1. **State the problem:** We need to find the region in the xy-plane that satisfies the system of inequalities: $$y \geq \frac{3}{2}x + 2$$ $$y \leq -2x - 5$$ 2. **Understand the inequalities:** - The first inequality represents the region above or on the line $y = \frac{3}{2}x + 2$. - The second inequality represents the region below or on the line $y = -2x - 5$. 3. **Find the intersection point of the two lines:** Set the right sides equal to find where the lines cross: $$\frac{3}{2}x + 2 = -2x - 5$$ 4. **Solve for $x$:** $$\frac{3}{2}x + 2 = -2x - 5$$ $$\frac{3}{2}x + 2x = -5 - 2$$ $$\left(\frac{3}{2} + 2\right)x = -7$$ $$\left(\frac{3}{2} + \frac{4}{2}\right)x = -7$$ $$\frac{7}{2}x = -7$$ $$x = \frac{-7}{\cancel{\frac{7}{2}}} \times \cancel{\frac{2}{7}} = -2$$ 5. **Find $y$ at $x = -2$:** $$y = \frac{3}{2}(-2) + 2 = -3 + 2 = -1$$ 6. **Intersection point is $(-2, -1)$**. 7. **Analyze the region:** - The region satisfying $y \geq \frac{3}{2}x + 2$ is above the line with positive slope. - The region satisfying $y \leq -2x - 5$ is below the line with negative slope. 8. **Check a test point to determine the feasible region:** Try $x=0$: - For $y \geq \frac{3}{2}(0) + 2 = 2$, so $y \geq 2$. - For $y \leq -2(0) - 5 = -5$, so $y \leq -5$. No $y$ satisfies both at $x=0$, so the feasible region is between the lines near their intersection. 9. **Conclusion:** The solution region is the triangular area bounded by the two lines intersecting at $(-2, -1)$, where $y$ is above the line $y=\frac{3}{2}x+2$ and below the line $y=-2x-5$. **Final answer:** The graph representing the system is the one showing the region above the line $y=\frac{3}{2}x+2$ and below the line $y=-2x-5$, intersecting at $(-2,-1)$.