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MATHEMATICS
L'BEARY
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UNIVERSITY OF ILLINOIS LIBRARY
NOV 22
ALGEBRA FOR BEGINNERS.
ALGEBRA
“OR BEGINNERS
BY ae | eke
ay vv at
‘abe 's! “HALL anv st Rs KNIGHT,
’ Autuors or “ELEMENTARY ALGEBRA FOR Scnootns,”’ ‘* Hiener
ALGEBRA,”’ ‘‘ ELEMENTARY TRIGONOMETRY,’ Ero. Ero.
REVISED AND ADAPTED TO AMERICAN SCHOOLS
BY
FRANK L. SEVENOAK, A.M., M.D.,.
PROFESSOR OF MATHEMATICS AND ASSISTANT PRINCIPAL IN
THE STEVENS SCHOOL, ACADEMIC DEPARTMENT OF
THE STEVENS INSTITUTE OF TECILNOLOGY
New Work
MACMILLAN & CO.
AND LONDON
1895
All rights reserved
CopyricuT, 1895,
By MACMILLAN & CO.
Norwood Wpress :
J. S. Cushing & Co. — Berwick & Smith.
Norwood, Mass., U.S.A.
_ f")
= fae fl
ated Ba ad Bs SS
H\4a JATNEMATICS LIBRARY
PREFACE.
Tur 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
tselass 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 :
Y “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
~“eproblems; and secondly, the strong opinion that all
%sreference to compound expressions and their resolution
nto 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
va. 351058
thang
& Mid =
‘
oy &
-
v1 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 found 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.
JUNE, 1895.
= : arn
'
OHAP. PAGE
I. DEFINITIONS. SUBSTITUTIONS . : : : : ; o
pis NEGATIVE QUANTITIES. ADDITION OF Likk TERMS : i
yi II. Simphe Brackets.» ADDITION. .. + + DS ee ce LL
" IV. SUBTRACTION . ; “. . ‘ . " ° 2 16
MISCELLANEOUS EXAMPLES I. . . . . ° ; 19
V. MULTIPLICATION ; ; ; - P c : : 21
wae, Vi, DIVISION, . : - ‘ a : : ‘ ° ‘ 31
« ¥
fp __NI1...REMOVAL AND INSERTION OF BRACKETS . F $ A 38
Seo MISCELLANEOUS EXAMPLES II, . : : : r 42
VIIl. BReEvIsION oF ELEMENTARY RULES . : : ‘ 3 44
IX. SiMPLE EQUATIONS . ; Se ‘ : : - 52
X. SyYMBOLICAL EXPRESSION . ‘ . : : : ‘ 59
XI. Propiems LEADING TO SIMPLE EQUATIONS . F A 64
>.
XII. Hicuest Common Factor, Lowest Common MULTIPLE
or Simple EXPRESSIONS. FRACTIONS INVOLVING
SIMPLE EXPRESSIONS ONLY. : : ; . : (fs
XIII. SimuULTANEOUS EQUATIONS : ‘ : P H ; 17
XIV. PROBLEMS LEADING TO SIMULTANEOUS EQuaTIons : 85
XV. INVOLUTION ; : ‘ ; : , ‘ ; 89
XVI. EvoOLUTION ; ; : : 5 : * : : 93
ry
XVII. RESOLUTION INTO FACTORS. CONVERSE USE OF FACTORS 101
MISCELLANEOUS EXAMPLeEs III. ‘ ; ; . Peg BE
XXV,
XXXVI.
CONTENTS.
PAGE
HIGHEST COMMON FACTOR OF COMPOUND EXPRESSIONS 116
MULTIPLICATION AND DIVISION OF FRACTIONS ; ; 122
LOWEST COMMON MULTIPLE OF CoMPOUND EXPRESSIONS 126
. ADDITION AND SUBTRACTION OF FRACTIONS . : fF Wea)
MISCELLANEOUS FRACTIONS . .. : ; : . 139
HARDER EQUATIONS . : 5 ; ; : ; . 146
HARDER PROBLEMS . ; ; ° . ° : . 153
Misce.taNrous Exampims IV. ee oe
QUADRATIC EQUATIONS . : ‘ ‘ ° ne LG2
PROBLEMS LEADING TO QUADRATIC EQUATIONS. Lis
MISCELLANEOUS EXAMPLES V, : ‘ 4 : alee
ALGEBRA.
CHAPTER L
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, aleebraical 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=1,” 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, +, 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 + and —. Thus 7a+5b—38c-—4#+2y 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 5a. A compound expression
consists of two or more terms. Compound expressions may be
€ H.A. A
2 ALGEBRA. [CHAP.
further distinguished. Thus an expression of two terms, as
3a — 26, is called a binomial expression ; one of three terms, as
2a—3b+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.
#3;In Arithmetic the product of 2 and 3 is written 2 x 3, whereas
in Algebra the product of a and 6 may be written in any of
the forms axb, a.6, or ab. The form ab is the most usual.
Thus, if a=2, b=3, the product ab=axb=2x3=6; but in
Arithmetic 23 means “twenty-three,” or 2x 10+3.
5, Each of the quantities multiplied together to form a pro-
duct is called a factor of the product. Thus 5, a, b are the
factors of the product 5ab.
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 5a), 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
Gubc, 6a may be appropriately called the coefficient of be. 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 multipled 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
axa is called the second power of a, and is written a? ;
LS URIS ere seas setter, POWE? OL Cau. wherne metic: fe 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 ‘fa 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’, 1a4 all have the same meaning.
I] DEFINITIONS. SUBSTITUTIONS. 2
8. The beginner must be careful to distinguish between
coefficient and index.
Lzxample 1, Whatis the difference in 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,
34:2 3X GS KAS
a=axaxa=4x4x4= G4.
Example 2. If b =5, distinguish between 4b? and 2b4+,
Here 4b =A b= 4x xX B= 1003
whereas OU Ue D0 = DOR Oo MOK D = 1250.
Hxample 3. If «x =1, find the value of 52%.
Here De = Mo woe x om Le heel =D;
Note. The beginner should observe that every power cf 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 inall. Thus
3x4=4x3,
In a similar way,
Ox4 xX Daat x ox Dat SD Mo
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 ab and ba each denote the product
of the two quantities represented by the letters a and b, and
have therefore the same value, Again, the expressions abc,
ach, bac, bca, 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.
Fractional coefficients which are greater than unity are
usually kept in the form of improper fractions.
Example 4. If a= 6, # = 7, z= 5, find the value of ‘as
Here Paws = 3x 6x7 x d= 273,
4 ALGEBRA. [crrap.
EXAMPLES I. a.
Ifa=5, (bes oe lee 3, y= 12"2-2 and thevalue of
Le ed. Ne ibe Oreos Ate. Ny eae
Geese Tye ab Se ce Ne a ARGs Oe
Lee Lp, Melons 18, 62°. TAS be. 15, Fe:
1b re 6, p =4,q=7, r=5, 2 =1, find the value of -
16 Sap. Lye opG: IG SCE, 1D eOTe: 90, S8aqa.
D1. por. OD SA0T eS DG 1 OTe: Pe le TANS | bas
DOs op. WA Hei 08> Dapon 29 556xro ith, een
lfih=5, k=3, 2 =4, y =], find the value of
1 1 The 1 pclos'f
\ps. Ve, La, 1 ike. 1 ys,
Bl. G6, 82, abe 88, fy 8A, hk, 85, Sy
Chee co yeeoe op ke yee og tis «Ag Shee
* 8 * OF “7125 sa ys
10, When several different quantities are multiplied together ~
a notation similar to that of Art. 7 is adopted. Thus aabbbbeddd
is written a?b’cd>. And conversely 7a*cd? has the same meaning
as 7xaxaxaxcxdxd.
Example 1. If c=3, d=, find the value of 16c*a?.
Here. 16ctP S16 x 34x? = (16 x5") 3 | 2000 81 62000:
Note. The beginner should observe that by a suitable combination
of the factors some labour has been avoided.
Example2, If p=4, q=9, r=6, s=5, find the value of oa
Slps*
ager 32q7r3_382x9x6?_ 32x 9x6x6x6 _3
Slp? = 81x4® S8lx4x4x4x4x4 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 O is usually called a zero factor.
For instance, if v=0 then ab’ry? contains a zero factor.
Therefore ab?xy? =0 when «=0, whatever be the values of a, b, 7.
Again, if c=0, then c?=0; therefore ab’c}=0, whatever values
aand b may have.
Note. Every power of 0 is 0.
ot
Tet DEFINITIONS. SUBSTITUTIONS.
EXAMPLES I. b.
feos pec =) 1 0, 2=7, find the value of
1, 3bp. O- Bax. 3, op. 4, Gaqz. 5, bpz.
6, 3079. pease Segre. 9, gz’. 10, 50? pa.
Vive tee oeop 7. - lomeeas 1S ares 15, Saiq’.
ieee = 2.90 = 0,-p = 3,'¢ = 45.97=.5, find the value of
3k? 513 m 3m? 16p?
16, p? he qr’ 18, Bye 19, 4/ $ 20. Iq" .
5m? 60 8r3 Imq 8lqtr?
als es a OO: 3q? 23. 25lq? 94, 4p? 25, 400p>
ma JN q® kr* 5m” ee
26, aT. G08, a9, S80,
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.
18. 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.
Hxample. When c=5, find the value of ct ~ 4c + 2c? — 8c?.
Here (oe OGD KO on
dee 4 = 2) «
Qe? 2 5) 2 6 DX = 250 +
BC = 8X07 KD
Hence the value of the expression
= 625 — 20 + 250-75 = 780.
14, The beginner must also note the distinction in meaning
between the swm and the product of two or more algebraical
quantities. For instance, ab is the product of the two quan-
tities a and 0b, and its value is obtained by multiplying them
together. But a+b is the sum of the two quantities a and 8,
and its value is obtained by adding them together.
6 ALGEBRA. [CHAP. 1.
Thus if a=11, 6=12,
the sum of a and b is 11412, that is, 23 ;
the product of a and b is 11x12, 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. Ifia=2,b=0, x=5, y =3, find the value of
5a’ — ab? + 2au°y + 8bay.
The expression = (5 x 2?) -0+ (2x 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
-,e . . .
quantities which are equal. Beginners should be particularly
careful not to employ the sign of equality in any vague and
mexact 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.
lig=4,b=21,0=3;f—o0.9 = 1, d= 0, find themalue of
1, 3/+5h- 7b. 2, Tco-9h+2a. 38, 49-5c-9b.
4, 3g-4h+7c. 5, 3f-29-6. 6, 9b-3c+4h.
7, 3a-9b+e. 8, 2f-39+5a. Q, 3c-—4a+7b.
10, 3/+5h-2c-4b+a. ll, 6h-7b-5a-7f+99.
12, 7c+5b —4a+Sh+3g. 13, 9b+a-394+4f+7h.
14, fgtgh-ab. 15, gb-3he+fb. 16, fht+hb-3he.
Ll]. f7*= 3a" 4.2¢7, 18, 0?-2h?+3a?, 19, 3b?-2b?+ 4h? - 2h.
CHAPTER II.
NEGATIVE QUANTITIES. ADDITION OF LIKE TERMS.
17, Iv his arithmetical work the student has been accus-
tomed to deal with numerical quantities connected by the signs
+ and — ; and in finding the value of an expression such as
17+72-—344+6-41 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, 13, to
which no sign is prefixed, is counted among the additive terms.
The same notions prevail in Algebra ; thus in using the expres-
sion 74+3b—4c—2d we understand 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.
G) 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,
(1) 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 absolute distance from the
starting point is the same; but by taking the positive and
negative e signs into account, we see that —30 is a distance from
the starting point equal in magnitude but opposite in direction
to the distance represented by +30. Thus the negative sign
nay here be taken as indicatir : l of direction.
may here be taken as indicating a reversal of direct
(iii) The freezing point of the Centigrade De ae 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 absolute value. In other words subtraction ts the reverse
of addition.
20, Derinition. When terms do not differ, or when they
differ only in their numerical coefficients, they are called like,
otherwise they are called unlike. Thus 3a, 7a; 5a°b, 2u*b ;
3a°b?, —4a°b? are pairs of like terms; and 4a, 3b ; “Te? 9a°b are
pairs of unlike terms.
Addition of Like Terms.
RuleI. The sum of a number of like terms is a like term.
Rule II, Jf all the terms are positive, add the coefficients.
u.] NEGATIVE QUANTITIES. ADDITION OF LIKE TERMS. 9
Example. Find the value of 8a + 5a.
Here we have to increase § 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+5a = 13a.
Similarly, 8a+5a+a+2a+6a = 22a,
Rule III. Jf all the terms are negative, add the coefficients
numerically and prefix the minus sign to the sum.
Example. To find the sum of -3a, -—5a, —7x, —2.
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 —3x, —5x, -—7x, -—x is — 162.
Rule IV. Jf the terms are not all of the same sign, add to-
gether separately the coefficients of all the positive terms and the
coefficients of all the negative terms; the difference of these two
results, preceded by the sign of the greater, will give the coefficient
of the sum required.
Example 1. The sum of 17% and —8z is 9x, 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, —lla, a.
The sum of the coefficients of the positive terms is 16.
PS ee rege cts ef cis yer es Puce vckies 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 sum.
Thus lla—27a+13a=-— 3a states that the algebraical sum
of lla, —27a, 13a is equal te — 8a,
22. The sum of two quantities numerically equal but with
opposite signs is zero. ‘Thus the sum of 5a and — du is 0.
10
ALGEBRA.
(CHAP. II.
EXAMPLES II.
Find the sum of
COTS Si Con
11,
13,
15,
A:
19,
2a, 3a, 6a, a, 4a.
Gb, 11b, 8b, 9b, 5b.
2p, p, 4p, Tp, 6p, 12p.
—2x, —62, —10a, - S82.
Uy 49) ou, OY ae
—2ly, —5y, -8y, - 18y.
-—4s, 3s, s, 2s, -2s, —-s.
32, 102, — 7x, 12x, Oa.
2ay, —4xy, — 32y, xy, Txy.
abe, - 8abe, 2abe, —5abce.
Find the value of
21,
23,
25.
27.
29.
— 9a? + lla? +3a2-—4a%. ¢
~ Lia? + 3a*>— 8a? — Fa? + 2a?
a?b? — 7a7b? + Sa2b? + 9a2b?.
2p°q? - 81 p3q? + 17 pq’.
9Jabed — llabcd — 4labed.
AY 2, oe, OL, O2.
6c, 7c, 3c, 16c, 18c, 101c.
d, 9d, 3d, 7d, 4d, 6d, 10d.
— 3b, -—13b, —19b, — 5b.
—l%c, —34¢, —9c, — 6c.
—4m, —138m, -17m, — 59m.
LLY Oy yoy, TY:
Sab, —6ab, 5ab, —4ab.
Spq, —8pq, 8pq, —4p¢.
— xyz, -Qxyz, Txy2z, — xyz.
3b? — 2b? +76? — 96°.
92° — 32° — Oa? — 923.
ae — lla®x + 3a2x — Qu?a.
Tmin —15m4n + 3m4n.
l3pqu — d5xpq —19qpz.
CHAPTER IM:
SIMPLE BRACKETS. ADDITION.
93, 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—b+c is equivalent to a+c-—), for in the first of the
two expressions 6 is taken from a, and ¢ added to the result ; in
_ the second ¢ is added to a, and 6 taken from the result. Similar
reasoning applies to all algebraical expressions. Hence we
may write the terms of an expression in any order we please.
Thus it appears that the expression a—6 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 0 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. vir.; 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+(18+5)=84+ 13+5=26.
Similarly a+(b+c) means that the sum of b and ¢ is to be
added to a. |
Thus a+(b+c)=a+b+e.
8+(13—5) 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+4+13 -5=16.
. Similarly a+(b-—c) means that to a we are to add 6, diminished
vy c. :
Thus at+(b-—c)=at+b-e.,
12 ALGEBRA. [oHap.
In like manner,
a+b—c+(d—e-f)=a+b—-c+d-—e-f.
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 without making any change in
the expression.
25, The expression a-(b+c) means that from a we are to
take the sum of band c. The result will be the same whether
b and ¢ are subtracted separately or in one sum. Thus
a—(b+c)=a-—b-c.
Again, a—(b—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 ¢ too much, and must therefore
add etoa—b. Thus
a—(b-—c)=a—b+e.
In like manner,
a—b—(c—d-—e)=a—b—c+d+te.
Accordingly the following rule may be enunciated :
Rule. 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.
Addition of Unlike Terms.
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 single like term. If, however, the terms are unlike they
cannot be collected; thus in finding the sum of two unlike
auantities 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+(—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 @ is the ¢ /.1e in
effect as to subtract 6; thus
a+(-—b)=a—6b;
that is, the algebraical sum of a and —b is expressed by a—b.
98, It will be observed that in Algebra the word swm is used
in a wider sense than in Arithmetic. Thus, in the language of
Arithmetic, a—6 signifies that 6b 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 —b without any
regard to the relative magnitudes of a and 0.
Example 1. Find the sum of 38a —-5b+2c, 2a+3b-d, -—4a+ 20.
The sum = (3a — 5b + 2c) + (2a +386 - d) +(—4a+ 2b)
= 38a -5b4+2c+2a4+3b-—d-4a+2b
= 8a+ 2a -4a—-5b4+3b+2b+2c-d
=a+2c-d,
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
may be in the same vertical columns: then add each column
beginning with that on the left.
3a — 5b + 2c The algebraical sum of the terms in the
2a +3b ayy first column is a, that of the terms in the
ep ee second column is Zero, The single terms
in the third and fourth columns are
a +2c-d brought down without change.
Hxample 2. Add together -5ab+6bc-TJac; 8ab+3ac-2ad;
~2ab+4act+5ad; be-3ub+4ad.
— 5ab + 6bce — Tac
Sab + 3ac —-2ad Here we first rearrange the ex-
“Ge LIES pressions so that like terms are in
r the same vertical columns, and then
~ 8ab+ be +4ad add up each column separately.
— 2ab + Tbe +7ad
EXAMPLES III. a.
Find the sum of
3a+2b-5c¢; —4a+b—Te; 4a-3b+6c.
32+2y+625; w2-3y-32; Qa+y—3z.
4p+3q+dr; -2p+8q—-8r; p-qt+r.
7a—5b+8c3; lla+2b-c; 16a+5b-2c.
8l-2m+5n; -614+7m+4n; -1—4m-8n.
5a-—7b+3c—4d; 6b-5c+3d; b4+2c-d.
2a+4b—5x; 2b-5a; -38a+2y; -G6b+Sa+y.
Wa-Sy-Tz; 4e+y; 62; b5a-38y+2z,
°. e «
°
ONaoorRrwhr
.
14 ALGEBRA. [CHAP,
9, a—-2b+7c¢+3; 2b-8c4+5; 38c+2a; a-8-Te.
10, 5-x-y; 7+2xe3; 3y-22; -4+a”-2y.
1], 25a-15b+c; 4c-10b+18a; a-—c+20b.
12, 2a-3b-2c+2x; 5x24+3b-7c; 9c-6x—2a.
18, 3a-—5c+2b-2d; b+2d-a; 5c+38f+3e —2a - 3d.
16a Pe ls POU EK Ps BOO ie eon
15, l7ab-13kl-Say; Tay; 12kl—5ab; 3xy—4kl-ab.
16, 2ax-38by-—2cz; 2y-ax+7cz3; axw-—4cz+7Tby; cz—Gby.
17, Saxv+cz-4by; Thy-S8ax-cz3; -3hby+9ax.
18, 3+5cd; 2fg-3st; 1-5ced3; -442st—fy.
19, 5cea+38fy-2+2s; -2fy+6-9s; -3s-4+4+2cx—-fy.
90, -—38ab+7cd-—5qr; 2ry+8qr—cd; 2cd-3qr+ab—2ry.
29, Different powers of the same letter are unlike terms ;
thus the result of adding together 2x”? and 3x7 cannot be ex-
pressed by a single term, but must be left in the form 2x? + 327.
Similarly the algebraical sum of 5a7b?, —3ab°, and —O* is
5a°b? — 3ab—b*, This expression is in its simplest form and
cannot be abridged.
Example. Find the sum of 6x? — 52, 2x7, 5a, —2a°, - 32, 2.
The sum = 62° — 54 +22? + 5a — 223 — 3x2+2
= 623 — 2Qu3 + Qu? — 8a? -5a+5x+2
= 47? — 7? +9,
This result is in descending powers of 2.
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 32° +7 + 6” - 52°; 27-8 - 9a;
Ada —2a° 4327; 32a°-9e—2?; w-a?-774+4.
303 — 522+ 6a+7 | In writing down the first expression
Tr ee we put in the first term the highest
ss power of xz, in the second term the
~ 20° + 3a? + daz next highest power, and so on till the
30° — x2? -—9x last term in which x does not appear.
ia dg ites peed The other expressions are arranged in
the same way, so that in each column
32° — 20° -Ta+3 we have like powers of the same letter.
III. |
ADDITION, 15
Example 2. Add together
8ab? - 2b? +a; 5a*b - ab? - 3a?
— 2b? + 3ab? ea?
5b?
3b?
— ab?+ 5a*b -3a3
+ 8a?
ab?+ 9a?b —-2a3
+ 3ab? + 14a7b + 4a?
; 8a2+5b?; 9a2b - 2a? + ab?
Here each expression contains
powers of two letters, and is
arranged according to descend-
ing powers of b, and ascending
powers of a.
EXAMPLES III. b.
Find the sum of the following expressions :
SON OORONH
v7 + 3xy-38y°?; -—3a?+ay+2y?; Q2x?-3xyt+y".
2427-2443; —247+5x4+4;
5a —a2+a-—-1; 2x? -Qr+5;
xz” — Qa -—6.
~— §2° + 5a — 4.
a®—a*b+5ab?+b?; -—a®—10ab?+6°; 2a°b+5ab?- b.
3a? —~9x?—-lle+7; 2u?-5x?4+2; 5a2+152°-TFas 8-9.
x°—52°+8a;3; Ta°+4a°+5e;
822-92; 2a°- Tx? — 4a.
4m? +2m2-5m+7; 3m?+6m?-2; -—5m?+3m; 2m-6.
ax®—4be?+cx; 8ba?-2cxr-d; bx*+2d; Qax?+d.
py — 99. It; = 2py" + 3qy —
5y® + 20y2+8y-1; -2y+5-
Gr; Tay-4r; 3py*.
7Ty?; —8y?-4+2y8-y.
2-—a+S8a?-a®; 2Qa?-3a2+2a-2; -—38a+4+7a3—5a?
1+2y - 3y?-5y?; | -1+4+2y?-
a7a3 —3a®x2-++-0; 52+ 7a®x? ;
xv — 4darty — Bary? 3 Baty + 2a3y3
y; 5y>+3y?+4.
4aPu? — a?x? — 5x.
-6xy?; 3°42 + 6ayt - y?.
a®—4a*b+6abe; a?b-l0abe+c?; 63+3a7b+abce.
ap’ -6bp?+Tep; 5-6cep+5bp?; 3-2ap®; 2p--7.
e7 —20° +11c8s; —2c7 —8c8+5c5; 4c8-10c>; 4c? -c®.
4h3—-7+3h4-2Qh; Th-3h?+2-ht; Qh4+2Qh?-5.
30° +4+2y?-5et+2; Tai-5y?+7x-5; 9a3+11-84%+4y?;
6x — y? - 182° — 7.
7+ Qry + 3y"; 327+ 2yz+y*;
ey + 22 + yx — 62? - 4y? — 22,
e274 32242923 22-3xy - 38yz;
CHAPTER IV.
SUBTRACTION.
31, THE simplest cases of Subtraction have already come
under the head of addition of dike terms, of which some are
negative. [Art. 20.]
Thus 5a-38a= 2a,
3a —Ta= —4a,
—da46a= — Qa.
Since subtraction is the reverse of addition,
+b—b=0;
. a=at+b—b.
Now subtract —b from the left-hand side and erase —b 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)=38a+5a
= 8a,
and —3a—(-—5a)= —3a+5a
= 2a.
Subtraction of Unlike Terms.
32. We may proceed as in the following example.
Kxample. Subtract 8a—2b—c from 4a—3b+5e.
The difference
The expression to be subtracted is
=4a—3b+5c—(8a—2b—c) | first enclosed in brackets with a minus
=4a—38b)+5c—8a+2b+¢ sign prefixed, then on removal of the
=4a—8a—3b42b+5c+¢ brackets the like terms are combined
=a—b+6e. by the rules already exclaimed in
Art, 20,
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, — 3b + 5e The like terms are written in
—3a+2b+e the same vertical column, and each
by addition, a— b+6c column is treated separately.
Rule. Change the sign of every term in the expression to be
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 5a?+ay take 2a? + 8axy - Ty.
5at+ xy In the first column we combine mentally 52?
and — 22°, the algebraic sum of which is 3x7, In
= the last column the sign of the term — 77 has to
3a" — Tay +7y? be changed before it is put down in the result.
2a? + Say — Ty?
Hxample 2. Subtract 3x?-2x from 1-2.
Terms containing different powers of the same letter being wnlike
must stand in different columns.
_ 73 co) 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
~ 23-327 +2e+1 and third columns each sign has to be changed.
3x? — 2x
The re-arrangement of terms 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
a+2b—c from 2a+3b+c.
2a—b+c from 3a—5b--c.
3a--y-—% from 2 —4y+4 32.
x+8y+8z from 10x —7Ty - 6z.
—m—3n+p from -—2m-+n- 3p.
3p -2q+r from 4p—7q+3r.
a—7b-3c from -—4a4+35)+4+8c.
-a-b-9c from -a+b-9e.
H.A. B
CONDO FPwONH
18 ALGEBRA. [CHAP.
Subtract
9, 38x-5y-7z from 2x+3y —4z.
10, -—4¢-2y+1lz from —~a#2+2y- 132.
ll, -2x-5y from 2+ 3y- 2z.
12, 3x-y-8z from 2+2y.
138, m-2n-p from m+2n.
14, 2p-3q-r from 2q-4r.
15, ab-2cd-ac from —ab—3cd+2ac.
