# How do you solve 5/(y-3) + 10/(y^2-y-6) = y/(y+2)?

Mar 12, 2018

$y = 10$

#### Explanation:

Move expression to the left side and change its sign

$\frac{5}{y - 3} + \frac{10}{{y}^{2} - y - 6} - \frac{y}{y + 2} = 0$

Write $- y$ as a sum or difference

$\frac{5}{y - 3} + \frac{10}{{y}^{2} + 2 y - 3 y - 6} - \frac{y}{y + 2} = 0$

Factor out $y$ and $- 3$ from the expression

$\frac{5}{y - 3} + \frac{10}{y \left(y + 2\right) - 3 \left(y + 2\right)} - \frac{y}{y + 2} = 0$

Factor out $y + 2$ from the expression

$\frac{5}{y - 3} + \frac{10}{\left(y + 2\right) \left(y - 3\right)} - \frac{y}{y + 2} = 0$

Write all numerators above the least common denominator

$\frac{5 \left(y + 2\right) + 10 - y \left(y - 3\right)}{\left(y + 2\right) \left(y - 3\right)} = 0$

Distribute $5$ and $- y$ through the parenthesis

$\frac{5 y + 10 + 10 - {y}^{2} + 3 y}{\left(y + 2\right) \left(y - 3\right)} = 0$

Collect the like terms

$\frac{8 y + 20 - {y}^{2}}{\left(y + 2\right) \left(y - 3\right)} = 0$

Use the commutative property to reorder the terms

$\frac{- {y}^{2} + 8 y + 20}{\left(y + 2\right) \left(y - 3\right)} = 0$

Write $8 y$ as a sum or difference

$\frac{- {y}^{2} + 10 y - 2 y + 20}{\left(y + 2\right) \left(y - 3\right)} = 0$

Factor out $- y$ and $- 2$ from the expression

$\frac{- y \left(y - 10\right) - 2 \left(y - 10\right)}{\left(y + 2\right) \left(y - 3\right)} = 0$

Factor out $- \left(y - 10\right)$ from the expression

$\frac{- \left(y - 10\right) \left(y + 2\right)}{\left(y + 2\right) \left(y - 3\right)} = 0$

Reduce the fraction with $y + 2$

$- \frac{y - 10}{y - 3} = 0$

Determine the sign of the fraction

$- \frac{y - 10}{y - 3} = 0$

Simplify

$\frac{10 - y}{y - 3} = 0$

When the quotient of expressions equals $0$, the numerator has to be $0$

$10 - y = 0$

Move the constant, $10$, to the right side and change its sign

$- y = - 10$

Change the signs on both sides of the equation

$y = 10$

Check if the solution is in the defined range

$y = 10 , y \ne 3 , y \ne - 2$

$\therefore y = 10$