## College Algebra (10th Edition)

$\dfrac{2x+10}{(x+2)(x-1)}; x \ne -2, 1$
The LCM of $x-1$ and $x+2$ is $(x-1)(x+2)$. Use the LCM as the expressions' LCD to make them similar: $=\dfrac{4\color{blue}{(x+2)}}{(x-1)\color{blue}{(x+2)}} -\dfrac{2\color{blue}{(x-1)}}{(x+2)\color{blue}{(x-1)}} \\=\dfrac{4x+8}{(x-1)(x+2)}-\dfrac{2x-2}{(x+2)(x-1)}$ The expressions are similar so subtract the numerators and copy the denominator to obtain: $=\dfrac{4x+8-(2x-2)}{(x+2)(x-1)} \\=\dfrac{4x+8-2x+2}{(x+2)(x-1)} \\=\dfrac{2x+10}{(x+2)(x-1)}; x \ne -2, 1$