Answer
$=\displaystyle \frac{x-2}{x-3}, \quad x\neq-2,3$
Work Step by Step
The fractions have a common denominator, we add the numerators...
$\displaystyle \frac{x^{2}-4x}{x^{2}-x-6}+\frac{4x-4}{x^{2}-x-6}=\frac{x^{2}-4x+4x-4}{x^{2}-x-6}$
$=\displaystyle \frac{x^{2}-4}{x^{2}-x-6}$
... factor what we can...
$\left\{\begin{array}{lll}
x^{2}-4 & =(x+2)(x-2) & \text{(diff. of squares)}\\
x^{2}-x-6 & =(x-3)(x+2) & \text{(factors of c whose sum is b)}
\end{array}\right.$
$=\displaystyle \frac{(x+2)(x-2)}{(x-3)(x+2)}$
... exclude values that yield 0 in the denominator:
$x\neq-2,3$
... and, cancel the common factor
$=\displaystyle \frac{x-2}{x-3}, \quad x\neq-2,3$
