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

Oct 5, 2015

$x = - 1$ and $x = - 10$

#### Explanation:

We have, at first, $\frac{x}{x + 2} - \frac{x + 2}{x - 2} = \frac{x + 3}{x - 2}$.
We pass $- \frac{x + 2}{x - 2}$ to the other side adding.
We now have$\frac{x}{x + 2} = \frac{x + 3}{x - 2} + \frac{x + 2}{x - 2}$.
As the denominator is common on both fractions of the right side, we add them.
We get $\frac{x}{x + 2} = \frac{2 x + 5}{x - 2}$.
We now cross multiply to get $x \left(x - 2\right) = \left(x + 2\right) \left(2 x + 5\right)$.
We multiply and get ${x}^{2} - 2 x = 2 {x}^{2} + 9 x + 10$.
We pass the left hand side working to the right, getting ${x}^{2} + 11 x + 10 = 0$ as an answer.
We factorise to get $\left(x + 1\right) \left(x + 10\right) = 0$.
Therefore, the answers are $x = - 1$ and $x = - 10$.
Hope it helps :D .