Chapter 3 - Linear Systems - 3-2 Solving Systems Algebraically - Lesson Check - Page 145: 4

$x = 5$ $y = -3$

Equations: $x + 2y = -1$ $x - y = 8$ Making the equations an equivalent system by multiplying the second equation by 2: $x + 2y = -1$ $2(x - y) = 2(8)$ $x + 2y = -1$ $2x - 2y = 16$ Eliminating $y$ to solve for $x$ $x + 2y = -1$ $+$ $2x - 2y = 16$ $3x = 15$ $x = 5$ Substituting $x$ into the second equation to solve for $y$ $x - y = 8$ $5 - y = 8$ $-y = 3$ $y = -3$