## Precalculus (6th Edition) Blitzer

$\displaystyle \frac{x-2}{x+1},\qquad x\neq-1,2,3$
Complex rational expressions have rational expressions in the numerator or/and in the denominator. Here, the numerator contains $\displaystyle \frac{3}{x-2}$ (exclusion from the domain: $x\neq 2.\ )$ To get rid of $\displaystyle \frac{3}{x-2}$, we multiply both the numerator and denominator with $(x-2).$ The denominator can not be 0, so we exclude x for which $x-\displaystyle \frac{3}{x-2}=0\qquad/\times(x-2),\qquad (x\neq 2)$ $x(x-2)-3=0$ $x^{2}-2x-3=0$ $(x-3)(x+1)=0\quad \Rightarrow x\neq-1,2,3$ $\displaystyle \frac{x-2}{x-2}\times\frac{x-3}{x-\frac{3}{x-2}}=\frac{(x-2)(x-3)}{x(x-2)-3}$ $=\displaystyle \frac{(x-2)(x-3)}{(x-3)(x+1)}\qquad$...a common factor. Reduce. $=\displaystyle \frac{x-2}{x+1},\qquad x\neq-1,2,3$