Charles Davies.

First lessons in algebra, embracing the elements of the science online

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sign of division, but which may frequently be reduced.

Take, for example, 12 a*b 2 cd 9 to be divided by 8a 2 bc\
which is placed under the form

I2a*b 2 cd ■
8a 2 bc 2 l



Quest. — 43. What is the first case named in which the division of
monomials will not be exact 1 What is the second 1 What is the third ]
If either of these cases occur, can the exact division be made 1 Under
what form will the quotient then remain ? May this fraction be often
reduced to a simpler form 1



40 FIRST LESSONS IN ALGEBRA.

this may be reduced by dividing the numerator and denomi-
nator by the common factors 4, a 2 , b, and c, which gives

I2a*b 2 cd 3a 2 bd



8a 2 bc 2 2c



44. Hence, for the reduction of a monomial fraction we
have the following

RULE.

I. Suppress the greatest factor common to the two co-
efficients,

II. Subtract the less of the two exponents of the same letter
from the greater, and write the letter affected with this differ-
ence in that term of the fraction corresponding with the greatest
exponent.

III. Write those letters which are not common, with their,,
respective exponents, in the term of the fraction which contains
them.

From this new rule, we find,

(1) (2)

48aWcd 3 4ao? ^ 37a b 3 c 5 d 37b 2 c

and



26a 2 b 3 c 2 de~~ 3bce ' 6a 3 b cH 2 ~~ 6a 2 d '

(3) (4)

7a 2 b 1 , 4a 2 b 2 2a

also — , , OTO = ■ ; : and



Ua 3 b 2 2ab * 6ab* 3b 2

7bc 2

5. Divide 49a 2 b 2 c 6 by \4cfibc*. Ans. — —

2a

6. Divide Samn by 3abc. Ans. — — .

* be

7. Divide \8o?b 2 mn 2 by \2a*¥cd. Ans, n *■* - .

r 2a 2 b 2 cd

Quest. — 44. Give the rule for the reduction of a monomial fraction



DIVISION. 41

7cfic*d

8. Divide 28a*b 6 c 1 d 8 by l6ab 9 cdFm. Ans. —■— — .

9. Divide 72a 3 c 2 b 2 by \2a 5 c*b 3 d. Ans.



10. Divide I00a 8 b 5 xmn by 25a 3 b*d. Ans.

11. Divide 96a 5 b 8 c 9 df by 75a 2 cxy. Ans.



a 2 c 2 bd
4a 5 bxmn

d '
32a 3 b 8 c 8 df
~~25xy~'



12. Divide 85m 2 n 3 fx 2 y 3 by I5am*nf. Ans. — — ——.

127

13. Divide I27d 3 x 2 y 2 by 16d 4 ;ry. -A** rsir*-*

I6dx 2 y 2

45. If we have an expression of the form

a a 2 a 3 a 4 a 5

— , or — , or — , or — , or — , &c,
a a 2 a 6 a> aP

and apply the rule for the exponents, we shall have



a ;_. n a 2 _ A a-



za l - l =a°, —=za 2 ~ 2 =a°, —=za 3 - 3 = a°, &c.
a 2, a 3



But since any quantity divided by itself is equal to 1, it
follows that

— =0<> = 1, ^-=a*-*=z<fi=l, &c,

a a*

or finally, if we designate the general exponent by m, we

have

a m

— =a m - m =a?=l ;

a m

that is, any power of which the exponent is is equal to 1.



Quest. — 45. What is aft equal to? What is bo equal tol What is
the power of any number equal to, when the exponent of the power is 1

4*



42 FIRST LESSONS IN ALGEBRA.

2. Divide 6a 2 b 2 c*d by 2a 2 b 2 d.
6a 2 b 2 c*d



2a 2 b 2 d



■=3a 2 - 2 J 2 - 2 c*cP- 1 =3e*.



