## College Algebra (6th Edition)

$x+y \leq 4$ $y +3x \leq 6$ .
First sentence: $\quad x+y \leq 4\Rightarrow y \leq -x+4$ Second sentence: $\quad y +(3x) \leq 6 \Rightarrow y \leq -3x+6$ Both inequalities have the $\leq$ inequality sign, so both border lines will be solid. 1st inequality: Graph $y=-x+4:$ y intercept=4, and for x=4, y=0 . Points on the line : (0,4) and (4, 0). Draw the line. Test point: (0,0). Is $0 \leq -(0)+4\quad ?.$ Yes. shade the region containing (0,0). 2nd inequality: Graph $y=-3x+6$ y intercept=$6$, and for x=$2$, y=0 . Points on the line : (0,$6$) and ($2$, 0) Test point: (0,0). Is $0 \leq -3(0)+6\quad ?.$ Yes. shade the region containing (0,0). The solution set is the region with both shadings. (grey on the screenshot)