## Algebra 2 Common Core

Infinite solutions (all points along the line $d = 3a - 1$)
Given: $9a - 3d = 3$ $-3a + d = -1$ Multiply the second equation by $3$: $-9a + 3d = -3$ Add equations: $9a - 3d + (-9a + 3d) = 3 + (-3)$ Expand: $9a - 9a - 3d + 3d = 3 - 3$ Simplify: $0 = 0$ Therefore infinite solutions. Solve to make $d$ the subject: Second: $-3a + d = -1$ Add $3a$: $d = 3a - 1$ Therefore infinite solutions, all points along the line $d = 3a - 1$