# How do you solve 1/(x+3) + 1/(x-3) = 10/ (x^2-9)?

Feb 13, 2016

$x = 5$

#### Explanation:

$\frac{1}{x + 3} + \frac{1}{x - 3} = \frac{10}{{x}^{2} - 9}$

First put everything with same denominator, recalling that $\left(x + 3\right) \left(x - 3\right) = {x}^{2} - 9 :$

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

We can eliminate the denominators, but keeping in mind that $x \ne \pm 3$

$x - 3 + x + 3 = 10$

$2 x = 10$

$x = 5$