## Precalculus (6th Edition) Blitzer

$=-\displaystyle \frac{3}{x(x+1)}, \quad x\neq-1,0$
The denominators are different. Find a common denominator. LCD=$x(x+1)$ $\displaystyle \frac{3}{x+1}-\frac{3}{x}=\frac{3x}{(x+1)\cdot x}-\frac{3(x+1)}{x(x+1)}$ $=\displaystyle \frac{3x-3x-3}{x(x+1)}$ $=-\displaystyle \frac{3}{x(x+1)}$ ... exclude values that yield 0 in the denominator: $x\neq-1,0$ ... and, cancel common factors (here, there are none) $=-\displaystyle \frac{3}{x(x+1)}, \quad x\neq-1,0$