The question asks us to solve the following system of inequalities graphically and identify a point within the solution set:

1. $y \leq x + 3$
2. $y < \frac{1}{2}x - 6$

Steps to Solve:

1. Graph each inequality:

   • For $y \leq x + 3$:

     • This is a straight line with slope $m = 1$ and $y$-intercept $b = 3$. Draw the line $y = x + 3$ as a solid line (since $\leq$ includes equality). Shade below the line.

   • For $y < \frac{1}{2}x - 6$:

     • This is a straight line with slope $m = \frac{1}{2}$ and $y$-intercept $b = -6$. Draw the line $y = \frac{1}{2}x - 6$ as a dashed line (since $<$ does not include equality). Shade below the line.

2. Find the overlap of the shaded regions:

   • The solution to the system is the region where the shaded areas of the two inequalities overlap.

3. Determine a point in the solution set:

   • Pick any point in the overlapping shaded region and check if it satisfies both inequalities.

Would you like me to graph this explicitly for clarity? Or proceed to identify a solution point? Let me know!

Related Questions:

1. How do you determine if a point satisfies a given inequality?
2. What does the slope and $y$-intercept represent in a linear inequality?
3. Why is one line dashed and the other solid in the graph?
4. What are the rules for shading when graphing linear inequalities?
5. How do you verify that a region is the solution to a system of inequalities?

Tip:

Always double-check the boundaries (solid vs. dashed lines) and test a sample point to confirm the solution region.