[tex]\begin{gathered} 3y>2x+12 \\ 2x+y\leq-5 \end{gathered}[/tex]
To graph a system of inequalities you need to find the coordinates of 2 points on each inequality (as they are lineal inequalities):
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[tex]3y>2x+12[/tex]
Solve for y: divide both sides of the inequality into 3:
[tex]\begin{gathered} \frac{3y}{3}>\frac{(2x+12)}{3} \\ \\ y>\frac{2}{3}x+\frac{12}{3} \\ \\ y>\frac{2}{3}x+4 \end{gathered}[/tex]
Find the points as in a equation:
If x is 0:
[tex]\begin{gathered} y>\frac{2}{3}(0)+4 \\ y>0+4 \\ y>4 \end{gathered}[/tex]
First point (0 , 4)
If x is 6
[tex]\begin{gathered} y>\frac{2}{3}(6)+4 \\ y>\frac{12}{3}+4 \\ y>\frac{12+12}{3} \\ y>\frac{24}{3} \\ y>8 \end{gathered}[/tex]
Second point ( 6, 8 )
As the inequality is y greather than (2/3x+4) you draw a dot line that go through the two points and shade the area over the line:
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Repeat the process to find two points in the second inequality:
[tex]\begin{gathered} 2x+y\leq-5 \\ \\ y\leq-2x-5 \end{gathered}[/tex]
If x is 0:
[tex]\begin{gathered} y\leq-2(0)-5 \\ y\leq0-5 \\ y\leq-5 \end{gathered}[/tex]
First point (0 , -5)
If x is -5
[tex]\begin{gathered} y\leq-2(-5)-5 \\ y\leq10-5 \\ y\leq5 \end{gathered}[/tex]
Second point ( -5 , 5)
As the inequality is y less than or equal to (-2x-5) you draw a line that go through the two points and shade the area under that line:
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Then, you get the next graph for the system of inequalities:
The solution is the area shaded by both inequalities.
