Answer:
Shown below
Step-by-step explanation:
The first system of inequality is the following:
[tex]\left\{ \begin{array}{c}y>2x+\frac{2}{3}\\y<2x+\frac{1}{3}\end{array}\right.[/tex]
To find the solution here, let's take one point, say, [tex](0,0)[/tex] and let's taste this point into both inequalities, so:
FIRST CASE:
First inequality:
[tex]y>2x+\frac{2}{3} \\ \\ 0>2(0)+\frac{2}{3} \\ \\ 0>\frac{2}{3} \ False![/tex]
The region is not the one where the point [tex](0,0)[/tex] lies
Second inequality:
[tex]y<2x+\frac{1}{3} \\ \\ 0<2(0)+\frac{1}{3} \\ \\ 0<\frac{1}{3} \ True![/tex]
The region is the one where the point [tex](0,0)[/tex] lies
So the solution in this first case has been plotted in the first figure. As you can see, there is no any solution there
SECOND CASE:
First inequality:
[tex]y<2x+\frac{2}{3} \\ \\ 0<2(0)+\frac{2}{3} \\ \\ 0<\frac{2}{3} \ True![/tex]
The region is the one where the point [tex](0,0)[/tex] lies
Second inequality:
[tex]y>2x+\frac{1}{3} \\ \\ 0>2(0)+\frac{1}{3} \\ \\ 0>\frac{1}{3} \ True![/tex]
The region is not the one where the point [tex](0,0)[/tex] lies
So the solution in this first case has been plotted in the second figure. As you can see, there is a solution there.
CONCLUSION: Notice that when reversing the signs on both inequalities the solution in the second case is the part of the plane where the first case didn't find shaded region.
