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

May 27, 2015

color(blue)(a^2 - b^2 = (a+b)(a-b)

$\left({x}^{2} - 9\right) = \left(x + 3\right) \left(x - 3\right)$

the expression given is:
$\frac{1}{x + 3} + \frac{1}{x - 3} = \frac{1}{{x}^{2} - 9}$

1/(x+3) + 1/(x-3) = 1/((x+3)(x-3)

(1.(x-3))/((x+3)(x-3)) + (1.(x+3))/((x-3)(x+3)) = 1/((x+3)(x-3)

$\frac{\left(x - 3\right) + \left(x + 3\right)}{\left(x + 3\right) \left(x - 3\right)} = \frac{1}{\left(x + 3\right) \left(x - 3\right)}$

$\frac{2 x}{\cancel{{x}^{2} - 9}} = \frac{1}{\cancel{{x}^{2} - 9}}$

$2 x = 1$

 x =color(blue)( 1/2 is the solution for the expression.