# How do you solve \frac { 9} { x ^ { 2} - 2x - 8} + \frac { x } { x + 2} = 2?

Jun 21, 2017

Solution: $x = 5 \mathmr{and} x = - 5$

#### Explanation:

$\frac{9}{{x}^{2} - 2 x - 8} + \frac{x}{x + 2} = 2$ or

$\frac{9}{\left(x - 4\right) \left(x + 2\right)} + \frac{x}{x + 2} = 2$ . Multiplying by $\left(x - 4\right) \left(x + 2\right)$ on both sides we get ,

$9 + x \left(x - 4\right) = 2 \left(x - 4\right) \left(x + 2\right)$ or

$9 + {x}^{2} - \cancel{4 x} = 2 {x}^{2} - \cancel{4 x} - 16$ or

$2 {x}^{2} - {x}^{2} - 16 - 9 = 0$ or

${x}^{2} - 25 = 0 \mathmr{and} \left(x + 5\right) \left(x - 5\right) = 0$. Either $x + 5 = 0 \therefore x = - 5$ or

$x - 5 = 0 \therefore x = 5$

Solution: $x = 5 \mathmr{and} x = - 5$ [Ans]