfect square, and its highest poller,
x\ has unity for its coefficient ^ we must always have the term
without X equal to the square of half the coefficient of 07. If,
therefore, the terms in x^ and x are given, the square may be
completed by adding the square of half the coefficient of x.

Example, Solve 07^+ 14a? = 32.

The square of half 14 is (7)'.

.-. 272+ 14a7 + (7)2 = 32+49;
that is, (a: + 7)- = 81;

.-. 07 + 7 = ±9;

/. 37= -7 + 9, or -7-9;
.-. 07 = 2, or - 16.

189. When an expression is a perfect square, the square terms
are always positive. Hencc, before completing the square th.e
coefficient of z^ should be mado equal to + 1.

^^^ ALGEBRA. [chap.

Example 1 . Solve 7a: « a:^ - 8.

Transpose »o an to Jiai^e the terms involving x on one side, and the
square term j^ositive.

Thus ar^-1x = 8.

Completing the square, a:* - 7a;+ r ] = 8 + ^^ ;

thatis. (ar-7y = ?^

2 -2'

• ^ 7.9.

• • a: = - + - :

2 2
.'. a; = 8, or - 1.

Example 2. Solve 4- ^ _ 33:^ + 5

3a:+l 3a: + l

Clearing of fractions, 12a: + 4-8 = 3ar* + 5;
bringing the teims involving x to one side, we obtai:)

3x2-12a; = -9.
Divide throughout by 3 ; then

a:2-4a: = -3;

.-. ar»-4a: + (-:)- = 4-3;
that is, (a; - 2)2 = 1 ;

.*. a:-2 = ±l;

.*. a: = 3, or 1,

EXAMPLES XXV. a.

Solve the equations :

1. 7{x^-7) = 6x\ 2. (a:+8)(a:-8) = 17. 3. (7 + a;)(7-a:) =24.

7. a:2 + 2a; = 8. 8. a;2 + 6a: = 40. 9. 3^^ + 35 = 120;.

10. ar» + a; = G. 11. a:2-156 = a:. 12. lla; + 12 = ar^.

13. a:2+4a,=,32. 14^ g^^^^^^ j^g^ a:^ ^ 15a: - 34 = 0.

XXV.]

QUADRATIC EQUATIONS.

165

Solve the equations :
16. i(2a;2 + 7)-(6-x2)=|(ar^ + 3).

18.

20.

a?+3 a;-2_ 5

x + 2 x-3 (a? + 2)(a;-3)*

3a: + l

x-l

= x + 2.

■irj a;f5_a: + 37

^'' x-2 2x- r

^^- 2x4-2 "^^ ^•

190. We have shown that the square may readily be com-
pleted when the coefficient of jp^ is unity. All cases may ho
reduced to this by dividing the equation throughout by the
coefficient of x^.

Example 1 . Solve 32 - 3a;2 = 1 Ox.

Transposing, 3x^ + 10a: = 32.

Divide throughout by 3, so as to make the coefficient of aP unity.

Thus

x^-^% = ^-,
S"^ 3

(5V 32 *'5

10.
3

that is,

f . 5V 121
("■^3-) =-9 '

:. a: = -|±y=2,or-5i.

o.ll 12
0?-+ ^x = -—.
o 5

Example 2. Solve 5x^ -h 1 la; = 12.
Dividing by 5,

Completing thesquare, x'+l^x+ (^ )' = ^ + J^J J

that IS, i" + ru)=100'

10 -10
11 .19 4 „

166 ALGEBRA. [chap,

191. We see then that the following steps are required for
solving jin affected quadratic equation :

{I) If necessary y simplify the equation so that the terms in
X- and X are on one side of the equation, and the term vdthout x
on the other.

(2) Make the coefficient of x*^ unity and positive hy dividing
throughout hy the coefficient of'x^,

(3) Add to each side of the equation the square of half the
coefficient of x.

(4) Take the square root of each side,

(5) Solve the restdting simple equations,

192. When the coefficients are literal the same method may
be used.

Example. Solve 7(a:+2rt)2 + 3a« = 5a(72; + 23a).

Simplifying, 7ar« + 28aa: + 28a* + Sa* = ?,oax -\- 1 loa^ ;
that is, 7ar»-7aa; = S4a2,

or a:'-aa;=12a^

(a\* «2

-j =12a2 -I- ;

., , . / aV 49a«

that IS, \^~2/ ~~T~ '

. a,_a_.7a.
• ^ 2"-"2'
.'. X = 4a, or - 3a.

193. In all the instances considered hitherto the quadratic
equations have had two roots. Sometimes, however, there is
only one solution. Thus if a'2-2a' + 1=0, then (j7-l)- = 0,
whence x=^\ is the only solution. Nevertheless, in this and
similar cases we find it convenient to say that the quadratic has
tioo equal roots,

EXAMPLES XXV. b.

Solve the equations :

1, 30^2 + 20: = 21. 2. 5ar^ = 8x+21. 3. 6aP-x-l=0,

4, 3~lla; = 4a;2. 5. 2la:2 = 2a: + 3. 6. 10 + 23a:+12a:2 = 0.