16, 3ab+6cd -3ac—5ld from 3ab+5cd —4ac -- 6bd.
17. -xyt+yz-—2x from 2xy+2.
18, -2pq-3qr+4rs from qr -4rs.
19, -mn+llnp-—8pm from — 1lnp.
20, x°y -2Qxy?+ 3xyz from 2x?y + 38xy? - xyz.
From
91, x?-38x?+2 take —2°+322-—-2.
99, -2x?—x? take 2° -2?-x.
93, a?+l—-B8abc take b?—2abe.
94, -8+4+6bc+b°c? take 4 —- 3be — 5b7c*.
25, 3p*—2p°q + Tpq? take p*gq — 8pq?+q’.
96. 7+x2-2 take 5-x+27+2°%.
QT, —44+2°y-ayz take —3-2x°y+llaryz.
98, —Sa’x?+5x?+15 take 9a?x? - 8x? -5.
99, p®+r-38pqr take 1°+¢9°+ 3pqr.
80, 1-32? take 2° -32?+1.
“3, 2430-72? take 322- 32-2.
82, x? +1la?+4 take 8a?- 52-3.
83, a®+5-2a? take 8a*+3a? -7.
04, 24+323-2?-8 take 244+ 32?-2#+2,
v85, 1-2x4+32? take 7x? —42?4-32+1.
36. xtyztyzx take — 3y?2u — Qary2? — xyz.
37, 4a%a?-38aat+a® take 3a°x?+7a72x3 — a.
88, l-w2+a°-at-az take zt-l+x-2".
89, —-Smn?+15m2n+n3 take m3 —n?+8mn? —7m?n.
40, 1-p? take 2p? - 3pq?- 2q'.
33, The following exercise contains miscellaneous examples
on the foregoing rules,
IV. | SUBTRACTION. 19
MISCELLANEOUS EXAMPLES I.
1, When x=2, y=3, 2=4, find the value of the sum of 52’,
-3xy, 2’. Also find the value of 327+ 324.
9, Add together 3ab+be-—ca, -ab+ca, ab-2be+5ca. From
the sum take 5ca+ be — ab.
3, Subtract the sum of x-y+3z and -2y—2z from the sum of
2a —-5y—3z and -3x+y+4+ 4.
4, Simplify (1) 3b —- 2b? - (2b — 30%).
(2) 8a-2b —(2b +a) —(a—5d).
5, Subtract 8c?+8c-—2 from c®-1.
6, When w=3, a=2, y =4, 2 =0, find the value of
Arey
9) Sata
(2) 3y
7, Add together 3a?-7a+5 and 2a*+5a-3, and diminish the
result by 3a7+2.
(1) 2a?- Bay +4az%,
8, Subtract 267-2 from -—2b+46, and increase the result by
S0— 1s
9, Find the sum of 32?-47+8, 24-33-47, and 2z?-2, and
subtract the result from 62743.
10, What expression must be added to 5a?-3a+12 to produce
9a?—7?
11. Find the sum of 22, — 2°, 3x7, 2; +52, -4, 32°, ~—52?, 8;
arrange the result in ascending powers of x.
12, From what expression must the sum of 5a7-2, 8a+a?, and
7 - 2a be subtracted to produce 8a?+a-—5?
18, When x=6, find the numerical value of the sum of 1-2 +2",
2a2—1, and x — x?
14, Find the value of 6ax + (2by — cz) — (2au - 8by + 4cz) — (cz + a2),
nauiema= 0,0 = 1,¢=2,0=8, y=3, 25 4.
15, Subtract the sum of 4° — 3x7, 2a?-7a, 8a-—2, 5- 32°, 2a°-7
from a3+a?+a+1,
aA) ALGEBRA. (CHAP. IV.
16, What expression must be taken from the sum of p*-—3p’,
2p +8, 2p7, 2p? — 3p4, in order to produce 477-3?
17. What is the result when -—32°+2z2-1lx2+5 is subtracted
from zero.
18, By how much does b+¢ exceed b-c?
19, Find the algebraic sum of three times the square of x, twice
the cube of x, —2?+a -22?, and 23-a—a?+1.
20, Take p?-—q? from 3pq—4q", and add the remainder to the
sum of 4pq — p? — 3q? and 2p* + 6q".
91, Subtract 3b°+2b6?-8 from zero, and add the result to
b4 — 2b? + 30.
99. By how much does the sum of —~m?+2m-—1, m?-3m,
2m? —2m?+5, 3m? + 4m?2+ 5m +8, fall short of 1lm* —8m?+3m?
93, Find the sum of 82° -4a3y?, Taty—axy*, 3a°y?+ 2a°y3 + dary},
y—4ayt+a%y?, 2° —y+a3y24+ vyt- ay? +3aty, and arrange the
result in descending powers of x.
24, To what expression must 3x -4a3+7a7+4 be added so as to
make zero? Give the answer in ascending powers of x.
95, Subtract 7a?-32-6 from unity, and x - 52? from zero, and
add the results.
96, When a=4, b=3, c=2, d=0, find the value of
(1) 8a? —2be —ad-+ 3b%ed. jp ee
; Ja
97, Find the sum of a, —3a?, 4a, —5a, 7, -18a, 4a7, —6, and
arrange the result in descending powers of a.
98, Add-together 44+327+23, 3-22-11, x?-—22?+7, and sub-
tract 22°+2?—7 from the result.
29, Ii a=5x-3y+2z, b= -2x+y-32z, c=x- 5y+6z, find the
value of a+b-—-c.
30, If «=2a?-5a+3, y= -8a?+a+8, z=5a?—6a-—5, find the
value of «-(y+2).
CHAPTER V.
MULTIPLICATION.
34, Mu.rreiicaTion in its primary sense signifies repeated
addition.
Thus 3x5=3 taken 5 times
one Sone
Here the multiplier contains 5 units, and the number of times
we take 3 is the same as the number of units in 5
Again axb=a taken b times
=a+a+ta+a+..., the number of terms being 0.
Also 3x5=5 x8; and so long as a and 6 denote positive whole
numbers it is easy to shew that
ax b=-b = at ;
that is, the index of a in the product is the sum of the indices
of a in the factors of the product.
Again, 5ba*=5aa, and 7a°=7aaa ;
ba? x 7a=b x X aadaa= 35a",
22 ALGEBRA. [ciaP.
When the expressions to be multiplied together contain
powers of different letters, a similar method is used.
Example. 5a®b? x 8a*bx? = 5aaabb x Saabuxx
= 40082)
Note. The beginner must be careful to observe that in this pro-
cess of multiplication the indices of one letter cannot combine in any
way with those of ancther. Thus the expression 40a°b’z? admits of
no further simplification.
37. Rule. Zo multiply 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.
Hxample 1. Find the product of 2”, x, and 2°.
‘The product)= a7 xa SG20 =e xe ee,
The product of three or more expressions is called the con-
tinued product.
Hxample 2. Find the continued product of 5x?y?, 8y°2, and 3x24.
The product 2 Hay? Sy poe nl our ay ee.
38, By definition, (a+b)m=m+m+me-+ ... takena+b times
=(m+m+m-+... taken a times),
together with (m+m+m-+... taken b times)
=am+bm.
Also (a—b)m=m+m+m-+ ... taken a—b times
=(m+m+m-+... taken a times),
diminished by’ (m+m+m+... taken b times)
=am—bm.
Similarly (a-—b+c)m=am—bm+em.
Thus it appears that the product of a compound expression by
a single factor ts the algebraic sum of the partial products of each
term of the compound expression by that factor.
Hxamples. 3(2a+3b- 4c) = 6a+9b —12c.
(4a? — Ty — 82?) x Bay? = 12a3y? — Qlay® — Way,
v.1 MULTIPLICATION. 29
EXAMPLES V. a.
Find the value of
BL oa a BER Go Ar eee one GEOG a Od
face < 7c 6, 9y? x 5y®. 7, 3m? xbm*, 8, 4a® x Gat.
Q, 3xx 4y. 10. ba x 60: ieee oa; TO apt og".
Domocaxdbax. 14, 3qrx4or. 15, abxab. 16, 3ac x 5ad.
eee ate, 19. 3x°y*x4y2 19, abe xab. 90, atx da°b*
Ql, a? x ab x 5ab4. OO Rie x Op rx [pr
93, Ga®y x ay x 9aty?. OA 1G? <3b" x De,
DD. Gay? x Tyz x az. 296, 3abcd x 5bca? x 4cabd.
Multiply
27, ab-—ac by are. 98, x*y—a®2+4y2 by a3y2?.
99, 5a? - 3b? by 8ab?ct. 30, ab -—5ab+6a by 3a*d.
Sigea-— 20? by 327. 32, 2ax?-b®y+3 by aray.
83. Tp*a-part+l by 2p. 384, m?+5mn-3n? by 4m?n.
35, zy? -3x?z-2 by 3yz. S00 oa Dy Zara.
39, Since (a —b)m=am— bm, [Art. 38. |
by putting c—d in the place of m, we have
(a—b)(c -d)=a(c— d) —b(c—d)
=(e—d)a—(ce—d)b
=(ace—ad)-— (be— bd)
=uac—ad — be+bd.
If we consider each term on the right-hand side of this result,
and the way in which it arises, we find that
(+a)x (+e) =+ae.
(—b)x(—d) = +d.
(—6b)x(+c) =—be.
(+a) x(-d)=-— ad,
These results enable us to state what is known as the Rule
of Signs in multiplication.
Rule of Signs. Zhe product of two terms with like signs ts
positive ; the product of two terms with unlike signs ts negative.
24 ALGEBRA, [cHaP,
40, The rule of signs, and especially the use of the negative
multiplier, will probably present some difficulty to the beginner.
Perhaps the following numerical instances may be useful in illus-
trating the interpretation that may be given to multiplication
by a negative quantity.
To multiply 3 by —4 we must do to 3 what is done to unity
to obtain — 4. Now —4 means that unity is taken 4 times and
the result made negative: therefore 3x (—4) implies that 3 is
to be taken 4 times and the product made negative.
But 3 taken 4 times gives +12;
oie 4)
Similarly —3x —4 indicates that —3 is to be taken 4 times,
and the sign changed ; the first operation gives —12, and the
second +12.
Thus (-—3)x(-4)=+4+12.
Hence, multiplication by a negative quantity indicates that we are
to proceed just as if the multiplier were posite, and then change
the sign of the product.
Hxample 1. Multiply 4a by —3b.
By the rule of signs the product is negative ; also 4a x 3b = 12ab;
*. da x (- 3b) = —12ab.
Example 2. Multiply -5ab®x by - ab?a.
Here the absolute value of the product is 5a7)°:*, and by the rule
of signs the product is positive ;
“. (—Sabix) x(—ab?x)= 5a7d's?.
Example 3. Find the continued product of 3a7b, -2a3b?, — abt,
This result, however, may be
a2 WD 3h =o Aap hos i
ie ae b m= oes written down at once : for
Pe Obi )” x ia a = OL . F >
3a7b x 2a30? x abt = 65’,
Thus ee complete pro- and by the rule of signs the re-
duct is 6a°b’. quired product is positive.
EHxample 4. Multiply 6a? - 5a2b -4al? by - 3ab>.
The product is the algebraical sum of the partial products formed
according to the rule enunciated in Art. 37 ;
thus (Ga? - 5a*b — 4ab?) x ( - 3ab?) = — 18a4h? + 15a7b3 + 12074,
v.] MULTIPLICATION, 25
EXAMPLES V. b.
Multiply together
eo = 2, 9. —3, 42. 3, —2*, -x. 4, —5m, 3m?
5, —4¢q, 3q”. 6, —4y°, —4y°. 7, —3m3, 3m3. 8, 424, — 424,
9, —3x, —4y. 10, —5a7,4~. 11, —3p?,-49°. 12, 3ab, —4abd.
Popeoa, — 0°, Zab.» 14, —a, -—0,' =c?. 15, 3a, — 2b, —4c?, —d.
16, —3ab, —4ac,3be. 17, -—2a?, -3a7b,-6. 18, —2p, -3¢q, 4s, -¢.
Multiply
19, -ab+ac-—be by -ab. 20, —38a*?-4ax+5z? by — ax’,
91, ac—ac?+c* by —ae. 22, —2ab+cd -efby —32°y%,
4], 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. Tfia= —4, find the value of a’,
Here a® = (-—4)? =(-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 posztive.
Example 2. Iia=-1, b=38,c= —2, find the value of — 3a‘bc3.
Here -3atbe? = -3x(-1)*x 3x (- 2)? We write. down at
= —~3x(+1)x38x(-8) once, (—1)*= +1, and
= 72, (-2)? = -8.
RENEE V..¢.
‘al 3 “a WV
faa —1,9b=0,°¢= 2,07 =, res find the value of
rot: OF 0G. 3, an. 4, (-a)?.
b, = 3c. 6, (-q)t tase 8, -ac.
9, ad. 10, —acn. TES 3a", 12. 4(—¥-
18, 2abe?. Thee Se. 15, -(a)* 16, -8a7q.
17, —a'n?, 18, ac’. 19, —a%c?, 20, ?g.
26 ALGEBRA. [cHAP.
li ea 3 esha) fk 5; y = —], find the value of
91, 3a-2y+4k. 99, -—4c—3x4+2y. 93, -4a+5y-x.
94, ac-—ds3cy- yk. 95, 2ay -ka+4k?. 96, a? -2c?4+3y?.
O17, -a-ayt3y". 98, ax-—yx-cy. 29, @-y-+y%
80, a-2?-2y. Bl, c*y2-2act+ch% 82, acy—y*t+2a?.
Multiplication of Compound Expressions.
49. To find the product of a+b and e+d.
From Art. 38, (¢1+b)m=am+bm ;
replacing m by e+d, we have
(a+b)(e+d)=a(e+d)+b(ce+d)
=(c+d)a+(ce+d)b
=ac+ad+be+ bd.
Similarly it may be shown that
(a—b\e+d)=ac+ad—be—bd;
(a+b\e—d)=ac—ad+be—bd;
(a—b)(e-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+ cm =am—bm+em ;
replacing m by x—y, we have
(a—b-+o\(e—y)=ale—y)—W(e-y) + (w—y)
=(ax — ay) - (bx — by)+(cx —cy)
=axn—ay — bx+by+cer—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 first expression by each term
of the second. When the terms multiplied together have like signs,
prefix to the product the sign +, when unlike prefie — 3; the
algebraical sum of the partial products so formed gives the com-
plete product.
45, It should be noticed that the product of a+b and x-y
is briefly expressed by (a+6)(#—-y), in which the brackets
indicate that the expression a+0 taken as a whole is to be
multiplied by the expression v—y taken as a whole. By the
v.] MULTIPLICATION. 27
above rule, the value of the product is the algebraical sum of
the partial products +ar, +bx, —ay, —by; the sign of each
product being determined by the rule of signs.
Hxample 1.
The product
Multiply «+8 by x+7.
=(e-S)e4 7)
=a 8a 72406
= “7+ 152+56.
The operation is more conveniently arranged as follows :
aie. We begin on the left and work
e+ 7 to the right, placing the second
e+ 8x result one place to the right, so
4+ Fa+56 that like terms may stand in the
ae Fag Sree same vertical column.
by addition, w7+15%+56.
Hxample 2. Multiply 2a-3y by 4a-—7y.
20 = DY
4a —Ty
8a? — 12xy
— l4ay + 21y?
by addition, 82? — 26ay + 21y?.
EXAMPLES V. d.
Find the product of
tp as 7, a5. Q, 2-3, +4, 3, a-6,a-7.
4, y-4, yt+4. 5h, 2+9, 2-8. 6, c-8, ct+8.
Tak), k— 5. 8, m-—9, m+12. 9, w2-12,2+11.
10.6 @=14, +1. 11 ep 10s p10. a Wine fe eer aro ae
13, 2-4, —x+4. 14, -y+3, -y-3 15, -a+4,-a45.
16, «2-10, -v7+8 17, -A+4, -k-7. 18, -y-7,-y-7.
19, 2a-5, 8a42. 90, «2-7, 2a+5. Ole 3a 4 Oe 8:
92, 3y-5, y+7. 23, 5m-4,7m-3. 94, Tp-2, 2p+7.
95, x-—3a, 2a+3a. 26, 3a-2b, 2a+38b. 27, 5c44d,5c—4d.
98, a-2x, 3at+2x. 29, Th+c, 7b-2e. 30, 2a-5c, 2a+4 5c.
Ol.. 3a—by, 4a+y. 82,. 2y*-3z, 2y+3z. 388, xy+2b, xy—2b.
04, 2u-3a,2x+3b. 85, 3x-4y, 2a+3b. 86, mn-—p, 2xy+3z.
28
46,
KHxample 1.
3x7 — 27 —5
2x — 5
62? — 4x”?-—10x
— 15x”?+ 10% +25
Ga 31942 +25.
ALGEBRA.
[ CHAP.
We shall now give a few examples of greater difficulty.
Find the product of 3a?-2a2-5 and 22-5, ~
Each term of the first expression is
multiplied by 2x, the first term of the
second expression; then each term of the
first expression is multiplied by —5; like
terms are placed in the same columns and
the results added.
Hxample 2. Multiply a-b+38c by a+ 20.
aa
b+ 3¢
& 2p :
a?- ab+3ac
2ab
= 20° +6be
a = ab + 38ac — 2b? + Bie
AT,
If the expressions are not arranged according to powers,
ascending or descending, of some common letter, a rearrange-
ment will be found convenient.
Hxample.
2a?— 3ab +4b?
= Ha?+ Sab £40?
— 10a4 + 15a*b -- 20a7b?
+ 6a°b-— 9a?b?+ 12ab?
8a2b? — 12ab?
+ 1604
— 10a4+ 21a) — 21a7b?
+ 1664.
Find the product of 2a7+ 4b? -3ab and 3ab —- 5a? + 4b".
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, 22-382-2, Qn-1.
3. 2y7%-sy+1, 3y—1.
Hh, 2a?-3a-6, a-2.
7. 327-2447], 20-7.
Q, xw?+x-2, w*-x4+2.
ll, 2a?- 3a-6, a?-a+2.
18, at+b-c, a-b+e.
4a?-a -2, 2a+3.
3a7+4e4+5, 4a —-5.
5b? -2b4+3, -—2b-3.
5c? -4c+8, -—2ce4+1.
w= 24 +5, 27-2745.
2k? —-3k=1, 3h? -—k=1.
a-2b-—38c, a—2b+3c.
v.] MULTIPLICATION, 29
15, 2-xyty, w+ayt+y’. 16, @-2a%+227, a?+2aa +227.
17, @-0?-38c?, -a?-b?-3c2. 18, w-3a?-a, 2?-3r+1.
19, a®-6a+5, a’?+6a—-5. 90, 2yt-4y?+1, 2y4-4y?-1.
91, 5m?+3-4m, 5-4m+3m. 22, 8a>-2a2-3a, 3a?+1- 5a.
98, 2x2+2a?- 327, 87+24+2e% 94, a? +? -a7l?, a*b?- a+b.
95, a +2°+8a274+8ara, a? + 3ax?- x? - 38a7n.
96, 5Spt-p?+4p?-2p+3, p?-2p43.
97, m®—2m*+3m3—4m?, 4m - 3m? + 2m.
98, at+1+6a?-4a?-4da, a®-1+4+38a-3a?.
99, a®+b?+c?+ab+ac—be, a-b-e.
80, 214+ Gary? + y* — 4a%y—4ay®, — at — y* — Cay? — dary? — 4a°y.
48, Although the result of multiplying together two binomial
factors, such as +8 and w-7, can always be 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
(e7+8)(@+7)=27+80+7x24+ 56
=7+154+56.
- (@ —8)(@—7) =a? — 8x —7# +56
=x? —152-+56.
(7+ 8)(4@—-7)=274+ 8x2 —Tx—56
) =2'+x—56.
(7 —8)'7+7)=x2°-—8x+ Tx —56
=? -x2- 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. The 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. [oHar. v.
The intermediate step in the work may be omitted, and the
products written down at once, as in the following examples :
(w7+2)\(v7+3)=2° + 5r+6.
(v—3)(e+4)=a27+4-12.
(v7+6)(@—9)=2?— 3x — 54.
(a —4y)(x2 — 10y) =x? —14ay + 40y”.
(7 —6y)(a@+ 4y) =x? — Qay — 24y?.
By an easy extension of these principles we may write down
the product of any two binomials.
Thus (2% + 3y)(% — y) = 2a" + 3.2xy — 2Qxey — 3y?
= 20° + xy — dy".
(8a — 4y)(2a + y) =6x? — 8xy + Bry — 4y?
= 6x? — Bay —4y’.
(v7+4)(~—4)=n7 +427 —4x- 16
=27°—16.
(2 + 5y) (2x — 5y) = 40? + 10xy — 10xy — 257?
= 44? — 25.
EXAMPLES V. f.
Write down the values of the following products :
1, (a+3)(a—-2). 9, (a-7)(a—-6). 8, (w—-4)(~%+5).
4, (b-6)(6+4). 5 Y-Hiy—D- 6, (a-1)(a-9).
7, (c-5)(¢+4). 8. (x-9)(a— 3). OS (yA) ye7)
10, (@-3)(a+8). ll, (%-5)(x— 8). 12, (a+7)(a—-7).
13, (4-6)(/-6). 14, (a-5)(a+5) 15, (¢+7)(¢+7)
16, (p+9)(p- 10). 17. (+5)(2=8). 18, (%-9)(%+9).
19, (w-8a)(v+2a). 20, (a-2b)(a+2b). QI, (a-—4y)(x~—-4y).
22, (a+4c)(a+4c). 23, (c-5d)(c-5d). 24, (p—2q)(p+2¢).
95, (2e-8)(82+2). 26, (Ba-1)(2Q4+1). 97, (5x—2)(52+4+2).
28, (8a+2a)(8a-2a), 29, (6u+a)(6x—-2a). 80, (7x+3y)(7x-y).
CHAPTER VI.
DIVISION.
49, Tur 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 ~+ 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 2 is 42, it follows that
when 4x is divided by x the quotient is 4,
or otherwise, 4u—a = 4,
Example 2, Divide 27a5 by 9a*.
ma : 9705 2Taaaan We remove from the divisor
The quotient = bt Oaae and dividend the factors com-
mon to both, just as in arith-
= 3a = 3a" metic.
Therefore 27a’ + 9a? = 3a?
Example 3. Divide 35a*b?c? by 7ab?c?.
3d5aaa . bb. ce
7a@. 6b. ce
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.
Cc
= 5aa.c= 5a.
The quotient =
32 ALGEBRA. [CHAP,
50. It is easy to prove that the rule of signs holds for
division.
Thus QOS eer
a a
a : a
: —a —a
Hence in division as well as multiplication
like signs produce +,
unlike signs produce —,
Rule. To divide one simple expression by another :
The index of each letter in the quotient vs obtained by subtracting
the index of that letter 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 that of the divisor.
Hxample 1. Divide 84a°x? by —- 12a4x.
The quotient = ( — 7) x a'-4z3-1 Or at once mentally,
= -— Tax? 84a°x? + (-12a4x) = — Jaz.
Example 2. —45a%b?a*+(~ 9a8bx?) = 5a®ba?.
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 Go? a? 9 = a;
3
but also roa J,
a
read be
This result will appear somewhat strange to the beginner, but
its full significance is explained in the Theory of Indices.
[See Hlementary Algebra, Chap. xxxt.]
Rule. Zo divide a compound expression by a single factor,
divide each term separately by that fuctor, and take the algebraic
sum of the partial quotients so obtained.
This follows at once from Art. 38.
Examples. (9x -12y+8z)+(-—3) = -—8x%+4y -2z.
(86a°L? — 24a7b° — 20a4b?) + 4a*b = 9ab — 6b* — 5a°b,
VI. ] DIVISION. 33
EXAMPLES VI. a.
Divide
Peace by x. 2. Ga’ by 3a. 3, oa’ by at.
4, 210 by 76°. 5, xy? by —axy. 6, —3axy*® by 3y.
7, 4p?q? by -2pq. 8, 1l5m®n by -—5m. 9, —2m? by —ln.
a0 tse" by —6a°. J], 35242 by -72. 19, ~7a*> by \—7.
lo. —28p'¢ by 28p*, 14, —Ta* by —2’. 15/ 2dxy2? by -— 322,
16, —12b%< by Gb?e® 17, -9k" by -k4. 18, 2h by —hi.
19, —45a*b%cl’ by 9a7b?c!, 90) Saryte? by —ax%yz?.
91. —168a7b?ca? by — Taba, 99, —35a%bSa? by — 7a*bta’.
23. 327-2e by x. 94, 5a®b- Tab? by ab.
25, 48p*q —24pq? by 8pq. 26, —152°+252* by —- 52°.
O17, w-xy-—xz by -x. 98, 10a?-5a*b+a by —a.
99, 403+36ax2—16«% by —4za. 30, 38a?—9a2b —Cab? by —3a.
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 be necessary.
Repeat these operations till all the terms from the dividend are
brought down.
Example 1. Divide x?+11%+30 by «+6.
Arrange the work thus :
+6) 22+ 11x% +30 (
divide x", the first term of the dividend, by x, the first term of the
divisor ; the quotient is x. Multiply the whole divisor by x, and put
the product «7+ 6x under the dividend. We then have
x+6)a*+1le+30(2#
o+ 6x
by subtraction, Ba +30
On repeating the process above explained, we find that the next
term in the quotient is +5.
HA; Cc
34 ALGEBRA. [CHAP.
The entire operation is more compactly written as follows :
x+6)x72+1l2+380(42+5
a+ 6x
ox + 30
5x +30
The reason for the rule is this: the dividend is separated
into as many parts as may be convenient, and the complete
quotient is found by taking the sum of all the partial quotients.
By the above process #?+112+30 is separated into two parts,
namely 27+ 6x, and 57+30, and each of these is divided by «+6;
thus we obtain the partial quotients + and +5.
Hxample 2. Divide 24a?- 65xy+2ly* by 8x -3y.
