1. **State the problem:** We are given the system of inequalities: $$3y > 2x + 12$$ $$2x + y \leq -5$$ We want to understand and graph the solution region. 2. **Rewrite inequalities in slope-intercept form:** For the first inequality: $$3y > 2x + 12$$ Divide both sides by 3: $$y > \frac{\cancel{3}y}{\cancel{3}} > \frac{2x}{3} + \frac{12}{3}$$ which simplifies to $$y > \frac{2}{3}x + 4$$ For the second inequality: $$2x + y \leq -5$$ Subtract $2x$ from both sides: $$y \leq -2x - 5$$ 3. **Interpret the inequalities:** - The first inequality $y > \frac{2}{3}x + 4$ means the solution region is above the line $y = \frac{2}{3}x + 4$ (not including the line because of strict inequality). - The second inequality $y \leq -2x - 5$ means the solution region is on or below the line $y = -2x - 5$. 4. **Graphing:** - Plot the line $y = \frac{2}{3}x + 4$ as a dashed line (since the inequality is strict). - Plot the line $y = -2x - 5$ as a solid line (since the inequality includes equality). - Shade the region above the first line and below the second line. 5. **Final answer:** The solution to the system is the set of points $(x,y)$ that satisfy $$y > \frac{2}{3}x + 4 \quad \text{and} \quad y \leq -2x - 5$$ This region is the intersection of the half-plane above the first line and the half-plane below the second line.