7. 15ar»-6a: = 9. 8. 4ar» - 17a; = 16. 9. 8a;2 - 19x - 15 = 0.

XXT.] QUADRATIC EQUATIONS. 167

Solve the equations :

10. I0x^ + Sx=h 11. 12a~»+7a: = 12. 12. 20a;3 - a: - 1 = 0.

13. a^+2ax = l5aK 14. 2ar» - Sa^ = 15aa:. 15. Sa;^ = it(2jfe - 5a:).

16. nhx+^b^ = Sx\ 17. 9a:2 - 143c» = 6ca:. 18. 2a^x^ = ax+h
19. (a; - S){x - 2) = 2(a:» - 4). 20. 5(a: + 1 )(3a; + 5) = 3(3ar2 + 1 la: + 10).
21. 3a:2+13+(a:-l)(2a;+l) = 2a;(2a; + 3).
oo 7a:-^_3a: oo 2_a:-l „. 3a;-l_2a:-9

25.?^ = ?-6. 26. .^+i= ^^

a; + 5 3 ^* 7a;+l 2 2(3 + 2a;)

27. 3(2a; + 3)^ + 2(2a; + 3)(2 - a;) = (a; - 2)2.

28. (3a?-7)2-(2a;-3)2 = (a:-4)(3a:+l).

[For additional examples see Elementary Algebra."]

194. Solution by Formula. From the preceding examples
it appeara that after suitable reduction and transposition
every quadratic equation can be written in the form

aa^+bx+c=Of

where a, 6, c may have any numerical values whatever. If
therefore we can solve this quadratic we can solve any.

Transposing, aar'+ba;=-c;

dividing by a, a^+-a:= —-.

a a

Complete the square by adding to each side ( ^ ) > thus

a \2aJ 4a^ a*
that is, (x+^)^*!z^ ;

extracting the square root,

2a 2a '
" " 25

168 ALGEBRA. [chap.

195. In the result a:= '"^- '^<^'"^>,

2a

it must be remembered that the expression JQi^ — Aac) is the
square root of the compound quantity 6^ -4ac, taken as a tohole.
We cannot simplify the solution unless we know the numerical
vahies of or, 6, c. It may sometimes happen that these values
do not make V- — 4ac a perfect square. In such a case the exact
numerical solution of the equation cannot be determined.

Example, Solve 5a^- 13a: - 11 = 0.

Here a = 5, 6 = - 13, c = - 1 1 ; therefore by the formula we have

^,_ (-13)±\/(-13)^-4.5(-li)

2.5

_ 13±\/l69 + 2^j
10

istv^Jso

""~10

Since 389 has not an exact square root this result cannot be
simplified ; thus the two roots are

13+^389 13-^389
10 ' 10 '

196. Solntion by Factors. Tliere is still one method of
obtaining the solution of a quadratic which will sometimes be
found shorter than either of the methods already given.

Consider the equation a^+-a:=2.

o

Clearing of fractions, 3jr^ + *7x — 6=0 (1);

by resolving the left-hand side into factors we have

(3:F-2)(.r + 3)=0.

Now if either of the factors 3.r - 2, :r+3 be zero, their product is
zero. Hence the quadratic equation is satisfied by either of the
suppositions

3jr-2 = 0, ora: + 3=0.

9

Thus the roots are -, — 3.

3'

It appears from this that w/icii a quadratic equation has been
simplified and brought to the form of equation (1), its solution
can always be readily obtained if the expression on the left-hand

xxv.] QUADRATIC KQUATIONS. 1G9

side can be resolved into factors. Each of these factors equated
to zero gives a simple equation, and a corresponding root of tlie
quadratic.

Example 1. Solve 2a^ -ax + 2bx = ah.

Transposing, so cbs to have all the terms on one side of the eqvxjUion^
we have

Now 2x^ - aa: + 26a; -ah = x{2.x - a) + 6(2a: - a)

= {1x-a){x + h).
Therefore (2a? -a){x + h) = 0\

whence 2a:-a = 0, or a: + 6 = 0,

.-. a; = -, or - 6.

Example 2. Solve 2{x^ - 6) = 3(a; - 4).

We have 2*2-12= 3x-12;
that is, 2x2 = 3a; (jj^

Transposing, 2x^ - 3a; = 0,

a;(2a?-3)=0.

.-. a; = 0, or2a?-3 = 0.
3
Thus the roots are 0, -.

Note. Tn equation ( 1 ) above we might have divided both sides by x
and obtained the simple equation 2a; = 3, whence a; = 1, which is one

of the solutions of the given equation. But the student must be
particularly careful to notice that whenever an x is removed by
division from every term of an equation it must not be neglected,
since the equation is satisfied by a; = 0, which is therefore one of
the roots.

197. Formation of Equations with given roots. It is

now easy to form an equation whose roots are known.

Example 1. Form the equation whose roots are 4 and —3.