Divide 2427 by 8x; this
» — 297) D472 — B5y PTa2( 3x -7
8a — dy) 24a" — Oday + 2ly"( 3x - Ty gives 32, the first term of the
me ES) quotient. Multiply the whole
- 56xy+21y? divisor by 38x, and place the
— 56ay + 21y? result under the dividend.
By subtraction we obtain
—56xy+2ly?. Divide the first term of this by 8z, and so obtain
—Ty, the second term of the quotient.
Hxample 3. Divide 16a*—46a?+39a-9 by Sa-3.
8a —3 ) 16a? — 46a? + 39a — 9 ( 2a? -5a+3
l6a°— 6a?
— 40a?+ 39a
— 40a? + 15a
94a —9
Y%4a—9
Thus the quotient is 2a? -5a+3.
EXAMPLES VI. b.
Divide
], a+2a+1 by a+. Q, U?+3b4+2 by 6+2.
3, 2?+42+3 by «+1. 4, y?+5y+6 by y+3.
5, 27?+5x2-6 by x-l. 6, 2°+2x2-8 by 2-2.
7, p?+38p-40 by pt8. 8. g@-4q-82 by q+4.
9, a?+5a-50 by a+10. 10, m*+7m-78 by m-—-S.
ll, #*+ax-30a* by x+6a. 12, a*+9ab—-36b? by a+120.
vI.] DIVISION. 35
Divide
18, -—2?+18x%-45 by x-15. 14, «* 4274441 by x--21.
15, 2au?-182-24 by 2443. 16, 5x74+16%+3 by «+3.
17, 6%7+52-21 by 22-3. 18, 12a*+au-6xz? by 3a- 22.
19, —52?+ay+6y? by ~x-y. 20, 6a? ac—35c? by 2a—-5e.
21, 12p?--'749q+12q? by 2p-12¢.
92, 4m?-49n? by 2m+7n.
238, 12a7--3lab+20b? by 4a—-5bd.
24, —2527+49y? by -5a+7y.
25, 21p?+1lpq—40q? by 3p+5q.
26, 82°+82?+4e+1 by 27+].
27, —2x3+1322-17x+10 by —x+5.
98, x? +ax?-3a°x-—6u? by w-2a.
29, Ga®y—a*y?-Txy?+12y* by 2a+4+3y.
30, 8x?-122?-14%+21 by 2a-3.,
52. The process of Art. 51 is applicable to cases in which
the divisor consists of more than two terms.
Hxample 1. Divide a*t-—2a*-—7a?+8u+12 by a?-a-6.
a*—-a-6)a*—2a?—7Ja?+8a+12(a2?-a-2
a*—a?— 6a?
-a-—a*+8a
—a?+a"+6a
— 2a7+2a+12
— 2a7+2a+12
Hxample 2. Divide 4x?-5x?+6x°-15-at-x by 3+22%--2,
lirst arrange each of the expressions in descending powers of 2.
2x7 -—2+3)62°— 244+ 403-5? -a2- 15 ( 822+ a?-2a-5
6x? — 324+ 9x3
24 —5a— 5a?
se a ode
—493— 827- x
—4a3+ 2o7?- 6a
*~ 1027+ 54-15
—10x7+5x-15
36 ALGEBRA. [CHAP.
Hxample 3. Divide 23a7-2a4-4a3412+2>-3lx by x2?-7x+5.
x? — Te +5 ) a° — Qat — 4a3 4+ 23a? — 31a +12 ( a? - 24+3
xe” —TJao+ 5a
~ 2x4 + 3a? + 18x? - 31a
— 2x4 +14a?—-10x
323+ 47-2174 12
32° —21x%4+15
hege® ae
Now 42? is not divisible by x’, so that the division cannot be
carried on any further ; thus the quotient is «?-2x+3, and there is
a remainder 4a7— 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 2a?+10-16a-39a?+15a4 by 2-4a—-5a?.
2-—4a-5a?)10-16a-39a?+ 2a°+ 15a4( 5+ 2a - 8a?
10 — 20a — 25a?
4a—14a?+ 2a°
4a— 8a*-~10a°
— 6a?+12a?+15a4
— 6a?+12a*?+15a4
EXAMPLES VI. c.
Divide
1, w-6a?+lla-6 by a?-4a+3.
9, x -40?4+24+6 by x?-x-2.
8. yty?-9yt12 by ¥°-3y+4+3.
4, 2lm?-m?+m-1 by 7m?+2m+1.
5, 6a*?-5a*-9a-2 by 2a7-3a-1.
6, 66-k-14k4+3 by 3h°+4k-1.
7, 6x?+1la?—-392-65 by 3x74 134413.
8,
122° — 8ax? — 27a7x%+ 18a? by 62? —13ax + 6a?
vi.] DIVISION. 37
Divide
Q, 162°+14a?y —129xy? - 15y? by 8x7 + 27xy + 3y?.
10, 2c? — 5c2d — 3ced? - 2d? by 7c?+3cd+d?.
1], 324-1003 +12¢?-1l2+6 by 32°-2x+3.
12, 30a*+1la’ —- 82a7-12a+48 by 8a7+2a-4.
13, v°--%?-8x-13 by 274+32+4+3.
14, a+38a?+6-10a? by a?-4a+3.
15, 21m?-—27m-26m?+20 by 38m+7m?-4.
16, 18x?+24a3 — 40a?x ~- 9ax? by 927+ 7a? - 18ax.
17, 3y'-4y?+10y?+3y-2 by y-y7+3y+2.
18, 5a?+1+10at- 4a? by 5a8-2a+1.
19, l2a4*+5x3-332?-32+16 by 4a°-x-5.
90, pt-6p?+18p?-10p+7 by p?-3p4+2.
D1, 28x*+69x"+2— 71x? - 35x? by 40746 —- 13a.
99, 5a°—7a4— 9a? —-1la?-388a+40 by -—5a?+17a—-10.
93, «x°>-8a® by 2?+2axr+ 4a. 24, yi+9y?+81 by y?-3y4+9.
95, «vt+4y* by x2 + 2ay + 2y". 96, 9at-—4a7+4 by 3a?-4a+2.
D7, a®+64 by at-—4a?+8.
98, 1624+ 362°4+81 by 4%7+62+9.
99, 4m? — 29m -36 + 8:n?-—7m?+6m* by m? —2m?+3m -4.
BO, 15a4 +22 — 320° - 302 +502? by 3-—4x + 52°.
Bl, 3a?+8ab+4b?+10ac+8bce4+ 8c? by a+2b4+ 3c.
62, 9u?-4y?+4yz2-2% by -—38x+2y—-z.
63, 4c7-12c-—d?+9 by 2c+d-3.
84, 9p? -16q?+380p +25 by -38p-—4q-5.
O06 Ce y tery —e+2°—y by -x-y.
86, xv +a2ty—2%y?+23-Qay? by 22 +ay—-y?.
37, a°b?+ab—9 —b4+ 3b? + 3b - at -— 3a? - 3a by 3-b+a’°,
88, z4+1 by v4+a?+e4+1. 39, 2a°+2 by a?+2a?+2a+1.
40, 2°-—6a4-82°-1 by 2? -2e-1.
CHAPTER VII.
REMOVAL AND INSERTION OF BRACKETS.
54, Quantities are sometimes enclosed within brackets to
indicate that they must all be operated upon in the same way.
Thus in the expression 2a--3b—(4a—2b) the brackets indicate
that the expression 4a—26 treated as a whole has to be sub-
tracted from 2a — 30.
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 brackets is preceded by the sign +,
the brackets can be removed without making any change in the
CXPTeSSON.
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 — 3b) — (3a + 4b) — (b - 2a).
The expression = 2a — 3b —- 38a-4b-—b+2a
= a-8b, by collecting like terms.
55, Sometimes it is convenient to enclose within brackets
part of an expression already enclosed within brackets. For
this purpose it is usual to employ brackets of different forms.
The brackets in common use are ( ), { }, [ ].
56, When there are two or more pairs 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.
Hxample. Simplify, by removing brackets, the expression
a — 2b — [4a — 6b - {3a -c + (2a — 4b +¢)}].
Removing the brackets one by one,
the expression = a ~ 2b —- [4a - 6b — {3a —c+2a -—4b+c}]
= a—2b—[4a - 6b-38a+c¢-2a+4b -c]
=a—-2b-4a+6b+38a—c+2a-4b+¢
=2a, by collecting like terms.
Note. At first the beginner will 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, @+2b+ (2a - 3b). 9, a+2b-(2a-3b).
8, 2a-—3b—-(2a+25). 4, a-2-(4-8a).
5, (x—3y) + (2u—-4y) —(x-8y). 6, @+2b-38¢-(b-a-4e).
7, («-8y+2z) -(2-4y+4+2z). 8, 4a -(2y +22) — (38x - dy).
9. 2a+(b—3a) — (4a — 8b) — (6b — 5a).
10, m-(n-p)—(2m-2p+28n) -(n-m+2p).
1], a-b+c-(a+c-—b)-(a+b+c) -(b+e-a).
12, 5a—-(Ty+3x) — (2y+7x) — (8a + 8y).
13, (p-49)-(¢-2p)+(2p — 9g) — (p - 29).
14, 2a? - (3y?— 22) — (a? 4y?).
15, (m?— 2n?) — (2n? — 3m?) — (3m? - 4n?),
16, (#—-2a) —(x% — 2b) —{2a—a%-(2b+2)}.
17, (a+3b) -(b—3a) —{a+2b -(2a-3)}.
bee 9 (9° + 2p") —{p" £39? — (2p? — 9’)}.
19, x-[y+{e-(y-x)}} 20, (a—b)-{a-b- (a+b) -(a—D)}.
21, p-[p-(¢+p)-{p—(2p-4)}].
92, 8e-y-[w-(2y-2)-{2e—(y—2)}}
93, 38a? —[6a? — {8b? — (9c? — 2a7)}].
Q4, [Ba — {2a —-(a—b)}] - [4a — {3a — (2a — b)}}.
57, *xat=a7*4=aq%=1. [See Note; Art: 50.)
ab or ae pCa Coes el = in
5
Now clear of fractions by multiplying by 5 x 7.x 4 or 140;
72a — 108 + 45a +405 = 280x — 2800 ;
2800 — 108 + 405 = 280a — 72a — 45a ;
3097 = 1632 ;
Zhe:
81, Tosolve equations whose coefficients are decimals, we may
express the decimals as common fractions, and proceed as before;
but it is often found more simple to work entirely in decimals.
Example. Solve ‘3752 ~1°875 = '12%+1-185.
Transposing, *375a - 12% =1'185+41°875 ;
collecting terms, (°375 — *12)~ =3°06 ;
that is, 255x = 3°06 ;
= 3°06
"255
= 12,
EXAMPLES IX, b,
Solve the equations :
1, (%+15)(a% - 3) —(x%-3)?=30-15(x—-1).
Q. 15 -3a =(2x2+1)(2x-1)— (2x —1)(2x2 +3).
2 Q1—a(2a+1)+2(a—-4)(a +2) =0.
4, 3(2+5)—3(2a —1)=32—4(x —5)?+ 422,
5, 3a? -- Ta — (2 +2)(%—2)=(2+1)(%—1) + (a2 -3)(2 +3).
6, (x-6)(2e-9) - (11 -2x)(7 - x) =5a —4—7(a -2).
x-1l, «x-9_ CAS -
Lh Oe tere SEP ger ane
“+8 5,x-6 62-2 , 3x+5_1
iT wctetear ek Ook Gish ie
58
ALGEBRA. ; [CHAP, IX.
Solve the equations :
1a
13.
15.
iy,
19.
21.
22,
20.
24.
25.
26.
27,
28,
29.
30,
dl.
33,
30,
37,
38.
ee oer 2. 12, 2+84+%27=7420.
2-0 -e=-de>> & Cee al
yaar ta NU) P(g aa
+5 ae2+1 _2+3 16 1i-62° 9-72 _5(¢%-1)
6 9 Wir ; 5 Oot ewe O
47 — 6x tan ed) 4—52 1-2 -13
5 en Oe see 18, 6 Fea
3¢-1 w-1_ 2xa- 31 1 Be SH OF Sa ek
ees nay PD pees 5) | 4
“Ww 2 3 hs Satay ver
3 5 8 5
P(x = —6) +—.
$(@-1) ~2(a—4) = 3(e-6) +=
= hs ]
= =(2ar — -(x-1)-2
S (w@- 4) ~ 3(2n-9)=3(e -1)
= (+4) — 5 (w—8) = s(8x ~ 5)-3(« ~ 6) - H(e-2)
] ] 1 ]
(3 87} eee Oe 1
(8 8a) — 51 x) + 5 = 62)
1 1 Li 1
2 4 20 — a) =— (5a — 1)-_(5a%- 18) +8.
Ca )- 7 x) is! ae ) go 3)+8
e+1 52+9 2+6 x-12
2 ones ae
~ l0x+1 x _l3e+4 S(x- 4)
5 - =
hn, te 18 4
sd ae 6 24+5
a SOS F "i == 1 =e
ae (2-3) - + 0) + 7 0
a+4 1] a 1
ESS e217) a9 M6 Sbay Ss lees
0 Be a pee Le
sy eet Li oe Speed | x
Se = eee ebe hes |.
na AG a al 5)
“Ta — 3°35 = 6°4 -— 32a. 33, *Da4+- 254 °141°25= 4a,
2 20e— lore 94 Ltbx. 34, ‘2x- 0lx+°005z7=11°7.
Be — 6x = "75x —11. ‘ 96, -4r—'83x2=°7-°3.
Find the value of x which makes the two expressions
(8% —-1)(4a-11) and 6(22-1)(@-38) equal.
What value of « will make the expression 77x — 3(2a — 1)(4a — 2)
equal to 337 — 8(38%—-1)(@+1)?
CHAPTER X.
SYMBOLICAL EXPRESSION.
62, 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 7 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 is
greater than 7 by ais xv+a,
88. 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 x- 17.
Similarly the defect of x from 17 is 17-2.
Hxample 2. If x is one part of 45 the other part is 45 — x,
Hxample 3. If x is one factor of 45 the other factor is 49
6;
Example 4. How far can a’man walk in @ 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 5. If $20 is divided equally among y persons, the share
of each is the total sum divided by the number of persons, or ¢ 20.
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—4e quarters,
and y half-dollars=2y quarters;
.. the sum left=(4”%+2y—z) quarters,
=25(4%+2y—z) cents.
EXAMPLES X, a,
By how much does xz exceed 5?
By how much is y less than 15?
What must be added to a to make 7 ?
What must be added to 6 to make b?
By what must 5 be multiplied to make a?
What is the quotient when 3 is divided by a?
By what must 6x be divided to get 2?
By how much does 6% exceed 2a ?
. The sum of two numbers is 2 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 x, what is the third ?
11, The product of two factors is 4x; if one of the factors is 4,
what is the other ?
12, The product of two numbers is p, and one of them is m3
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 @ ; 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 48x; what is the value of
each number ?
18, How many numbers each equal to y must be taken to make
ldxy?
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?
.
COABDO POOH
x SYMBOLICAL EXPRESSION. 61
91, Ifthere are x numbers each equal to p, what is their product ?
99, If there are n books each worth y dollars, what is the total
cost ?
93, If mn books of equal value cost x dollars, what does each
cost ?
94, How many books each worth two dollars can be bought
for y dollars ?
95, If apples are sold at x for a dime, what will be the cost in
cents of y apples ?
96, What is the price in cents of m oranges at six cents a score ?
97, If I spend n dimes out of a sum of $5, how many dimes
have I left?
98, What is the daily wage in dimes of a man who earns $12
in p weeks, working 6 days a week ?
99, How many days must a man work in order to earn $6 at
the rate of y dimes a day ?
80, 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 50x cents ?
382, 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 g 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 »?
The two numbers consecutive to n aren+1 andn+2;
.. the sum=n+(n+4+1)+(n+2)
=3n+4+3.
And the product =n(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 +5 years old ; therefore his father
will then be 2(z+5), or 2x+10 years old; his present age must
therefore be 27+ 10-5 or 2%+5 years.
62 ALGEBRA. [onar.
KHxample 3. A and B are playing for money; A begins with $p
and B with g dimes. B wins $a; 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—« dollars, that is 10(p—a) dimes,
B has q dimes + dollars, that is g+ 10x dimes.
Thus the required equation is 10(p—x) =8(q+10z).
Hxample 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
es eae SA errs Be creea 5 hour ;
2 a
UAT Ieee ah eee Ch We an: hours.
Similarly the train travels b miles in = hours.
5
a
“, the whole time occupied is 7+ © hours.
25
Example 5. Wow many men will be required to do in p hours
what g men do in np hours?
np hours is the time occupied by q men ;
ne Lk HOUT Aa eas etee eee qx np men;
thatiis,-7 HOULSs 6), deeeeee ceanes = £2 een.
P
Therefore the required number of men is qn.
EXAMPLES X. b.
1. Write down three consecutive numbers of which a is the least.
9, Write down four consecutive numbers of which 0 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 is 2a7+1.
7, How old is a man who will be x years old in 15 years ?
8, How old was a man 2 years ago if his present age is m years?
9, In 2x yearsa man will be y years old, what is his present age ?
x. ] SYMBOLICAL EXPRESSION. 63
10, How old is a man who in # years will be twice as old as
his son now aged 20 years ?
11. In 5 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 algebraical
symbols.
13, A has 25 dollars and B has 138 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 2 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 ?
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 « men do a work in 5z hours, how many men will be
required to do the same work in y hours ?
20, How long will it take p men to mow g acres of corn, if each
man mows 7 acres a day ?
91, Write down a number which, when divided by a, gives a
quotient ) and remainder c.
99. What is the remainder if x divided by y gives a quotient z?
93, What is the quotient if when m is divided by n there is a
remainder r?
94, If a bill is shared equally among n persons, and each pays
75 cents, how many dollars does the bill amount to ?
95, A man has $x in his purse, he pays away 25 dimes, and
receives y cents ; express in dimes the sum he has left.
96, How many dollars does a man save in a year, if he.earns
$« a week and spends y quarters a calendar month ?
97, What is the total cost of 6% nuts and 4% plums, when x
plums cost a dime and plums are three times as expensive as
nuts ? bs
98. 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 ?
CHAPTER XI.
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 :
Represent the unknown quantity by a symbol, as 2, 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 LX.
Example I, Find two numbers whose sum is 28, and whose
difference is 4.
Let x be the smaller number, then x +4 is the greater.
Their sum is x+(x%+4), which is to be equal to 28,
Hence xet+u+4=28;
Ue = 243
w= 12,
and x+4= 16,
so 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
$10 more than B, and B $8 more than C.
Let « represent the number of dollars that C has ; then B has
x+8 dollars,;and A has «+8+10 dollars.
Hence e+ (#+8) + (#+8+10) =47 ;
¢+%4+84+2748410=47,
34221 3
Bs | ;
so that C has $7, B $15, A $25,
CHAP, XI.] PROBLEMS LEADING TO SIMPLE EQUATIONS. 65
EXAMPLES XI, a,
], Six times a number increased by 11 is equal to 65; find it.
2, Find a number which when imultiplied by 11 and then
diminished by 18 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.
“ 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 $30 more
than A.
10, Divide $66 between A, B, and C so that B may have $8
more than A, and C $14 more than B.
11. A, B, and C have $72 among them; C has twice as much
as B, and B has $4 less than A ; find the share of each.
12, How must a sum of 78 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 —-~z is the less.
Three times the greater part is 3x, and its excess over 100 is
32 — 100.
Eight times the less is 8(60- x), and its defect from 200 is
200 — 8(60 - z).
Whence the symbolical statement of the question is
3x — 100 = 200 — 8(60 — x) ;
3x — 100 = 200 — 480+ 82,
480 -— 100 — 200 = 8a - 32,
Da = 180%
x = 36, the greater part,
and 60 — ax = 24, the less.
H.A. E
66 ALGEBRA. [omar
Example IV, A is 4 years older than B, and half A’s age
exceeds one-sixth of 5’s age by 8 years; find their ages,
Let x be the number of years in B’s age, then A’s age is 2 +4 years.
One-half of A’s age is represented by $(~+4) years, and one-sixth
of B’s age by ta years.
Hence 3(xv+4)-4r=8;
multiplying by 6 3x+12-—x%=48;
Vi oe
hoped Noe
Thus B’s age is 18 years, and A’s 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.
90. If A, B, and C have $76 between them, and A’s money is
double of B’s and C’s one-sixth of B’s, what is the share of each?
91, Divide $511 between A, B, and C, so that B’s share shall be
one-third of A’s, and C’s share three-fourths of A’s and B’s together.
29, B is 16 years younger than A, and one-half 5’s age is equal
to one-third of A’s ; how old are they ?
93, A is 8 years younger than J, and 24 years older than C;
one-sixth of A’s age, one-half of 5’s, and one-third of C’s together
amount to 38 years ; find their ages.
94, Find two consecutive numbers whose product exceeds the
square of the smaller by 7. [See Art. 84, Ex. 1.]
95, 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, and B has six dimes; after B has won
from A acertain sum, A has then five-sixths of what B has; how
much did B win ?
XI.]} PROBLEMS LEADING TO SIMPLE EQUATIONS. 67
Suppose that B wins # dimes, A has then 60—«a dimes, and B
has 6+” dimes.
Hence 60—»=2(6+4+2);
360-—6x—= 30+ 52,
Lins sa0%
7=380.
Therefore B 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’s age be x years, then A’s age is 2% years.
Ten years ago their ages were respectively «—10 and 27%—10
years ; thus we have
2x—10=4(x%—10) ;
2x—10=4a—40,
2x%—= 30 ;
a 15,
- so that Bis 15 years old, A 380 years.
EXAMPLES XI. b.
1, A has $12 and Bhas $8; after B has lost a certain sum to A
his money is only three-sevenths of A’s; how much did A win ?
9, Aand B begin to play each with $15; if they play till B’s
money is four-elevenths of A’s, what does B lose ?
8, Aand 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, Ahad 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. A is 20 years older than B, 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 15 years ?
8, Ais twice as old as‘’B; 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 ALGEBRA. [cuap.
Example III. A person spent $56.40 in buying 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 « be the number of geese, then 108—~ is the number of ducks,
Since each goose costs 7 dimes, x geese cost 7x dimes.
And since each duck costs 3 dimes, 108—w ducks cost 8(108—~)
dimes,
Therefore the amount spent is
7x+8(108—«a) dimes.
But the question states that the amount is also $56.40, that is 564
dimes.
Hence 72+8(108—2) =564 ;
74+ 824 —3x=564,
4%=240,
*, =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’s share”’’ or ‘‘let x=the
ducks,’’ or any statement so vague and inexact.
It will sometimes 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 9% cents,
and 16 eggs cost 16x cents ;
oe 9x%—25=—75—162,
’ 25x%= 100 5
eA
Thus the price of an egg is 4 cents, and the number of eggs
=100+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 15 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 ?
138, 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 50 coins, which are either quar-
ters or dimes ; how many are there of each ?
15, A sum of $11.45 was paid in half-dollars, quarters, and
dimes ; the number of half-dollars used was four times the number
of quarters and ten times 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 $15.10, and in all gets 24 lbs. ; how
much of each did he buy ?
17, A man sold a horse fora sum of money which was greater
by $68 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.
990, Asum 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 $45, and C and A have $53; what is the share of each ?
91, 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 ?
99, By buying eggs at 15 for a quarter and selling them at a
dozen for 15 cents a man lost $1.50; find the number of eggs.
93, I bought a certain number of apples at four for a cent, and
three-fifths of that number at three for acent; by selling them at
sixteen for five cents I gained 4 cents ; how many apples did I buy ?
94, If 8 lbs. of tea and 24 lbs. of sugar cost $7.20, and if 3 lbs.
of tea cost as much as 45 lbs. of sugar, find the price of each per
pound.
70 ALGEBRA. [CHAP. XI.
95, Four dozen of port and three 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.
96, Aman 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.
97, 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.
98, A sum of money is divided between three persons, A, B,
and CO, 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.
99, The difference between two numbers is 15, and the quotient
arising from dividing the greater by the less is 4; find the numbers.
380, 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 ?
381, Find two numbers which are to each other as 2 to 3, and
whose sum is 100.
382, 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 3 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.
338, The sum of $34.50 was given to some men, women, and
children, each man receiving $2, each woman $1, and each child
50 eents. 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 8 cents and the
remainder at 2 for a cent, gaining 10 cents; how many did he
buy ?
35,
Example 2. The highest common factor of a?b4, a*b°c?, atb’c is
a*b*; for a? is the highest power of a that will divide a’, a?, at;
b+ is the highest power of b that will divide b+, b°, b7; 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
coefficient to the algebraical highest common factor.
Hxample. The highest common factor of 2latz*y, 35a7xty, 28a%ay
is Ja*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 XII. a.
Find the highest common factor of
eeoceD aur, Doc Ary, ,. 0c’, 50°C. PT ns hea Verne
Dewar cn auc.) Ga.eaco,Oabe. {, 6a*ytz, Qay. 8, lby®, Sry fe,
9, 12a%bc?, 18ab?c, Oe iecyr a: Zl aae.
1]. 8ax, 6a*y, 10ab?x?, Lee oes OF cary”,
18, 14bc?, 63ba?, 56b7c. 14, lary, 60a5y72*, 25022",
15, Ixy’, Slayz, 34a°yz «16, Tia°bic?, B3a%%c%, ab ct,
72 ALGEBRA. [CHAP.
Lowest Common Multiple of Simple Expressions,
91, Derrryirion. 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 multiple.
92. In the case of simple expressions the lowest common
multiple can be written down by inspection.
Hxample 1. The lowest common multiple of a4, a3, a?, a® is a®,
Example 2. The lowest common multiple of ab‘, ab®, a?b7 is
a®b’; for a? is the lowest power of a that is divisible by each of the
quantities a®, a, a7; and b7 is the lowest power of b that is divisible
by each of the quantities b*, b°, b’.
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 2latz*y, 35a°aty,
28a°®xy is 420ata4y ; 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
Reo, oY, 2, abt abe, Bee ory er.
A 4a, Baber. 5, 4a4bc?, 5ab?. 6. 2ab, 4ay.
