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

The equation has two solutions, $\frac{-6}{5}$ and $-1$.
$\frac{s}{3s+2}+\frac{s+3}{2s-4}=\frac{-2s}{3s^2-4s-4}$ $\frac{s}{3s+2}+\frac{s+3}{2(s-2)}=\frac{-2s}{(s-2)(3s+2)}$ $2(s-2)(3s+2)(\frac{s}{3s+2}+\frac{s+3}{2(s-2)})=2(s-2)(3s+2)\frac{-2s}{(s-2)(3s+2)}$ $2s(s-2)+(3s+2)(s+3)=-4s$ $2s^2-4s+3s^2+11s+6+4s=0$ $5s^2+11s+6=0$ $5s^2+5s+6s+6=0$ $(s+1)(5s+6)=0$ $s+1=0$ or $5s+6=0$ $s=-1$ or $s=-\frac{6}{5}$ ($\frac{-6}{5}=-1.2$) Check: $\frac{-1}{3(-1)+2}+\frac{-1+3}{2(-1)-4}=\frac{-1}{-1}+\frac{2}{-6}=\frac{2}{3}; \frac{(-2)(-1)}{3(-1)^2-4(-1)-4}=\frac{2}{3+4-4}=\frac{2}{3}$ $\frac{-1.2}{3\times(-1.2)+2}+\frac{-1.2+3}{2\times(-1.2)-4}=\frac{1.2}{1.6}+\frac{1.8}{-6.4}=\frac{3}{4}-\frac{9}{32}=\frac{15}{32}; \frac{-2(-1.2)}{3(-1.2)^2-4(-1.2)-4}=\frac{2.4}{5.12}=\frac{9}{32}$ The equation has two solutions, $\frac{-6}{5}$ and $-1$.