1. **State the problem:** We need to solve the system of inequalities graphically: $$y < -\frac{1}{3}x - 3$$ $$y \leq x + 5$$ and find a point in the solution set. 2. **Understand the inequalities:** - The first inequality is a strict inequality $y < -\frac{1}{3}x - 3$, meaning the solution region is below the line $y = -\frac{1}{3}x - 3$ (not including the line). - The second inequality is $y \leq x + 5$, meaning the solution region is on or below the line $y = x + 5$. 3. **Graph the boundary lines:** - For $y = -\frac{1}{3}x - 3$, the line has slope $-\frac{1}{3}$ and y-intercept $-3$. - For $y = x + 5$, the line has slope $1$ and y-intercept $5$. 4. **Determine the solution region:** - Shade below the line $y = -\frac{1}{3}x - 3$ (not including the line). - Shade on or below the line $y = x + 5$. 5. **Find the intersection point of the boundary lines:** Set $$-\frac{1}{3}x - 3 = x + 5$$ Multiply both sides by 3 to clear the fraction: $$3\left(-\frac{1}{3}x - 3\right) = 3(x + 5)$$ $$\cancel{3}\left(-\frac{1}{\cancel{3}}x\right) - 9 = 3x + 15$$ $$-x - 9 = 3x + 15$$ Add $x$ to both sides: $$-9 = 4x + 15$$ Subtract 15 from both sides: $$-9 - 15 = 4x$$ $$-24 = 4x$$ Divide both sides by 4: $$\frac{-24}{\cancel{4}} = \frac{4x}{\cancel{4}}$$ $$-6 = x$$ 6. **Find $y$ coordinate:** Substitute $x = -6$ into $y = x + 5$: $$y = -6 + 5 = -1$$ So the lines intersect at $(-6, -1)$. 7. **Choose a test point in the overlapping shaded region:** Try $(-7, -2)$: - Check $y < -\frac{1}{3}x - 3$: $$-2 < -\frac{1}{3}(-7) - 3 = \frac{7}{3} - 3 = \frac{7}{3} - \frac{9}{3} = -\frac{2}{3}$$ Is $-2 < -\frac{2}{3}$? Yes. - Check $y \leq x + 5$: $$-2 \leq -7 + 5 = -2$$ Yes, equality holds. Therefore, $(-7, -2)$ is in the solution set. **Final answer:** A point in the solution set is $\boxed{(-7, -2)}$.