7, mn, ni, lm. OP yy oor eee: 9, xy, 8yz, 4zx.
LOmee oT Ag pt Pg: ell ploa ys barye. 1357 9ab?, 2107e:
13. 27a?; 816%, 18076": [Ae har Gey ait 6
Lo mewod 0, 2007 yo0z. 16. 72p79°r4, 108p%97r.
Find both the highest common factor and the lowest common
multiple of
17, 2ab?, 3a2b3, 4a4b. 18, 1523 y?, 5a?yz°. 19, 2a4, 8a2h3c7,
OD eS ia coe. 21, 32a*b®c, 48a7bc°.
99, 5lim®p?, pn, 34mnp'. 93, 49a4, 56b4c, 2lac’.,
94, 66a°*b%cat, 55ab’ayrz, 12a? yz’.
XII] ELEMENTARY FRACTIONS. 73
Elementary Fractions,
94, Derrixition. If a quantity w be divided into b equal
parts, and a@ of these parts be taken, the result is called the
Jraction ; of x. If « be the unit, the fraction of w is called
simply “the fraction @ ”+. go that the fraction
ply b
b
represents a
equal parts, b 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 WSchools,
Chapter xv. :
Rule, Zo reduce a fraction to rts lowest terms: divide
numerator and denominator by every factor which is common to
them both, that 1s by their highest common factor.
Dividing numerator and denominator of a fraction by a com-
mon factor is called cancelling that factor.
ESOS ae Lee
35a°b%c _ 5atb _
4
Tab-c 1 oe
EXAMPLES XII, c,
Reduce to lowest terms :
2.3 oe 3
2203 3/2 af 232
d. “Tea ; 10. eye H. yee La. iba
a Beaty Eee ag Ses ag Be
74 ALGEBRA. [CHAP.
Multiplication and Division of Fractions.
96. Rule, Zo multiply algebraical fractions: as in Arith-
metic, multiply together all the numerators for a new numerator,
and all the denominators for a new denonunator.
Hxample | 2a x 5a" v, BD? (QO Dae oUe ae
é . 3b 2a7b on ~ 3b x 902 x on oq,”
by cancelling like factors in numerator and denominator.
8a°b a
Do Oras 0.
all the factors cancelling each other.
97, Rule, Zo divide one fraction by another: invert the
divisor and proceed as in multiplication.
Ta? Boke, 28a%e?
4a3y? 5ab? 15b%xry?
= Ta? x Beta ——— 15b*xy?
42°y2" Bab? 28a?
Hxample 2.
Example.
all the other factors cancelling each other,
EXAMPLES XII. d.
Simplify the following expressions :
ay, a°b? ab . 4c*d FLEE lie ANE
1. ab ‘ye 2. Qed abe 3: 3y°z * date
Gara? 140% 3ab? 15b7c* Tee eee
= ; SK pees fic ® :
e Tab? c l2ax 5. 5b’c— 9a2b* 6, Bbc? ‘ l4be
7 am ,, 2cd? Dmy, g 4a*b , 3p7q? . PY
' By *3ab 4am? Yay * Babb? a2 ee
9 20% ,, 100? «te i 10 Uae iW Oe 34/3 ye
' Bat de®” 308" "gad ged” oe by
Sore") Sax? xy? 15b2_ 14d? , 81d?
x . cr °
Ll. aby * Bate * by 12. Fc * abe BIA
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.
XI. J ELEMENTARY FRACTIONS. 75
Hxample. Express with lowest common denominator the fractions
age anak
Say’ 6xyz Qyz
The lowest common multiple of the denominators is 6ayz. Multi-
plying the numerator of each fraction by the factor which is required
to make its denominator 6xyz, we have the equivalent fractions
20 eee 3cx
6xyz 6xyz Bayz
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 :
Tee oer Be a 4, =
Bay ae 6. ay He lh @ = 8, a =
9. ? & 105); Qa. Lge ae -
Bowe Bowe wWoaet
Addition and Subtraction of Fractions.
99, Rule. To add or subtract fractions: express all the
fractions with their lowest common denominator ; form the algebrat-
cal sum of the numerators, and retain the common denominator.
5x 712
. Simplify 27 4?x-/™,
Example 1. Simplify 5 sie 7
The least common denominator is 12.
202+92—l4a_15x”_ 5x
The expression = 12 =i puepie
» ae 3a ab ab
Example 2. Simplify Pe Fy
76 ALGEBRA. [CHAP. XII.
ey
azc2 3ca®
. axe — C2 : Wek :
The expression = ee and admits of no further simplification.
CG: Ce
Example 3. Simplify
Note. The beginner must be careful to distinguish between
erasing equal terms with different signs, as in Example 2, and
cancelling 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 as a whole.
6ax —cy
cae
and not the whole numerator. Similarly a cannot be cancelled
because it only divides 6ax% and not the whole numerator. The
fraction is therefore in its simplest form.
Thus in c cannot be cancelled because it only divides cy
When no denominator is expressed the denominator 1 may
be understood.
te OL Ooty a0
Example 4. ee ee ee Be
xcample 3x dy elias vy,
If a fraction is not in its lowest terms it should be simplified
before combining it with other fractions.
vi : at ay _ax x _sav—2Qe
xample 5 5 aay a 5
EXAMPLES XII. f.
Simplify the following expressions :
ee ee ee oe
5 5-5 DD Gey TB rap Bh Bom
9, ate 10. os 11, aoe 12, re
Ry Sa ee 14, Pees 15, 18-2
Be ee uae ae
19.6 2 20 en Bes Sheen
93, oe dx, xy OA, a= a? ae
~ dy t By 3a ab? 6be
CHAPTER] XLT,
SIMULTANEOUS EQUATIONS,
100, ConstpEr the equation 22+ 5y=28, which contains two
unknown quantities.
By transposition we get
By =23 — 2x ;
that is, a= abs nd, RR (1).
From this it appears that for every value we choose to give
to # there will be one corresponding value of y. Thus we shall
be able to find as many pairs of values as we please which
satisfy the given equation,
For instance, if v=1, then from (1) we obtain 2
Again, if = —2, then ya ; and so on.
But if also we have a second equation containing the same un-
known quantities, such as 8x+4y=24,
we have from this y 7 Ree (2).
If now we seek values of w and y which satisfy both equa-
tions, the values of y in (1) and (2) must be identical.
Therefore pe ee era,
5 4
Multiplying across 92-—8r=120—-152;
(LZ 5
xv=A4,
Substituting this value in the first equation, we have
8 + 5y = 23 ;
w. SY=15;
con ah
and c=4,
Thus, if both equations are to be satisfied by the same values
of # and y, there is only one solution possible.
78 ALGEBRA. [CHAP.
101, Derrryition. When two or more equations are satisfied
by the same values of the unknown quantities they are called
simultaneous 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 svmultaneous equation ; 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
only of the unknown quantities.
1038. 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 BOA TY =O) senna ote shee eee (1);
BiH 2y = 16.20 oes ue oe ee ee ee (2).
To eliminate x we multiply (1) by 5 and (2) by 8, so as to make
the coefficients of 2 in both equations equal. This gives
15x + 35y = 135,
. ldx+ by = 48 ;
subtracting, 297 = 877;
i MN oa;
To find x, substitute this value of y in ether of the given
equations.
Thus from (1), Noe Spd Ri Ear
meas
and. y = tf
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
5a+6=16;
a = 2, as before.
XIII. ] SIMULTANEOUS EQUATIONS. 79
Example 2. Solve PRS ick LIE ech ESE ee eee (1),
aoe ELEM Ae agile pe cor
Here it will be more convenient to eliminate y.
Multiplying (1) by 2, l4x+4y = 94,
and from (2) 5a —4y=13
adding, 19% = 95 ;
Goa
Substitute this value in (1),
30 + 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 Lat NO ah cise aeantiaate ks soe nite a asst ie Ag
QE OY aieniians wie NE he cote en stens (2).
Here we can eliminate x by substituting in (2) its value obtained
from (1). Thus
24 — T5y +1) = By;
48 — 35y —7 = by ;5x
4] =41y;
noe fit
and from (1) La.
104, Any one of the methods given above will be found
sufficient ; but there are certain arithmetical artifices which
will rnorincn shorten the work.
Hxample. Solve DSA cO UY eat anon Sar Enki eigs ei tst ete reeset ‘ae
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 252ax —207y = 198,
252% —-220y = 68;
subtracting, 13y = 130;
that is, y = 10,
and therefore from (1), x= 9,
[CHAP.
80 ALGEBRA.
EXAMPLES XIII. a.
Solve the equations:
l, x«+y=19, 9. ety = 23, oO ety=ail;
See TE Li Yeo. ey = —9,
4, x+y = 24, he, ey, G6, w—y= 25,
x-y= 0. et 8/0. cry = 13,
7, 3a2+5y = 50, 8, x+5y= 18, 9, 4%+ y=10,
44+ 3y = 41. 3a+2y = 41. 5u+Ty = 47.
10, 7 -6y = 25, ll, 5e+4y=7, 12, 3¢—Ty=1
5a+4y = 51. 4x + 5y = 2. 4x+ y= 53.
18, 7x+5y = 45, 14, 4¢+5y =4, Lijelig= fy at433
Qu - 3y = 4. 5x — 3y = 79. 2x — 3y = 13.
16, 4x-3y=0, 17, -2x+3y = 22, 18, 72+3y = 65,
7a —4y = 36. 5a +2y = 0. iz —8y = 32.
19, 1l3e-y=14, 20. 9ux-8y = 14, Q1, 14%+13y = 35,
2a —7Ty = 9. 15x - 14y = 20. 21x+19y = 56.
0D. bf ay 21, Doaoot = sor, 04, bdea7y= it,
21x —9y = 75. 10x” = Ty — 15. etn a bap
95, 138x-9y=46, 26, 6x-5y=11, 27, lly-llx=66,
llz-12y=17. 28a+21y = 7. 7x +8y = 3.
98, 6y-5x =11, 29, 32+10=5y, 80, 4y =47+4+32,
4a = Ty — 22. Ty = 4¢+13. 5a = 30- ldy.
ol, 12+138y 57, Cols lly ieee Glee
13x4+ lly = 17, 292 — 39y = 17. 4lz+37y = 17.
105, We add a few cases in which, before proceeding to
solve, it will be necessary to simplify the equations.
Example, Solve 65(%+2y)—(38a+ ly) =14 crrcceccccceccceneeees (1);
Ja Oy Sie — dy) a 38 ase ee (2),
From (1), 52+10y - 3x -lly=14;
Dae aap ct We ee vargas eaenecne toes (3).
From (2), 7x —9y -3x+12y = 38;
Ape 38 Ure, ee (4).
From (3), 6a — 3y = 42,
By addition, 10z = 80; whence x = 8,
From (8) we obtain y = 2.
xu. ] 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 —4 ie = =e Lek hy a eee Cy
i
By +4 — lig. _ aie a
5 3(2e Weare Ieee cartes veer ecnecises es Cay
Clear of fractions. Thus
from (1), 42a —2y +10 = 287-21;
VA Ga Le NO has cnt casiceaee ate arse (3).
From (2), 9y +12 --10%+25 = ldy ;
LOS UG Spencer yes clea castes steeds ev (4).
Eliminating y from (38) and (4), we find that
oe
13
Eliminating x from (3) and (4), we find that
y 2 20!
26
107. Simultaneous equations may often be conveniently
ee 1 uh ee
solved by considering — and ~ as the unknown quantities.
we o
Example. Solve 2 se Sat LO come nat teestie Sin aennces Rese tee ce ‘id Ws
Ae ey
EOE TER else en: Ne ee OP (2)
te Ms
Multiply (1) by 2 and (2) by 3; thus
16_18_y
oY
30 | 18 = 9A >
og wes
adding, a0 8 ;
%
multiplying up, 46 = 232 ;
fig A
and by substituting in(1), 4 =3.
HLA. F
82 ALGEBRA. [cuar.
EXAMPLES XIII. b.
Solve the equations :
1, 2x-y=4, 9, 4e-y=1, 3. £+2y = 13,
Bed Ep eed ee US
Did en ne 35
HA a eye Bx a
4, eae 5, 2 =O, 6, eae G
%+3y = 2 4x -3y=1 4x + 5y = 0.
Ty bare Ay, 8, «x-y=9, Q, x+y=-2,
4x By _ Bd 08 e1Y_o
i ae ye any 46
1 3 ] ]
Ee =I, é Sop oD) = 20); ‘ Rig 20) =
10; 5(v +3) 0 11 pe ~ 2y 0 12 ate al 0
sey = 4h s(y-+8) = 2. Sm = Dy
13. 3(v@-y)+2%a+y)=15, 38(at+y)+2(x-y) = 25.
14, 3(@+y-5)=22(y—-x), 3(v-—y-7)+2(r4+y-— 2) =0.
15, 4(2x-y-6)=38(8%-2y-5), 2x-y+1)+4x = 3y+4.
16. 7(2a—y)+5(8y—4x7)+30=0, 5(y-—2+3) = 6(y— 22).
+4 y-4_ oe le ya lis ot oy
17. yma ee 18. 25
peak 90, 24° =39, Ole eae
cy “x sy eae ats
bigs 708 Big SoEE
aay ay “x y
108, In order to solve simultaneous equations which contain
two unknown quantities we have seen that we must have two
equations.
Rule.
be solved by the refes already given.
Similarly we find that in .erder to solve simul-
taneous equations which contain three unknown quantities we
must have three equations.
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
The remaining unknown is
then found by substituting in any one of the given equations.
XIII. | SIMULTANEOUS EQUATIONS. 83
Example. Solve PRO Orees etme coe Pee bess saa seecdaceeerts (1),
Miri as Vice go ycame 1 uae ae fo ane eee Retry (2),
DOE EA Ber SN es ys as bog teeceaiy orsniehaanes (8).
Choose y as the unknown to be eliminated.
Multiply (2) by 5, 20%+10y-15z=0;
Multiply (1) by 2, 14%+10y-—142= - 16;
by subtraction, O75 —@ 1 On see testare sees keoe' beuvce, (E)e
Multiply (2) by 2, 8x+4y-62=0;
from (3), 5a — 4y + 42 = 35 5
by addition, 13x — 2z = 35.
Multiply (4) by 2, 122% —2z = 32;
by subtraction, Sr Ds
From (4) we find a= 2,
and from (2), y=-3.
109. Some modification of the foregoing rule may often be
used with advantage.
Example. Sol ees ba tO.
xample olve 5 re id
ja teen as
379 2
From the equation 5 Sle a+ 1,
we have a 2) deride eral oA ee Peccceeeseae eomeeh (1).
Also from the equation 57 is +2,
we have rh Mio) ah re aE Nec tence an odds Aras avosalloas ieee (2).
And from the equation at5e 13,
we have Ri aa eae) ee ee Tatar ceestbycaetes tas (3).
Eliminating z from (2) and (3), we have
21x +4y = 282 ;
and from (1} 12x -4y = 48 ;
whence #=10, y=18. Also by substitution in (2) we obtain z= 14,
84
ALGEBRA.
EXAMPLES XIII.
Solve the equations :
Ub,
co
Lt,
12,
13,
14.
15,
16,
3x —2y+2= 4, Dh.
2x + 3y —2= 3,
Lt Y+tZa S,
xv + 2y +32 = 32, 4,
da — dy + 6z.=.27,
7x +8y —9z = 14,
7«—4y-3z= 0, 6,
5x — 3y + 22 = 12,
34+2y-5z= 0.
3y — 62 —5a = 4, 8,
22 -3dx- y=8,
x —2y+22+2=0.
10,
dar (zy) = 11,
Qe sey) =2, Haty) = 78-2),
:
3
(CHAP. XIII.
ox + 4y — 6z = 16,
4eu+ y-— z= 24,
x —d3y—225
LNs Rp Sema 4¥,
62 + dy + 2z = 84,
da + 4y — 5z = 13.
4x+3y- z= 9,
9G = Omer}
wAYy — 32 =) 2.
by +22 +52 = 21,
82 — 32 9 = 8,
22 +22 -3y = 39.
(z-4x) =y.
w= 4y +32.
GCHAPTER: XLV;
PROBLEMS LEADING TO SIMULTANEOUS EQUATIONS.
1190, Ivy 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 yt en atta saree a ae eres § Gh):
Also oe =o,
or GY = 40 S232 head eee toe (2).
By addition 22=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 « cents,
a eet COMLCE aie iae Ys aaa ele
Then from the question we have
1g Oy = la bW aes aes weNierds ables bia 8 Ly,
PAVERS GD ERO 115 Diya eer ey ee (2).
Multiplying (1) by 5 and (2) by 38, we have
752+ 50y=7750,
752+ 89y =7365.
Subtracting, weLiv— ooo.
Y= 35.
And from (1), 15% +850 = 1550 ;
whence tba 1200 5
MzaS0):
.*. the cost of a pound of teais 80 cents, .
and the cost of a pound of coffee is 35 cents.
86 ALGEBRA. [CHAP.
Hxample 3. In a bag containing black and white balls, half the
number of white is 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 7 +y balls.
We have the following equations :
pag AI HO Neh RAM, l
2 3 es. ( )
2A Y) =H OY EE vacenceatessesessescescststeaae (2).
Substituting from (1) in 2, we obtain
AY 4 y= By +4;
whence Uai2:
and from {1), x= 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
4x100+3x10+5. Similarly a number whose digits beginning
from the left are x, y, z
=x hundreds+y tens+z units
= 100% + 10y +2.
Hxample. 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 ;
tind 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 10x+y, and the number
formed by reversing the digits will be represented by 10y+2.
Hence we have the two equations
LODE 83 (20) ee coetans cout oe eee (1),
_and 10% dD = LOG 12s as ssi vacks ees creek wares (2).
From (1), Ley
from (2), yee),
From these equations we obtain x = 2, y = 7.
Thus the number is 27.
XIv.] PROBLEMS LEADING TO SIMULTANEOUS EQUATIONS. 87
EXAMPLES XIV.
1, Find two numbers whose sum is 54, and whose difference is
12
foal
9, Thesum of two numbers is 97 and their difference is 51; find
the numbers.
8. 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.
5, ‘Four sheep and seven cows are worth $131, while three cows
and five sheep are worth $66. 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 A’s age exceeds three times B’s age by 2 years ; if
the sum of their ages is 61 years, how old are they ?
8, Half of A’s age exceeds a quarter of B’s age by 1 year, and
three-quarters of B’s age exceeds A’s by 11 years; find the age of
each ? .
9, Ineight hours C walks 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 dees 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 $15 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 ?
18, 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 18 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, . 31, (—2m2n’)6, 32, (—a3y2V’,
mG) me CBR on CH we
To Square a Binomial.
115, By multiplication we have
(a+6)?=(a+b)(a+b)
07 200 Bs on. 5 ats ate eeoeee (1);
(a—b)?=(a—b)(a—b)
= P= 200 4-07. aa sche steals ae
Xv. ] INVOLUTION. 91
These formulz may be enunciated verbally as follows :
Rule 1. The square of the sum of two quantities ts equal to
the sum of their squares increased by twice their product.
Rule 2. The square of the difference of two quantities is equal
to the sum of their squares diminished by twice their product.
Hxample 1. (2+ 2y)? = 274+2.%.2y+(2y)?
= x7 + 4ay +4y*.
Example 2. (2a3—8b?)?= (2a3)?—2 . 2a3 . 3b?+ (8b?)2
= 4q°— 12a3b?+ 964.
To Square a Multinomial.
116, By the preceding article
(a+b+c)?={(a+b)+cP
=(a+b)?+2(a+b)e4+c?
= a*+ b?+ 7+ 2ab+2ac +4 2be.
In the same way we may prove
(a—b+cP=a?+b?+ 0? —2ab+ 2ac— be.
(a+b+c4+d)?=040+C4+d*+2ab4+2ac+ 2ad+ 2be+ 2bd + 2cd.
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 square terms are always positive.
The same laws hold whatever be the number of terms in the
expression to be squared.
Rule. Zo find the square of any multinomial: to the sum of
the squares of the several terms add twice the product (with the
proper sign) of each term into each of the terms that follow tt.
Ez. 1. («¢-—2y—382z)?=x? + 4y? + 92? --2.%.2y—-2.4.3824+2.2y.32
= 207 + 4y? + 927 — day — 6az+ 12yz.
Hix. 2. (1420 -—32?)?=1 4+ 40?+4 9at+2.1.27-2.1.382?-2.27.32?
=1+427+ 924+ 4a - 6x? — 1223
=]+4a — 2x? — 124° 4 9x4,
by collecting like terms and rearranging.
92 ALGEBRA. [CHAP. xv.
EXAMPLES XV. b.
Write down the square of each of the following expressions :
apa. 27; Dae = 2y: 3, a+30. 4, 2a-3b.
5, 38a+b. 6, x-—5dy. 7. 2n+7n. 8, 9-2.
9, 2-ab. 10, abc+l. ll, ab-cd. 12, 2ab+ay.
Tope = 22 14, 34+2pq. 15, x?-3a. 16, 2a-+ab.
17, atb-e. 18, a-b-e. 19, 2a+b+¢.
20, 2u-y-2 91, «x+3y—-2z. 90, we ea +1,
93, 3x+2p-q. 94, 1-22-32. 95, 2-8u+22
26, xty+a—). 27, m—n+p—4q. 28, 2a+8b+%—2y.
To Cube a Binomial.
117. By actual multiplication, we have
(a+b) =(a+b)a+b)(a+6)
= 03 +3a°b + 3ab? + b*.
Also (a — 6b)? =a? —3a7b + 3ab? — Bb’,
By observing the law of formation of the terms in these
results we can write down the cube of any binomial.
Example 1. (2a +y)3 = (2x)? + 3(2x)?y + 3(2ax)y? + y?
= 8a? + 122°y + 6xy? + y?.
Example 2. (8% - 2a?) = (38x) — 3(3x)?(2a*) + 3(8a)(2a?)? — (2a?)3
= 272° — 542°a? + 36xa4 - 8a,
EXAMPLES XV. c.
Write down the cube of each of the following expressions :
Le 2, m-n. 3, a- 2b. 4, 2c+d.
5, x+3y. 6, «x+y. Ley: 8. 5442.
Re rd be 10, 22?-+y7. ll, 2a%-30% 12, 4y?-3
CHAPTER XVI.
EVOLUTION.
118, Derinirion. 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 even 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.
Hxample. J9a?x® = +3a2°,
In the present chapter, however, we shall confine our attention to
the positive root.
Examples. ab! = 032, because (a%b2)? = a4.
/ — x9 = — x, because (- 2°)® = - 2,
Nc = ct, because (c4)> = c%,
A/81a!2 = 3x3, because (323)! = S1x!2,
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 by Arithmetic, and
prefix the proper sign.
(2) Divide the exponent of every factor of the expression by the
index of the proposed root.
Hxamples. —642° = — 427,
/ 16a8 = 2a2,
/81a1° = 9x°
25c8 Bc?
94 ALGEBRA. [oHAP.
EXAMPLES XVI, a,
Write down the square root of each of the following expressions :
{bap ese 2.6 20a Us 3, 49c7d%, A abe
De Shaw ek 6, 162%. fis 0 avian Seopa
4x6 ae isa. 144
9, 16a4 10, 36° ag “O5 iby a2
Write down the cube root of each of the following expressions :
13, x*y®. 14, -a%®, 15, 82%, 16) 2/24
pe bss 8a%h? 1250727) 64a77b?
ive. 27° 18, “ys 19, ky [ae 20, ae nee
Write down the value of each of the following expressions ;
Dinan ey. DDN arta O38 N= ete
24, N6ta®. 25, Na?tb™, 26, Npq”.
97, N= x35y56, 298. N8laty®?. 29, §/320>bMe%,
121. By the formule in Art. 115 we are able to write down
the square of any binomial.
Thus (20 + 3y)? =4u? + 12xy +97".
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 25a? — 40xzy + 16y’.
The expression = (5a)?- 2. 20xy+ (4y)?
= (5x)? — 2(5a)(4y) + (4y)?
= (5x - 4y)?.
Thus the required square root is 5x - 4y.
64a? 32a
Example 2, Find the square root of OOF UL Rae
The expression = 3 i + (2)? + 2( 27 )
Thus the required square root is ee +2,
XVI. ] 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 student is advised, here and elsewhere, to employ methods
of inspection in preference to rules.
To Find the Square Root of a Compound Expression.
123, Since the square of a+b is a?+2ab+b%, we have to dis-
cover a process by which a and 0, the terms of the root, can be
found when a?+2ab+0? 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 +? or (2a+b) x 6.
Now the first term 2ab of the remainder is the product of
2a and b. Thus to obtain 6 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+.
The work may be arranged as follows :
a’ +2ab+b? (a+b
ae
Qat+b 2ab +b?
2ab+ 6?
Example Find the square root of 9x? — 42xy + 49y.
9x? — 42xy + 49y? ( 8a - Ty
9x?
6xu — Ty | —42xy + 49y? ©
| — 42xy +49y?
Explanation. The square root of 92? is 3%, and this is the first
term of the root.
By doubling this we optain 6x, which is the first term of the
divisor. Divide —42zy, the first term of the remainder, by 6x 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
remainder, 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
2527a? — 1220? + 1624 + 4a4 — 24270,
Rearrange in descending powers of x.
16a4 — 2443 + 25x70? — 12x03 + 4a4 ( 4a? - 8xa + 2a?
16x4
82? — 3x0 — 2420, + 25270?
—24e2a+ 9x02
82? — 6za + 2a? 1622a? — 12xa? + 4a4
16z7a? — 1220? + 4a4
Explanation. When we have obtained two terms in the root,
4x? — 32a, we have a remainder
1627a? - 12203 + 4a’.
Double the terms of the root already found and place the result,
8x2 —6xa, as the first part of the divisor. Divide 16xz°a?, the first
term of the remainder, by 8x, 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
4a? +b? +c¢?+4ab — 4ac — 2be.
Arrange the terms in descending powers of a, and let the other
letters be arranged alphabetically ; then
the expression = 4a? + 4ab —4ac + b?- 2b¢4+ c?
= 4a?+4a(b-c)+(b-c)?
