## Algebra 2 (1st Edition)

$$=\frac{3\left(x-3\right)}{3x+7}$$
Simplifying the fraction using the rules of fractions, we find: $$\frac{3}{\left(x+5\right)\left(\frac{2}{x-3}+\frac{1}{x+5}\right)} \\ \frac{3}{\frac{3x+7}{\left(x-3\right)\left(x+5\right)}\left(x+5\right)} \\ \frac{3}{\frac{3x+7}{x-3}} \\ =\frac{3\left(x-3\right)}{3x+7}$$