Chapter 4 - Section 4.2 - Solving Systems of Linear Equations by the Substitution Method - Exercise Set - Page 300: 3

{(2.1)}

x+3y=5 x+3y-3y=5-3y x=5-3y Plug this value for x into the other equation. 4(5-3y)+5y=13 4(5)-4(3y)+5y=13 20-12y+5y=13 20-7y=13 -7y=13-20 -7y=-7 y=$\frac{-7}{-7}$ y=1 Plug this back into the first equation to find x x=5-3(1) x=5-3 x=2 Therefore the solution is {(2,1)}