The inequalities provided in the image are:

1. $-2y < 3x + 12$
2. $x \geq -3$

Let's break these down step by step:

First inequality: $-2y < 3x + 12$

• To graph this, we need to first convert the inequality into slope-intercept form, which is $y < mx + b$.
• Divide both sides by $-2$ (note that when dividing by a negative number, the inequality sign reverses): $y > -\frac{3}{2}x - 6$
• The line corresponding to this inequality is $y = -\frac{3}{2}x - 6$. Since the inequality is "greater than" (not equal), we will draw a dashed line to indicate that points on the line are not included in the solution.
• The region above the line is the solution region for this inequality.

Second inequality: $x \geq -3$

• This represents a vertical line at $x = -3$, with shading to the right of the line (since $x \geq -3$).

Graphing Procedure:

1. Graph the dashed line $y = -\frac{3}{2}x - 6$.
2. Shade the region above the dashed line.
3. Draw a solid vertical line at $x = -3$ and shade to the right of it.

Solution Set $S$:

• The solution set is the region where the shading from both inequalities overlaps. This is the area that satisfies both conditions.

Let me know if you'd like more help with this or further steps! Would you like to go into more detail about graphing inequalities, or need clarification?

Here are some related questions:

1. How do you graph a system of linear inequalities?
2. What does the inequality $y \geq mx + b$ represent?
3. How do you determine if a point lies within the solution region of an inequality system?
4. What does a dashed line in an inequality graph indicate?
5. How would you graph the inequality $y \leq -x + 3$?

Tip: When graphing inequalities, always start by sketching the boundary line and determine whether to use a dashed or solid line, then decide which side to shade!