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

The solution is $v=-2$ or $v=-6$.
$\frac{v}{3}+\frac{v}{v+5}=\frac{-4}{v+5}$ Add: $\frac{v(v+5)}{3(v+5)}+\frac{3v}{3(v+5)}=\frac{-12}{3(v+5)}$ $v^2+5v+3v=-12$ Simplify: $v^2+8v+12=0$ Factor: $(v+2)(v+6)=0$ $v=-2$ or $v=-6$ Check: $\frac{-2}{3}+\frac{-2}{-2+5}=\frac{-4}{-2+5}$ $\frac{-4}{3}=\frac{-4}{3}$ $\frac{-6}{3}+\frac{-6}{-6+5}=\frac{-4}{-6+5}$ $4=4$ The solution is $v=-2$ or $v=-6$.