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

The solution of the system is $(\displaystyle \frac{80}{3},\frac{40}{3})$
Substitute $2y$ for $x$ in the first equation $x+y=40$ $(2y)+y=40$ $3y=40$ $y=\displaystyle \frac{40}{3}$ minutes Substitute back into $x=2y$ $x=2\displaystyle \cdot\frac{40}{3}=\frac{80}{3}$ minutes The solution of the system is $(\displaystyle \frac{80}{3},\frac{40}{3})$