## Precalculus: Mathematics for Calculus, 7th Edition

$\frac{-2x-1}{(x-3)(x+2)}$
$\frac{x}{x^{2}-x-6}-\frac{1}{x+2}-\frac{2}{x-3}$ 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: $=\frac{-2x-1}{(x-3)(x+2)}$