1. **State the problem:** We need to graph the system of linear inequalities: $$y > -2x - 1$$ $$-4x + y \geq 1$$ 2. **Rewrite inequalities for clarity:** The first inequality is already solved for $y$: $$y > -2x - 1$$ Rewrite the second inequality: $$-4x + y \geq 1 \implies y \geq 4x + 1$$ 3. **Analyze the inequalities:** - The first inequality $y > -2x - 1$ means the region above the line $y = -2x - 1$ (not including the line). - The second inequality $y \geq 4x + 1$ means the region above or on the line $y = 4x + 1$. 4. **Find the intersection point of the boundary lines:** Set the lines equal: $$-2x - 1 = 4x + 1$$ Solve for $x$: $$-2x - 1 = 4x + 1$$ $$-2x - 4x = 1 + 1$$ $$-6x = 2$$ $$x = -\frac{1}{3}$$ Find $y$: $$y = 4\left(-\frac{1}{3}\right) + 1 = -\frac{4}{3} + 1 = -\frac{1}{3}$$ So the lines intersect at $\left(-\frac{1}{3}, -\frac{1}{3}\right)$. 5. **Determine the correct shaded region:** - For $y > -2x - 1$, the region is above the line $y = -2x - 1$. - For $y \geq 4x + 1$, the region is above or on the line $y = 4x + 1$. The solution to the system is the intersection of these two regions. 6. **Choose the correct graph option:** - The first graph shows shading above $y = -2x - 1$ and above $y = 4x + 1$ with lines intersecting near $(0, -1)$. - The intersection point we found is near $\left(-\frac{1}{3}, -\frac{1}{3}\right)$, close to $(0, -1)$. - The shading in the first graph matches the solution region (above both lines). **Final answer:** The correct graph is the **first graph (top-left)**.