91Ó°ÊÓ

Rewrite each inequality in terms of y: \[ \begin{aligned} 7y &\leq -6x + 84 &\Rightarrow y &\leq -\frac{6}{7}x + 12 \\ 11y &\geq 12x - 18 &\Rightarrow y &\leq \frac{11}{12}x - \frac{3}{2} \\ 7y &\geq 6x - 28 &\Rightarrow y &\geq \frac{6}{7}x - 4 \\ \end{aligned} \] The fourth inequality, \(y \geq 0\), is already in slope-intercept form.

Sketch each of the inequality lines on the coordinate plane: 1. Line 1: \(y = -\frac{6}{7}x + 12\). Plot the points \((0, 12)\) and \(\left(7,6\right)\) and draw a solid line through them, since the inequality sign is \(\leq\). 2. Line 2: \(y = \frac{11}{12}x - \frac{3}{2}\). Plot the points \(\left(0,-\frac{3}{2}\right)\) and \((2, 1)\) and draw a solid line through them, since the inequality sign is \(\leq\). 3. Line 3: \(y = \frac{6}{7}x - 4\). Plot the points \((0, -4)\) and \((7, 2)\) and draw a solid line through them, since the inequality sign is \(\geq\). 4. Plot the \(y \geq 0\) inequality by drawing a solid line along the x-axis. For each line, shade the area that satisfies the inequality. The region where all shaded areas overlap is the solution set.

By examining the plot, we can see that the area where all shaded regions overlap is a polygon located in the first quadrant. Since this polygon is enclosed by lines, the solution set is bounded. The final solution is a bounded region in the first quadrant where all inequalities are true.