6. Graph the feasible region \[ \left\{\begin{array}{c} 2 x+y \geq 2 \\ x-3 y \leq 6 \\ -4 x+y<-3 \end{array}\right. \] Report the Corners

Step 1 ：First, we need to graph each inequality on the same coordinate plane.

Step 2 ：For the inequality \(2x + y \geq 2\), we first graph the line \(2x + y = 2\). This line intersects the y-axis at \(y = 2\) and the x-axis at \(x = 1\). We then shade the region above the line because the inequality is greater than or equal to.

Step 3 ：For the inequality \(x - 3y \leq 6\), we first graph the line \(x - 3y = 6\). This line intersects the y-axis at \(y = -2\) and the x-axis at \(x = 6\). We then shade the region below the line because the inequality is less than or equal to.

Step 4 ：For the inequality \(-4x + y < -3\), we first graph the line \(-4x + y = -3\). This line intersects the y-axis at \(y = -3\) and the x-axis at \(x = 0.75\). We then shade the region above the line because the inequality is less than.

Step 5 ：The feasible region is the area where all three shaded regions overlap.

Step 6 ：The corners of the feasible region are the points where two or more of the lines intersect. To find these points, we solve the equations of the lines simultaneously.

Step 7 ：For example, to find the intersection of the lines \(2x + y = 2\) and \(x - 3y = 6\), we can solve these two equations simultaneously: Subtract the second equation from the first to get: \(5y = -4\), so \(y = -4/5\).

Step 8 ：Substitute \(y = -4/5\) into the first equation to get: \(2x - 4/5 = 2\), so \(x = 9/5\).

Step 9 ：So one corner of the feasible region is at \(\boxed{(9/5, -4/5)}\).

Step 10 ：Repeat this process for the other pairs of lines to find the remaining corners.