Answer
See below:
Work Step by Step
Consider the provided inequalities:
\[2x+y<3\]and\[x-y>2\]
Replace each inequality symbol with an equal sign:
\[2x+y=3\]and\[x-y=2\]
Now, draw the graph of the equation\[2x+y=3\].
Put \[y=0\] for x-intercept and \[x=0\] for y-intercept in the equation\[2x+y=3\].
So, the x-intercept is \[\frac{3}{2}\] and y-intercept is 3.
Therefore, the line is passing through \[\left( \frac{3}{2},0 \right)\]and\[\left( 0,3 \right)\].
Now, consider a test point\[\left( 0,0 \right)\], which lies in the half-plane.
Substitute \[x=0\] and \[y=0\]in\[2x+y<3\].
\[\begin{align}
& 0+0<3 \\
& 0<3
\end{align}\]
Since \[\left( 0,0 \right)\]satisfies the above inequality \[2x+y<3\], the test point \[\left( 0,0 \right)\] is a part of the solution set.
The solution should be the half plane, which is containing the point \[\left( 0,0 \right)\].
Since,\[2x+y<3\] is not contain an equal sign, the line should be dotted.
Now, draw the graph of the equation\[x-y=2\].
Put \[y=0\] for x-intercept and \[x=0\] for y-intercept in the equation \[x-y=2\].
So, the x-intercept is 2 and y-intercept is \[-2\].
Therefore, the line is passing through \[\left( 2,0 \right)\]and \[\left( 0,-2 \right)\].
Now, consider a test point \[\left( 0,0 \right)\], which lies in the half-plane.
Substitute \[x=0\] and \[y=0\]in \[x-y>2\].
\[\begin{align}
& 0-0>2 \\
& 0>2
\end{align}\]
\[\left( 0,0 \right)\]does notsatisfy the above inequality \[x-y>2\].
So, the test point \[\left( 0,0 \right)\] is not a part of the solution set.
The solution should be the other half of the plane, which is not containing the point \[\left( 0,0 \right)\].
\[x-y>2\]is not contain an equal sign.
So, the line should be dotted.
Therefore, the graph of the linear inequality, \[2x+y<3\] and \[x-y>2\] is provided below:
