# How do you solve (y^2-4)/(y+3) = 2- (y-2)/(y+3)?

May 26, 2015

First multiply all the terms by $\left(y + 3\right)$ to get:

${y}^{2} - 4 = 2 \left(y + 3\right) - \left(y - 2\right)$

$= 2 y + 6 - y + 2$

$= y + 8$

Subtract $\left(y + 8\right)$ from both sides to get:

${y}^{2} - y - 12 = 0$

This factors as $\left(y - 4\right) \left(y + 3\right) = 0$

So ${y}^{2} - y - 12 = 0$ has solutions $y = 4$ and $y = - 3$

Of these, $y = - 3$ is not a solution of the original problem, because it causes the denominators of two of the terms to be zero.

So the unique solution of the original problem is $y = 4$.