## Precalculus: Mathematics for Calculus, 7th Edition

$\frac{x-3}{x+2}$
$\frac{x^2+2x-15}{x^2-25}\times\frac{x-5}{x+2}$ Simplify $\frac{x^2+2x-15}{x^2-25}$: $=\frac{(x+5)(x-3)}{(x-5)(x+5)}$ $=\frac{(x-3)}{(x-5)}$ So it becomes: $=\frac{x-3}{x-5}\times\frac{x-5}{x+2}$ Multiply the two fractions (note: $x-5$ cancels out.): $=\frac{x-3}{x+2}$