= (2a)7+2. Qa(b-c)+(b-c)?;
whence tke square root is 2a+(b-c). [Art. 121.)
XVI. ] EVOLUTION. 97
EXAMPLES XVI. b.
By inspection or otherwise, find the square root of each of the
following expressions :
1, @-8a+16. Q, «?+142+49.
8, 6444824 922. 4. 25-30m+4 9m?.
5, 36n4-84n?+49. 6, Sl+144y?+64y'%,
T, 2° — 62% ytz4 + 9y8z8, 8, 4a2b4 — 12ab2c 4+ 919,
ee ons 6 9a? , 24ac , 16c?
9, ra 3xy? + 9y?. 10; yet bd +—a9-
9a? 2562 16a? , 49y4
Daa 0 all) a achat bd 2x4
Nl. psp 2+ Gar 12. Foye" Tea oY
18, 1624-3223 +242?- 82+].
14, 25 -30a+ 29a? - 12a°+ 4a4.
15, 9a®-12a° - 2a4+4a?+1.
16, 25p*- 30p? + 121 — 101p?+ 66p.
17, 8v?+14+4a4- 42.
18, 201a?- 108a? + 100 + 36a4 — 180a,
19, a?+0?+c¢?+2ab —2ac — 2be.
90, yet t+ 22x? + Py? — Qa?yz + Qary?z — Qarye.
307 1
4— q+ =
Q1, at—2ab+ > - 5 +i5
93, Imi+ 4 8 OO mtn
94, 9at+ 144074 12ax?+ 4a? — 720° - 4802.
95, x8 —4ax7 4+ dao + 62° — 1404+ 40° + 9a? -6r4+1.
96, a?+9b?+c?- 6ab+6be —2ac.
27, Mp crepe 47 +4,
n* n? n?
9a? b? 6a 2b.
—-§4+—-— pe ze
oF az 0b
H.A. G
98 ALGEBRA. [OnAr.
{ 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?+3a%b+3ab?+b3, we have
to discover a process by which a and b, the terms of the root,
can be found when a?+3a%)+3ab?+ 03 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 + 3ab? + D3 or (3a°?+3ab +b") x b.
Now the first term of the remainder is the product of 3a?
and 6. Thus 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. Three times the product of this first term a, and the new
term 0.
3. The square of 6.
The work may be arranged as follows :
a +3ab+ 3ab?+b? (a+b
he
3(aPr = 3a? 3a°b + 3ab? +3
3xaxb= +4+3ab
Qe = +0?
3a7+3ab+b? 3a°b + 8ab?+ 63
Lxample 1. Find the cube root of 8x? — 36z°y + 54ary? — 27y?,
8x3 — 36x?y + B4ay? — 27? ( Qa — 3y
823
B23)" = 109°
3x 2x x (- 3y) = — 1ory
(—3y)' = + 9y"
1227 - 18xy + 9y? —
— 86x7y + 54ay? — 27y?
— 36x%y + 54ay? — 277
va
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eS 6+ =(¢—)
RB LOE + UST - =(¢—) x (xp p#Z) x8
iS LZ — X8OI — X06 — LEFT +4298 — LOb + COP — eTL = (ZH — eB) XE
& =
XF9 —pUOG + —USP—-| XII + ek FS— LVI
erg + ‘ =(xp—)
eXFG — = (xp—) x (2B) XE
eX0S +5L09 + 6XSF - PXSL = o(zlG) XE
9S
2 -— XP— XLS) LZ — CSOT — X06 — eVOS8 + 5X09 t+ oT SP — 978
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XVI. ]
100 ALGEBRA. [CHAP. XVL
EXAMPLES XVI. c.
Find the cube root of each of the following expressions :
], @+12a?+4+ 48a + 64. 9, 82? +120?+6xr+1.
9 6423 — 1442+ 108a — 27. 4, 8p*-36p!+54p? - 27.
5, m?—18m?+ 108m - 216. 6, x°+ 6xty? + 12a°y4 + 8y®.
7, 1-3c+6c? — 7c? + 6c4 — 3c° + c%,
8; 8+36m + 66m? + 63m? + 33m4 + 9m +m,
Q, 216 -108k+ 342k? — 109K? + 171K4 — 27K + 27K8.
10, 48y°+ 108y + 6044 — 90y? - 27 + 8y® — 80z°.
11, 64+192k4 + 240k? + 160K? + 60K4 + 1245 + h°,
12, x? —62°y — 3272+ 12xy? + lQxyz + 3x27 — 8y3 — 12yz - Byz? — 2.
[For additional examples see Hlementary 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.
Hxample. 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.
at+b
614125 ( 80+5 = 85
512000
3a7 =.3 K (80) = 19200) 1102125
3xaxb=3x80x5= 1200
A= Bx Os 25
20425
In Arithmetic the ciphers are usually omitted, and there are
other modifications of the algebraical rules.
CHAPTER XVII.
RESOLUTION INTO FACTORS.
128. Derritiox. 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? — 6ab have a common
factor 3a;
3a? -- 6ab = 8a(a — 2b).
Example 2. 5a*ba? — 15aba? — 20032? = 5bx?(a*a — 3a — 407).
EXAMPLES XVII. a.
Resolve into factors :
], 2+ax. Oy 262 =. a0. Ce Os. 4, ai--a?b.
5, 3m2-6mn. 6, p?+2p?q. T, 2-52, 8, yxy.
Q, 5a? -—25a7b. 10, 12%+4827y. ll, 10c? — 25c4d.
12, 27-162z. 13, x°y?+82y. 14, 17x2-5la.
15, 2a?-a?+a. 16, 32°+6a?x? - 3a°x.
17, 7p?-7Tp? + 14p%. 18, 40°+6a7b? — 262.
19, xy? — 2?y?+ 2Qary. 920, 26a°b°+39a%b2,
102 ALGEBRA. [onar.
Expressions in which the Terms can be so grouped as to
contain a Compound Factor that is Common.
180, The method is shown in the following examples.
Example 1. Resolve into factors «2—ax+bxe—ab.
Since the first two terms contain a common factor x, and the
last two terms a common factor b, we have
x? ~— ax +ba—ab = (x?—ax)-+(ba-ab)
a(x ~a)+b(~-a)
(a —a) taken w times plus (w-—a) taken 0 times
= (x-—a) taken (x+b) times
= (~-—a)(~+b).
II
I
| EHxample 2. Resolve into factors 6a? — 9ax + 4ba - Gab.
6x? — 9ax+4bx - bab = (6x? — 9ax) + (4ba — 6ab)
= 3a(2a —- 3a) + 2b(2x - 3a)
= (24 — 3a)(3a” + 2b).
Example 3. Resolve into factors 12a? + ba? — 4ab — 3ax.
12a? + ba? — 4ab — 38ax? = (12a? — 4ab) — (Bax? — bx?)
¢ = 4a(3a —b)- x°(8a —- b)
= (3a —b)(4a - x).
EXAMPLES XVII. b.
Resolve into factors :
1, x?+ayt+ue+y% 2, w—xz+auy-y2.
8, a?+2a+ab+2b. 4, a®?+ac+4a+4e.
By Qa+2xe+an+x2, 6, 3q-3p+pq-p.
7, am—bm-an+bn. 8, ab—by-ay+y’.
9, pataqr-pr-7. 10, 2ma+na+2Qmy + ny.
ll, awv-2ay - ba +2by. 12, 2a?+3ab —2ac — 3le.
18, ac?+b+bc?+a. 14, ac*—2a -bc?4+ 2b.
15, a@-a?+a-l1. 16, 22°+3+4+2a+4 32.
17. ax—aby+2ax—2by. 18. axy+bexy—az—bez.
Trinomial Expressions.
181, 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 (BED GAS) Sa FBG ALD, ets scse veces (1),
(a —5)\ae— et HO Arty Le cain cea ina ies cath pae (2),
CoD BN OD cciaces voden se Catees eee
CoD Reto a 2b LO ans cein ew saddens (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 2.
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 —5 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 # in the trinomial ; e.g. in (4) the
sum of —5 and +3 gives —2, the coefficient of w in the tri-
nomial.
The application of these laws will be easily understood from
the following examples.
Example 1. Resolve into factors v2+ lla +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.
x? + 1la+24 =(x+8)(~+4+3).
Example 2. Resolve into factors x? - 10x” +24,
The second terms of the factors must be such that their product
is +24, and theirsum —10. Hence they must both be negative, and
it is easy to see that they must be -6 and —4.
x? —10x”+24 = (a -—6)(a - 4).
EHxample 3. x?-18x%+81 = (x-9)(x-9)
= (x - 9).
Example 4. + + 10a? + 25 = (2? + 5)(y?+5)
ea (ro Ve
Example 5. Resolve into factors x? - llaa+10a?.
The second terms of the factors must be such that their product
is +10a7, and their sum —lla. Hence they must be —-10a and —a.
—llax+10a? = (x - 10a)(x — a).
Note. In eed: 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.
EXAMPLES XVIL ec.
Resolve into factors:
ik +3e+
2. 2. y+5y+6. 3. x tiy+12.
4. a?—3a+2. 5. a?—6a+8. 6, B-55+6.
7, +135+42. 8. 6-135+40. 9, =-132+36.
10, 2°-15r+56. ll, 2?-157454. 12, 24+152+44.
13, 126436. 14, a@84+15a+56. 15, a?—120+27.
16, 2°+92+20. 17, z2-10r+9. 18, 2°-162+64.
19, »°-23y+102. 90, y®-24y+95. 21. 2+ 54y +729.
99. a-+10ab+21F%. 23, a? +12ab+11F. 24, a? - 23064 132%.
95, m*+8m?+7. 26. 429m*n?+14n*. 27, 6-S5rix.
98, 54-l5a+a?. 29. isp 30, 216-35a+a*.
1382. Next consider a case where the third term of the tri-
nomial is negative.
Example 1. Resolve into factors z*+2z—- 35.
The second terms of the factors must be such that their product
is —35, and their algebraical sum +2. Hence they must have
opposite signs, and the greater of them must be positive in order
to give its sign to their sum.
The required terms are therefore +7 and —5. >
x? + 2a —35 = (x+7)(x-5).
Example 2. Resolve into factors z* — 3z - 54.
The second terms of the factors must be such that their product
is —54, and their algebraical sum —3. Hence they must have
opposite signs, and the greater of them must be negative in order
to give its sign to their sum.
The required terms are therefore —9 and +6.
x? -3z2-54 =(x-9)x+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 z*y* + 23zy — 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
ry"? + 23zy — 420 = (zy + 35)(zy — 12).
xviIL] RESOLUTION INTO FACTORS. 105
EXAMPLES XVIL @
Resolve into factors:
1, 242-2 2. -z-6. 3, 2-2-2.
4. y+4y-12 5. y+4y-21. 6. y°-5y-36.
7, @+Sa-3. 8. 2-14-30. 9, wia- 132.
10. --—12%-4. lL &+145-51. 12. F+10-39.
13, m*-—m—56. 14, m-im-S. 15, minm- 5.
16. ~-—Sp-6&. 17. #+3p-108. 18. #+p-110.
19, 2+27-4. 90, = -—7z-120. 91. =-2z-132
92. xt+139¢-48. 23. t4dzry-S9. 24. ¥+izy-Ke.
% aia FT 96. a+ab-—240F. o7, 14-i-&.
28. 3-%-¥. 29. 9645-1. 90, 72+5-F
133. We proceed now to the resolution into factors of tri-
nomial expressions when the coefiident of the highest power is
not unity.
Again, referring to Chap. v. Art. 48, we may write down the
following results -
se et Oe A, (1),
— 2) x—4}=32°— 1448.0... (2),
ons (x—4)=322— eoige hee eee, (3),
—9\(x+-4)=32"+410e—8........---------- (4)
The converse bea fe presents more ORS than the cases
we have yet considered.
Before endeavouring to give a general method of procedure
it will be worth while to examime im detail two of the identities
given above.
Consider the result 32° — 14r+8=(3r—2yYxr—4)}
The first term 32° is the product of 3r and x.
The third term +8....................- —land -4+4
The middle term —14r is the result of adding together the
two products 3r x —4 and xx —2
Again, consider the result 3x.°*—10r—8=(3r+2\2-—4)
The first term 3 is the product of 3r and x.
The third term —8..................... +2and -+
The middle term — 1 ts the result of adding together the
two products 3rx —4and 2x2; 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 coefficients of the expres-
sion to be resolved.
Example. Resolve into factors 7x? - 19x - 6.
Write down (7z 38)(@ 2) for a first trial, noticing that 3 and 2
must have opposite signs. These factors give 7x? and —6 for the
first and third terms. But since 7 x2—3x1=11, the combination
fails to give the correct coefficient of the middle term.
Next try (7x 2)(a 3).
Since 7x38-2x1=19, these factors will be correct if we insert
the signs so that the negative shall predominate.
Thus 7a? — 192-6 = (7x +2)(x — 8).
[Verify by mental multiplication. ]
1385, 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 14x%?+ 29% -15 ....... ee, (1)
]474— 203-15 se eee (2).
In each case we may write down (72 ~3)(2x% 45) 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,
Ati A ZY UNS NEGA UY Cr...:-cas+s ccs aecrest aeee eye
Therefore 1427+ 29a —15 = (7a — 3)(2a +5).
7
XVII. J RESOLUTION INTO FACTORS. 107
Example 2. Resolve into factors 547+ 172 +6 .....cccccceccseeee eee (1);
arid PEA Oe tan tis eae coo (2).
In (1) we notice that the factors which give 6 are both positive.
EDC ys Sis oan sen ie a aC Oe a asc negative.
And therefore for (1) we may write (5v+ )(x+ ).
And, since 5x 3+1x2=17, we see that
5a? +172+6 = (5a+2)(x+3).
52? — 17246 = (5a —2)(x- 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 XVII. e.
Resolve into factors :
1, 2a7+3a+1. 9, 3a7+4a+4+1. 3, 407+5a4+1.
4, 2a?+5a4+2. 5, 3a7+10a+3. 6, 2a?+7a+3.
7, 5a7+7a+4+2. 8, 2a7+9a+10. Q, 2a?+7a+6.
10, 2%7+92%+4+4. 11, 2%?+52 - 3. 12, 327+52-2.
13, 39?+y-2. 14, 3y?-7y-6. 15, 2y?+9y-5.
16, 207-5b-3. 17, 60?+7b-3. 18, 26°+b-15.
19, 4m?+5m-6. 90, 4m?-4m-38. Q1, 6m?-7m-3.
99, 427-8xry—-—5y*. 98, 6x?-Tayt+2y*, 94, 6a?-1382y+2y?.
95, 12a?-17ab+6b?. 26, Ga?-5ab-6b7. 97, Ga?+35ab— 607.
28, 2-3y- 2y*. 29, 3+23y -8y?. 80, 8+18y—5y2.
9], 44172-1522. 39. 6-13a+6a%. 33, 28-31b—502
When an Expression is the Difference of Two Squares.
186. By multiplying a+b by a—b we obtain the identity
(a+b)(a—b) =a?—}?,
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 difference of two squares
may at once be resolved into factors.
108 ALGEBRA. [CHAP,
Example. Resolve into factors 25x? - 16y?.
25a? — 16y? = (5x)? — (4y)?.
Therefore the first factor is the sum of 5% and 4y,
and the second factor is the difference of 52 and 4y.
2527 — 16y? = (5a + 4y)(5ax — 4y).
The intermediate steps may usually be omitted.
Example. 1 — 49c® = (1+ 7c3)(1 — 7°).
The difference of the squares of two numerical quantities is
sometimes conveniently found by the aid of the formula
a? —b?=(a+b)(a—b).
Example. (329)? — (171)? = (329 + 171)(329 - 171)
= 500 x 158
= 79000.
EXAMPLES XVII. f.
Resolve into factors :
1, a’-9. 9, a? -49. 3, a’-8l. 4, a’?-100.
Boa = 25. 6, «144. 7. 64=:2. 8, 81-42".
Q, 4y?-1. 10, y?-9a% Il, 4y°-25. 12, 9y?-49z%.
18, 4m?-81. 14, 36a?-1. 15, k?-640?.
16, 9a?- 250". L]7, (121 1697, 18, 121-3627.
LON, 25 =e 90, a?b? — a7y". 91, 49a*— 1006.
992, 64a? - 492. 93, 4p?q?-81. OA rat hea a D
95, 2x®- 4a, 06, x7 = 2027. 1 0 = po.
98, 16a!6- 90%. 99, 2xa’-4, 380, a%b8c4 — 92.
Find by factors the value of
Ql, (89)? — (31). 82, (51)? - (49). 33, (1001)?-1.
34, (82)-(18)?. 35, (275)?- (225), 36, (936)? — (64).
When an Expression is the Sum or Difference of Two Cubes.
187. If we divide a?+b3 by a+b the quotient is a?—ab+6?;
and if we divide a?—b3 by a—b the quotient is a?+ab+b”.
We have therefore the following identities :
a®+b3=(a+b)(a?—ab+b?) ;
a®—b3=(a—b)(a?+ab+b?).
These results enable us to resolve into factors any expression
which can be written as the sum or the difference of two cubes.
XVII. | RESOLUTION INTO FACTORS. 109
Hxample 1. 8a? — 27y? = (2x)? - (By)3
= (2a — 3y)(4u? + Gay + 9y?).
Note. The middle term 6zy is the product of 2a and 3y.
Example 2. 64a? +1 = (4a)? +(1)3
= (4a + 1)(16a2 — 4a +1).
We may usually omit the intermediate step and write down
the factors at once.
Examples. 343a® - 272° = (7a? — 3x)(49a4 + 21a? + 92").
8x9 +729 = (223+ 9)(4a% — 1823 +81).
EXAMPLES XVII. g.
Resolve into factors :
ec = U2, 9. a +b°. Bala: 4, 1-7.
5, S8a?+1. 6, x? — 823. Le 0 270°. 8, 2-1.
9, 1-8a?. 10, 08-8. A ant bee 12, 64-p*.
13, 12503+1. 14, 216-23 15, xy? +343.
16, 1000a°+1. ie Silas 7. 18, a*b°®c? — 27.
19, Sx? - 343. 90, x? +2167". 91, x8 -— 272.
92, m—1000n°. 93, a? —729b%, 94, 125a%+512b%.
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 + 2b)? — 162°.
The sum of a+ 20) and 4x is a+2b+ 4a,
and their difference is a+2b-—- 4a.
(a +2b)? — 16x? = (a+2b +4x)(a+2b—4zx).
If the factors contain like terms they should be collected so
as to give the result in its simplest form.
Example 2. (8”+7y)? — (2x -3y)?
= {(32+ Ty) + (2a - 8y)} {(3a-+Ty) - (2x — 3y)}
= (8a¢+7Ty + 2x - 3y)(8x+ Ty — 2a + 3y)
= (5¢+4y)(~+10y),
110 ALGEBRA. [ CHAP.
139. By suitably grouping together the terms, compound
expressions can often be expressed as the difference of two
squares, and so be resolved into factors.
Hxample 1. Resolve into factors 9a? — c?+4cu - 4a”,
9a? — c? + 4ca — 4a? = 9a? — (c? — 4ca + 477)
= (3a)? — (c — 22)?
= (8a+c¢ — 2x)(3a —-c+2z2).
Example 2. Resolve into factors 2bd — a? — c? + b?+ d?+2ac.
Here the terms 2d and 2ac suggest. the proper preliminary
arrangement of the expression. Thus
2bd — a? —c? + b? + d*+2ac = b?+ 2bd + d? — a? + 2ac — c?
= b?+ 2bd + d? — (a? — Qac +c?)
= (b+d)?-(a-c)?
=(b+d+a-c)(b+d-atec).
140. The following case is important.
Example. Resolve into factors 24+ 22y? + y4.
x4 ee al ab y4 = (a*+ Dy224 2 +y') ms raf
= (22+ 92)? = (ay)?
= (2? -+y? + ay)(a?+y? — xy)
= (x? +ay+y?)(a?-xy+y?).
141, Sometimes an expression may be resolved into more
than two factors.
gf
£1 1
Example 1. Resolve into factors 16a‘ -— 814.
16a4 — 81b4 = (402+ 9b2)(4a2 — 9b?)
= (4a? + 9b*)(2a + 3b)(2a — 3b).
Example 2. Resolve into factors 2° — y°,
af — yf = (a? + y°)(28 — y?)
= (e+ y)(a?— ay + yu —y)(a?+ay ty),
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 of two squares.
XVII. |
RESOLUTION INTO FACTORS.
111
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.
Huample.
Resolve into factors 28x4y + 64a°y — 60x7y.
28u4y + 642%y — 60x?y = 4x?y(7x? + 16x — 15)
= 4a°y(7x — 5)(@ +3).
EXAMPLES XVII. h.
Resolve into two or more factors :
1, (w+y)-2. 2, (w-y)-2.
4, (a+3c)?-1. 5, (2¢-1)?-a?.
7, 4a?-(b-1)?. 8, 9-(a4+2).
10, (18%+y)?-(17"-y). its
12, 4a? - (2a - 3b). 13.
14, (x+y)? (m-n)?. 15.
16, a@-2ax%+a?- 4b. live
18, 1-a?—2ab-—b?. 19.
90, c?-a?-b?+2ab. O11.
99, x*+y*—2t-at4 Qary? —2Qa72?, 93,
94, at+a?+l. 95, a‘b*—16.
27, 16a4b?- 6%. 98, 64m7 — mn,
80, «@7b>-81a?d. 81, 400a?x — x,
83, 2160° + ad, 34, 25027 +2.
386, aa? — ax? -240ax. 37,
88, mt+4m?n2p? + 4ntp+. 39,
40. Gay? + 1l5a°y? - 36ay". 41,
42, 9824 — Tay? — yy? 48.
44, 2? -2u7-4%+4+2. 45,
46, a’2x? — 8a2y? — 4b7x3 + 32077".
47, 2p-3q+4p? -9q°. 48,
49, 24a7b? — 30ab? — 3604. 50,
HI, 2*+427+4+ 16. bo:
58, a*t—18a7b? + b+. 54,
8, (a+2b)? -—c?.
6, @-(b+c).
Q, (2a-—3b)?-c3.
(6a + 3)? — (5a — 4)”.
x — (2b -- 8c)".
(3x + 2y)? — (2a - 3y)?.
x? + a? + Qaa — 22,
12ay +25 — 4x7 - 9y?.
x? — Qa +1 —m?—-4mn —4n2.
(m+n+p)?—-(m-n+p).
96, 25624 -81y%.
99, at—aty?.
ou. L—=129y*.
95, 1029—3x%
aca? + bex — adx — bd.
82243 — 20°. .
2m8n*t — Tmin’ — 4n8,
a°b? — a? — 6? +1.
(a@+bpP+1.
119+ 10m — m?.
240? + aSy/4 — a7l0y8,
at + y+ — Ta2y?,
8 +o4+1.
[For additional examples see Elementary Algebra. ]
112 ALGEBRA. [crraP.
Converse Use of Factors.
148, 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 formulze 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.
Hxample 1. Multiply 2a+3b-c¢ by 2a-3d+¢.
These expressions may be arranged thus:
2a+(8b-—c) and 2a-—(3b-c),
Hence the product = {2a + (3b —c)} {2a — (3b —c)}
= (2a)? - (8b -c)?
= 4a? — (9b? — 6bc +c?)
= 4a? — 9b? + 6be - c?.
EHxample 2. Find the product of
x+2, 2-2, w?-Qa+4, u?+2r+4.
Taking the first factor with the third, and the second with the
fourth,
the product = {(a + 2)(x? — 2a +4)} {(a - 2)(u?+2x%+4)}
= (a + 8)(a — 8)
= 26 — 64.
Example 3. Divide the product of 2u?+a”-6 and 62?-5x+1
by 3x7+5x-2.
Denoting the division by means of a fraction,
(2a?+ x2 — 6)(6x? —5x”+1)
327 +52 —2
(2a — 8)(a +2)(3x —1)(2a—1)
(8% —1)(%+2)
= (2% —3)(2x- 1),
by cancelling factors which are common to numerator and denomin-
ator.
the required quotient =
XVII. ] CONVERSE USE OF FACTORS. TS
Example 4. Prove the identity
17(5x + 3a)? — 2(40a + 27a)(5x + 3a) = 25x? - 9a.
Since each term of the first expression contains the factor 5x + 3a,
the first side = (5% + 3a){17(5x + 8a) — 2(40x + 27a)}
= (5x + 8a)(85x + 5la — 80x — 54a)
= (5% + 3a)(5x — 3a)
= 252? — 9a?.
EXAMPLES XVII. k.,
Employ factors to obtain the product of
1, a-b+c, a-b-e. 2, 2w—yt2, W+y+z
8, 14+2x-x7, 1-2x-2?, 4, ¢c?+3c4+2, c?-3c-2.
5, a+b-—c+d, at+b+c-d. Go Po rey, p-g— ary.
7, a’®—4a2d + 8ab? —-8b°, a*+4a°b + 8ab? + 8b?
Find the continued product of
8, (a—b), (a+b), (a? +b?)
9, (icy, (L+2)°, (ia? )".
10, a-4a+3, a?-a-2, a?+5a+6.
ll, 3-y, 3+y, 9-3y+y’, 94+3y+y’.
12, l+ct+e?, l-c+e, 1-c?+c4
18, Divide a3(a+2)(a*-a-56) by a?+7a.
14, Divide the product of z7+a”-2 and 27+4x%+3 by 2?+5x+6.
15, Divide 32°(2+4)(z?-9) by 2?+a”-12.
16, Divide the product of 2x%7+lla-21 and 38a?-20a-7 by
a” — 49.
17, Divide (2a?-a-3)(8a?-a-2) by 6a7-5a—-6.
18, Divide x6-7x?-8 by (~#+1)(u?+2u +4).
Prove the following identities :
19, (a+b)?-(a-b)?(a+b) = 4ab(a+b).
90, ct-—dt—(c—d)(c+d) = 2cd(c? - d?).
91, (m-n)(m+n) — m+ nt = 2mn(m? — v7).
22, (a+y)t-8ay(a+y)? = (a+y)(a+y").
