Eramples. 2. Divide a +5a'r +100°r* +- 10a®z + bart tr.

Ans. a* + 3 a 1 + 3 ars + . 3. Divide 12 x 13 r* — 34 x' + 35 x* by 4.m.; - 78.

Ans. 3r + 2r2 5r. 4. Divide z' — t - H2 + 2r - 1 by x? + j.

Ans. It m3 + x2 + I. 5. Divide yo — 3y*x® + 3yor* + zo by ys – 3yor + 3yr* — 23.

Ans. y' + 3yor + 3yrs + II 6. Divide as 5a'x + Joao.ro 10a’r + 5art -- XS by a' Zar + 3*.

Ans. a- За'r + Зах? .

Miscellancous Examples. 1. Divide as I by a — x. Ans, a' ta'r -- a* ra t ar + r. 2. Divide 4.23 76ax: + 35a’r + 105a2 by 2r 3a.

Ans. 2x 35ar - 35a?. 3. Divide 23 + y by r + y.

Ans. x2 y? - yr. 4. Divide 2a* 32 by a Ans. 2a3 + 4a2 + 8a + 16. 5. Divide as + air •a2x2 7a-r! + 6x by a? 7.

Ans, a' t a? - 6r. Divide x + 3r2y + 3ry? t ys by za + 2xy + ye.

Ans. x + y. 7. Divide a' — 3a?c + 4ac - 26 by a? — 2ac + c.

acc Ans. a 6+

a 2ac + c 8. Divide 6.x* — 96 by 3r – 6. Ans. 2x3 + 4x2 + 80 + 16. 9. Divide 1 50 + 10x2 - 10.23 + 5rt Is by 1 -2x + xs.

Ans. I 30 + 3.2% - 13.

2.

FRACTIONS. Algebraic Fractions are subject to the same rules as are the fractions of commun arithmetic, and their reduction requires only the application of the principles laid down in the preceding rules.

PROBLEM I.
To reduce a mised quantity to an improper fraction.

Rule. Multiply the integer by the denominator of the fraction, and to the product annex the numerator with its proper sign ; under this same place the former denominator, and the result will give the improper fraction required.

Example 1.

2.0 Reduce 3a + to an improper fraction. Here the integral part x denom, of the fraction + the numerato

3a X 5a+ 202 = 15a + 2x. Hence

is the fraction required.

5a

15 a

+ 2x

The one to an improper fraction

C

2. Reduce 51

30 apk 4.2 Ans.

ба? 23 3. Reduce 57

to a common denominator.

330 + 3* Ans.

7 20 4. Reduce hab + to a common denominator.

За

40 5. Reduce 362

to a common denominator. 5.3

a 6. Reduce a It

to a common denominator.

C

a

PROBLEM II.
To reduce an improper fraction to a whole or mixed quantity.

Rule. Observe which terms of the numerator are divisible by the denominator without a remainder, the quotient will give the integral part ; to this annex the remaining terms of the numerator, with

the denominator under them, with their proper signs, and the result will be the mixed quantity required.

Erample 1. Reduce a2 + ab + be

to a whole or mixed quantity. Here az + ab

62
=at the integral part; and- the fractional.

a,
7.8
is a tbt is the mixed quantity required.
15a2 + 20 Зc

23 2. Reduce

30

Ans. 3a + 5a

5a 5a

sa 3. Reduce

2013 12a + 40 3c 4. Reduce

42 10x2

2/8 5. Reduce

a

42%

Ans. 2x

+ 3.20

PROBLEM III.
To reduce fractions to a common denominator.

Rule. Multiply each numerator into every denominator 'ut its own for the new numerators, and multiply all the denominators for a common denominator. Sa mukiplied by 7 = 353 ; and 28 - 3 subtracted from 35x leaves 53378

i

a

Examples and , to equivalent having a common denominator.

.

с

1. Reduce

с

7

d Here a Xcx d = acd

New

Hence the fractions bi xbxd=l'd

Numerators. required are схих

acdb2d, c'd

and xcx d = bod Com. Denom.

lcd lcd

écd

C 3 cd

6x +9
Ans. and

5x +2. 3r

31 5. Reduce 41+2r 372

2r and

to a com. denominator. 5 4a

3ს
48 alx° +2 ali
Ans.

45 biza

40 ar

and 60 ab 60 al

60 ab 6. Reduce

1
and
4.ro 1 + 2

to a common denominator. 2x

2atal

Зас 7. Reduce

X2

and 27 26

to a common denominator. X +

7x2

PROBLEM IV.

To reduce a praction to its lowest terms.

Rule. Observe what quantity will divide all the terms of both mumerator and denominator without any remainder : divide them by this quantity, and the fraction is reduced to its lowest terms.

Examples. 1. Reduce 14 2 + 7ar + 2x

to a common denominator.

35 Nute.-The co-efficient of every term of both numerator and denominator

is divisible by 7, and x enters also joto every terın; + 7x will therefore divide both nomerator and denominator without a remainder. For 14./9 + 7ar + 21x2

= 22a + a + 3x ; 35 ro

2a + a +3 And

=5x. Hence, the frac. in its lowest terms, is 77

5x

72

2. Reduce

0

to its lowest terms. 12

a?

1 Ans.

a to

a

b

1 8. Reduce

Ans,
a US
. 69

a + ab + 1,2 Scholium. When, therefore, we have occasion to reduce algebraic fractions to their lowest terms, by finding the greutest cuinmon meusure of the numerator and the denominator, as in conimon arithmetic, it is evident that the rule which we have given for the solution of the foregoing problem answers every practi. cal purpose, since it is only requisite to discover what quantity will divide both numerator and denominator, and that the quantity must therefore be the greatest common measure.

PROBLEM V.

(43.) To add fractional quantities. Rule. Reduce the fractions to a common denominator, as in Problem III. Add all the numerators together, and below their sum set down the common denominator, and it will give the sum of the frac

tions required.

+3, 3r_1

and

2r + 3

4.0 2. Add

and

together.
5
2.1

7
Ans.
68ro + 1471 – 35

703
31 5.7
4r

4072
3. Add
and

Ans.
7 2
11

154
3a2 2a

31

Ans.

105a + 28a"), + 306 4. Add

26
5
7a

70ab
2r + 1 4r + 2

169x+77 5. Add

and

Ans.
3
5

105
5ar + b 4a2 + 2b

and 6. Add

37a +116

Ans.
36
56

150
X + 1
2r

11x + 377 + 15 and

Ans.
r +2' 3
5

15x + 30
2r 5
1

4.3% -72-31
8. Add
and

Ans.
3
2.1.

Or
2.x2
Ans..

2x + 1
and
11

XP

7. Add

9. Add

(44.) To subtract fractional quantities.

Rule. Reduce the fractions to a common denominator as in Problem III ; and then subtract the numerators from each other, and under the difference, write the common denominator.

Eramples. 3.7

14.0 1. Subtract

(rom 5

15

C

Here 32 x 15 = 45% .:: 70x 457

25.7 and 147 x 5 = 70.7

is the Also 5 * 15 = 75

75

75

3 difference required.

2a 4.2 2. From

take 30

50 Here (Z - a) 5c = 50% 5ас

5cx sac bal 12br and (2am-fx)8l=6ab 12lir Also 36 50 = 15 lc

15bc

15bc 501 5ac 6ab + 12bx

the difference required.