1. **State the problem:** We need to graph the system of inequalities: $$x + y > -3$$ $$y > \frac{3}{2}x - 1$$ and identify the region that satisfies both. 2. **Rewrite inequalities in slope-intercept form:** For the first inequality: $$x + y > -3 \implies y > -x - 3$$ The second inequality is already in slope-intercept form: $$y > \frac{3}{2}x - 1$$ 3. **Understand the inequalities:** - For $$y > -x - 3$$, the solution is the region above the line $$y = -x - 3$$. - For $$y > \frac{3}{2}x - 1$$, the solution is the region above the line $$y = \frac{3}{2}x - 1$$. 4. **Graphing:** - Plot the line $$y = -x - 3$$ (dashed because inequality is strict). - Plot the line $$y = \frac{3}{2}x - 1$$ (dashed as well). - Shade the region above both lines. 5. **Find the intersection of the boundary lines:** Solve: $$-x - 3 = \frac{3}{2}x - 1$$ Add $$x$$ to both sides: $$-3 = \frac{3}{2}x + x - 1 = \frac{5}{2}x - 1$$ Add 1 to both sides: $$-2 = \frac{5}{2}x$$ Multiply both sides by $$\frac{2}{5}$$: $$x = -\frac{4}{5}$$ Find $$y$$: $$y = -x - 3 = -\left(-\frac{4}{5}\right) - 3 = \frac{4}{5} - 3 = -\frac{11}{5}$$ So the lines intersect at $$\left(-\frac{4}{5}, -\frac{11}{5}\right)$$. 6. **Final answer:** The solution region is the intersection of the half-planes above both lines, i.e., the area above $$y = -x - 3$$ and above $$y = \frac{3}{2}x - 1$$.