## Precalculus (6th Edition) Blitzer

$\displaystyle \frac{1-x}{(x+6)(x-3)}, \qquad x\neq-6,-5,3$
Factor all denominators of the complex rational equation. (For $x^{2}+2x-15$, we find factors of $-15$ that add to $+2$. These are $+5$ and $-3.$) $...=\displaystyle \frac{\frac{6}{(x+5)(x-3)}-\frac{1}{x-3}}{\frac{1}{x+5}+1}=...$ Complex rational expressions have rational expressions in the numerator or/and in the denominator. Here, multiplying both the numerator and denominator with $(x+5)(x-3)$ would get rid of these fractions. We have to exclude any value that yields 0 in any denominator. Excluded: $x=-5,3$. The denominator of the expression itself can not be zero, so we will exclude any values for which it is 0: $\displaystyle \frac{1}{x+5}+1=0\quad/\times(x+5)$ $1+(x+5)=0$ $x+6=0\qquad -6$ is also excluded. $...=\displaystyle \frac{\frac{6}{(x+5)(x-3)}-\frac{1}{x-3}}{\frac{1}{x+5}+1}\times\frac{(x+5)(x-3)}{(x+5)(x-3)}$ $=\displaystyle \frac{6-1(x+5)}{(x-3)+(x+5)(x-3)}$ $=\displaystyle \frac{1-x}{x-3+x^{2}+2x-15}$ $=\displaystyle \frac{1-x}{x^{2}+3x-18}\qquad$ factors of -18 with sum 3 are $+6$ and $-3$ $=\displaystyle \frac{1-x}{(x+6)(x-3)}, \qquad x\neq-6,-5,3$