Here a; = 4, or a; = - 3 ;

/. a;-4 = 0, or a; + 3 = 0;
both of these statements are included in

(a;-4)(a; + 3) = 0,
or a2-a;-12 = 0,

which is the required equation.

170 ALGEBRA. [chap.

Eocample 2. Form the equation whose roots are a and - - .

Here

b
x = ay or ^ = -^'f

the equation is

(a:-a)(a: + |)=0;

that is.

ix-a){Sx-hh)=0,

or

3ar'-3aa; + 6x-a6 = 0.

EXAMPLES ZXV. c.

Solve by formula the equations :

1. a:3 + 2a:-3 = 0. 2. a:*-2a;-l=0. 3. ar»^3a: = 5.

4. 3x2-2a;=l. 5. 2a:--9a; = 4. 6. 3ar» + 7a: = 6.

7. 4ar»-14 = 3x. 8. 6ar» - 3 - 7a: = 0. 9. 12ar^+ 10 = 23a:.

Solve by resolution into factors :

10. a;2-9a: = 90. 11. a;^- 11a: = 152. 12, a:2_85=i2a:.

13. 2a;2-3a: = 2, 14. 3a:2 + 5a; + 2 = 0. ^5^ 4ar«-14 = a:.

16. 6af2-lla: + 2 = 0. 17. stP-a^^O. 18. ar»-7aa: = 8a«

19. 12a:2- 236a: +1062 = 0. 20. 3aar» + 26a: = 7a:.

21. 24a:2 + 22ca: = 21c2. 22. ar^ - 2a: + 46 = 26a;.

Solve the equations :

23. 2a:(a: + 9) = (a:+l)(5-a;).

24. (2a;-l)2-ll = 5a: + (a:-3)2.

25. 6(a:-2)2+13(l-a:)(a:-2) + 6ar« = 6(2a:-l).

26. -4---^ = i- 27.

x-6 a:-5 2a:-l a;+l x

28. .1^_? = -A_. 29. "^ + 3_^^ 4(a:-6) _3
*a;-4a;a:-5* 'a: -3 a: + 6

30. 3a: = -i- + 2. 31. "" ^ ""^'^

x-\-l *""• 3 a; 3(a: + 4)

32- ^AH^^'^- 33. ^5-5=6.

XXV.] QUADRATIC EQUATIONS. 171

Solve the equations :

86. ^l+#z8=i7 27. i_,.^+ 3 _l

a:+"8 3ar-l 12" 2(x-i) ip-l 4"

Sa ^<£z^-^ = q^Lr£). 39. (p-3)x + ?/ = (i> + ?).

a — o - a a:

40. -^+ « =2. 41. («-!)» = (*- 'Vx.

[For additional examples see Elementary Algebra.']

198. SimnltaneouB Quadratic Equations. If from either

of two equations which involve a: and v the value of one of the
unknowns can be expressed in terms of the other, then by sub-
stitution in the second equation we obtain a quadratic which
may be solved by any one of the meUiods explained in this
chapter.

Example. Solve the simultaneous equations

5x+7y=l, 4x2 + 3a:y-2y2 = l0.

From the first equation, x = ~ ^^ , and therefore by substitution

5

in the second equation, we have

4(l-7y)'' ^ 3y(l-7y) _ gys = lo ;

ifid O

whence 4 - 5Cy + 196y2 + 15y - I05y^ - 50y^ = 260 ;
that is, 4l2^ - 41y - 246 =

y2-y-6 =

/. (y-3)(y + 2) =
.-. y = 3, or -2.

From the first equation, we see that if y=3, then x=—4, and if
y=— 2, then x=3.

Homogeneous Equations of the Same Degree.

199. The most convenient method of solution is to substitute
y=mx in each of the given equations. 15y division we eliminate
X and obtain a quadratic to determine the values of m.

172 ALGEBRA. [chap. xxv.

Examjile. Solve the simultaneons equations

5ic2 + 3y= = 32, x^-xy^2y^=\iS.

Put y = mx and substitute in each equation. Thus

ar«(5 + 3m2) = 32 (1),

and ar5(l-»i + 2m2) = 16 (2).

By division, _5 + 3m3 32 ^ ^ ;

that is, wi»-2m-3=0;

.*. (m-3)(m + l) = 0;
.'. m = 3, or - 1.

(1) Take m = 3 and substitute in either (1) or (2).

From (1), 32a:2 = 32 ; whence x = ± 1.

/. y = mx = Sx=± 3.

(2) Take m = - 1 and substitute in (1). Thus

Ss^ = 32 ; whence x=±2.
.*. y = mx = - a: = + 2.

EXAMPLES XXV. d.

Solve the simultani'ous equations :

1. a; + 3y = 9, 2. 3a;-4y = 2, 3. 2x + y = 5,

5a^ -xy = 2,

6. 2x-5y =1,
a;--8y2=l.

9. «:-f=3,
xy-y^ = 4:.
12. 2a;-?=l,

y

3y-? = 8.

X

15. 2a;2 + 5y2 = 143,
Sa:y + 3y'-=195.
+ 3a:y-y2 = 37i,
5a;2 + 3a;y + 52r = 265.

a^ = 6.

xy = 2.

