In this example the work is shortened by adding the first two fractions, then by adding this sum and the third fraction, then this latter sum and the fourth fraction.

23.

+

24.

25.

(a-2)(a-1) (1− a) (3− a) (3a) (a−2)

1

+

1

+

1

(a - b) (a — c)' (b −c)(b − a)' (e− a)(c—b)

+ (e − a) (c − b )

(a - b)(a — c) (a−b)(b−c)' (c − a) (c —b)

x

64

40 a2b2 - 16 b4

a1- 16 b4

-

CHAPTER XV

EQUATIONS

160. Example 1. Solve

2x+3 4 3
=- +

x(x-3) x X 3

Multiplying by x(x-3), the L. C. M. of the denominators,

The first and third terms are meaningless; as the value 3 for x causes a denominator to vanish, it cannot be asserted that 3 is a solution of the equation.

This value of x does not make a denominator zero. It will satisfy the equation, and hence it is a solution.