## Intermediate Algebra (12th Edition)

$x=10$ $y=14$
The first equation is already solved for $y$, then we have to substitute this expression into the second equation: $y=1.4x$ Second equation: $0.5x+1.5(1.4x)=26.0$ $2.6x=26.0$ $x=10$ $y=1.4(10)=14$ We have to check, if the solution is right. Substitute $x$ with 10 and $y$ with 14: $14=1.4(10)$ $\checkmark$ $0.5(10)+1.5(14)=5+21=26.0$ $\checkmark$