## Algebra 2 Common Core

$x = 5$, $y = 4$, or $(5,4)$
Given: $20x + 5y = 120$ $10x + 7.5y = 80$ Multiply both sides of the second equation by $2$: $20x + 15x = 160$ Subtract the first equation, $20x + 5y = 120$, from the second, $20x + 15x = 160$. $20x + 15y - (20x + 5y) = 160 - (120) \\20x - 20x + 15y - 5y = 160 - 120 \\10y = 40$ Divide both sides by $10$: $y = 4$ Substitute $y = 4$ into the first equation, $20x + 5y = 120$: $20x+5y=120 \\20x + 5(4) = 120 \\20x + 20 = 120$ Subtract $20$ from both sides: $20x = 100$ Divide both sides by $20$: $x = 5$ Thus, $x = 5$ and $y = 4$