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CHAPTER V.

Fractions.

56. It is unnecessary to repeat here the propositions relating to fractions which were proved in Arithmetic, Chap. II. of this work. The student will see that, by substituting general symbols for the particular figures there used, the reasoning will equally hold. We shall work out a few examples to show the method of dealing with them in algebra.

23
2 202 + 3

12 Ex. 1. Simplify the fraction

2cm + 2 x 15 By inspection (Art. 30) we see that 2 – 3 is a factor of

x a numerator and denominator. We have then2013

- 12 #2 (C – 3) + x (x – 3) + 4 (20 3) 2c2 + 2a 15

(oc 3) (x + 5) (2x2 + x + 4) (3C – 3) 2c2 + x + 4

Ans. (x – 3) (x + 5)

1

1 Ex. 2. Find the value of

+ a + 7

a

6 a” + 62 1 1

(a - b) + (a + b) + a + b

a

7 a + 62 (a + b) (a - b) a“ + 62

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bc Ex. 3. Find the value of

(a b) (a – c) * 76 - a) (6 c) ab (c a) (c 6)

The second denominator has a factor, (b a), which differs from a factor, (a b), of the first denominator in sign only. We shall therefore change the sign of the second fraction, and also of its first factor. This will not alter its value.

And, similarly, we find that by changing the signs of each of the factors of the third denominator we shall have them in a form corresponding to factors of the first and second denominators. The sign of the third fraction will not be changed, as the sign of the denominator will, on the whole, be unchanged. The given expression then will stand thusbc

ab (a - b) (a – c)

c) * (a - b) (6 c) * (a c) (6 c) bc (b c) ac (a c) + ab (a 6)

(a - b) (a c) (6 c) bc (b − c) – aʼc + ac® + aob

ab (a - b) (a c) (6 c)

; or, re-arranging, a(b c)

a (62
ca) + bc (6 c)

then, dividing nume(a - b) (a c) (6 c) rator and denominator by b C, as a (b + c) + bc (a - b) (a c)

1. (a - b) (a – c) (a - b) (a c) Ex. 4. Simplify, 4 ax 3 ax + aci?

4 a-x + 3 ax + 23 ?

ar a + The given expressiona*(a + x) – 4 aʻxc + 3 ax” – 243 a(a 2) + 4 aʼx + 3 ax? + 23

a + 20 ai +, aʻx 4 a-x + 3 ax2 – 203 - aʻx + 4 aʻx + 3 axa + 23

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Ex. XIV. Simplify the following expressions:

x2 5x + 4 23 3 x + 2 1.

X2 + 2a 24' 23 + 4 x – 5 2.* 6 x2 + 29 x + 35 2 203 + 72 9 14 2 + 39 x + 10' 5 23 3 22 4x + 2

a4 2 ab+ b4 24 a3 28 a b + 6 ab? 773 3.

a' - 4a ́b + 4 ab? 73 6 a + 11 ab 21 62
2c(yo – 2) - xy (2 y + yz - ?) + y(y + 2) 2* + x*y* +y*

2ʻy2 2* (y + x)} – xy(2 y2 + 3 yz + x) + yo (y + 2)206 200 1 1 1

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