1. **State the problem:** Solve the system of inequalities: $$y \leq 2x + 3$$ $$y < -x + 1$$ 2. **Understand the inequalities:** - The first inequality represents all points on or below the line $y = 2x + 3$. - The second inequality represents all points strictly below the line $y = -x + 1$. 3. **Find the intersection of the boundary lines:** Set $2x + 3 = -x + 1$ to find where the lines intersect. $$2x + 3 = -x + 1$$ $$2x + x = 1 - 3$$ $$3x = -2$$ $$x = \frac{-2}{3}$$ Substitute $x = \frac{-2}{3}$ into $y = 2x + 3$: $$y = 2 \times \frac{-2}{3} + 3 = \frac{-4}{3} + 3 = \frac{5}{3}$$ So, the lines intersect at $$\left(\frac{-2}{3}, \frac{5}{3}\right)$$. 4. **Determine the feasible region:** - For $y \leq 2x + 3$, the region is below or on the line. - For $y < -x + 1$, the region is strictly below the line. The solution is the set of points that satisfy both inequalities simultaneously, i.e., the intersection of these two regions. 5. **Check a test point in the intersection:** Try $x=0$: - For $y \leq 2(0) + 3 = 3$, so $y \leq 3$. - For $y < -0 + 1 = 1$, so $y < 1$. The intersection at $x=0$ is $y < 1$ (since $y$ must satisfy both). For example, $y=0$ satisfies both. 6. **Summary:** The solution region is all points below or on the line $y = 2x + 3$ and strictly below the line $y = -x + 1$. The boundary lines intersect at $$\left(\frac{-2}{3}, \frac{5}{3}\right)$$.