3. Divide 8a*b 3 c±d 5 by 4a 2 b 3 c*d 5 . Ans. 2cF.

4. Divide 16a 6 5 8 ^ by 8a G b B d. Ans. 2d*.

5. Divide 32m 3 n 3 oc 2 y 2 by 4m 3 n 3 xy>. Ans. 8xy.

6. Divide 96a*b 5 d Q c 9 by 2U*¥d 5 c*. Ans. 4bd 3 .

SIGNS IN DIVISION.

46. The object of division, is to find a third quantity
called the quotient, which, multiplied by the divisor, shall
produce the dividend.

Since, in multiplication, the product of two terms having
the same sign is affected with the sign + , and the product
of two terms having contrary signs is affected with the
sign — , we may conclude,

1st. That when the term of the dividend has the sign -f->
and that of the divisor the sign of + , the term of the quo-
tient must have the sign + .

2nd. When the term of the dividend has the sign + , and
that of the divisor the sign — -, the term of the quotient
must have the sign — , because it is only the sign — ,
which, multiplied with the sign — , can produce the sign +
of the dividend.



Quest. — 46. What will the quotient, multiplied by the divisor, be
equal to ? If the multiplicand and multiplier have like signs, what will
be the sign of the product? If they have contrary signs, what will be
the sign of the product ? When the term of the dividend and the term
of the divisor have the same sign, what will be the sign of the quotient ?
When they have different signs, what will be the sign of the quotient ?



DIVISION.



43



3rd. When the term of the dividend has the sign — -, and
that of the divisor the sign +> the quotient must have the
sign — . Again we say for brevity, that,

+ divided by -fj and — divided by — , give + ;
— divided by -f, and + divided by — , give — .



EXAMPLES.



1. Divide 4ax by —2a. Ans. —2x.

Here it is plain that the answer must be —2x ; for,
—2a x — 2x=z -\-4ax, the divisor



2. Divide 36a 3 * 3 by — I2a 2 x.

3. Divide — 58a 3 b 5 c 2 d 2 by 29a 2 b*c.

4. Divide — 84a±¥d 3 by — 42a 2 b 2 d.

5. Divide 64c*d 5 x 3 by I6c 4 dx.

6. Divide —88b±x 5 y e by —24b 3 cdx 5 .

7. Divide lla^y 3 ^ by -\\a±y 3 z\

8. Divide 84a 4 b 2 c 2 d by — 42a*b 2 c 2 d.

9. Divide - 60« 7 #W by — 12c 8 i 7 c 5 ^.

10. Divide — 88a 6 b'c 6 by 8a 5 6 G c 6 .

11. Divide I6x 2 by —8a?.

12. Divide —\ba 2 xy 3 by 3«y.

13. Divide — 84ab 3 x by — 126 2 .

14. Divide -96a±b 2 c 3 by 12a 3 5c.

15. Divide — 144a*b 8 c 1 d 5 by — 36a 4 6W.

16. Divide 256a 3 bc 2 x 3 by — 16a 2 cx 2 .

17. Divide — 300a 5 6 4 c 3 a: 2 by 30a*b 3 c 2 x.

18. Divide 500<z 8 6 9 c 6 by — 100a 7 & 8 c 4 .



-4n,?. +5-



j4ft^. — 3a*.

-4fts. —2abcd 2 .

Ans. 2a 2 b 3 d 2 .

Ans. 4d 4 x 2 .

Ans +H^- 6
A7lS ' + 3cd '

-4fts. —7.

4ft*. -2.

1_

abed'

Ans. —l\ab.

Ans. —2x.

Ans. —5axy 2 .

Ans. 7abx.

Ans. — 8abc 2 .

Ans. 4a 5 b 2 cd*.

Ans. — 16abcx.

Ans. — lOabcx.

Ans. — babe 2 .



