Answer
$x+y \leq 4$
$y +3x \leq 6$
.
Work Step by Step
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)
