Find the feasible regions of the systems of inequalities. Also determine the corners of each feasible region (If there are no corners write "no corners").
$$\begin{array}{c}x+y>4\\ 2x-3y\ge 8\end{array}$$

Step 1 ：Graph the inequality $x+y>4$. This inequality can be rewritten as $y>-x+4$. The boundary line $y=-x+4$ is a straight line with a slope of -1 and a y-intercept of 4. Since the inequality is 'greater than', we shade the region above the line.

Step 2 ：Graph the inequality $2x-3y\ge 8$. This inequality can be rewritten as $y\le \frac{2}{3}x-\frac{8}{3}$. The boundary line $y=\frac{2}{3}x-\frac{8}{3}$ is a straight line with a slope of $\frac{2}{3}$ and a y-intercept of $-\frac{8}{3}$. Since the inequality is 'less than or equal to', we shade the region below the line.

Step 3 ：The feasible region is the area where the shaded regions of both inequalities overlap.

Step 4 ：Set $-x+4=\frac{2}{3}x-\frac{8}{3}$ to find the x-coordinate of the intersection point. Adding x to both sides gives $4=\frac{5}{3}x-\frac{8}{3}$. Multiplying every term by 3 gives $12=5x-8$. Adding 8 to both sides gives $20=5x$. Dividing every term by 5 gives $x=4$.

Step 5 ：Substitute $x=4$ into $y=-x+4$ to find the y-coordinate of the intersection point: $y=-4+4=0$.

Step 6 ：So, the corner of the feasible region is $\boxed{{\displaystyle (4,0)}}$.

Step 7 ：Both inequalities are satisfied at the point $(4,0)$, so this is indeed a corner of the feasible region.