44 FIRST LESSONS IN ALGEBRA.

19. Divide — 64<z 5 6 8 c 7 by — 8aWc 6 . Ans. Babe.

20. Divide +96a 5 ^ 9 by — 24aW. Ans. —4ab 2 d 8 .

21. Divide 72a 5 b 3 d i by — 8a 4 b 2 d. Ans. — 9abd 3 .



V



a — a?
Quotient.



Division of Polynomials. v

FIRST EXAMPLE.

47. Divide a 2 — 2ax+x 2 by a— a?.

[t is found most convenient, Dividend. Divisor.

in division in algebra, to place a 2 — 2ax-\-x 2

the divisor Hi the right of the a 2 — ax
dividend, and the quotient di- — ax-\-x 2

rectly under the divisor. — ax-\-x 2

We first divide the term a 2 of the dividend by the term a
of the divisor : the partial quotient is a, which we place
under the divisor. We then multiply the divisor by a, and
subtract the product a 2 — ax from the dividend, and to the
remainder bring down x 2 . We then divide the first term of
the remainder, — ax by «, the quotient is — x. We then
multiply the divisor by — x, and, subtracting as before, we
find nothing remains. Hence, a—x is the exact quotient.

In this example, we have written the terms of the dividend
and divisor in such a manner that the exponents of the same
letter shall go on diminishing from left to right. This is
what is called arranging the dividend and divisor with
reference to a certain letter. By this preparation, the first
term on the left of the dividend, and the first on the left of
the divisor, are always the two which must be divided by
each other in order to obtain a term of the quotient.

Quest. — 47. What do you understand by arranging a polynomial with
reference to a particular letter ?



division. 45

48. Hence, for the division of polynomials we have the
following

RULE.

I. Arrange the dividend and divisor with reference to a cer-
tain letter, and then divide the first term on the left of the
dividend by the first term on the left of the divisor, the result
is the first term of the quotient ; multiply the divisor by this
term, and subtract the product from the dividend.

II. Then divide the first term of the remainder by the first
term of the divisor, which gives the second term of the quotient ;
multiply the divisor by the second term, and subtract the pro-
duct from the result of the first operation. Continue the same
process until you obtain for a remainder ; in which case the
division is said to be exact.

SECOND EXAMPLE.

Let it be required to divide

b\a 2 b 2 +\0a*— 48a 3 b — I5b*+4ab 3 by 4ab—5a 2 +3b 2 .
We here arrange with reference to a.

Dividend. Divisor.



10a 4 — 48a 3 b+5la 2 b 2 + 4ab 3 — l5b*

4_10a4_ 8a 3 b— 6a 2 b 2

—40a 3 b+57a 2 b 2 + 4ab 3 —l5b*
— 40a 3 b + 32a 2 b 2 + 24ab 3



— 5a 2 +4ab+3b 2

—2a 2 +8ab — 5b 2

Quotient.



25a 2 b 2 —20ab 3 — l5b*
25a 2 b 2 —20ab 3 — l5b*



Quest. — 48. Give the general rule for the division of polynomials 1
If the first term of the arranged dividend is not divisible by the first term
of the arranged divisor, is the exact division possible 1 If the first term
of any partial dividend is not divisible by the first term of the divisor, is
the exact division possible 1



46 FIRST LESSONS IN ALGEBRA.

Remark. — When the first term of the arranged dividend
is not exactly divisible by that of the arranged divisor, the
complete division is impossible ; that is to say, there is not
a polynomial which, multiplied by the divisor, will produce
the dividend. And in general, we shall find that a division
is impossible, when the first term of one of the partial
dividends is not divisible by the first term of the divisor.

GENERAL EXAMPLES.

