## Precalculus (6th Edition) Blitzer

$=\displaystyle \frac{x-2}{x+2}, \quad x\neq-2,3$
The fractions have a common denominator, we subtract the numerators... $\displaystyle \frac{x^{2}-4x}{x^{2}-x-6}-\frac{x-6}{x^{2}-x-6}=\frac{x^{2}-4x-(x-6)}{x^{2}-x-6}$ $=\displaystyle \frac{x^{2}-5x+6}{x^{2}-x-6}$ ... factor what we can; for trinomials $x^{2}+bx+c,$ we search for two factors of c whose sum is b $=\displaystyle \frac{(x-2)(x-3)}{(x-3)(x+2)}$ ... exclude values that yield 0 in the denominator: $x\neq-2,3$ ... and, cancel the common factor $=\displaystyle \frac{x-2}{x+2}, \quad x\neq-2,3$