Question

Simplify the sum. State any restrictions on the variables.

Answer 1

The sum is $\frac{x+2}{x+3}$.

Solution:

Given equation:

$\frac{x-2}{x+3}+\frac{10 x}{x^{2}-9}$

        $=\frac{x-2}{x+3}+\frac{10 x}{x^{2}-3^2}$

Using algebraic expression: $a^2-b^2=(a-b)(a+b)$

         $=\frac{x-2}{x+3}+\frac{10 x}{(x-3)(x+3)}$

To make the denominator same, multiply and divide the first term by (x –3)

         $=\frac{(x-2)(x-3)}{(x+3)(x-3)}+\frac{10 x}{(x-3)(x+3)}$

         $=\frac{(x-2)(x-3)+10x}{(x+3)(x-3)}$

          $=\frac{x^2-2x-3x+6+10x}{(x+3)(x-3)}$

         $=\frac{x^2+5x+6}{(x+3)(x-3)}$

         $=\frac{(x+3)(x+2)}{(x+3)(x-3)}$

Cancel the common factors.

         $=\frac{x+2}{x+3}$

$\frac{x-2}{x+3}+\frac{10 x}{x^{2}-9}=\frac{x+2}{x+3}$ ; but x ≠ –3

If you substitute x = –3, the value is indeterminate.

Hence the sum is $\frac{x+2}{x+3}$.