Chapter 4 - Section 4.2 - Solving Systems of Linear Equations by the Substitution Method - Exercise Set - Page 300: 6

{(-5,5)}

x+2y=5 x+2y-2y=5-2y x=5-2y Plug this value for x into the other equation. 2x-y=-15 2(5-2y)-y=-15 2(5)-2(2y-y)=-15 10-5y=-15 -5y=-15-10 -5y=-25 y=$\frac{-25}{-5}$ y=5 Plug this back into the first equation to find x. x=5-2y x=5-2(5) x=5-10 x=-5 Therefore the solution is {(-5,5)}