# How do you simplify and find the excluded value of (6-y)/(y^2-2y-24)?

Jun 13, 2017

factor the denominator then reduce

#### Explanation:

${y}^{2} - 2 y - 24$ factors to $\left(y + 4\right) \left(y - 6\right)$.
Now factor a -1 out of the numerator $6 - y = - 1 \left(y - 6\right)$

The factored problem is now [-1(y-6)]/[(y+4)(y-6)

reduce to get $- \frac{1}{y + 4}$

Jun 13, 2017

$= - \frac{1}{\left(y + 4\right)}$

excluded value when $y = - 4$

#### Explanation:

$\frac{6 - y}{{y}^{2} - 2 y - 24} = \frac{6 - y}{\left(y - 6\right) \left(y + 4\right)}$

$= \frac{6 - y}{- \left(- y + 6\right) \left(y + 4\right)} = \frac{\cancel{6 - y}}{- \cancel{\left(6 - y\right)} \left(y + 4\right)}$

$= \frac{1}{- \left(y + 4\right)} = - \frac{1}{\left(y + 4\right)}$

excluded value when $\left(y + 4\right) = 0 \to y = - 4$ therefore $y = - 4$