93, 3ab(a—b)?+(a—b)* = (a —- b)(a? — b*).
H.A. vee
114 ALGEBRA. (CHAP.
MISCELLANEOUS EXAMPLES III.
], Find the product of 10z?-12-3% and 2%-4+432%.
Qo Leal. b= 1 ea 2, 0.0, tindsthe walle of
a?—b? b*-cd , cb - 0?
a2+b? 2b?+cd Babe°
3, Simplify 2[4a - {2y+(2e-y)-(«+y)}].
4, Solve the equations :
(1) z—-3 3-9 12x (2) 3u — 4y = 25,
5 3 15 ie bt 2y =i.
5, Write down the square of 223-—a+5.
6, Find the H.C.F. and L.C.M. of 3a2b°c, 12a4b?c?, 15a%b°c. -
7, Divide a*+4b* by a?—2ab+26?.
8, ind in dollars the price of 5x articles at 8a cents each.
9, Find the square root of 24-82? + 24a? — 32x+16.
QO, lh.a=5, b=3.¢ =]; ind the value of
(a—b) , (b-c) , (ac)
a+b b+e a+c¢
Ti eeccire (78 +5) -~73=13 -f(a i AI.
12, A is twice as old as B; twenty years ago he was three
times as old. Find their ages.
18, Simplify (1 - 2a) -{3-(4-5x)}+{6- @7 -8z)}.
14, The product of two expressions is
6x4 + Say + 6x?y? + Bay? + 6y4,
and one of them is 247+ 3ay+2y?; find the other.
15, How old is a boy who 2x years ago was half as old as his
father now aged 40?
16, Find the lowest common multiple of 2a?, 3ab, 5a%bc, 6ab*c, 7a”.
17, Find the factors of
(1) 2?-axy-72y*. (2) 6x?-18x%+4+6.
18, Find two numbers which differ by 11, and such that one-
third of the greater exceeds one-fourth of the less by 7.
XVII. ] MISCELLANEOUS EXAMPLES III. 115
ifo@al, b=—)) c= 2, d.=0, find.the value of
at+b c+d, ad— be een
a-b ce-d bd+ac a?+b™
- : 3 ae | ad be
90, Simplify 8x —y—{2v-ly-7-(%-4)+(2-}e)l
91, Solve the equations :
(1) (3a -8)(8x% +2) -— (4% -—11)(2%+1) =(%-3)(x%4+7) 5
x — ; 25 ene
(2) SU Siva 2, aty-5=ay-2).
99, + 7c4 — 20c?.
9
a
°
4m —9m?, 6m? —5m?-6m, 6m4+ 5m — 6m?.
3a42x3 — Sax? + 40723, 3a5x? - llatz? + 6a%x,
Satz? + 16a%23 — 120223,
118 ALGEBRA. [CHAP.
147, The highest common factor should always be determined
by inspection when possible, but it will sometimes happen that
expressions cannot be 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
aleebraical process of finding the highest common factor; for
the proof of the rules the reader may consult the Hlementary
Algebra, Arts. 102, 108. We may here conveniently enunciate
two principles, which the student should bear in mind in reading
the examples which follow.
I. Jf an expression contain a certain factor, any multiple of
the acpression is divisible by that factor.
Il. If two expressions have a common factor, ct will divide
their sum and their difference ; and also the sum and the difference
of any multiples of them.
Example. Yind the highest common factor of
423 — 3a?- 244-9 and 8x3 — 2x? - 53x — 39.
x | 4a? — 8a? — 24a -—9 8a? — 2a? — 53a — 39 | 2
4a3 -- 5a? — 21a 823 — 62? — 48a -—18
Pee Na 272 — 32-9 2 Gaeta OA Mal?
Qn — 6x 47 —. 62-18
3 32-9 x- 3
32-9
Therefore the H.C.F. is x—-3.
Explanation. 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 columns as above. When the first remainder 42? — 5a — 21
is made the divisor we put the quotient x to the left of the dividend.
Again, when the second remainder 2a?-—3x-—9 is in turn made the
divisor, the quotient 2 is placed to the right; and so am. As in
Arithmetic, the last divisor x-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.
XVIII. ] HIGHEST COMMON FACTOR. 119
Hxample. Find the highest common factor of
2424 — 2a3 — 60a? - 32a” and 1824 — 62? — 39x? — 182.
We have 2424 — 223 — 602? — 32a = 2a (122° — x? — 30x — 16),
and 1824 — 6x3 — 39x? — 18x = 32 (623 — 2x? — 13x - 6).
Also 2x and 3x have the common factor x. Removing the simple
factors 2a and 3x, and reserving their common factor x, we continue
as in Art. 148.
2% | 6a? — 2a? —- 1382-6 1227 -— x?-30x-16|2
622 —8x?-— 8x 122° — 4a? — 26x — 12
2 6a? -— 52-6 3u7— 4r— 4) 2
62?— 8x-—8 3x7 + Qe
38x24+2 —- 6x%- 4/-2
— 6a- 4
Therefore the H.C. F. is 7(8%+2).
150, So far the process of Arithmetic has been found exactly
applicable to the algebraical expressions we have considered.
But in 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, L. & IL]
Example 1. Find the highest common factor of
3x? — 1327+ 23% -21 and 62° +2? - 447421,
3x3 — 13274+23x2-21| 623+ x2?-44r421]2
6x? — 26x? + 46x — 42
27x? — 902 + 63
Here on making 27z?-90x+63 a divisor, we find that it is not
contained in 3x3 — 132?+23x%-21 with an integral quotient. But
noticing that 27x? — 90x + 63 may be written in the form 9(3z?-10x+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(3x7-10x+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
32? - 10x+4+-7.
120 ALGEBRA. (CHAP.
Resuming the work, we have
x | 323 — 13224 234-21 3x2-10e+7 |x
32° -—10a?+ Tx Ste | ae
-] — 32°+162-21 —- 3x4+7|-1
— 327+10a- 7 — 3x+7
2)6x-14
Auaias fi
Therefore the highest common factor is 3x —7.
The factor 2 has been removed on the same grounds as the factor 9
above.
151, Sometimes the process is more convenient when the
expressions are arranged in ascending powers.
Hxample. Find the highest common factor of
Oo 40160? = 0gr oo ee ee (1),
and AiG LOG Sa" 2. rete ere ee Oo}
As the expressions stand we cannot begin to divide one by the
other without using a fractional quotient. The difficulty may be
obviated by introducing 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 :
4— Ja- 19a?- 8a? 12 - 16a — 64a? - 36a? | 3
5 12 — 21a — 57a? — 24a
4|20-35a— 95a2— 40a a|5a— Ta? — 12a"
20-28a-— 48a? 5-— Ja —12a2|5
— Ja- 47a*- 40a3 5+ 5a
at 12a —12a?|- 124
7a ~ 35a-+ 2350? +200a° —12a -—12a?
3da— 49a2-— 84a?
2840? | 284a? + 284a3
l+a
Therefore the H.C.F. is 1+a.
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 term of
4—7Ta-—19a?—8a? is not divisible by the first term of 5—7a-— 12a?
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 XVIII. b.
Find the highest common factor of
243 + 322+ a+6, 22+ 4242443.
2y? — 9y?+9y-—7, y> — 5y? + 5y —4.
2a? + 8a? — 5a -—20, 6a? -4a?-1524+10.
a’? +3a2—-16a4+12, a®+a?-10a+8.
62° — a? — 7a —-2, 2a? —Ta?+x2+6.
g—3q+2, g@-59q?+7q-3.
at+a®—20?+a-3, 5a°+3a?-17a+6.
3y4 — 3y® — 15y?-9y, 4y° — 16y4 — 44y? — 24y/?.
1524 — 152? + 1022-102, 302° + 12024 + 2027 + 8022,
2mt+7m?+10m?2+ 35m, 4m*+ 14m? — 4m? -— 6m 4 28,
3act — 9a? + 12a?-12%, 6a? -6x7-15%+6.
2a° —4a4*- 6a, a’ +a*—- 3a - 3a?
x? +442 -—922—-15, x2? —2Qlx — 36.
14, 9at+2a72?+a4, 3a*- 8a%x + 5a7a? — 2a’.
15, 2-3a+5a?-2a3, 2-5a+8a?~ 3a’.
16, 3a?-52°-15a4- 40°, 6a — 7x? -— 2923 — 1224,
-
DOONIOARwWNH EH
ee
oONnNrH ©
[For additional examples see Hlementary Algebra. |
GHA PPE Rex
FRACTIONS.
153, Tue principles explained in Chapter xvi11. 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 we multi-
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 fraction is said to be in its lowest terms.
9407 ¢7a2
18a°2? — 120223
Hxample 1. Reduce to lowest terms
~ 2Aaen*
6a7x7(3a — 22)
=. ac
3a — 2a
The expression =
6x? — Say
Jay — 12
On a5 A
The expression = aA _ 2%
by(3u—4y) 3y
Note. The beginner should be careful not to begin cancelling
until he has expressed both numerator and denominator in the most
convenient form, by resolution into factors where necessary.
Hxample 2. Reduce to lowest terms
EXAMPLES XIX, a.
Reduce to lowest terms :
a 9 ae —2a_ 9 3ab +b?
622 —32y * 4a? — 8a? * 6a*b? + 2ab
1.
CHAP. XIX. ] FRACTIONS. hes
Reduce to lowest terms :
0 eae Baryz Fetes ox? —y" 6 222y7 — 8
* Bay + 10x72 * 62°y — Qary? * Bay + 6x
i, pies Gen Gilin OaC eens et = 30
pa ie 12. * Bea? — 10c2a + 5c" * Bat + 30a”
2a +b)? a? + 63 2c? + Bcd — 3d?
Pe el ee ee
4a° — ab? a? — ab — 2b? c7 + 6cd + 9d?
x? —4¢ —21 a — I —15 92 +4 -3
Re ag pe ee pe es
3274+10c+3 327 — 122-15 227+ 1lla#+12
3a? — 24 4a? —"Bary? 18a? + 6a2x + 2aa?
iy | ere Ab et See ele ae ee es
4a” + 4a —24 227 4+- ay — 1oy" 2 Oe
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.
3a? — 1327+ 23% — 21
15a? — 3827 — 2x%+21
Example. Reduce to lowest terms
The u.c.¥F. of numerator and denominator is 3a — 7.
Dividing numerator and denominator by 32-7, we obtain as
respective quotients 2? -2x+3 and 5x7-x-38.
oa? —l3at+23e—21 (82 —7\(x2-2e+3) 272-2243
15a? — 3827-22+21 (8%-—7)(5z*?7-x2-3) 522-2 -3
Thus
156, If either numerator or denominator can readily be
resolved into factors we may use the following method.
x? + 3x7 — 4x
72 — 1827+ 62+5
The numerator = 2(x?+ 3x - 4) =a(x%+4)(x-1).
EHxample. Reduce to lowest terms
Of these factors the only one which can be a common divisor is
zx-—1. Hence, arranging the denominator so as to shew z-1 asa
factor,
a(a+4)(a—-1)
Fax —1)—1lx(z—1)—d5(x—-1)
ee ee le ale ts)
~ (@—-1)(7a? — lla -—5)” 7a2?-lla—5
the fraction =
124 ALGEBRA. [coHaP.
EXAMPLES XIX. b.
Reduce to lowest terms :
1 x? — 4? +24 —2 9 oe+at+2
© “Bat4 72242 * @—402+5a—-6
9 yey ly 3) 4 m>—m?—2Qm
* 3y2+4y?+4y+1 * m—m?-—m—-2
5 a? — Qab?+ 21 b3 ; 6 Qa? — aa = 90° ‘
* a? — 4a7b —2lab? * 323 — 1l0ax? — Ja2x — 4a?
7 5a? —4a-—] : 9 +27 — 12cd? = 9d? :
* 923 — 32741 * Qc? + 6c2d — 28cd? — 24d3
9 aA —21e+8_ 10 ap Oy ly = oy".
* 8a4—21a? +1 * y+ Ty’ + 3y?- lly
ll Rea ings 12 2 —5ax — 4a? + 323
2-a +923 4+ 4a + 922+ 423 — Bat
[For additional examples see Hlementary Algebra. ]
Multiplication and Division of Fractions.
157, Rule. To multiply together two or more fractions:
multiply the numerators for a new numerator, and the denome-
nators for a new denominator.
Ce Caae
Thus B* dba
Sunilarly, as del sl
BaF bdf’
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 common
to numerator and denominator.
2a? + 3a y 4a? —6a
dae) 18a 15)
a(2a +3) 2a(2a — 3)
4a 6(2a + 3)
_2a-3
oom
by cancelling those factors which are common to both numerator and
denominator.
EHxample. Simplify
The expression =
158, Rule. Zo divide one fraction by another: invert the
divisor, and proceed as in multiplication.
XIX.] FRACTIONS. 125
d ad
Thus sa a oe ah
by di bd -e* 06
Fe Vy PC) ae
Example. Simplify Cea aed pae A ASS aD er
ax — az 9u7-4a? 38ax+2a?
: 6x? — ax — 2a? x-a 3ax + 2a?
Th Ss = : x )
e expression en ara eo a
: _ (8% ~2a)(2a +a) u-% ee C(O -+ 2a)
. a(x — a) (84+2a)\(8a2-2a) Weta
=,
since all the factors cancel each other.
EXAMPLES XIX. c.
Simplify
7 zea pe, 9. ee 2b — Bab®
te ARG UG e724 2-12ab a*b?-4
9 2c? + 3cd , e+d 4 5y - lOyees ] -Qy
* 4c? - 9d? ° 2cd - 3d? * 12y? + 6y? © Qy + y?
5 x? —4 Nate 6 b2 uy oes 16a?
* 9244744 °> 424+2 1 3b=—da~ = $2295 ~
7 Eee a 8 aU heey ey = 24
* @?+5e+4 a74+3%+4+2 ; pees y+ 6y+9°
G74 2) ee 40 = 2) a*—3a—2_ 3a?-8a-3
Drees gyee gr eaee 10, x pee
Ay 6? + 125 . 2O0 tn 12 38m?—m—-2 , 4m?+m—-6
* 5b?+24b-5 63—5b2+25d * 3m2?+8m+4° m+2
13 merely Os 5p +6 ee 2p-- 15
“ “p=9 p?-5p p-4
64a72—=1 4#7=49 . -w=7
14, ie eat Ee ee Sa
15 407+ 44-15 x+8 ; 2a? + 5x
Piet -46 (abe eis 6
16 a? + 8ab — 9b? a? — Tab +12? a+ ab fab?
, ard 27b ai? — b? a? — 3ab — 4b?
17 — 16a? 4 x*+ae—2007 , x? - pee Cat
; ae —ax—3002 ax2+ 9a2x +2003 x? + 8ax + l5a2
(a—b)?-c? a?+ab+ac . (a+b)?-¢?
18, Gee ele (a—c)?- 0? 7 (a+b+c)?
[For additional examples see Llementary Algebra. ]
CHAPTER XX.
LOWEST CoMMON MULTIPLE.
159, Derinition. The lowest common multiple of two
or more algebraical expressions is the expression of lowest di-
mensions which is divisible by each of them without remainder.
The lowest common multiple of compound expressions which
are given as the product of factors, or which can be easily
resolved into factors, can be readily found by inspection.
Example 1. The lowest common multiple of 6x°(a—2)?, 8a®(a — x),
and l2ax(a— 2x) is 24a%x?(a — x)°.
Hor 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.
Haample 2. Find the lowest common multiple of
38a7+9ab, 2a*- 18ab?, a® + 6a2b + 9ab?.
Resolving each expression into its factors, we have
3a? + 9ab = 8a(a+3d),
2a? — 18ab? = 2a(a+3b)(a — 3d),
a? + 6a*b + Yab? = a(a + 3b)(a+4+ 3b)
= a(a + 3b).
Therefore the L.C.M. is 6a(a+3b)?(a— 3b).
Example 3. Find the lowest common multiple of
(y2—xyzy, ya — x3), x44 Qad-+ a2,
Resolving each ee into its factors, we have
(yo —ay2)= {yee 2) =P 2-2),
yy? (202 — 203) = y?ar( 2? — x?) = wy?(z — x)(z+2),
ZA 4 Dare? 4 9022? = 2( 2? + Daz + 2?) = 22(z-+-00),
Therefore the L.C..M. is wy?2*(z+a)?(z-—a)%,
CHAP. XX.] LOWEST COMMON MULTIPLE. 127
EXAMPLES XX, a.
Find the lowest common multiple of
1, a’, a-a’. 9, x7, 2? — 323, 3. 4m?, 6m —8m?2.
4, 627, 2*+327. 5, b+), b-b. 6, «2-4, 2°+8.
I, 9a?b-b, 6a? +2a. 8, B-k+1, B-1.
YY. m?-5m+6, m?+5m-14. 10, y?+8y?, y?-9y’.
ll. 2?-9x2+14, x?+4e-12, 12, 2+27y*, x?+xy- by’.
13, 067+9b+20, b?+6- 20. 14, c?-3cx-182?, c?-8ex+ 122%.
15, a*?-4a-5, a?-8a4+15, a?- 2a?- 3a,
16. 2a?-4ay-16y?, 2?-6ary+8y?, 3x? - 127°.
17, 3a3-12a?x, 4x7+1l6arv+16a7, 18, a’c—ac®, (a2c¢+ac?)*.
19, (a%x—2ax*)?, (Qax — 42)2, 90, (2a-a7)3, 4a?-4a?+a4,
91, 2u?-x-3, (2x -38)?, 4z7-9.
99, 2x2-Tx-4, 6x?-Tx-5, x2? - 8274+ 16x.
28, 10x°y%(x?-y*), ldy*(a—y)?, l2a%y(a — y)(a? — y?),
94, 2a? +x-6, 7x?+1lxa-6, (7x7- 32).
95, 6a? — Tax -3ax", 10a?x - llax?- 6x, 10a? -2lax — 102".
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.
Hxample. Find the lowest common multiple of
2a + a? — 20a? — 7x +24 and 2a4+ 3x° — 138x2-7a+15.
The highest common factor is z?+2z.- 3.
By division, we obtain
Qa + a3 — 20a? — Ta 4+ 24 = (x? + 2a — 3) (Qa? — Ba - 8).
Qart + Ba3 — 13x? — Ta +15 = (2? + Qa — 3)(22? — w — 5),
Therefore the L.C.M. is (xz? + 2a — 3)(2a? - 3x” —8)(2u?- a” -5).
128 ALGEBRA. [CHAP, Xx.
EXAMPLES XX. b.
Find the lowest common multiple of
], «-Qa?-132-10 and x - a? 10x- 8.
2, y'+3y?-- 38y-9 and y*®+ 3y? - 8y— 24.
8, m+3m?—-m—3 and m?+6m*+1l1m+6.
4, 2Qat-2e3+274+3x2-6 and 4a4-2a°+ 3x -9,
. Find the highest common factor and the lowest common
multiple of (%—«?)®, (~?-a°)?, a? — a4
6, Find the lowest common multiple of (a*- ax), (a?+az)’,
(aa — x7).
7, Find the highest common factor and lowest common multiple
of Ga?+5¢-6 and 6a?+x%-12; and 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?+5ab+6b*, a*—4b?, a? — 3ab?+2b°,
9, Find the lowest common multiple of 1-a?-a4+2° and
1422 +2? — at — 2,
10. Find the highest common factor of (a? —4ab?)?, (a?+2a°b)°,
(a*x + 2abx)?.
11, Find the highest common factor and the lowest common
multiple of (3a? —2ax)?, 2a7xa(9a? — 4a"), 6a? — 18a7x? + bax’.
12, Find the lowest common mrltiple of #+a°y+ay?, wy -y', .
amy + ay? + xy,
[For additional examples see Hiementary Algebra. ]
CHAPTER XXI.
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 dowes¢ common denominator.
Rule. Zo 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 by the denominator of the
first fraction, and multiply the numerator of this fraction by the
quotient so obtained; and do the same with all the other given
fractions.
Hxample. Express with lowest common denominator
2a(x — a) 32e(a2 — a?)
The lowest common denominator is Gax(x%-a)(w+a).
We must therefore multiply the numerators by 32(v+a) and 2a
respectively.
Hence the equivalent fractions are
15x7(a~+ a) er 8a?
6ax(«-a)\(x+a) 6ax(x—-a)(v+a)
162, We may now enunciate the rule for the addition or
subtraction of fractions.
Rule. Zo add or subtract fractions: reduce them to the
lowest common denominator; find the algebraical sum of the
numerators, and retain the conmmon denominator,
1 a,c _ad+be
zou ue he
and a_¢_ad—be
6d bd
130 ALGEBRA. (citar.
168. We begin with examples in further illustration of
those already discussed in Chapter x11.
: 2e+a 5a?-4ax
Hxample 1. Find the val fee ae
xample ind the value of + 2-5
The lowest common denominator is 9a.
Therefore the expression = da(2x + a) + Sat — dam
9a?
= 6Gax + 3a?2+4+ 52? -4ax tt 3a? + 2ax + 5a?
i 9a? > 9a*
x—-2y 3y-a 34-240
ee eR ae
The lowest common denominator is axy.
Thus the expression = COA at eae
axy
ac —2ay + 3xy -ax—38xy + 2ay
= waxy
= 0,
since the terms in the numerator destroy each other.
Hxample 2. Find the value of
Note. To ensure accuracy the beginner is recommended to use
brackets as in the first line of work above.
EXAMPLES XXTI. a.
Tind the value of
Be-1 2+3 2e-1
1 a-2 a-l at+5 9
Sure oe eOe lpia GEE ees
peor iy LO MueuNaG caret.
5 eg! 2-2 6 22-5 2-4 a2 —dAg
we ye yp RS) Mt Mel eu: ener
Ye ee atx at2ex x-5a.
1. ye | wet xy the 2a 3a. | 64
i 3 2 joe a PAs
9. 2a°—-5a a pee reads 10 See. Ey eee, 4x?
a a" a? y oe Bry
R20 30 aC, ab-be a 2a®-ab
TSE Re Ope a Se tea)
13. ay-axy+4e_ 1 _ a. 14, Sa era
Quy Ee a" be ca
XXI.] ADDITION AND SUBTRACTION OF FRACTIONS. Tal
164, We shall now consider the addition and subtraction
of fractions whose denominators are compound expressions.
The lowest common multiple of the denominators should always
be written down by inspection when possible.
22-30 2x-a
Hxample 1. Simplify 5 :
x-20 x-a
The lowest common denominator is (% —2a)(z—- a).
Hence, multiplying the numerators by x-a@ and x-2a respec-
tively, we have
the expression = es ee)
(2 — 2a)(x — a)
_ 2a? — 5am + 8a? — (2a — Sax + 2a?)
he (a —2a)(%— a)
Qu? — Fax + 3a? -— 27+ 5ax - 2a?
(x —2a)(x—-a@)
a
> (a —2a)(x — a)
Note. In finding the value of such an expression as
— (2% -—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 ae + ame
The lowest common denominator is (a -- 4)(~% +4).
(82% +2)(x+4) +(x —5)(x - 4)
(7 -4)(~+ 4)?
_ 3x7 +14e+8+2?—9x+20
(x —4)(~ +4)?
_ 447+ 5a +28
(x—4)(a +4)?
Hence the expression =
165, If a fraction is not in its lowest terms, it should be
simplified before it is combined with other fractions.
132 ALGEBRA. [cHaP.
B be Wa are v?+5ey—4y?_ xy — 3y?
cample. Simplify 2 16,2 a wrap,
2 — Ay? =
The expression =~ ey. ay = ED
a*—l6y? (w+ 4y)(a — 3y)
Ua Oly tee oe
x” — 167? x+4y
+ bay Ay yee ay)
x? — 16y?
_v+5ay —4y—- xyt+4y?
x? — 16y
_e+4ay wetd4y) &
2 —l6y? (a+4y)(a—4y) x—4y
EXAMPLES XXI, b.
Find the value of
] 1 1 1 i 1
1. a—2*a-3 2. asd ae a 3. b-2 642
a b Gt be a+3 a-3
4, x-a «x-—b 5, ate 'a-@ 6. a-3 a+3
z he 3a 1 a Leta? ata
x uw? C ye
: : _9
Example 2. Simplify = a
Hoe
Here the reduction may be simply effected by multiplying the
fractions above and below by 6a, which is the L.C.M. of the
denominators.
18 + 2a? — 12a
a?+3a-18
_2(a?-6a+9) 2(a-3)
\(a¢6)(a—38) “a+6
Thus the expression =
a2 +0? a? 0?
Example 3. Simplify ere
a0 aD
TI des (Ga Oe ae 4a°b*
ae numerator (a?+ B5(a2 02) = (2 ED a2 BY
similarly the denominator = ne
: 4a?b? dab
H the fraction = ae
ence the fraction (®t B\ ew) * @Lbanb
a 4a7b? (a+ b)(a—b)
(a@2@+e)\(a2—b)~ 4abd
= 20d
a2+b?
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.
Haample. Find the value of : ;
ae eee
oe
The expression ek ae Eten ed inl
7 3 ceo — x)
Q2-24+2% 2-2
Si
iS 1 1
8-44-3432 5-a@
2-2 2-2
2+2
62
EXAMPLES XXII, a,
Find the value of
1 a
1, pry Us oa o - is 4, —.
+2 b= = ee | =
e Zz d a> l-a
aes en pel P_y
ae te NOE d 4g
5, ea 6. y a le a 1 8. ab Tr
ie OL xy Bard Pa p
a+? -5 y-34+2 aoe =
9, 10. . ak, ea
ee eee noe
a2 a y n
oan p-2-—8, ee
12, ———. Bo ——— ss 14, —
x-44_l% Dens BF coat
243 b+3 at abt R
c+d c-d qa - 222 2+3 «+3
ae TG l-ab a 4
15. c+ad c-d 16. 1 a(a—b) 17. x-3 2-3
c-d'ct+d L—ab z—1
1 1 eek
18, 1 * tomes 19. C+ ¥ 20, a c
1+- x — 4--=
a a ad
142 ALGEBRA [CHAP.
Find the value of
]
21, —— 22, ——— 23, ——.
