Chapter 4 - Section 4.1 - Solving Systems of Linear Equations in Two Variables - Exercise Set - Page 212: 56

Ordered Pair Solution $(4,10)$

We have $$1)\frac{y}{5}=\frac{8-x}{2}$$ $$2)x=\frac{2y-8}{3}$$ Substituting (2) in (1) $$2y = 5*(8-\frac{2y-8}{3})$$ $$2y = 40-\frac{10}{3}y+\frac{40}{3}$$ $$\frac{16}{3}y = \frac{160}{3}$$ $$y=10$$ Solving for $x$ $$x=\frac{2*10-8}{3} = 4$$ Ordered Pair Solution $(4,10)$