1. Divide 18# 2 by 9x. Ans. 2x.

2. Divide I0x 2 y 2 by —5x 2 y. Ans. —2y.

3. Divide — 9ax 2 y 2 by 9x 2 y. Ans. —ay.

4. Divide — 8x 2 by —2x. Ans. + 4a\

5. Divide \0ab-\-\5ac by 5a. Ans. 25+ 3c.

6. Divide 30ax—54x by 6x. Ans. 5a— 9.

7. Divide 10x 2 y—l5y 2 —5y by 5y. Ans. 2x 2 - 3y—l.

8. Divide \2a-\-3ax— 18ax 2 by 3a. Ans. 4 + x— 6x 2 .

9. Divide 6ax 2 -{-9a 2 x-\-a 2 x 2 by ax. Ans. 6x-\-9a-\-ax.

10. Divide a 2 -\-2ax-\-x 2 by a-\-x. Ans. a-\-x.

11. Divide a 3 — 3a 2 y-\-3ay 2 — y 3 by a— y.

Ans. a 2 — 2ay-\-y 2 .

12. Divide 2Aa 2 b — \2a 3 cb 2 — 6ab by —6ab.

Ans. — 4a-\-2a 2 cb-{-l.

13. Divide 6x*— 96 by 3a;— 6. Ans. 2# 3 + 4a: 2 +8tf+16.

14. Divide . . . a 5 — 5a i x+l0a 3 x 2 — \0a 2 x 3 +5ax*— x 5
by a 2 — 2ax-\-x 2 . Ans. a 3 — 3a 2 x-\-3ax 2 — x 3 .

15. Divide 48x 3 —76ax 2 — 64a 2 x+105a 3 by 2x — 3a.

Ans. 24x 2 —2ax—35a 2 f



DIVISION* 4t

16. Divide y 6 — 3y% 2 +3y 2 a 4 — x 6 by y 3 — 3y 2 x+3yx 2 — x 3 .

Ans. y 3 +3y 2 #+3y# 2 +# 3 .

17. Divide 64aW— 25a 2 b 8 by 8a 2 b 3 +5ab\

Ans. 8a 2 b 3 —5ab*.

18. Divide 6a 3 +23a 2 b+22ab 2 +5b 3 by 3a 2 +4ab+b 2 .

Ans. 2a-{-5b.

19. Divide 6ax 6 +6ax 2 y 6 +<±2a 2 x 2 by ax+5ax.

Ans. x 5 -\- xy Q -\-l ax .

20. Divide . . — lba± +37 a 2 bd-2§a 2 cf-20b 2 d 2 +Ubcdf
— 8c 2 / 2 by 3a 2 — bbd+cf. Ans. — ba 2 -\-Ud—8cf.

21. Divide x* -\- x 2 y 2 -\- y* by x 2 —xy-\-y 2 .

Ans. x 2 -\-xy-\-y 2 .

22. Divide # 4 — y 4 by #— y. .Arcs. a; 3 +^ 2 y+ a? y 2 +y 3 -

23. Divide 3a 4 — $a 2 b 2 +3a 2 c 2 +5¥ — 3b 2 c 2 by a 2 — b 2 .

Ans. 3a 2 — 5b 2 + 3c 2 .

24. Divide . . 6x 6 — 5x 5 y 2 — Qx A y^-{-Qx 3 y 2 -{-\bx 3 y 3 — 9# 2 y 4
+ 10* 2 y 5 -fl5y 5 by 3# 3 +2tf 2 y 2 +3y 2 ,

Ans. 2x 3 — 3x 2 y 2 -$-5y 3 .

25. Divide . — c 2 +16a 2 # 2 — 7abc— \a 2 bx— 6a 2 b 2 + 6acx
by Sax— Gab— c. Ans. 2ax+ab + c.

26. Divide .... 3# 4 +4* 3 y-4o; 2 -4a: 2 y 2 +16a:y-15
by 2xy+x 2 — 3. Ans. 3x 2 —2xy + 5.

27. Divide # 5 +32y 5 by x-\-2y.

Ans. # 4 — 2# 3 y+4# 2 y 2 — 8#y 3 +16y 4 .

