## Algebra 1: Common Core (15th Edition)

The equation has two solutions, $\frac{-3}{2}$ and $4$.
$\frac{3}{m-1}=\frac{2m}{m+4}$ $3(m+4)=2m(m-1)$ (Cross Products Property) $3m+12=2m^2-2m$ $2m^2-5m-12=0$ $2m^2-8m+3m-12=0$ $2m(m-4)+3(m-4)=0$ $(2m+3)(m-4)=0$ $2m+3=0$ or $m-4=0$ $m=-\frac{3}{2}$ or $m=4$ Check: $3/(4-1)=1; (2\times4)/(4+4)=1$ $3/(\frac{-3}{2}-1)=3\times\frac{-2}{5}=\frac{-6}{5}; (2\times\frac{-3}{2})/(\frac{-3}{2}+4)=(-3)\times\frac{2}{5}=-6/5$ The equation has two solutions, $\frac{-3}{2}$ and $4$.