The system of linear inequalities of the given points is required.
The system of inequality is
[tex]x-3y\leq -9[/tex]
[tex]3x-y>2[/tex]
Points of the first line
[tex](0,3)[/tex]
[tex](3,4)[/tex]
The equation of the line is
[tex]y-3=\dfrac{4-3}{3-0}(x-0)\\\Rightarrow y-3=\dfrac{1}{3}x\\\Rightarrow y-\dfrac{1}{3}x=3\\\Rightarrow 3y-x=9\\\Rightarrow x-3y=-9[/tex]
Let us take a point above this line [tex](2,4)[/tex]
It is also a solid line.
[tex]2-3\times 4=-10<-9[/tex]
So, the inequality is [tex]x-3y\leq -9[/tex]
Points of the second line
[tex](0,-2)[/tex]
[tex](1,1)[/tex]
The equation of the second line is
[tex]y+2=\dfrac{1+2}{1-0}(x-0)\\\Rightarrow y+2=3x\\\Rightarrow 3x-y=2[/tex]
Let us take a point towards the right of the line [tex](2,2)[/tex]
[tex]3\times 2-2=4>2[/tex]
Since, the line is dashed the second inequality will be [tex]3x-y>2[/tex]
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