1. The problem involves solving and graphing the system of inequalities: $$3x + 2y \leq 5$$ $$2x + 3y \geq 5$$ $$2x + 3y \leq 5$$ $$3x + 2y \geq 5$$ 2. Each inequality represents a half-plane bounded by a line. The goal is to find the region that satisfies all inequalities simultaneously. 3. Let's analyze the first inequality: $$3x + 2y \leq 5$$. - The boundary line is $$3x + 2y = 5$$. - To graph, find intercepts: - When $$x=0$$, $$2y=5 \Rightarrow y=\frac{5}{2}$$. - When $$y=0$$, $$3x=5 \Rightarrow x=\frac{5}{3}$$. 4. The slope of the line is $$-\frac{3}{2}$$ (from rearranging to $$y = -\frac{3}{2}x + \frac{5}{2}$$). 5. The shaded region is below or on the line because of the $\leq$ sign. 6. The second inequality is $$2x + 3y \geq 5$$. - Boundary line: $$2x + 3y = 5$$. - Intercepts: - $$x=0 \Rightarrow 3y=5 \Rightarrow y=\frac{5}{3}$$. - $$y=0 \Rightarrow 2x=5 \Rightarrow x=\frac{5}{2}$$. - Slope is $$-\frac{2}{3}$$. - The shaded region is above or on the line. 7. The third inequality $$2x + 3y \leq 5$$ is the opposite of the second, so the shaded region is below or on the line $$2x + 3y = 5$$. 8. The fourth inequality $$3x + 2y \geq 5$$ is the opposite of the first, so the shaded region is above or on the line $$3x + 2y = 5$$. 9. Since the second and third inequalities contradict each other (one $\geq$, one $\leq$ for the same line), and similarly for the first and fourth, the system has no solution where all inequalities hold simultaneously. 10. The graph description matches the line $$3x + 2y = 5$$ with slope $$-\frac{3}{2}$$ and intercept $$\frac{5}{2}$$, shading below and to the left. Final answer: The system of inequalities has no common solution region because the inequalities contradict each other.