]-—= L+H lee
x x y
Rew
D4, 2 “ ~ 95. a 26. “ ~
es l= ee eae |
ie loa dae
Lite, 2-y ip
22] 8a —2
i Sera 28, ae
De ia Si Oe ee
lee [eae
x—-1 3a — 2c
175, Sometimes it is convenient to express a single fraction
as a group of fractions.
Day — lay loy? bay 10ry? | be
Example.
102: ~ 10x%y? 10x%y* 102%?
Lael ae,
ee eae
2 pe aeaete
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.
+7 (#+2)+5_ 44. 5
Hxample |
x+2 x+2 x+2
Example 2. 3x-2 _ 3(@ +5) — 15-2 _ 3(@4 +o)— 17 17 ae LZ ate
2+5 x+5 x+5 x+5
Example 3. Shew that 2Qu*— Te—1 =22-—1= meal
she © 2-3
By actual division, x -3)2x4?-TJa-—1(2xr-1
200 — 62
- x-l
—- £+3
-4
Thus the quotient is 22-1, and the remainder ~4,
Peas cs
Therefore vast adel Me 2a—-1- Lies
«“-3 x-3
XXxII.] 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 j —_ = 2% — 6x? + 182° - tees
By division 1+322 ) 2e ( 2a — 6a? +1825
2x + 623
Fe3
— 62? — 1825
1825
182° + 54x7
— 54x47
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
Qa ; 16229
g PEs Ont G8 1905 <4 a
aS ree ch me aia
The terms in the quotient may be fractional; thus if 2?
is divided p x*—a', the first four terms of the quotient are
ied ae ea
ae sua and the remainder is —
A ei gw
CO ae pe
178, The following exercise contains miscellaneous examples
which illustrate most of the processes connected with fractions.
EXAMPLES XXII, b,
Simplify the following fractions :
l 1-23 9 12a? +x2—-1 ,1+6%+92x?
* 14294 2424 ge * 1-824 1622" 1622-1 ~
a+b. 4ab a+b 2a
3. i teas. = at 4. o—ab—26? a?—4o°
5 a—1 xt+e2+1 6, (w+y)?_(#- (w-y)
gz-l x+e+) CV-Y “bY
144 ALGEBRA. [crrap.
Simplify the following fractions :
7 aba" — aca + bay — cy 9 “I ee! )- a
ax*+xy—-ax—y ‘ g\a-x% a+3x/ a+3u
2 3 5
9. 6848 bat alle +08, Oneal:
Se 02+ a* 1
3a.) a Cee
10. 3a+1 “ORE H a? + 23 Sore
1 1 eee
=, z—1
12, ae a 18, 92 _ 3(a? — 2ia+ 54)
cd(a? + b?) + ab(c? + d?) ( x ee
1d. cd(a? — b*) + ab(c? — d2) 16. lta a a
16 ] ., 1
: —8a22+4a? 23 — aa? — 4074+ 403
2
Bae oO 2(422)
17 2 eae mW 0 ngs
e = 3 * ° =
204 a’+8 lig 3
a a
19 223+ 42-32 | 5a?-8x-21 , 2x?-3x-9
* 8542+ 24a -—35° w+ 7a? -S8a ° 722+5la—-40
Cat Pr Te p= Dad ee
MS IISA Cae al 1rd
21 ye ee ee
; “a-y xt+yl \x-y “t+y 2a7y + 2ry"
a?—(b-c)? b?-(c-a)? , 2 -(a—b/y
22, (ctaf—B atbpie (b+c)?-a
23, (= +¥)( : )- ye a el
YING) ae Pay ae x+y
94. y a1) te a*—4a+3, a? — 4 — tale
a+a-6 a2 —-4a4+4 weet ae
Waste at sea ~4ax
25. (55 Toa z) (sa saa) 6a*(x - 2a)(% — 3a)’
XXII. ] MISCELLANEOUS FRACTIONS.
Simplify the following fractions :
i] ] i 2a?
20. 6a —6 6a+6 30243 8a4+3°
97 4ah? Ae 1 a) a x 1
* 2a4+32b4 8a+16b 4a7+1602 8(2b-a)
3b7-+6 .2b=7, 26°=3b ;
A Rae EB at
28. Geb t 195 28 43"
99, att
ND I HI
x x y y Z 2
Dga 2
LT we FES een :
20 (ger) xy 9] m m+m
ee “- Pigede ie 1 ns
x) \a a m4 m
ab ac ] ] if
32. (sea) A aie a
[ba bes ha
re b Cc
eli pends 6
99 Ltac = 1 -—ae Eee a)
ny | sree cose aa eee) | a
l+ac l-—ac
1 ]
(30-- wee oe 1
34. ul to 3a Riera iy) TPR)
a a
145
CHAPTER XXIIL
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.
6x-3 32-2
2+] e+5°
Clearing of fractions, we have
(6a — 3)(v7+5) = (8a —2)(24+7),
6x7 + 27x — 15 =62?+17x-14;
Example 1. Solve
1022 1%
1
ea=—.,.
10
Note. By asimple 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 ‘‘ multiplying up,’’ as it is sometimes called.
87+23 5a+2 2%4+3
Example 2. Sol ie = eat;
xanrple olve 50 te hae 1
Multiplying by 20, we have
8a +23 _ 20(5% + 2)
cxaisd Ge = 82x + 12- 20.
: a 20(52 +2)
By t t BY fee ee
y transposition, Nera
Multiplying across, 93x7+124=20(57+2),
84= 725
¢= 12,
CHAP, XXIII. ] HARDER EQUATIONS. 147
180. When two or more fractions have the same denomin-
ator, they should be taken together and simplified.
24-52 Sx-49 28
E. Lax kee OLY 6 ane = — 13.
aample NGS ary tap tame eae
By transposition, we have
8a — 28 — (24 — 5x),
ee eect a ea Ook
4— : x-2
Gree dae
4-¢ 2-2
Multiplying across, we have
3a — 5a?-6+102 = 16-424 20x — 52? ;
that is, — 32 =22;
Bix _ 22
=
Example 2. Solve a EN ea am “2-7
we 108 wae Boe ESO.
This equation rent 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
(10) Pl (e = 6) oe Ge ye (ae 9) 2
Sanyo © Eee yh een ay
whence we have
2 2 2 2
i el
Pa (ae ae es Seay, =9t
A A 1 1 i ]
h = : ‘
al la #210) eEbiao7 2-9
Transposing ! st Vis Rie a ec ae
: x-10 x-7 «x-9 x-6
3 3
(a —10)(z7—7)° (%—9)(a—6)
Hence, since the numerators are equal, the denominators must be
equal ;
that is, (a —10)(a-—7)=(a%-9)(x%-6),
x? —17a+70=27-15%+543
TG es da
ee “& 8.
148
ALGEBRA.
EXAMPLES XXIII. a.
Solve the following equations :
ape! 9 ie Gabe
5e-9 42-10 © 62-17 42-13
7_3-4a 1 ee,
9 4-52 4, Fhe ees ae
Bu-8_5x+14 F 8a-1_4a-3
pad “aToo °° 6x42 32-1
220-12 _ 4, 3e+7 9n- 22 38a-5_
8a-5 4248 8. On 5 One]
8x-19 1_3e-4 10, 2#t2 1, 6e=1,
42-10 2 2x41 = 3(¢ 21) ee ook
e-5 22] oe =1e 52
2 3x42 10 5
Bu-17 , 2¢-11_28_3e-7
13 -—4a 14 42 Ai
4x-—3 1-92 _4%+3 1
ald eh tee en gOd ea
eee eae al
xt+l “+2 %x4+3 3x2+6
Be aes 2 es
zx-4 3x-18 47-16 xw-6
Liye eee ae
“X+6 3a+12 2%+10 6(@+4)
2-1 #£-5_ “2-3 0-7
e-2 x-6 «x-4 “2-8
1 és ] 1 ]
oO ee l7. 2 llete— 15:
§x-64_ 4%-55_2x—-IT x-
6
2-18 @-14 2-6 -#=7 -
[crrA.
XXIII. ] HARDER EQUATIONS. 149
Solve the following equations :
99 5a+31_ 22+9_ 42-6, 2e-13
we t6 (#45 2-5 2-6
12¢+1 Bo bia 2a
Sa-1 1-922 1432 ~
es | ev 9 x—3
a = hi eee
24. x*7-9 x-3 xL+3
28,
[For additional examples see Hlementary Algebra. ]
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.
Haample 1. Solve (x+a)(x+6)-—c(a+c) =(x-—c)(x+c)+ab.
Multiplying out, we have
x*+axn+ba+ab-—ac—c?=27-c?+ab;
whence ax +bx = ac,
(a+b)x=ac;
ac
Tete :
a+b
6b _a-b
Hxample 2. Solve
Simplifying the left side, we have
a(x —b)—b(x-a)_a—b
(e-a)(x—-b) 2aw-—c
a 1
(w-a)(x—b) xw-c
Multiplying across, x?-—cwv =2?-ax-bxtab,
ax+ba—cx = ab,
(at+b-c)u=ab;
ab
150 ALGEBRA.
Example 3. Solve the simultaneous equations :
OL — DY A Cae, eset howto ea eee
DEO =P Moe Eanes me ee
To eliminate y, multiply (1) by g and (2) by 6;
thus agqx — bay = cq,
bpx + bay = br.
By addition, (aq+bp)x =cq+br ;
ey SE br
aq +bp
We might obtain y by substituting this value of x in ether of the
equations (1) or (2); but y is more conveniently found by eliminat-
ing x, as follows.
Multiplying (1) by p and (2) by a, we have
apx — bpy = cp,
apx+aqy=ar.
By subtraction, (aq+ bp)y = ar-cp;
_ar-cp
ay+ bp
EXAMPLES XXIII, b,
Solve the following equations :
10, x+(2-a)(x—b)+a?+b0%? =b+27-a(b-1).
ae ek ae 2 ys. ps 2 2
vik Qu a_ 3x b_ 3a 8b 19. a-x b-2x a’ +b
b a _ ab a—-b at ~ at oF
ax—-b bxe-c_a-cx x+a-b_x+b-c
13. ¢ a a bai 14 xt+tb+e x+at+b
ty a eee aN ey
LO) nee q)*- pP(P a) + pa(Z 1) 0
1, av+b?=a?- ba. 9, 2*-a? = (Qa-x),
3. a@(a—x)+aba = b*(x—-b). 4, (b4+1)\(x%+a) =(b-1)(a-a)
5, a(«+b)-b?=a?-b(a-2), 6. ca Ct=ad2+c.
7, a(e—a)+b(a—b)+e(x-c) = 2(ab+be+ca).
a? b? Qe) ee
A es psy : = Fa eee,
tbe aby Oxas i aa
XXIII. ] HARDER EQUATIONS. Tol
Solve the following simultaneous equations :
6. x-y=a4+), 17, ce-dy=+a, Toe exer by,
y
ax + by = 0. w+ y= 2c. UC,
tes ab _ EN hah
19, pe ne 90. x y 0, yA a-y bv
i Yn oe ¥-9, BELG 25)
pon - ute a+b
z_y_a,b etY _€-Y_¢,
= Ss C= Ye PY So he
a(a+2) = b(b-y) Dp ae 2 [on AGS
94 2a—-b_2yta_su+y 95 axtby_1_ a?
i a b a+2b * betay 2 bx+ay
Irrational or Surd Equations.
182. Derintrion. If the root of a quantity cannot be ex-
actly obtained the indicated root is called a surd.
Thus ./2, 9/5, /a3, Na?+0? are surds.
A surd 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 e-
mentary Algebra, Chap. xxxi1. 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 2 Je —n/4e-11 = 1.
Transposing, 2/2 —-1=/4x—11.
Square both sides; then 4xv-4,/x+1=42—11,
4,/x = 12,
ee oy
e='9:
152 ALGEBRA. [CHAP. XXIII.
Example 2. Solve 24 Vxe—5 = 13.
Transposing, Na-5=11.
Here we must cube both sides; thus 2-5=1331;
whence x = 1336.
Hxample 3. Solve
6Je-ll_2Je+1
3,/x
Multiplying across, we have
(6,/%-11)(,/7+6) =
Jxz+6-
3r/2(2J/a+1)3
that is, 62-11 ,/2+36/x - 66 = 6%4+38,/2,
—11,/x+36,/x-3,/x = 66,
Da Mace tits
Vx =3;
Me eee
EXAMPLES XXIII, c,
Solve the equations :
L Neoe= 1. 0. NS — 2a = 7. 3. Ne=7 =e.
4, oJe+1=3. ieee fhe = ae
7 1-52 =3N 1-2. 8, 2N5x2-3-7/x=0.
9, N4a?— 112-7 = 2-3. 10. 38V1-7x+422 = 5-62.
ll. 14+ V2°—-322+72—-ll=z, 12) Nee wee)
13, V4a+134+2,/a = 13. 14, 34+ V12z2—33 = 2A/32.
15 Je-1_J/¢-3 16 Net _ J/e+5
ie. ir ie ‘ 32-8 3\/2=7
; ]
17 Mepey ae 19, 2v%=7-9, 1
Jxa-1 Wee Ja 14 4,/x-13
3 3
3 Sane 1
Ee poecae, ly —-3=—_——.
19. 1+474+2,/x AEs 90. Va+nNa-8 pe
Q1. N4a+7-NaetrlsNe-3. 99, Nde-3-Nat+3= Ve-4.
CHAPTER XXIV.
HARDER PROBLEMS.
184, Is 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 numerator of a fraction is increased by 2 and
the denominator by 1, it becomes equal to 8; and if the numerator
and denominator are each diminished by 1, it becomes equal to 4:
find the fraction.
Let a be the numerator of the fraction, y the denominator ; then
- . xv
She fraction is =.
From the first supposition,
H+2_5
yt ae) peor reerer eee eeeoeroee eeerreeoesses (Ly
from the second,
rd alt
— = Fever sevescnes Peorereese ee evcoen 2, s
ae (2)
From the first equation, 8x#-d5y=-11,
and from the second, 22—- y=1;
whence x= 8, y=15.
Thus the fraction is =
Hxample 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 5 minute-divisions in 2 minutes,
a
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+13, or 33 divisions more than the hour-
hand.
154 ALGEBRA. [crar.
a
Hence soSe + 33,
11
—— i 33 ;
12"
Viera.
Thus the time is 36 minutes past 4.
If the question be asked as follows: ‘‘ At what times 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-band. In this case the minute-hand
gains 20 - 13, or 7 divisions.
Hence x = wat 7;
which gives Sehr i
,
Therefore the times are i past 4, and 36’ past 4.
Example 3. A grocer buys 15 Ibs. of figs and 28 Ibs. of currants
for $2.60; 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 2, y denote the number of cents in the price of a pound of
figs and currants respectively ; then the outlay is
154%-+28y cents.
Therefore LODE 28 Y= 200 Fe eeitn se lawienaeatenes tee (3;
The loss upon the figs is “6 x 15a cents, and the gain upon the
currants is x 28y cents ; therefore the total gain is
a) Te cents ;
5 2
pp eey 2h Tay
EAD ;
that is, BY = Daa 00g ag ete h en eee een (2).
From (1) and (2) we find that «=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, ¢ miles apart, and walk in the same direction. A travels
at the rate of p miles an hour, and B at the rate of g 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 a —c miles.
A walking at the rate of p miles an hour will travel 2 miles in
x—-—C
* hours; and B will travel a—c miles in hours; these two
)
times being equal, we have
a _m-¢
fo) ee
Gx = px—pe 5
whence peed San
Ped
Therefore A has travelled ae miles.
9) coat
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 zy miles.
On the first supposition the speed per hour is 7 +6 miles, and the
time taken is y—4 hours. In this case the distance traversed will
be represented by (x+6)(y- 4) miles.
On the second supposition the distance traversed will be repre-
sented by (x—6)(y+6) miles.
All these expressions for the distance must be equal ;
xy = («+6)(y —4) = (w@-6)(y+6).
From these equations we have
xy = xy + by — 4x - 24,
or Oran aerlee aes itn. gnaw an tanhihrare eee: Cl ys
and xy = xy — by + 6x — 36,
or Goes Ope VOM ate ae seein en'n.s 8016 Riss anions (2).
From (1) and (2) we obtain x = 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 4,
and if the denominator is increased by 9 it reduces to 4: find the
fraction.
156 ALGEBRA. [cHAP.,
9, Find a fraction such that it reduces to 2 if 7 be subtracted from
its denominator, and reduces to 2 on subtracting 3 from its numerator.
3. If unity is taken from the denominator of a fraction it reduces
to 4; if 3 is added to the numerator it reduces to +: required the
fraction.
4, Find a fraction which becomes # on adding 5 to the numerator
and subtracting 1 from the denominator, and reduces to 4 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 3; if 6 is taken from the denominator the fraction
reduces to 2: 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 time 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 ?
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, ) Hind: the WH, ©. ot
323 —1le?+a%+4+15 and 5a4— 7x3 — 20a? - lla -3.
23, Express in the simplest form
See
he at 3
C1y ee eee e eee ee
LEY Ee e-l1 «+1 x +
y x
24, — u4+427’ (2) Bote!
1, 2+3
ates
pa:
98, What value of a will make the product of 3-8a and 8a+4
equal to the product of G6a+11 and 3-4a?
XV] MISCELLANEOUS EXAMPLES IV. 161
99, Find the L.C.M. of 2? -—2?-3x-—9 and 2° - 2z?-5x2—- 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,
81, Find the value of
6a? — Bay — Gy? . 3x? — xy — 4 2, 9a" — Gay — Sy"
2x? +ay—y? Qa? — Bayt dy?” La —Bayt+y?
82. Resolve each of the following expressions into four factors :
(1) 4a4—17a?b? + 404; (2) a8 —256y8.
383, Find the expression of highest dimensions which will divide
24a4b — 2a%b? — 9ab* and 18a° + a4b? —- 6a°b*? without remainder.
34, Find the square root of
(I) w(a+1)(e+2)(4+3)+1;
(2) (2a?+13a+15)(a?+ 4a —5)(2a2+ a - 3).
35, Simplify
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 $10000. If the average price
per acre were $50, the sum obtained would be 10 per cent. higher:
find how much of the land is pasture, and how much arable.
CHAPTER XXV.
QUADRATIC EQUATIONS.
185, Derinirion. 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 be a pure quadratic.
Thus 2x?7—5xz=3 is an affected quadratic,
and 5x2=20 is a pure quadratic.
Pure Quadratic Equations.
186, A pure quadratic may be considered as a simple equa-
tion in which the square of the unknown quantity is to be found.
Example. Solve ——— Hiatal — ei.
v?—-27 2-11
Multiplying across, 9x?—99=25x?— 675 ;
Hak Veni Jaa yh
ios ob 5
and taking the square root of these equals, we have
L= +6.
[In regard to the double sign see Art. 119. ]
187. In extracting the square root of the two sides of the
equation z?=36, it might seem that we ought to prefix the
double sign to the quantities on both sides, and write +z2=+6.
But an examination of the various cases shows this to be un-
necessary. For +x=+6 gives the four cases:
t+z4=+4+6, +x=—-6, —x7=+6, —x=—6,
and these are all included in the two already given, namely
x=+6,2=—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 z?=36 is an instance of the simplest form
of quadratic equations. The equation (#—3)?=25 may be
solved in a similar way; for taking the square root of both
sides, we have two simple equations,
Bao 5,
Taking the upper sign, #2—3=+5, whence 7=8;
taking the lower sign, x—3=—5, whence r= — 2.
.. the solution is = 6,06 2.
Now the given equation (7—3)?=25
may be written x* — 6x +(3)?=25,
or x? —62=16.
Hence, by retracing our steps, we learn that the equation
w= 6216
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,
a+ 2ar+a°=(¢@+a)’,
and uv —2ax+a?=(«4-ay;
so that if a trinomial is a perfect square, and zits highest power,
x*, has unity for its coefficient, we must always have the term
without w equal to the square of half the coefficient of w. If,
therefore, the terms in xz? and z are given, the square may be
completed by adding the square of half the coefficient of vr.
Example. Solve 2?+14x = 32.
The square of half 14 is (7)?.
that is, (fF 7 j= 81 ;
189, When an expression is a perfect square, the square terms
are always positive. Hence, before completing the square the
coefficient of x” should be made cqual to +1,
164 ALGEBRA. [orar.
Hxamplel. Solve Fx =2?-8.
Transpose so as to have the terms involving x on one side, and the
square term positive.
Thus ele = 8.
Completing the square, 2? -—7a+ & = oe :
that is, Ga ;
2 +
4 bis iin 2
pee),
odio >
Ze
$ _ 8x45
38a+1 3x2+1-
Example 2. Solve 4-
Clearing of fractions, 12%+4-8 = 327+5;
bringing the terms involving 2 to one side, we obtain
3x7 — 12% =-9.
Divide throughout by 3; then
x?-4e=-3;
x? -444+(2)?=4-3;
that is, (w= Que
e-2=+1;
Gis. OL ad
EXAMPLES XXV. a.
Solve the equations :
Ll.) W(=7) = 627. 2. (24 8)(e—-8)=17. Be (i eal aye
eee oe eerie ee an”
eae Drea: 8, 2?+6a = 40. = Baa gale 4a
10) 927 £25. Lie? 1G =a: Loe Leia
18, 27+47=32. 14, 974+36=27. 15, 2«7+152-34=0.
XXV.] QUADRATIC EQUATIONS. 165
Solve the equations :
] a9 J eb) 2 2 9 tA _ “+37
16, ee ee) Ne ae +3). Lis ane ei
+3, %-2_ 5 Deaar Le near ge
18. e2+2 “2-3 (#+2)(x—3) 19. Qa +2 eri
2 Se eee
90, 5 aay x+2,
190. We have shown that the square may readily be com-
pleted when the coefficient of #2 is unity. All cases may be
reduced to this by dividing the equation throughout by the
coefficient of 2°.
Example 1. Solve 32-3827=10z.
Transposing, 3x? + 10% = 32.
Divide throughout by 3, so as to make the coefficient of x nity.
oy ANU Sy
TI Co .
lus 3 op 3
5\2 OF
Completing the square, 27 + Put 3 a4 ;
ae oy 4h.
that is, (« oe rg
5) 11
C+—= *-— 3
oS
e
a3 =2, or —5%
Example 2. Solve 527+1llx=i2.
Dividing by 5, eo AZ aae
Completing thesquare, 4 1 Gar = + “i -
that is, (2+ i = ae ;
ete
166 ALGEBRA. [cIAP.
191. We see then that the following steps are required for
solving an affected quadratic equation :
(1) If necessary, simplify the equation so that the terms in
x“ and x are on one side of the equation, and the term without x
on the other.
(2) Make the coefficient of x* unity and positive by dividing
throughout by 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 resulting simple equations.
192. When the coefficients are hteral the same method may
be used.
Hxample. Solve 7(#+2a)*+ 8a? = 5a(7x + 23a).
Simplifying, 1x7 + 28ax + 28a? + 8a? = 35ax + 11da? ;
that is, 7x? — Tax =84a?,
or Tae Ce OO.
Completing the square, 2*-ax+ OF = 12a + ;
2 2
that is, (« oe B = ae 3
C— x = +14 R
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 #?—-227+1=0, then (7—1)’=0,
whence w=1 is the only solution. Nevertheless, in this and
similar cases we find it convenient to say that the quadratic has
two equal roots.
EXAMPLES XXV., b,
Solve the equations :
1, 3a2+2a = 21. 9, 5a2=S8a42]. 8 6a2-x-1=0.
43-1 le =e"; He 2a = Dares. 6, 10+232+4 122? = 0.
7, 152?-6x = 9. §, 4a0°-17a = 15. Q, 8a?-197-15=0.
XXV. J QUADRATIC EQUATIONS, 167
Solve the equations :
HH) 1077 oe = 1. Pee Ta = 12: 12, 2027—2-—1=0.
toe pean loa. 14° 29*—S8at= lbax. 15, 3a*=k(2k— 5a).
nGseloee 20e ans 17, 9a" — 14h = box, 18. 2e72* = ar).
19, (w7-3)(~—-2) = 2(a7-4). 90, 5(%+1)(8%+4+5) = 3(82?+ lla +10).
D1, 322+134+ (%-1)(2%4+1) = 2x(2x 4+ 3).
Tx-3_ 3x 2 2-] 3a-1_2a-9
22, a O° 23, 375577 24, eto ei 4
62-5 2 x-4 1 11
== Oo. See Shed oot Se ec
25. z+5 3 26, je+1 2 2(3+ 22)
27, 3(2a+ 3)? + 2(2x% +3)(2- a) = (w- 2)",
98, (38a —7)? — (2a - 3)? = (w—4)(8a+1).
[For additional examples see Elementary Algebra. |
194, Solution by Formula, From the preceding examples
it appears that after suitable reduction and_ transposition
every quadratic equation can be written in the form
an +br+c=0,
where a, 6, c may have any numerical values whatever. If
therefore we can solve this quadratic we can solve any.
Transposing, ax’ +be=—c¢;
anon b c
dividing by a, Cg,
a a
Complete the square by adding to each side cal ; thus
20
faa nee bee,
a
2a Aa? a’
: b\2 6b%—4ae
that is (« ey = s
; oe 4a?
extracting the square root,
+ Ge
2a 2a
bie Ot (Ot 4ac)
Qa ;
168 ALGEBRA. [CHAP.
—b+ ,/(b’ — 4ac)
2a 4
it must be remembered that the expression ,/()?—4ac) is the
square root of the compound quantity b? —4ac, taken as a whole.
We cannot sumphfy the solution unless we know the numerical
values of a, b,c. It may sometimes happen that these values
do not make b?—4ac a perfect square. In such a case the exact
numerical solution of the equation cannot be determined.
195. Inthe result w=
Hxample. Solve 52?-13x-11=0.
Here a=5, b=-13, c=~—11; therefore by the formula we have
(—13) + \(S19P—4 7) B11)
oa
2.5
_13 + V169 +22)
10
_ 13+ /389
10
Since 389 has not an exact square root this result cannot be
simplified ; thus the two roots are
13+ ,/389 13— ,/389
102s 10
196. Solution by Factors. There 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 2+ la =
Clearing of fractions, $3274 72-60 assent -ceecesnessennaee CLs
by resolving the left-hand side into factors we have
(3x2 — 2)(a+3)=0.
