## Precalculus (6th Edition) Blitzer

$-\displaystyle \frac{x-14}{7},\quad x\neq-2,2.$
Start by recognizing a difference of squares in the denominator of the the denominator. $\displaystyle \frac{\frac{3}{x-2}-\frac{4}{x+2}}{\frac{7}{x^{2}-4}}=\frac{\frac{3}{x-2}-\frac{4}{x+2}}{\frac{7}{(x-2)(x+2)}}$ Complex rational expressions have rational expressions in the numerator or/and in the denominator. Here, multiplying both the numerator and denominator with $(x-2)(x+2)$ would get rid of these fractions. Currently, we note that we have to exclude $\pm 2$ as they yield zero in corresponding denominators.. The denominator of the complex rational expression, $\displaystyle \frac{7}{(x-2)(x+2)}$ can not be zero as it has a nonzero numerator. No new exclusions. $\displaystyle \frac{\frac{3}{x-2}-\frac{4}{x+2}}{\frac{7}{(x-2)(x+2)}}\times\frac{(x-2)(x+2)}{(x-2)(x+2)}=\frac{3(x+2)-4(x-2)}{7}$ $=\displaystyle \frac{3x+6-4x+8}{7}$ $=\displaystyle \frac{-x+14}{7}$ $=-\displaystyle \frac{x-14}{7},\quad x\neq-2,2.$