Question

Subtract$\frac{-13q-8}{q^2+2q-24}-\frac{q+2}{4-q}$.

Short answer

The difference is$\frac{q-1}{q+6}$.

Step-by-step solution

Given Information  

The Given expression is$\frac{-13q-8}{q^2+2q-24}-\frac{q+2}{4-q}$.

Find the LCD of denominators.     

$q{}^2+2q-24=(q-4)(q+6)$

$4-q=-(q-4)$

Rewrite the expression in factored form.

$\frac{-13q-8}{(q-4)(q+6)}+\frac{q+2}{q-4}$

Rewrite each fraction as an equivalent fraction with the LCD. 

$$\begin{gathered} =\frac{-13q-8}{(q-4)(q+6)}+\frac{(q+2)(q+6)}{(q-4)(q+6)} \\ =\frac{-13q-8}{(q-4)(q+6)}+\frac{q^2+8q+12}{(q-4)(q+6)} \end{gathered}$$

Take common denominator and solve.

$$\begin{gathered} =\frac{-13q-8+q^2+8q+12}{(q-4)(q+6)} \\ =\frac{q^2-5q+4}{(q-4)(q+6)} \end{gathered}$$

Factor numerator.

$=\frac{(q-1)(q-4)}{(q-4)(q+6)}$

Cancel common factors.

$=\frac{q-1}{q+6}$

Thus, the difference is$\frac{q-1}{q+6}$.