1. **State the problem:** We need to graph the solution to the system of inequalities: $$-5x + 2y \leq 2$$ $$4x + 3y > -9$$ and find one point in the solution set. 2. **Rewrite inequalities in slope-intercept form:** For the first inequality: $$-5x + 2y \leq 2$$ Add $5x$ to both sides: $$2y \leq 5x + 2$$ Divide both sides by 2: $$y \leq \frac{\cancel{2}}{\cancel{2}} \cdot \frac{5x}{2} + \frac{2}{2}$$ which simplifies to: $$y \leq \frac{5}{2}x + 1$$ For the second inequality: $$4x + 3y > -9$$ Subtract $4x$ from both sides: $$3y > -4x - 9$$ Divide both sides by 3: $$y > \frac{\cancel{3}}{\cancel{3}} \cdot \frac{-4x}{3} - \frac{9}{3}$$ which simplifies to: $$y > -\frac{4}{3}x - 3$$ 3. **Graph the boundary lines:** - For $y \leq \frac{5}{2}x + 1$, graph the line $y = \frac{5}{2}x + 1$ with a solid line because the inequality includes equality ($\leq$). - For $y > -\frac{4}{3}x - 3$, graph the line $y = -\frac{4}{3}x - 3$ with a dashed line because the inequality is strict ($>$). 4. **Shade the solution regions:** - Shade below or on the line $y = \frac{5}{2}x + 1$. - Shade above the line $y = -\frac{4}{3}x - 3$. The solution to the system is the region where these shaded areas overlap. 5. **Check the point (0,0):** - Substitute into the first inequality: $$-5(0) + 2(0) = 0 \leq 2$$ which is true. - Substitute into the second inequality: $$4(0) + 3(0) = 0 > -9$$ which is true. Therefore, the point (0,0) lies in the solution set. **Final answer:** The solution region is the intersection of the half-planes defined by $y \leq \frac{5}{2}x + 1$ and $y > -\frac{4}{3}x - 3$. The point (0,0) is one point in the solution set.