## Algebra 1

$u=0,2$
Given:$\dfrac{u+1}{u+2}=\dfrac{-1}{u-3}+\dfrac{u-1}{(u^2-u-6)}$ Need to find least common denominator(LCD). LCD: $(u+1)(u+2)(u^2-u-6)$ $\dfrac{u+1}{u+2}=\dfrac{-1}{u-3}+\dfrac{u-1}{(u^2-u-6)}$ $\dfrac{(u+1)(u-3)+(u+2)}{u^2-u-6}=\dfrac{u-1}{(u^2-u-6)}$ $(u+1)(u-3)+(u+2)=(u-1)$ $u^2-2u-3+u+2-u+1=0$ $u(u-2)=0$ Hence, $u=0,2$