28. Divide 3a 4 -26a 3 £-14a& 3 +37a 2 & 2 by 2b 2 -5ab
_|_3 a 2 Ans. a 2 — lab.



48 FIRST LESSONS IN ALGEBRA*



CHAPTER II.

Algebraic Fractions.

49. Algebraic fractions should be considered in the same
point of view as arithmetical fractions, such as ^, U ; that
is, we must conceive that the unit has been divided into as
many equal parts as there are units in the denominator, and
that one of these parts is taken as many times as there are
units in the numerator. Hence, addition, subtraction, mul-
tiplication, and division, are performed according to the rules
established for arithmetical fractions.

It will not, therefore, be necessary to demonstrate those
rules, and in .their application we must follow the procedures
indicated for the calculus of entire algebraic quantities.

50. Every quantity which is not expressed under a
fractional form is called an entire algebraic quantity.

51. An algebraic expression, composed partly of an
entire quantity and partly of a fraction, is called a mixed
quantity.



Quest. — 49. How are algebraic fractions to be considered] What
does the denominator show ? What does the numerator show ? How
then are the operations in fractions to be performed 1 50. What is an
entire quantity \ 51. What is a mixed quantity?



ALGEBRAIC FRACTIONS. 49



To reduce a fraction to its si?nplest terms.

52. The rule for reducing a monomial fraction to its low-
est terms has already been given (Art. 44).

With respect to polynomial fractions, the following are
cases which are easily reduced.

1. Take, for example, the expression

a 2 — b 2



a 2 —2ab-\-b 2 '
This fraction can take the form

(a+b) (a—b)



(a -by

(Art. 39 and 40). Suppressing the factor a—b, which
is common to the two terms, we obtain

a — b'
2. Again, take the expression

5a 3 — \0a 2 b+5ab 2
8a 3 — 8a 2 b

This expression can be decomposed thus :

5a(a?—2ab + b 2 )
8a 2 (a-b) '

ba{a-b) 2



or,



8a 2 (a—b)'



Quest. — 52. How do you reduce a fraction to its simplest terms 1
5






50 FIRST LESSONS IN ALGEBRA.

Suppressing the common factors a (a— b), the result is

5(a- b)
8a

Hence, to reduce any fraction to its simplest terms, we sup-
press or cancel every factor common to the numerator and
denominator.

Note. — Find the factors of the numerator and denomina-
tor as explained in (Art. 41).

EXAMPLES.

i „ a 3a 2 +6a 2 Z> 2 . . .

1. Reduce — - = - to its simplest terms.

12a 4 +6a d c J

Ans.



4a 2 +2ac 2 '

■ _ . 15a 5 c-f25a 9 d ,

2. Reduce ^ F , nfx to its simplest terms.

25a 2 + 30a 2 £

3a 3 c+5a 7 d

^ n

3. Reduce 8 to its simplest terms.

j 17



3 c 7

J n . 60cW* . . .

4. Reduce ■ ^ , 7 ^ Q - to its simplest terms.

12c 5 d 8 /9 r 5c

Ans.



87«*J*— 81aW 3o 3 -9J2

5 - RedUC6 63745=36^' ^-Y4^4a-

96a 3 Z> 2 c

6. Reduce ■■ - to its simplest terms. Ans. —8.

m m - 245 5 -36a£ 4 45-6a

7. Reduce — — - — -—rrz' Ans.



48a 4 6 4 — 66a 5 6 6 ' ' 8a 4 — lla 5 6 2 .



ALGEBRAIC FRACTIONS. 51

> CASE II.

53. To reduce a mixed quantity to the form of a fraction.

RULE.

Multiply the entire part by the denominator of the fraction ;
then connect this product with the terms of the numerator by
the rules for addition, and under the result place the given
denominator.

EXAMPLES.



1. Reduce d>\ to the form of a fraction

6x7=42: 42 + 1=43: hence, 6j=~



43

7*



2. Reduce x i- to the form of a fraction.

x

a 2 —x 2 x 2 —(a 2 —x 2 ) 2x 2 — a 2



x —



X X



ax~\~ x

3. Reduce a?—- — to the form of a fraction.

2a

ax—x 2
Ans.



