# How do you solve 2/(y^2-y-2)+18/(y^2-2y-3)=(y+20)/(y^2-y-2) and check for extraneous solutions?

Nov 24, 2016

First of all, the LCD (Least Common Denominator) is $\left(y - 3\right) \left(y + 1\right) \left(y - 2\right)$.

$\frac{2}{\left(y - 2\right) \left(y + 1\right)} + \frac{18}{\left(y - 3\right) \left(y + 1\right)} = \frac{y + 20}{\left(y - 2\right) \left(y + 1\right)}$

$\frac{2 \left(y - 3\right)}{\left(y - 2\right) \left(y + 1\right) \left(y - 3\right)} + \frac{18 \left(y - 2\right)}{\left(y - 3\right) \left(y + 1\right) \left(y - 2\right)} = \frac{\left(y + 20\right) \left(y - 3\right)}{\left(y - 2\right) \left(y - 3\right) \left(y + 1\right)}$

$2 y - 6 + 18 y - 36 = {y}^{2} + 20 y - 3 y - 60$

$0 = {y}^{2} - 3 y - 18$

$0 = \left(y - 6\right) \left(y + 3\right)$

$y = 6 \mathmr{and} - 3$

We check our original restrictions ($y \ne 3 , - 1 , 2$), and find that they don't interfere with our solutions.

Hopefully this helps!