4.

ar'+4y» = !

3,
29.

5.

3x +y =9,
3ary-y2 = 9.

7.

a:y = 24.

8.

?-A=i,

10a;y=l.

10.

-2-y-''

3+y-i

11.

? + y = 3.
2^3 '

8 3^j
a? y

13.

3a;2 + 7y' =
2^2 + 7a:y =

= 55,
= 60.

14.

16a:y-3a;2:r
7ary + 3y2 =

77,
110.

16.

X' + 2xy +
3jc2-9a:y-

2y- =

17,
119.

17.

21
5a;

CHAPTER XXVI.
Problems LEADING to Quadratic Equations.

200. We shall now discuss some problems which give rise
to quadratic equations.

Example 1. A train travels 300 miles at a uniform rate ; if the
speeil had been 5 miles an hour more, the journey would have taken
two hours less : find the rate of the train.

Suppose the train travels at the rate of x miles per hour, then the

time occupied is ^ — - hours.

X

On the other supposition the time is hours ;

UJ •!" o

. 300 300 2-

a; + 5 X

v.^hence :x^ + ^x- 750 = 0,

or (a; + 30)(«-25) = 0,

.*. a: = 25, or -30.

Hence the train travels 25 miles per hour, the negative value
being inadmissible.

[For an explanation of the meaning of the negative value see
Elemental^ Algebra.']

Example 2. A man buys a number of articles for 1^2.40, and sells
for ^2.52 all but two at 2 cents apiece more than they cost; how
many did he buy ?

Let X be the number of articles bought ; then the cost price of
each is — cents, and the sale price is -^ cents.

X 05-2

252 240 o
.'. =2;

that is, 221. 120^1

x—2 X

174 ALGEBRA. [chap.

After simplification, 6ar + 240 = a:^ - 2x,

or ar=-8a:-240 = 0;

that is, (a:-20)(x+12) = 0;

.'. a; = 20, or - 12.

Thus the number required is 20.

Example 3. A cistern can be filled by two pipes in 33^ minutes ;
if the larger pipe takes 15 minutes less than the smaller to fill the
cistern, find in what time it will be filled by each pipe singly.

Suppose that the two pipes running singly would fill the cistern

in X and a; - 15 minutes ; then they will fill - and of the cistern

X a; — 15

respectively in one minute, and therefore when running together they

will fill ( + — ) of the cistern in one minute.

\x a:- 15/

1 3

But they fill -— , or - -- of the cistern in one minute.

Hence 1+_1_ = J-,

X a -15 100

100(2a; - 15) = 3ar(a: - 15),

3a:»- 245a: +1500=0,

(x-75)(3a:-20) = 0;

X = 76, or 68.

Thus the smaller pipe takes 75 minutes, the larger 60 minutes.

The other solution 6§ is inadmissible.

201. Sometimes it will be found convenient to use more than
one unknown.

Example. Nine times the side of one square exceeds the Peri-
meter of a second square by one foot, and six times the area of the
second square exceeds twenty-nine times the area of the first by one
square foot : find the length of a side of each square.

Let X feet and y feet represent the sides of the two squares ; then
the perimeter of the second square is 4i/ fcct ; thus

9x - 4i/ = 1.

The areas of the two squares arc x^ and ^ square feet ; thus

6y«-29a:2 = l,

XXVI.] PROBLEMS LEADING TO QUADRATIC EQUATIONS. 175

9a: — 1
From the first equation, y = — — — .

By substitution in the second equation,

that is, Ila:2_54a;-5 = o,

or (a;~6)(Il2;+l) = 0;

whence a; = 5, the negative value being inadmissible.

Also, y = 5^ = 11.

4

Thus the lengths are 5 ft. and 11 ft.

EXAMPLES XXVI.

1. Find a number which is less than its square by 72.

2. Divide IG into two parts such that the sum of their squares
is 130.

3. Find two numbers diflfering by 5 such that the sum of their
squares is equal to 233.

4. Find a number which when increased by 13 is 68 times the
reciprocal of the number.

5. Find two numbers differing by 7 such that their product
is 330.

6. The breadth of a rectangle is five yards shorter than the
length, and the area is 374 square yards : find the sides.

7. One side of a rectangle is 7 yards longer than the other, and
its diagonal is 13 yards : find the area.

8. Find two consecutive numbers the difference of whose reci-
procals is if^.

9. Find two consecutive even numbers the difference of whose
reciprocals is x^.

10. The difference of the reciprocals of two consecutive odd
numbers is ^f ^ : find them.

11. A farmer bought a certain number of sheep for $315 ;
through disease he lost 10, but by selling the remainder at 75 cents
each more than he gave for them, he gained f 75 : how many did
he buy ?

12. By walking three-quartersof a mile more than his ordinary
pace per hour, a man finds that he takes 1 \ hours less than usual to
W£dk 29| miles : what is the ordinary rate ?