2a

2x—l

4. Reduce b-\ to the form of a fraction.

3x

17*— 7

Ans. — .

3a?

5. Reduce 1 to the form of a fraction.

a

2a—x-{-l
Ans. .



Quest.— 53. How do you reduce a mixed quantity to the form of a
fraction 1



52 FIRST LESSONS IN ALGEBRA.

x 3

6. Reduce l+2# — to the form of a fraction.

10;r 2 +4;r+3
ggu Arts. .

3c+4

7. Reduce 2a+b — to the form of a fraction.

iea+Sb — 3c— 4
An*. 8 .

(5a x ~—~ ab

8. Reduce 6ax+b - to the form of a fraction.

4a

18a 2 x+5ab

Ans. .

4a

84-6a 2 b 2 x*

9. Reduce 8+3ab - 7 ; — to the form of a frac-

96«5a: 4 +30a 2 ^ 4 — 8

tion. Ans. r~ . ; .

I2abx*

3J2__g c 4

10. Reduce 9H — — to the form of a fraction.

a—b 2

9«_65 2 -8c 4
Ans. u .

CASE III.

54. To reduce a fraction to an entire or mixed quantity.



RULE.

Divide the numerator by the denominator for the entire part,
and place the remainder, if any, over the denominator for the
fractional part.



Quest. — 54. How do you reduce a fraction to an entire or mixed
quantity ?






ALGEBRAIC FRACTIONS. 53

EXAMPLES.



1. Reduce — - — to an entire number.

8

8)8966(



1120 ... 6 rem.

Hence, 1120f= Ans.



2. Reduce to a mixed quantity.



Ans. a .

x



3. Reduce to an entire or mixed quantity.



Ans. a—x.



„ , ab—2a 2 . ,

4. Reduce - to a mixed quantity.



2a 2
Ans. a



b *

5. Reduce to an entire quantity. Ans. a-\-x.

a — x

6. Reduce — to an entire quantity.

x — y

Ans. x 2 -{-xy-\-y 2 .

1 _. '' 10* 2 -5;z+3 : .

7 Reduce to a mixed quantity.

DX

3

Ans. 2x— 1+— •
5a;

. „ , 36a 3 — 72x+32a 2 x 2 ,

8. Reduce to a mixed quantity.

A m ' m 32a 2 x
Ans. 4a; 2 — 8 -| — .



54 FIRST LESSONS IN ALGEBRA.



CASE IV.



55. To reduce fractions having different denominators
to equivalent fractions having a common denominator.

RULE.

Multiply each numerator into all the denominators except its
own, for the new numerators, and all the denominators together
for a common denominator.

EXAMPLES.

1. Reduce \, J, and £, to a common denominator.

1x3x5 = 15 the new numerator of the 1st.
7x2x5=70 „ „ „ 2nd.

4x3x2=24 „ „ „ 3rd.

and 2 X 3 X 5 = 30 the common denominator.

Therefore, Jf , |~g, and f£, are the equivalent fractions.

Note. — It is plain that this reduction does not alter the
values of the several fractions, since the numerator and
denominator of each are multiplied by the same number.

2. Reduce — and — to equivalent fractions having
a common denominator.

the new numerators.



aXc—ac > .
bxb = b 2 >



)Xl
and bxc—bc the common denominator.

Quest. — 55. How do you reduce fractions to a common denominator!



ALGEBRAIC FRACTIONS. 55

Hence, — and r— are the equivalent fractions.
be be

3. Reduce — and to fractions having a com-

, ac ab4-b 2

mon denominator. Ans. — and — ; .

be be

4. Reduce — , — , and d, to fractions having a

Zu oc

Qcx ^ab 6acd

common denominator. Ans. —r — , -3 — , and



6ac 6ac 6ac

* r> ! 3 2# . , 2# _ .

