# How do you solve (y+4)/(y-2)+6/(y-2)=1/(y+3)?

Apr 12, 2018

Solution: $y = - 4 , y = - 8$

#### Explanation:

$\frac{y + 4}{y - 2} + \frac{6}{y - 2} = \frac{1}{y + 3}$. Multiplying by $\left(y - 2\right) \left(y + 3\right)$

on both sides we get, $\left(y + 4\right) \left(y + 3\right) + 6 \left(y + 3\right) = y - 2$ or

${y}^{2} + 7 y + 12 + 6 y + 18 - y + 2 = 0$ or

${y}^{2} + 12 y + 32 = 0$ or

${y}^{2} + 8 y + 4 y + 32 = 0$ or

$y \left(y + 8\right) + 4 \left(y + 8\right) = 0$ or

$\left(y + 8\right) \left(y + 4\right) = 0 \therefore$ either $y + 8 = 0 \therefore y = - 8$

or $y + 4 = 0 \therefore y = - 4$

Solution: $y = - 4 , y = - 8$ [Ans]