# How do you subtract 1/(x-6) - 3/(x+6) + (3x)/(36-x^2)?

Jun 25, 2016

$\frac{1}{x - 6} - \frac{3}{x + 6} + \frac{3 x}{36 - {x}^{2}} = \frac{24 - 5 x}{{x}^{2} - 36}$

#### Explanation:

$\frac{1}{x - 6} - \frac{3}{x + 6} + \frac{3 x}{36 - {x}^{2}}$

= $\frac{1}{x - 6} - \frac{3}{x + 6} - \frac{3 x}{{x}^{2} - 36}$

As GCD of $\left(x - 6\right)$, $\left(x + 6\right)$ and $\left({x}^{2} - 36\right)$ is $\left({x}^{2} - 36\right) = \left(x - 6\right) \left(x + 6\right)$, above can be written as

$\frac{\left(x + 6\right) - 3 \left(x - 6\right) - 3 x}{{x}^{2} - 36}$

= $\frac{x + 6 - 3 x + 18 - 3 x}{{x}^{2} - 36}$

= $\frac{24 - 5 x}{{x}^{2} - 36}$