Solve each system of inequalities by graphing.

$\begin{array}{c}\text{3}y\le 2x-8\\ y\ge \frac{2}{3}x-1\end{array}$

The steps to graph the inequality are provided below.

1. If the inequality contains greater than or less than sign then the boundary of the line will be dashed. If the inequality contains signs of greater than or equal to or less than or equal to then the boundary of the line will be solid.

2. Select a point (known as test point) from the plane that does not lie on the boundary on the line and substitute it in the inequality.

3. If the inequality is true then shade the region that contains the test point otherwise shade the other region when inequality is false.

Consider the inequality provided below.

$3y\le 2x-8$

The inequality contains the sign of less than or equal to.

Therefore, the boundary line will be solid.

Next, consider the inequality $y\ge \frac{2}{3}x-1$

The inequality contains the sign of greater than or equal to.

Therefore, the boundary line will be solid.

Graph the inequalities $3y\le 2x-8$and $y\ge \frac{2}{3}x-1$ on same plane and shade the region.

The corresponding equation is $3y=2x-8$.

Take a test point that does not lie on the boundary of the line, say $\left(0,0\right)$.

Substitute the point $\left(0,0\right)$ in the inequality and check whether it’s true or not.

This is false.

Therefore, shade the region not containing the point $\left(0,0\right)$.

Draw the line $y=\frac{2}{3}x-1$.

Take a test point that does not lie on the boundary of the line, say $\left(0,0\right)$.

Substitute the point $\left(0,0\right)$ in the inequality and check whether it’s true or not.

This is true.

Therefore, shade the region containing the point $\left(0,0\right)$.

Thus, the shaded regions are provided below.

The region 1 corresponds to inequality $3y\le 2x-8$.

The region 2 corresponds to inequality $y\ge \frac{2}{3}x-1$.

There is no region common to both the inequalities $3y\le 2x-8$ and $y\ge \frac{2}{3}x-1$.

Hence, the solution to the system of inequalities is $\varnothing$.