## Algebra 1

$r=-14$
Given:$\frac{2r}{r-4}-\frac{4}{r+5}=2$ $\dfrac{2r}{r-4}=2+\dfrac{4}{r+5}$ $\dfrac{2r}{r-4}=\dfrac{2r+10+4}{r+5}$ Apply cross products. $2r(r+5)=(2r+14)(r-4)$ $2r^2+10r=2r^2-8r+14r-56$ $10r+8r-14r+56=0$ Hence, $r=-14$