5. Reduce — , — , and a-\ , to fractions having

a common denominator.

9a Sax . 12a 2 +24x

Ans. — - r — , -— — , and



12a ' 12a ' 12a

y>2 /y2 I /yj2

6. Reduce — , — , and — — — , to fractions

<c> o a~\~ x

having a common denominator.

3a+3a? 2a 3 +2a 2 ;r 6a 2 +6x 2

6a+6#' 6a + 6# 6a +6x

ft Gax a - ■ ■ x

7. Reduce — , — - — , and = — to a common

So oc a

denominator.

5acd I8abdx ., 15a 2 bc — Ibbcx 2

Ans. - ^ 7 - , — ; ? » and



ISbcd ' 15fod ' 15fod

8. Reduce — , , and r-i t0 a common de-

5a c a-\-b

nominator.

ac 2 -\-c 2 b 5a 3 —5ab 2 . 5ac 2

Ans. —-z — — , — — — , and



&a 2 c+babc ' 5a 2 c+5abc . 5a 2 c-f-5a5c



56 FIRST LESSONS IN ALGEBRA.



56. To add fractional quantities together,

RULE.

Reduce the fractions, if necessary, to a common denomina-
tor ; then add the numerators together, and place their sum
over the common denominator. ~~

EXAMPLES.

1. Add §, f, and | together.

By reducing to a common denominator, we have

6x3x5=90 1st numerator.

4x2x5 = 40 2nd numerator.

2x3x2 = 12 3rd numerator.

2x3x5 = 30 the denominator.

Hence, the fractions become

90 40 12 142



30 ' 30 ' 30"" 30 '
which, by reducing to the lowest terms become 4^-|.

2. Find the sum of — , —r-, and — F ..
o d f

Here axdxf—adf\

cxbx f=cbf > the new numerators.

ex bxd—ebd J
And b x d xf= bdf the common denominator.

adf cbf ebd adf-\-cbf-\-ebd

Hence ' TdT + W + Tdf=-b^ thesum -



Quest. — 56. How do you add fractions.



ALGEBRAIC FRACTIONS. 57



3. To a ; — add b-\



A ,- , 2abx—3ex 2

Ans. a +0-1



be
4. Add — , — and — together. Ans. a?+



- A ,i *— 2 , 4» . \ 19a: — 14

5. Add — — and — together. Ans. — — .

m .,, , a?— 2 ■ ■ 2a?— 3 , 10a;— 17

6. Add a?H — — to 3a?H — .. Ans. 4x-\ - — .

5x x I a

7. It is required to add 4a?, -— — , and together.

. , 5x 3 -]-ax-\-a 2
Ans. 4a? -f -



2ax

hi
_ ___



8. It is required to add — , — , and — - — together.

49a?+12



Ans. 2a?+-



60



9. It is required to add 4a?, — , and 2+— together.

44a?+90

Ans. 4a? +• — ,

45

2a? 8a;

10. It is required to add 3a? -f— and a? — — together.

o y

o 2 3a?
Ans. 3a?-



45

11. Required the sum of «c— ■— - and 1 — =-,

oa a

Sa 2 cd—6bd+ Sad—Sac
Ans , __



58 FIRST LESSONS IN ALGEBRA.



57. To subtract one fractional quantity from another.

RULE.

I. Reduce the fractions to a common denominator,

II. Subtract the numerator of the fraction to be subtracted
from the numerator of the other fraction, and place the dif-
ference over the common denominator,

EXAMPLES.

3 2

1. What is the difference between -— - and — .

7 8

j*_ 2 _24 14_10__ j>_
T~T~5^~5^~56~28* nS *



2. Find the difference of the fractions -—7— and — .

2b 3c

tt<vh„ (#— a)x3c=3cx— 3ac ) ,
•Here, .* ' v the numerators

(2a— 4;r)x2&=4a&— 8ta? J

And, 2bx3c=z6bc the common denominator.

