## Algebra 1

C) $x-y > -3$
(0,3) and (-3,0) $m = (3-0)/(0--3)$ $m = 3/3 = 1$ $y=mx+b$ $3 = 0*1 + b$ $3 = 0 + b$ $3 = b$ $y=x + 3$ (0,0) is part of the solution set, so we use the $\lt$ sign $y \lt x+3$ $y \lt x+3$ $y*-1 \lt (x+3)*-1$ $-y \gt -x-3$ $-y+x \gt -x-3+x$ $x-y \gt -3$