## Precalculus (6th Edition) Blitzer

$\displaystyle \frac{2(10x-3)}{(5x-2)(5x+2)}, \quad x\neq-\frac{2}{5}, \displaystyle \frac{2}{5}$
Factor each denominator. Recognize a difference of squares in the second. $\displaystyle \frac{3}{5x+2}+\frac{5x}{25x^{2}-4}=\frac{3}{5x+2}+\frac{5x}{(5x)^{2}-2^{2}}$ $=\displaystyle \frac{3}{5x+2}+\frac{5x}{(5x-2)(5x+2)} \qquad ...$LCD=$(5x-2)(5x+2)$ $=\displaystyle \frac{3}{5x+2}\times\frac{5x-2}{5x-2}+\frac{5x}{(5x-2)(5x+2)}$ $=\displaystyle \frac{15x-63}{(5x-2)(5x+2)}+\frac{5x}{(5x-2)(5x+2)}$ $=\displaystyle \frac{15x-6+5x}{(5x-2)(5x+2)}$ $=\displaystyle \frac{20x-6}{(5x-2)(5x+2)}$ $=\displaystyle \frac{2(10x-3)}{(5x-2)(5x+2)}\quad x\neq-\frac{2}{5}, \displaystyle \frac{2}{5}$