Answer
$\frac{-2x-1}{(x-3)(x+2)}$
Work Step by Step
$\frac{x}{x^{2}-x-6}-\frac{1}{x+2}-\frac{2}{x-3}$
Factor $x^{2}-x-6$ and replace in the denominator:
$=\frac{x}{(x-3)(x+2)}-\frac{1}{x+2}-\frac{2}{x-3}$
Find the lowest common denominator (i.e. $(x-3)(x+2)$) and adjust the fractions accordingly:
$=\frac{x}{(x-3)(x+2)}-\frac{1\times (x-3)}{(x+2)\times (x-3)}-\frac{2\times (x+2)}{(x-3)\times (x+2)}$
$=\frac{x}{(x-3)(x+2)}-\frac{x-3}{(x+2)(x-3)}-\frac{2(x+2)}{(x-3)(x+2)}$
Combine the fractions:
$=\frac{x-(x-3)-2(x+2)}{(x-3)(x+2)}$
$=\frac{x-x+3-2x-4}{(x-3)(x+2)}$
Collect like terms:
$=\frac{-2x-1}{(x-3)(x+2)}$