176 ALC

13. A cistern can be filled I
ntes less tlian by the smaller. '
cistern Is tilled in H minutes:
could be filled by eacb of tbe pi

14. A man buys a dozen eg
been a cent per dozen cheaper I
twelve cenu : what is the price

15. Tlie large wheel of a ca
ference than the small wheel,
mile : find the circumference o<

16. A boy was sent out to I
at« two, and Ills master bad in
cent per dozen more than tbe q
the boy buy ?

17. A lawn 46 feet long an
width round it ; if the area of

18. By Belling one more ap]
a woman finds that she gets a c
she nqw get per dozen ?

19. Four times tlie side of <
of a second sqaara by 12 feet, t
is IcBS than Ave tlmea the area i
the length of a aide of each aqua

20. Find a number of two di
product of its digits thn quotie
the number the order of the dig.

21. A p.r.on b.,. »m. 61
been t5 less he would have reci
bia money i at what price did b

22. Tlie area of each of twi
length of one is S feet more thai
of their breadths la 3 feet : find

S3. There are three numhen
the lirst by 6 and less than tbi
tlirce is 2S0 timea the greatest, 1

24, Find four consecutive ii
two greatest is represented by a

25. Two trains A anil B sta
P and which are 200 milefl a
B reocliea /* in 43 hours after tli

178 ALGEBRA.

13. Ada together a-{b + c-{a + b)} + c, 2(3a + 2ft) - 4(6 + 2a) - c,
and 3(26 - a) - 2(36 - a) + e.

14. Find what value of x will make the product of a; +3 and
2a; +3 exceed the product of x+ 1 and 2a; + 1 by 14.

15. Divide 6» + 8 - 125c3 + 306c by 6 - 6c + 2.

16. Simplify

13a6V 87c«cP^266cd.

2ycV 2a^b* ' 4ab^ '

(•)^^-

,ox 2aa; _ / o'xc' o'6a; o*6 V \
^ ' 6"^ \2a6V lia=*6^ 2a36W'

17. How old will a man be in m years who n years ago was
p times as old as his son then aged x years ?

18. I boaght a certain number of pears at three for a cent,
and two-thirds of that number at four for a cent ; by selling them
at twenty-iive for 12 cents, I gained 18 cents. How many pears did
Ibuy ?

19. Solve the equations :

17=3x^4_x±2^g__g 7x_tl4
^ ^ 6 3 3 •

(2) g+y. ^x-Sy ^g ^+1. = 1.
^^24 ' 14 18

20. Divide aW + (2ac - 6^) x* + c^ by ax^-^c- hx^.

21. If a hor&es are worth 6 cows, and c cows are worth d sheep,
find the value of a horse when a sheep is worth $2.

22. I^^ind the highest common factor of 3a*^6'c, V2a^bc^y ISa^ft* ;
and the lowest common multiple of 4a6"^c', 12a'6, ISac*-*.

Also find the value of -^ - JL + ^ _ «'«*

2a62c 362 52^ Ga'^b^^

23. A gentleman divided ^49 amongst 160 children. Each girl
had 50 cents, and each boy 25 cents. How many boys were there ?

24. If r=5a + 46-0c, X=r-3a-96 + 7c, r=20a+76-5c
^=13a-66+0c, calculate the value of V-{X-\-Y) + Z.

MISCELLANEOUS EXAMPLES V. 179

25. Solve the equations :

.jv 3(6-5a;) ^63a;_3a;_

36

5 ~50 ~2 125*

(2) l(a:+y) = a:+l, ^{y - x) = 2x - 1.

26. Find the factors of

(1) a2-a-182; (2) Sx^+Ux-Q.

27. If a? = 4, y = 5, and z = 3, find the vahie of

\/{5(jr* - 2®) -x^}+ VMxix" - z2) - 1}.

28. The product of two expressions is {x + 2//)^ + (3a; + z)^, and one
of them is 4a; + 2y + z ; find the other.

29. When A and B sit down to play, B has two-thirds as much
money as A ; after a time A wins $15, and then he has twice as
much money as B. How much had each at first ?

30. Find the square root of 16a« + 4a + 4-16a3 + a2- 8a*.

31, Find the value of

24{.-l(»-3)}{.-2(. + 2)}{.-?(x-|)}.

and subtract the result from (a; + 2)(a;-3)(a: + 4).

32. Find the square root of

a^ 2ar» lla:^ 9

4"X""T6~"*" "^ie*

33, If 2a = 36 = c = 4rf = 1, find the value of

a 3c 9a-

34. Separate into their simplest factors :

(1) a^~xy-Qi/; (2) x^-4xy^-xh/ + 4y^.

. 35. Solve the equations :

(1) (a;-l)(a:-2)(a;-6) = (x-3)^

36. A farmer sells to one person 9 horses and 7 cows for $375,
and to another 6 horses and 13 cows at the same prices and for the
same sum : what was the price of each ?

180 ALGEBRA.

37. When tt = 3, 6 = 2, c =~ 7, find the value of

a a-c b + 2c
(2) 4c + {c-(3c-26)+26}.

