Given the system of inequalities:
2x + 3y < -6
-2x + 3y < 6
Let's solve the system by graphing.
To graph, rewrite the inequalities in slope-intercept form:
y = mx + b
Inequality 1:'
Subtract 2x from both sides:
2x - 2x + 3y < -2x - 6
3y < -2x - 6
Divide all terms by 3:
[tex]\begin{gathered} \frac{3y}{3}<-\frac{2x}{3}-\frac{6}{3} \\ \\ y<-\frac{2}{3}x-2 \end{gathered}[/tex]
Inequality 2:
Add 2x to both sides:
-2x + 2x + 3y < 2x + 6
3y < 2x + 6
Divde all terms by 3:
[tex]\begin{gathered} \frac{3y}{3}<\frac{2x}{3}+\frac{6}{2} \\ \\ y<\frac{2}{3}x+2 \end{gathered}[/tex]
Now, let's plot 3 points from each inequlality and connect using a straight edge.
Inequality 1:
When x = -3
Substitute -3 for x and solve for y:
[tex]\begin{gathered} y<-\frac{2}{3}(-3)-2 \\ \\ y<2-2 \\ \\ y<0 \end{gathered}[/tex]
When x = 0:
[tex]\begin{gathered} y<-\frac{2}{3}(0)-2 \\ \\ y<-2 \end{gathered}[/tex]
When x = 3:
[tex]\begin{gathered} y<-\frac{2}{3}(3)-2 \\ \\ y<-2-2 \\ \\ y<-4 \end{gathered}[/tex]
From inequality 1, we have the points:
(x, y) ==> (-3, 0), (0, -2), (3, -4)
For inequlity 2:
When x = -3:
[tex]\begin{gathered} y<\frac{2}{3}(-3)+2 \\ \\ y<-2+2 \\ \\ y<0 \end{gathered}[/tex]
When x = 0:
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