# How do you solve (x-4) / (x+2) + 2 /( x-2) = 17 / (x^2-4)?

Apr 3, 2017

$x = - 1$ or $x = 5$

#### Explanation:

Start from the left hand side:

$\frac{x - 4}{x + 2} + \frac{2}{x - 2} = \frac{\left(x - 4\right) \left(x - 2\right) + 2 \left(x + 2\right)}{\left(x + 2\right) \left(x - 2\right)} = \frac{{x}^{2} - 6 x + 8 + 2 x + 4}{{x}^{2} - 4} = \frac{{x}^{2} - 4 x + 12}{{x}^{2} - 4}$

which is equal to the right hand side $\frac{17}{{x}^{2} - 4}$.

So we have
${x}^{2} - 4 x + 12 = 17$
${x}^{2} - 4 x - 5 = 0$
$\left(x + 1\right) \left(x - 5\right) = 0$
$x = - 1$ or $x = 5$