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"). \[ \begin{array}{c} x+y>4 \\ 2 x-3 y \geq 8 \end{array} \]

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 \(2x - 3y \geq 8\). This inequality can be rewritten as \(y \leq \frac{2}{3}x - \frac{8}{3}\). 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 \(\boxed{(4, 0)}\).

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