38. Solve the equations :

^ ' x^2x Sx 3'

(2) 6x + 3//=120, 10a; = Cy + 90.

39. Find the highest common factor of 5a:2+2a::^-15a:-6 and
Tar* -4x'2-21x + 12.

40. A coach travels between two places in 5 hours ; if its speed
were increased by 3 miles an hour, it would take 3J hours for the
journey : what is the distance between the places ?

41. From a:(a; + o-6)(a:-a + 6) take {x-a){x-b){x + a + b),

42. Find the value of
2a^* + 3c:-3^3ar^-10a: + 3 . 6ar2-5a:+l

Qc^-Qx V + 3a; + 2 ' 3ar* + 7ar* + 2a;'

43. Divide a^-^i^ + Sxy-l by x + y-l, and extract the square

root of a.-* - ar» + -llar» + 2a; + t

12 9

44. A man can walk from ^ to i? and back in a certain time at
the rate of 4 miles an hour. If he walks at the rate of 3 miles an
hour from A to 7i, and at the rate of 5 miles an hour from B to A,
he requires 10 minutes longer for the double journey. What is the
distance from ^ to ^ ?

45. Find the highest common factor of

1x* - lOax^ + 3a-x^ - ia'^x + 4a*, 8.r* - 1.3aar» + 5aV - Sa'a: + 3a*.

46. Solve the equations :

^'^ '=-(^-'-^')=i<2^-=''-3=

(2) x-2y-\-z-=0, dx-Sy + 3z = 0, 2a; + Sy + 5s = 36.

47. Find the lowest common multiple of

6a:2-a:-l, Sx- + 'Jx + 2, 2ar» + 3a:-2.

MISCELLANEOUS EXAMPLES V. 181

48. I'he expression ax + Zh is equal to 30 when x is 3, and to 42
when a; is 7 : what is its value when a; is 1 ; and for what value of x
is it equal to zero ?

49. Find the lowest common multiple of

^a^ + ab), 12{a62 _ fP)^ is{a^ - 52),

50. Extract the square root of

43^ 12a; ^ or 24a . 16a»

+ 25- +-T-.

a^ a X a^

51. Reduce to lowest terms

l2x* + 4x^-2Sx^-^x-9
8a;*-14a;'^-9

52. Solve the equations :

(1) x-?^=y-4£r5y, y-?^ = 2(:.-l)-3^g-7;

(2) 2a:-y + 3z=l, 4a;+3y-2z=13, 6a:-4y + s = 20.

53. Simplify

>l» a + h _ 2a arb-a^ _

(2)

6 a + b d^b- b^
1 I

a:^ + 8a:+15 a;'+lla; + 30

54, The sum of the two digits of a number is 9 ; if the digits are
reversed the new number is four-sevenths of what it was before.
Find the number.

55. Solve the equations :

(1) 4x-l{5y-i)=ly 3y-2a; 1 1

(2) 3a; + 4y-ll=0, 5y-6z= -8, 7z-8a;-13 = C.

56. Find the value of

2 1 _ 3a; _ a

a + x a-x x'^-d^ (a + a;)^'

57. Resolve into factors :

(1) afi-2a^+x^i (2) a« + a*-n3-l.

182 ALGEBRA.

58. T^y^o persons started at the same time to go from A to B.
One rocle at the rate of 7i miles per hour ami arrived half an hour
later than the other who travelled by train at the rate of 30 miles
per hour. What is the distance between A and B ?

59. Find the square root of

4qi^_x_ 16a:* 9y» 6gy 16a:*
9y* 2 15yz l&a 5z« 26z«'

60. Find the factor of highest dimensions which will exactly
divide each of the expressions

2c*+c»d-c»d»-7cd»-4£i*, Sc^+c'd- 2c-d»-9cflP-W*.

p+2_ p _p*-2p*.

61. Simplify (1) '^-_^^--Z£.^;

62. Find the highest common factor of

6a:*-2a:»+9a:* + 9a:-4 and 9a?*+80a:2-9.
What value of x will make both these expressions vanish ?

63. Solve the equations :

(1) ^+^=2;

(2) -2^+3^1-5^15^0.

x-i x+2 X'2

64 Simplify 6^-5^-63^ _ ISar'+Sa?/- 12//*
u». « F / 14ar»-23ary + 3y* 35ar^ + 47a:y+6jr''

66. Find the value of

x-2a a: + 2a 16a6 ,^.j^^^ ^ ^ 4aft

a: + 26 a:-2Z> 46a-ar»' a + h

66. An egg-dealer bought a certain number of eggs at 16 cents
per score, and five times the number at 75 cents per hundred ; he
sold the whole at 10 cents per dozen, gaining $3.24 by the transac-
tion. How many eggs did he buy ?

