Graph the feasible region for the given system of inequalities. \[ \begin{array}{r} x+4 y \leq 12 \\ 3 x+4 y \geq 12 \end{array} \] Use the graphing tool to graph the system. Graph the region that represents the correct solution only once.

Step 1 ：Rearrange the inequality \(x + 4y \leq 12\) to the form \(y \leq -\frac{1}{4}x + 3\). This is a line with a slope of -1/4 and a y-intercept of 3. The inequality \(y \leq -\frac{1}{4}x + 3\) means that the feasible region is below the line.

Step 2 ：Rearrange the inequality \(3x + 4y \geq 12\) to the form \(y \geq -\frac{3}{4}x + 3\). This is a line with a slope of -3/4 and a y-intercept of 3. The inequality \(y \geq -\frac{3}{4}x + 3\) means that the feasible region is above the line.

Step 3 ：The feasible region for the system of inequalities is the region that is both below the line \(y = -\frac{1}{4}x + 3\) and above the line \(y = -\frac{3}{4}x + 3\). This region is a triangle with vertices at the points where the lines intersect the x-axis and y-axis.

Step 4 ：To find these intersection points, set \(y = 0\) in both equations and solve for \(x\), and set \(x = 0\) in both equations and solve for \(y\).

Step 5 ：For \(x + 4y = 12\), when \(y = 0\), \(x = 12\). When \(x = 0\), \(y = 3\).

Step 6 ：For \(3x + 4y = 12\), when \(y = 0\), \(x = 4\). When \(x = 0\), \(y = 3\).

Step 7 ：So, the vertices of the feasible region are at the points (12, 0), (4, 0), and (0, 3).