__ 3cx — 3ac Aab — 8bx 3cx — 3ac — 4ab-\-8bx

Hence ' —637 6fc—= 633 • Ans -

3. Required the difference of — - and — . Ans.



7 5 ' 35 *

4. Required the difference of 5y and -~. Ans. — ~-.

o 8

5. Required the difference of — and — . Ans. -^^-.



Quest. — 57. How do you subtract fractions ?



ALGEBRAIC FRACTIONS.



59



6. Required the differ ence between — - — and -j.



Ans.



7. Required the difference of — — — and



dx-\-ad — be

Yd '

2*+7



Ans



5b 8

24x+8a—l0bx—35b

40b '



8. Required the difference of 3a?+— and x .

cx-\-bx — ab



Ans. 2x-\-



bc



CASE VII.

58. To multiply fractional quantities together.

RULE.

If the quantities to be multiplied are mixed, reduce them to
a fractional form ; then multiply the numerators together for
a numerator and the denominators together for a denominator.

EXAMPLES.



\ZJ



1. Multiply



of T by 81



We first reduce the com- 1

pound fraction to the sim- 6

pie one ^ anc ^ then the
mixed number to the equiva-
lent fraction 2 ^ ; after
which, we multiply the Hence,
numerators and denomina-
tors together.



Operation.

f 3 3

° f T = 42'

3 3

JL 25 _ 75 _ 25

42 X 3 ~~ 126 — 42'

Ans. g.



60 FIRST LESSONS IN ALGEBRA.

_ ,., , . . i he m c ~. bx a 2 -\-bx

2. Multiply a-\ by -r. First, a -\ = — ■ .

ad a a

TT a 2 4- bx c a 2 c4-bcx

Hence, .... X-r= % — . Ans.

a a ad

„_,.., , „ 3# n 3a 9ax

3. Required the product of — and — . Ans. ., ■.

5 o po

2x 3x 2

4. Required the product of — and



5 2a

3x*



Ans.



5a



5. Find the continued product of — , and — -.

r « c 2b

Ans. 9ax.

6. It is required to find the product of b-\ and — .

ab-\-bx
Ans. .

x

x 2 yz #2 I £2

7. Required the product of — 7 and -7— — .

x*-b±
Ans.



b 2 c-{-bc 2



8. Required the product of x-\ , and =-.

a a-\-b

ax 2 — ax-\-x 2 — 1



Ans.



a 2 -\-ab



ax a — x

9. Required the product of a-\ by



a — x J x -\-x 2
a±—a 2 x 2



Ans.



ax-\-ax2 — x 2 — a: 3 '



Quest. — 58. How do you multiply fractions together ?



ALGEBRAIC FRACTIONS. 61

CASE VIII.

59. To divide one fractional quantity by another.

RULE.

Reduce the mixed quantities, if there are any, to a fractional
form; then invert the terms of the divisor, and multiply the
fractions together as in the last case,

EXAMPLES.

x.. ., 10 , 5

1. Divide. . . . — by — .

If the divisor were 5, the Operation.

quotient would be y 1 ^. But, 5 __ 1

since the divisor is -§• of 5, 8 "~" 8

the true quotient must be 8 10 10

times -^q, for the eighth of 24^ ~ 120

a number will be contained 10 80 2

in the dividend 8 times more 120 120 3*

than the number itself. In

this operation we have actually multiplied the numerator
of the dividend by 8 and the denominator by 5 ; that is, we
have inverted the terms of the divisor and multiplied the fractions
together.

x.. ., ° 1 f

2. Divide . . a— — by — .

2c g

b 2ac — b



2c 2c



b f 2ac—b g 2acg—bg

Hence, a-— -f-^-=— - X^= — % s ' Ans '

2c g 2c f 2cf



Quest. — 59. How do you divide one fraction by another 1


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