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

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Answer 1 · 2015-05-26T20:20:17

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

$y^2-4 = 2(y+3) - (y-2)$

$=2y+6-y+2$

$=y+8$

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

$y^2 - y - 12 = 0$

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

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 $y = 4$ .

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