Now if either of the factors 32 —2, «+3 be zero, their product is
zero. Hence the quadratic equation is satisfied by either of the
suppositions
a7 —-2=0, or r+3=0.
> Ef)
It appears from this that when 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
Thus the roots are
XXv.] QUADRATIC EQUATIONS. 169
side can be resolved into factors. Hach of these factors equated
to zero gives a simple equation, and a corresponding root of the
quadratic.
Example 1. Solve 2x°-ax+2bx = ab.
Transposing, so as to have all the terms on one side of the equation,
we have
2x7 —ax+2bxe—-ab=0,
Now 22:7 — ax + 2ba -—ab = x(2% -a)+b(2x%-a)
= (2%-a)(x+b).
Therefore (Qe-a)\(v+l)=0;
whence 22-—-a=0, or 7+b=0,
ane, or —b.
Y
Example 2. Solve 2(x?-6) = 3(a”—-4).
We have 2e2 \2 = oe = 12 >
that is, CA laa Woe ee RAPE APT tee NORA (1).
Transposing, 20% Bu =),
x(2%—-3)=0
v= O,or 27 = 3 =0
Thus the roots are 0, 2
Note. In equation (1) above we might have divided both sides by x
a]
3 . > oO . F,
and obtained the simple equation 2a = 3, whence x = 5, 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 x=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 : v= Ay or asia) *
G=4'3Q, Of T+ a= 0 3
both of these statements are included in
(x —4)(x+3) =0,
or a*-x2-12=0,
which is the required equation,
170
ALGEBRA.
[CHAP.
Example 2. Form the equation whose roots are a and - : :
Here
the equation is
that is,
or 320?
(x —a)(8a+ 6) = 0,
—8axr+ba-—ab=0.
EXAMPLES XXV. c.
Solve by formula the equations :
a oi Uk —24-1=0. 3. x7 ore:
Ae Oar =e eel 5, Qa?-9x = 4. 6. 327+7x2=6.
7 4a2—14 = 82. 8. 6a2-3-—Ta=0. 9, 1222410 = 23m.
Solve by resolution into factors :
10.30 —92 =90; Teale] 1625 19) ar 8a = 1
13, 2_ 37 = 2, 14) 3224+52-+42=0"" 15, 427° ite
16.6 Ser Sle 2 0) 17 a2 — 0.0; 182° 2? = Tax =8e%
19, 122? — 2362 + 100? = 0. 90, 3ax?+2ba = Tax.
91, 2422+ 22cx = 21c?. 09. a — 22 44h = 2he,
Solve the equations :
93, 2a(a+9) =(e%4+1)(5--z).
94, (2x-1)?-11 = 5x+(x-3)2.
95, G(x -—2)?+138(1 - x)(a - 2) + 6a? = 6(2x -1).
= 4 3 2 2.
aan 9, EASA =
26, x-6 x-5 27. Sx-1l x+1 2
10 oe +3 4(x~6)_
a a ese se ea Oo ae ie
] Gon Wines
eee sat as
a Sseeeig S13 Mead}
Tee OG hs TE) e415
Sakon teal apres ee ao.
32, io ales wre
Xxv.] QUADRATIC EQUATIONS. 171
Solve the equations :
eee et" oot
34, Gry Vel Oa” 35. Bar’ — 4 aap
Ce eee — 3 Pel D) u ei?
Hi og oT el eA
3, SEM a G2 99, (p— abe + 2 = Wn +9.
b P 5b c\?
2 aN a
40, Spans 41. i 1) (: i) ®
[For additional examples see Hiementary Algebra. |
198. Simultaneous Quadratic Equations. If from either
of two equations which involve w and y 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 methods explained in this
chapter.
Example. Solve the simultaneous equations
5xt+Ty=1, 447+3axy —2y?=10.
From the first equation, x = a , and therefore by substitution
in the second equation, we have
a(S ures ye 272=10;
25
whence 4 — 56y + 196y? + 15y — 105y? — 50y? = 250 ;
that is, 4ly*—4ly —246=0 ;
| y’-y-6=0;
(y—8)(y+2)=05
y=3, or —2.
From the first equation, we see that if y=3, then x= —4, and if
y=—2, then x=.
Homogeneous Equations of the Same Degree.
199, The most convenient method of solution is to substitute
y=me in each of the given equations. By division we eliminate
x and obtain a quadratic to determine the values of m.
ve ALGEBRA. (CHAP. XXV.
Example. Solve the simultaneous equations
527 + 3y?= 32, “x? -a2y +2y?=16.
Put y= mez and substitute in each equation. Thus
roms MaRS (Ed beciea ors Pare, a Pe SM eT A, oe | (qj
and col 22) = LG eae eee ee 2).
By division, Maweisil ayia 2
l1-m+2m* 16
that is, m* -2m-3=0;
(m—3)(m+1)=0;
= 3,.0F «2.1,
(1) Take m=3 and Syustitnte in either (1) or (2).
From (1), 32x? = 32: whence x= + 1.
WA = a a es
(2) Take m= —1 and substitute in (1). Thus
827=32; whence v= + 2.
Y=MNE= Lf. 2.
EXAMPLES XXV. d.
Solve the simultaneous equations :
1, w+3y=9, 9, 38x-—4y=2, oe 2r+y=5,
vy bs iy =e De — wy = 2.
ee Te aaa 5 ot er vans, 6, 28-215
x? + 4y? = 29. 3xy — y= 9. mete Lb
co on ae ies) ai pot
hs aa 1; 8, Batry fs ’ 9, x oe
xy = 24 10zy=1 xy —yr=4
eel eat Pees be
ieee . — A Lie Dates itm
10.5 5 Le ong ae ;
3.Y ee) seein
Prv=] peas a | — = 8,
22 ee ey oe
10; 327-79? 255... 14, léry—-Se*= 77, 15. a2 ey a
2c? ey OU. Tay + 3y? = 110. Say + 3y7 = 195.
16, 27+ 2ay + 2y? = 17, 17, 2la?+3ay - y? =37)1,
3a? — Oxy — y= 119. 5a? + day + Sy? = 265.
CHAPTER XXXVI.
PROBLEMS LEADING TO QUADRATIC EQUATIONS.
200. Wr shall now discuss some problems which give rise
to quadratic equations.
Hxample 1. A train travels 300 miles at a uniform rate; if the
speed 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
: See ay
time occupied is —— hours.
i
ry
On the other supposition the time is a5 hours ;
5
300 _300_o.
Se Ce ee ’
ce oe ie
whence e+ 5a —750 = 0,
or (7+ 30)(% — 25) = 0,
* oa = 25,06) — 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
Elementary Algebra. |
EHxrample 2. Aman buys a number of articles for $2.40, and sells
for $2.52 all but two at 2 cents apiece more than they cost; how
many did he buy ?
Let « be the number of articles bought; then the cost price of
each is ae cents, and the sale price is 5 cents.
HY Spas
252 240_..
fo ee f
that is, Oe me Eas
174 ALGEBRA. [cHAP.
After simplification, 6x +240 = a? - 2a,
or x? —8x2-240=0;
that is, (7 —20)(4+12)=0;
Lia UP Orie,
Thus the number required is 20.
Example 3. A cistern can be filled by two pipes in 334 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 «-15 minutes; then they will fill ba i yet 5 of the cistern
2
a
respectively in one minute, and therefore when running together they
will fill (+ i of the cistern in one minute.
But they fill a , or a of the cistern in one minute.
1 1 3
H = eee niet
a fen anOG
100(2% - 15) = 38x(x—- 15),
3a? — 245x + 1500 =0,
(a —75)(3a —20)=0;
x ='75, or 62.
Thus the smaller pipe takes 75 minutes, the larger 60 minutes,
The other solution 62 is inadmissible.
901. 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 4y feet ; thus
9x. 4y-= 1,
The areas of the two squares are x” and y” square feet; thus
6y? - 2927 = 1,
XXVI.] PROBLEMS LEADING TO QUADRATIC EQUATIONS. 175
From the first equation, Ye a _
By substitution in the second equation,
Se Woe oon 1 :
8
that is, llz?-—547%-5=0,
or (2—d5)(llz+1)=0;
whence x =5, the negative value being inadmissible.
Also, y = a L STi
Thus the lengths are 5 ft. and 11 ft.
EXAMPLES XXVI,
], Find a number which is less than its square by 72.
9. Divide 16 into two parts such that the sum of their squares
is 130,
3, Find two numbers differing 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 sz.
9, Find two consecutive even numbers the difference of whose
reciprocals is <3.
10. The difference of the reciprocals of two consecutive odd
numbers is 725: 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 $75: how many did
he buy ?
12, By walking three-quarters of a mile more than his ordinary
pace per hour, a man finds that he takes 1} hours less than usual to
walk 293 miles; what is the ordinary rate ?
176 ALGEBRA. (CHAP. XXVI.
13, + 16p*- 33p°+14p? by p?+7p.
3, Find the sum of a-2(b-3c), 3{a-2(b+c)}, 2{b -2(a —2b)}.
4, Simplify by removing brackets 7[8a-4{a-—b+3(a+b)}}.
5, Solve the equations :
x+4,«a-4
(1) agentes =4;
6. A is three times as old as B; two years ago he was five times
as old as B was four years ago : what is A’s age ? “i
7. Find the product of 2a -3b-(a-—2b-—c) and b-2c-—(a-c).
Gide] b= 0, ¢=-11) do =2, ei 2) find the value of
+4+0+E+4+0R—-8+a24+l2+C°+d?-e.
9, Remove brackets from the expressions :
(1) a -[5b-{a—(3c -3b)+2c- (a - eel
(2) Qa - 3{b-4(c - d)}]-[a—4fb - 6(c--d)}].
10, If the price of 5 acres of land is $a, what is the price of x
acres ? and how many acres can be bought for $b ?
11, Divide at—4 by a?—2a42.
12, There are 150 coins in a bag which are either half-dollars
or quarters. If the value of the coins is $58.50, find the number
of each kind.
H.A. M
178 ALGEBRA.
13. Add together a—{b+c-(a+b)!+c, 2(3a+2b) -4(b+2a)-c,
and 3(2b — a) -2(3b-a)+e.
14, Find what value of x will make the product of +3 and
2%+3 exceed the product of x+1 and 2x+1 by 14.
15, Divide b?+8-125c?+30bc by b-5c+2.
16, *Simplify
(1) 13ab*c? ‘ 87c°d? , 26bed ,
29c4d ~ 2a2bt * 4ab3 ’
202 ee a®bx , atb%a4 )
“b2 \2Qab 2c? Ga2b — Qakb4x3/°
17, How old will a man be in m years who n years ago was
p times as old as his son then aged 2 years ?
(2)
18, I bought 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-five for 12 cents, I gained 18 cents, How many pears did
I buy ?
19, Solve the equations:
a a _ 4%42 74+14
(1) — alee
e
=5—6%+
9) X+Y 8z—5Y _o ada bate
a ey az : cAlce
90. Divide a2xa§+ (2ac—b?)x#+c? by axt+c—b2?.
91, Ifa horses are worth b cows, and c cows are worth d sheep,
find the value of a horse when a sheep is worth $2.
99, Find the highest common factor of 8a7b3c, 12a%bc?, 15a2b5 ;
and the lowest common multiple of 4ab2c3, 12a%b, 18ac?.
aah oy IPOS MUS
2ab2c 362 b2d =~ 6. a2?
93, A gentleman divided $49 amongst 150 children. Each girl
had 50 cents, and each boy 25 cents. How many boys were there ?
94, Ii V=5a+4b—-6c, X=—8a—9b+7c, Y=20a+7b—85c,
Z=18a—5b+4+9c, calculate the value of V—(X+ Y)+Z.
Also find the value o
MISCELLANEOUS EXAMPLES V, 179
95, Solve the equations :
l 3(6 — 5x) 63a _ 3a _ 3G
() a) a KE Oy Rae
2) plety)=e+1, Hy-n)=20-1.
96, Find the factors of
(1) a?-a-182; (2) 8a?+13x-6.
Oi 4 eye AN ey, ye the value of
Ni{5(y? — 22) — ah + V/Bfa(a? — 2?) — 1h.
98, The product of two expressions is («+ 2y)?+ (3%+z)°, and one
of them is 4%+2y+2; find the other.
99, When 4 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 6. How much had each at first?
80, Find the square root of 16a°+4a+4 -16a?+a?-8at.
|
81, Find the value of
| 1 ae Teed pte
2a {2 p(e3)} {a —3le42)} (e-F(2-5)},
and subtract the result from (a+ 2)(%—38)(~+4).
382, Find the square root of
a4 293 lla? 9
Gs. PAG eG
BO ela ciieiaD =.c= es = 1, find the value of
arc _ nt Jatd — Vaid + o
me
a
34, Separate into he simplest factors :
(1) 2?—xy—-6y’; (2) 2° —4xy?- ay +4y%.
85, Solve the equations :
(1) (a-—1)(a%-2)(%-6) = (w- 3);
~ ome:
(2) 2 5 3
2a! By =0, ms ys 13.
36, A farmer ere 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 a=3, b=2, c=—7, find the value of
(1) 3Bb%c | Bbc? _ Ga? ;
a a-c b+2c¢
(2) 4c+{c-(3¢-—2b) + 2b}.
88, Solve the equations :
1,1 1
(2) 542+3y = 120, 10.c = Sy + 90.
39, Find the highest common factor of 5a°+2z?-—15a2-—6 and
7x? — 447 — 21a +12.
40, A coach travels between two places in 5 hours; if its speed
were increased by 3 miles an hour, it would take 34 hours for the
journey : what is the distance between the places ?
41, From 2(a+a-b)(c-—a+b) take (%-a)(x%—-b)(vt+a+tDd).
42, Find the value of
2a*+5a~3 3 3a? —102+3, 62? -—52+1
a3 — 9x 2+ 3042 © Bae +7a%+2a
43, Divide 2°+y?+3xy-1 by x+y-—-1, and extract the square
root of «4 - 3.° +5 4 204%
44, Aman can walk from A to B 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 B, 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 A to B?
45, Find the highest common factor of
7x4 — l0ax® + 8a72? - 4da®x+4at, 8rt— 13az3+ 5a2x? - 3a%a + 3a'.
46. Solve the equations :
(2) w-2y4+z2=0,° 9%-8y+32=0, 22+3y+5z2 = 36.
47, Find the lowest common multiple of
6a?-a2-1, 3a7+7x4+2, 227+32—-2.
MISCELLANEOUS EXAMPLES V., 181
48, The expression ax+3b is equal to 30 when 2 is 3, and to 42
when 2 is 7: what is its value when 2 is 1; and for what value of x
is it equal to zero?
49, Find the lowest common multiple of
4(a?+ab), 12(ab?-b*), 18(a?—b?).
50, Extract the square root of
4a? _ 128, on | 24a , 16a?
da 8G: x Figs
51, Reduce to lowest terms
1224 + 423 — 23x? — 9a — 9
8x4 — 1427-9
52. Solve the equations :
OU CY ett OY Qu —Ty 3a —7
1 a et se Fh a ee PP ames | Vireo :
Gy 2— Fer em eg y 6 ial ae
(2) 2e-y+32=1, 4¢+3y-2z2=13, 6x-4y+2=20.
53. Simplify
(1) GHD 2a 8 obo
b a+b ab—b?’
ee 1
24+ 8a+15 x?+1le+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.
(2)
55, Solve the equations :
(1) 4ar- (5y-4)=1, es 1-1,
(2) Sx+dy-11l=0, 5y—-6z=-—8, 7z2-8x%-—13=0.
56, Find the value of :
2 ] 3x a
atx a-x x -a@ (a+a)*
57, Resolve into factors :
A) 28 -2a* a; (2) af+at—ai—l1.
182 ALGEBRA.
58, Two persons started at the same time to go from A to B.
One rode at the rate of 74 miles per hour and 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
Ae Loge ays 4 Gay 16a?
9y2 2 l5yz 1622 52? 2527"
60. Find the factor of highest dimensions which will exactly
divide each of the expressions
2c4+ c8d — c2d? --7cd?—4d4, 3c4+ c8d - 2c?d* — 9cd? — 5d4.
: : 2 3 — Qn?
Simplif Tye et ee ee
G1, Simplify (1) —;| Fail Mee
DACiie aler)
62, Find the highest common factor of
624 — 2a3+927+9"—4 and 924+ 80x? — 9.
What value of x will make both these expressions vanish ?
63. Solve the equations :
9 293
(1) x-3 x-6
2¢ Se 1 ba 15>
yi ~ SAN
(2) rE NE eS x=2
e
9
6x? — Say —6y? — 15x? + 8ay — 121?
4x? — Bay +3y? 35a?+47xy + by”
64, Simplify i
65, Find the value of
x-2a_x2+2a Il6ab Shane
e+2b 2-2b 462-2? ..
4ab
a+b
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=-1, b=-2, c=-3, d=—4, find the value of
2a? +ab?—8abe_a-b+c-d_a ” Ze
a? — b?-—abc be — 2ad brad
68, Solve the simultaneous equations :
Dee TY ee al eZ
<=) >
2 14 8 4
ey YO 5(y +1)
MAA ite aE
sau 710 Pe ay
69, Simplify the fractions:
(iy cae ey)
x-%Z x-Yy (%-x)(y- 2)
wig ut el
5 B a b a
eee) ee
b a/\b a a2 hb? ab
70, From a certain sum of money one-third part was taken and
From the sum thus increased one-fourth part
$50 put in its stead.
If the amount was now $120,
was taken and $70 put in its stead.
find the original sum.
71, Find the lowest common multiple of
(a4 —a%c?)?, 4a®-Sate?+4a2ct, a+ 3a3c + 3a%c?+ ac’.
72, Solve the equations:
; ‘135% —°225_ 36 -09%-°18 ,
(1) *l5a+ 03 es, z= 9 ;
ll - 5a, da —-33
6
l0x+4, 7-20? _
(2) SAY EVER S as
73, Simplify
x?—44¢—-21 y C+ Ga? -247a , x? - 202491
ae era
e+17 x? —a2-12
74, What must be the value of x in order that
(a+2a)?
a? + 70ax + 3.x?
may be equal to 14 when a is equal to 67 ?
184 ALGEBRA.
75, Find the highest common factor of 1674+36a?+81 and
8x? +27; and find the lowest common multiple of 82°+27,
16a*+ 3627+81, and 62?-5x-6.
76, Solve the equations :
(1) a(a—a)-—b(a-—b)=(a+b)(x-a-—b).
(2) (a+b)z--ay = a?, (a? + b*)x -— aby = 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
ey vty 2a-8b 3b
(1) oe eA ee eth gh (2) 2Qa—6b 2a
a?—y? atty?? ~" 2a=3b 3b
ory? 72 — 2a 2a —6b
79, Find the H.C.F. of 212? - 26a?+ 8a and 6a723 — a7a? — 2a°x.
Also the L.C.M. of z?-2, ax?+2a2—-3a, x?-—72?+6z2.
80, Simplify the expression
(a+b+c)(a-b+c)-{(atcP—-b— (a+b +c")}.
81, Solve the equations :
T+u QWw-y 5Sy-7 , 4a-3 <
ee nD : sis oF;
(1) 5 p 3y — 9, 5 “b 6 Ov
(2) Se hie anh)
3u—-2Z x2+1 2243
82, Extract the square root of
aes 3003 ey:
#4 V2 +4) 4+9(2 +4) ~4(Z 42) 45.
Ope ee pe 3 ae JP
83, Find the value of
4a+6b | 6a-4b _4a°+65? , 4b?- 6a? , 200!
a+b a—b a? — b? a+b? at—bt
84, A bag contains 180 gold and silver coins of the value alto-
gether of $144. Each gold coin is worth as many cents as there are
silver coins, and each silver coin as many cents as there are gold
coins. low many coins are there of each kind ?
MISCELLANEOUS EXAMPLES V. 185
85. Solve the equations:
ibe ob Soaiom ke
glo a+4 247?
(2) V22+6-Nx—1=2.
86, Find the factors of
(1) 20a?+2lab-27b?; (2) a3-3z?--9x+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.
68, Simplify the expression
{(a+b)(at+b+c)+c7{(a+b)? - 7}
{(a+bP-C}{at+b+ey
89, Find the square root of
(2x + 1)(2a + 3)(2% + 5)(2%+7) + 16.
90, Simplify
10a? paket fe 2
(1+a?)(l-4a2) 1-2a 1+a?™
91, Solve the equations:
liye PER SNE ae
ine ain me oe
Oh ye mae ee
99, Resolve into factors:
(1) 622+52-6. (2) 9at-82ax°y?+9y4.
a4 — 154? 4+- 28a - 12
Red
ea Ema Be
to its lowest terms.
94, Simplify the fraction
(a+b)? a+ 2b+a , (a+ b)z ak
(a-a)(xtat+b) Wwe-a) w#+be-a?-ah 2
186 ALGEBRA.
95, A person being asked his age replied, ‘‘Ten years ago I was
five times as old as my son, but twenty years hence he will be half
my age.” What is his age?
96, Find the value of
{ 2a, Lee ee eee
(a-a)(ate) a-a wtasy az
Oy. When ia yoo ee g d=-1, find the value of
a’ — 03 —(a—b)> - 11(80 +20)(2e-F).
a“
98, Find the square root of
a* + b4 — a®b — ab? + 9a"
99, Solve the equations:
(1) 3x 24
= =92- 9 2 = 493.
a OeeT 4 (2) 327+ 227 = 493
LOO DEY Negi onan ined
101, What value of x will make the sum of pear and Hane
equal to 2?
102, Aman drives toacertain 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 74 minutes longer than in going: how long is each road ?
103, Find the product of
(1) 32°?-4ay+ Ty’, 3a*+4ay+ Ty? ;
(2) #?-2y7, 2?-Qay+2Qy?, x?+2y?, 2? +2ry+2y?.
104, Extract the square root of
309
1-50? 4-208 + Sat - S054 a
MISCELLANEOUS EXAMPLES V. 187
105, Find the highest factor common to
x(62? — 8y?) — y(3a?-4y?) and Qay(2y —x) +423 — 2y3,
106, The 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?
2
the yeeaaes (a+1) = 3, prove that a® neo.
(eR?
108, Solve the equations :
(1) (x+7)(y-3)+2-(y+8)(x-1) =5x-1ly+35=0;
(2) pastes) =
3 64-32 3(x-2)
109, If x=b+c, y=c-a, z=a-b), find the value of
x+y? +27 — Qay — Quea+ Qyz.
110, Express in the simplest form
en Leer Pee tee . a
02+ 3e+1 6a2+bet+1] 1222+7x24+1 2022+9xe41
11]. Resolve into factors :
(1) 4a7b? — (a2 +b? -c?)*;
(2) ab(m?+1)+m(a? +b”).
112, Simplify the fractions :
] a-x
——— at Nee
(1) —: @) —
ete CH= the
14.21 ie
3-2 x
1138, Solve the equations :
75-2 , 802+21_ 23
1 : = 55
(1) 3(e41) 5(8a42) £41
(2) Ve+12— Ja =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 34 hours after the arrival
of the express. Find the rate at which each train travels.
115, Simplify
a’ — 64a , [ 2a27+5a4+2. ee
a*—4 °2a?+9a+4° \a?+4a° a?+a—2 ;
116. Resolve into four factors
4(ab + cd)? — (a? +b? - c? - d’)?.
LL When Oa 4 OH ec :, d=-1, find the numerical
value
4c? — ala — 2b — d) —‘Vb4e + 11b3a?.
118, Find the value of
119, Solve the equations :
120, A has 19 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 walking at the same rate as & for the remainder of the journey
he arrives half an hour sooner than he expected. Find how long the
journey occupied each man.
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Of the authors’ ‘ Higher Algebra,’ as of their ‘ Elementary Algebra,’ we un-
hesitatingly assert that it is by far the best work of the kind with which we are
acquainted. It supplies a want much felt by teachers.’’— The Athenwum.
MACMILLAN & CO.,
GORE PELE Asay ENG Ey ra VVeey OR IK,
ELEMENTARY TRIGONOMETRY
BY
H. S. HALL, B.A., and S. R. KNIGHT, B.A.
”
Authors of ‘Algebra for Beginners,’ “Elementary Algebra for Schools,” ete.
Cloth. $1.10.
“I consider the work as a remarkably clean and clear presentation of the principles
of Plane Trigonometry. For the beginner, it is a book that will lead him step by step
to grasp its subject matter in a most satisfactory manner.” — E. MILLER, Unzversity
of Kansas.
“The book is an excellent one. The treatment of the fundamental relations of
angles and their functions is clear and easy, the arrangement of the topics such as
cannot but commend itself to the experienced teacher. It is, more than any other
work on the subject that I just now recall, one which should, I think, give pleasure
to the student.” — JoHN J. SCHOBINGER, 7he Harvard Schood.
WORKS BY REV. J. B. LOCK.
TRIGONOMETRY FOR BEGINNERS.
AS FAR AS THE SOLUTION OF TRIANGLES,
16mo. 75cents.
‘(A very concise and complete little treatise on this somewhat difficult subject for
boys; not too childishly simple in its explanations; an incentive to thinking, not a
substitute for it. The schoolboy is encouraged, not insulted. The illustrations are
clear. Abundant examples are given at every stage, with answers at the end of the
book, the general correctness of which we have taken pains to prove. The definitions
are good, the arrangement of the work clear and easy, the book itself well printed.”
— Fournal of Education.
ELEMENTARY TRIGONOMETRY.
6th edition. (In this edition the chapter on Logarithms has been carefully revised.)
NGMO.) 51.10.
‘“‘ The work contains a very large collection of good (and not too hard) examples.
Mr. Lock is to be congratulated, when so many Trigonometries are in the field, on
having produced so good a book; for he has not merely availed himself of the labors
of his predecessors, but by the treatment of a well-worn subject has invested the
study of it with interest.” — ature.
MACMILLAN & CO.,
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