#### Answer

$= \frac{5x-4}{(x+2)(x+5)(x-5)}$

#### Work Step by Step

$\frac{2}{x^{2}+7x+10} + \frac{3}{x^{2}-25}$
1. Factor the denominator of the first fraction
$ = \frac{2}{x^{2}+5x+2x+10} + \frac{3}{x^{2}-25}$
$ = \frac{2}{x(x+5)+2(x+5)} + \frac{3}{x^{2}-25}$
$ = \frac{2}{(x+2)(x+5)} + \frac{3}{x^{2}-25}$
2. Factor the denominator of the second fraction:
$ = \frac{2}{(x+2)(x+5)} + \frac{3}{x^{2}-5x+5x-25}$
$ = \frac{2}{(x+2)(x+5)} + \frac{3}{x(x-5)+5(x-5)}$
$ = \frac{2}{(x+2)(x+5)} + \frac{3}{(x-5)(x+5)}$
3. Find a common denominator:
Multiply the first fraction by $(x-5)$ and the second fraction by $(x+2)$ to obtain a common denominator of $(x+2)(x+5)(x-5)$
$ = \frac{2(x-5)}{(x+2)(x+5)(x-5)} + \frac{3(x+2)}{(x-5)(x+5)(x+2)}$
4. Add the fractions
$= \frac{2(x-5)+3(x+2)}{(x+2)(x+5)(x-5)}$
$= \frac{2x-10+3x+6}{(x+2)(x+5)(x-5)}$
$= \frac{5x-4}{(x+2)(x+5)(x-5)}$