## Algebra 2 Common Core

$r = 4$, $s = 1$
Given: $r + 3s = 7$ $2r - s = 7$ Multiply the first equation, $r + 3s = 7$, by $2$: $2r + 6s = 14$ Subtract the second equation, $2r - s = 7$, from the first, $2r + 6s = 14$: $2r + 6s - (2r + -s) = 14 - (7)$ $2r - 2r + 6s - (-s) = 7$ $6s + s = 7$ $7s = 7$ Divide both sides by $7$: $s = 1$ Substitute $s = 1$ into the first equation, $r + 3s = 7$. $r + 3s = 7$ $r + 3(1) = 7$ $r + 3 = 7$ Subtract $3$ from both sides: $r = 4$ $r = 4$, $s = 1$