Answer
$\frac{3}{s+5}-\frac{2}{s-3}=\frac{s-19}{s^2+2s-15}$
Work Step by Step
$\frac{3}{s+5}-\frac{2}{s-3}=\frac{3}{s+5}\frac{s-3}{s-3}-\frac{2}{s-3}\frac{s+5}{s+5}=\frac{3s-9}{s^2-3s+5s-15}-\frac{2s+10}{s^2+5s-3s-15}=\frac{3s-9-2s-10}{s^2+2s-15}=\frac{s-19}{s^2+2s-15}$