MISCELLANEOUS EXAMPLES V. 183

67. If a = -l, 6 = -2, e = -3, </ = -4, find the value of

2a^ + al^ - Sahc _ a - h + c-d _a 2c
a^-b^-abc bc-2ad b ad'

68* Solve the simultaneous equations :

ar-2 _ a: + y _a?-y-l_ y+12
2 "IT" 8 ~4~*

a; + 7 .y-5_, 5(y+l)

"3-^"ir"^ "" ~7~'

69. Simplify the fractions :

(1) x-y^x-z_ {y-z? .

x-z x-y (z- a;)(y - x) *

af_&^ 1-1

y o'^ b a

\6 a)\b a J a^ h'^ ah

70. From a certain sum of money one-third part was taken and
$50 put in its stead. From the sum thus increased one-fourth part
was taken and $70 put in its stead. If the amount was now $120,
find the original sum.

71. Find the lowest common multiple of

{a*-a^c^)\ 4a«-8aV + 4aV, a* + 3a»c + 3aV + oc».

72. Solve the equations :

•135a; --225 "36 OOaj-lS

(1) •15x +

•03 "2 -y

(2) 10a? + 4 7-2ar » _ 11-53: 4a; -3^
^ ' 21 14(a;-f)'" 16 6 *

73. Simplify

iK^-Ax-2\ ^ a:^+ 6a;'-24 7a;^ a;g-20a;+9 1
a;+17 ^-x-\2 x^

74. What must be the value of a; in order that

(o + 2a;)«
a^ + lOax + Zx^

may be equal to 11 when a is equal to 67 ?

184 ALGEBRA.

75. Find the highest common factor of 163P* + 36a:' + 81 and
8ar* + 27 ; and find the lowest common multiple of 8a;' +27,
lCir*4-36x2 + 81, and GaP-bx-e.

76. Solve the equations :

(1) a{x-a)-h{x-h) = {a + h){x-a'li),

(2) (a + h)x " aij = a\ (a^ + h'^)x - ahy = a\

77. A farmer bought a certain number of sheep for $30. He
sold all but five of them for $27, and made a profit of 20 per cent,
on those he sold : find how many he bought.

78. Find the value of

x-y _x + y 2a - 36 3ft

l\\ a^ + y ^-y . /o\ 2a- G6 2a
^ ' x'-y^TT + y^' ^'^ 2a -.36 36
3^ + y^ x^-i^ 2a 2a-66

79. Find the H.C. F. of 2 lar» - 26ar» + 8a? and QaV - a^x^ - 2a^x,
Also the L.C.M. of x^-x, aa:2^2aa:-3a, ar» - 7ar» + 6a;.

80. Simplify the expression

(a + 6 + c)(a-6 + c)-{(a + c)«-6»-(a2 + 62 + c2)}.

81. Solve the equations :

(1) 7+£_2£^=3y-5. ^+*-^=lS-Bx;

(2) * <5 ^

Sx-2 x+l 22+3

82. Extract the square root of

y* X* \y^ x^J \y^ st^J \y xl

83. Find the value of
4a + 66^6a-46 4a2+ 662 ^46^-602 2O6*

+ ;— - -. r:;-+ .. . » +

a + 6 a-6 a^-62 a^ + V^ a*-b^

84. A bacj contains 180 gold and silver coins of the value alto-
gether of $144. l^acli gold coin is worth as many cents as there are
silver coins, and each silver coin as many cents as there are gold
coins. How many coins are there of each kind ?

MISCELLANEOUS EXAMPLES V. 185

85. Solve the equations :

(1) ^^ + » - 13 >
^ ar+lO x + 4: a:+7'

(2) 'j2x+6-'s/x^=2,

86. Find the factors of

(1) 20a2+21a6-276»; (2) a:»-3ar»-9aj+27.

87. If the length of a field were diminished and its breadth
increased by 12 yards, it would be square. If its length were
increased and its breadth diminished by 12 yards, its area would be
15049 square yards. Find the area of the field.

88. Simplify the expression

{{a + b)(a + h +^ + c ^}{(a + 6)»-c^}
{{a + bf-c^}{a + h + c}

89. Find the square root of

{2x 4- 1 ){2a: + S){2x + 6)(2ar + 7) + 1ft.

90. Simplify

IQqg 12

(l + o2)(l-4a2)"l-2a T+^

91. Solve the equations :

a: + 6 3 x

(1)

4' y-2 2'

(2) l»2a:-'^^^"'^^='4a?+8'9.
'6

92. Resolve into factors :

(1) 6x2 + 5a;-6. (2) 9ar*-82aV+9y*-

93. Eeduce ^:i^^.t^'":^'
to its lowest terms.

94. Simplify the fraction

(a + 6)' _ a-\-2h-\-x (a-\-h)x 1

(a?-a)(a; + a + 6) 2(x-a) a^ + bx-a'-ab 2'

186 ALGEBRA.

95* A person being asked his age replied, "Ten years ago I was
five times as old as my son, bat twenty years hence he will be half
my age." What is his age ?

