## Intermediate Algebra (12th Edition)

$x=-5$ $y=\frac{-10}{3}$
The second equation is already solved for $x$, we have to substitute this expression into the first equation: $\frac{3}{2}y=3y+5$ $3y=6y+10$ $-10=3y$ $\frac{-10}{3}=y$ $x=\frac{3}{2}y=\frac{3}{2}\times\frac{-10}{3}=-5$ We need to check, if it is the right solution, we substitute x with $-5$ and y with $\frac{3}{2}$: $-5=3\frac{-10}{3}+5$ $\checkmark$ $-5=\frac{3}{2}\times\frac{-10}{3}$ $\checkmark$