## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\dfrac{2(x+5)}{(x-1)(x+2)}$
Since the least common multiple (LCM) of the denominator is $(x-1)(x+2)$, then we have $$\frac{4}{x-1}-\frac{2}{x+2}=\frac{4(x+2)}{(x-1)(x+2)}-\frac{2(x-1)}{(x-1)(x+2)} .$$ Now, since the denominators are the same, then we have $$\frac{4(x+2)}{(x-1)(x+2)}-\frac{2(x-1)}{(x-1)(x+2)}=\frac{4x+8-2x+2}{(x-1)(x+2)}\\ =\frac{2x+10}{(x-1)(x+2)}=\frac{2(x+5)}{(x-1)(x+2)} .$$