## Precalculus (6th Edition) Blitzer

$\displaystyle \frac{x-3}{x+2}, \qquad x\neq-1,\ 3,\ -2$
Factor all denominators of the complex rational expression. (For $x^{2}-2x-3$, we find factors of $-3$ that add to $-2$. These are $+1$ and $-3.$) $\displaystyle \frac{\frac{1}{x+1}}{\frac{1}{x^{2}-2x-3}+\frac{1}{x-3}}=\frac{\frac{1}{x+1}}{\frac{1}{(x+1)(x-3)}+\frac{1}{x-3}}$ Complex rational expressions have rational expressions in the numerator or/and in the denominator. Here, multiplying both the numerator and denominator with $(x+1)(x-3)$ would get rid of these fractions. We have to exclude any value that yields 0 in any denominator. Excluded: $x=-1,3$. The denominator of the expression itself can not be zero, so we will exclude any values for which it is 0: $\displaystyle \frac{1}{(x+1)(x-3)}+\frac{1}{x-3}=0\qquad/\times(x+1)(x-3)$ $1+1(x+1)=0$ $1+x+1=$ $x+2=0$ Excluded: $x=-1,\ 3,\ -2$ $=\displaystyle \frac{\frac{1}{x+1}}{\frac{1}{(x+1)(x-3)}+\frac{1}{x-3}}\times\frac{(x+1)(x-3)}{(x+1)(x-3)}$ ... the denominator was simplified while solving the above equation. $=\displaystyle \frac{x-3}{x+2}, \qquad x\neq-1,\ 3,\ -2$