96, Find the value of

f 2a ^ 1 _ 1 U

\la-x){a + x) x-a x + aj ^ ^ . a'

97. When o = 4, 6 =-2, c = ?, d = - 1, find the value of

98. Find the square root of

o4+54_a»fe-a6'+?^

4

99t Solve the equations :

(1) j£^- 2»_ 2; (2) 3a:»+22a; = 493.
x — \ 2a; — 1

100. SimpUfy =^(^ + 3y)-3(x-fy) + 2y' ^

2(a;+y-l)-(l + a;)

101. What value of x will make the sum of ^"^^ and -^-■*"— ?
• 2(o-6) 3(a + 6)

equul to 2 ?

102. A man drives to a certain place at the rate of 8 miles an hour ;
returning by a road 3 miles longer at the rate of 9 miles an hour he
takes 7i minutes longer than in going : how long ia each road ?

103. Find the product of

(1) 3a:2-4a:y + 7y', 3a;' + 4a:y + 7y* ;

(2) x^-2y\ x^'-2xy+2y\ ar» + 2y2, a?+2xy+2y^.

104. Extract the square root of

2r 16 2

MISCELLANEOUS EXAMPLES V. 187

105. Find the highest factor common to
a?(6a;3-8y2)-y(3ic'-4y2) and 2xy{2y-x) + 4sfi-2y^,

106. ^he sum of the digits of a number is 9, and if five times the
digit in the tens' place be added to twice the digit in the units'
place, the number will be inverted. What is the number ?

107. If («+-)' = 3, prove that o^ + i^ = 0.

108. Solve the equations :

(1) (a; + 7)(y-3) + 2-(y+3)(a;-l) = 5a;-lly + 35 = 0;

(2) 1-7^"^ - ^^'^

3 6J-3a: 3(a;-2)

109, IfaJ = 6 + c, y = c~-ay z = a-b, find the value of

a^+y^+z^-2xy-2xz+2yz.

110, Express in the simplest form

2a:2 + 3a;+l 6ar' + 5a;+l 12a:''' + 7a; + 1 20ar»+9a; + l

111. Resolve into factors :

(1) 4a26a-(a2 + 62-c2)2;

(2) db{m^ + l) + m{a^ + h^).

112. Simplify the fractions :

x + = a^ -ax- ^ '

l+|±i 1-^

3-a: X

113. Solve the equations :

li\ 75-a? 80a?+21 , 23 g.

^ ' ,3(a;+l) 6(:^a; + 2) «+! '

(2) \/« + 12 - ^/« = 6,

188 ALGEBRA.

114. Ten minutes after the departure of an express train a slow
train is started, travelling on an average 20 miles less per hour,
which reaches a station 250 miles distant 3.^ hours after the arrival
of the express. Find the rate at which each train travels.

115. Simplify

a»-&4o ^r 2a« + 5a-f2 _^ r a» - 4a ^ o^-f 7a- 8 \"|
a«-4 'L2a^+9a+^' Xa^+^a' a^ + a-2Jj'

116. Bcsolve into four factors

4(a6 + cd)2 - (a» + fc^ _ c« - cP)«.

117. When o = 4, 6 =-2, c=^, d=-l, find the numerical

value

M\c^ - a(a - 26 - d) - V6*c + 1 Wd?.

118. Find the value of

(1) a- i- ; (2) 1

6 + '^^r- 2ar-3+ ^

a+ — . 1-

a-o a:-6

119. Solve the equations :

(1) 150ar»=299a:+2; (2) aa;+6y = ay-6a: = a^+ja,

120. -^ h&8 Id miles to walk. At the end of a quarter of an
hour he is overtaken by B who walks half a mile per hour faster ;
by walkins at the same rate as B for the remainder of the journey
he arrives naif an hour sooner than he expected. Find how long the
journey occupied each man.

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" In this work the subject is carried somewhat farther than is customarv in
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able aids to the right understanding of the more transcendental methods."

Introductory to the Above.

ELEMENTARY SYNTHETIC GEOMETRY

OF THB

Point, Line, and Circle in the Plane.

16mo. $1.10.

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AMERICAN EDITION OF

Charles Smith's Elementary Algebra.

FOR THE USE OF

PREPARATORY SCHOOLS, HIGH SCHOOLS,
ACADEMIES, SEMINARIES, Etc.

BY

JRVING STRINGHAM, PH.D.,

Profbssok op Mathkmatics, and Dean op ths Collbgb Faculties
IN THE University op Calipornia!

BRIEFER EDITION (408 iMges) $1.10

Tbis edition is tbe same as Chapters l.-XXVI. of the
COMPLETE EDITION (584 paffes) $1.20

'< I have always liked Charles Smith's Alge-
bra, and the new edttion contains a good many
improvements, and seems to me an excellent
work. The use of the book in schools prepar-
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consulted me."

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A TREATISE ON ALGEBRA.

By CHARLES SMITH, M.A.

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HIGHER ALGEBRA FOR SCHOOLS.

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ELEMENTARY TRIGONOMETRY

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WORKS BY REV. J. B. LOCK.

TRIGONOMETRY FOR BEGINNERS.

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1 6mo. 75 cents.

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ELEMENTARY TRIGONOMETRY.

6th edition. (In this edition the chapter on Logarithms has been carefully revised.)

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