## Algebra 2 Common Core

$x = 0$, $y = 3$, or $(0,3)$
Given: $3x + 2y = 6$ $3x + 3 = y$ Subtract the second equation, $3x + 3 = y$, from the first, $3x + 2y = 6$: $3x + 2y - (3x + 3) = 6 - (y) \\3x + 2y - 3x - 3 = 6 - y$ Combine like terms: $2y - 3 = 6 - y$ Add $y$ to both sides of the equation: $3y - 3 = 6$ Add $3$ to both sides of the equation: $3y = 9$ Divide both sides of the equation by $3$: $y = 3$ Substitute $y=3$ into first equation: $3x+2y=6 \\3x + 2(3) = 6 \\3x+6=6$ Subtract $6$ from both sides: $3x = 0$ Divide both sides by 3: $x = 0$ Thus, $x = 0$, $y = 3$