1. **State the problem:** We are given two inequalities: $$y \leq \frac{7}{4}x + 3$$ $$y < -\frac{3}{2}x + 5$$ We want to understand the solution region that satisfies both inequalities on the coordinate plane. 2. **Understand the inequalities:** - The first inequality represents all points on or below the line with slope $\frac{7}{4}$ and y-intercept 3. - The second inequality represents all points strictly below the line with slope $-\frac{3}{2}$ and y-intercept 5. 3. **Find the intersection point of the two lines:** Set the right sides equal to find where the lines cross: $$\frac{7}{4}x + 3 = -\frac{3}{2}x + 5$$ 4. **Solve for $x$:** $$\frac{7}{4}x + \frac{3}{2}x = 5 - 3$$ Convert $\frac{3}{2}x$ to $\frac{6}{4}x$ for common denominator: $$\frac{7}{4}x + \frac{6}{4}x = 2$$ $$\frac{13}{4}x = 2$$ Multiply both sides by $\cancel{\frac{4}{13}}$: $$x = 2 \times \frac{4}{13} = \frac{8}{13}$$ 5. **Find $y$ at $x=\frac{8}{13}$:** Use the first line: $$y = \frac{7}{4} \times \frac{8}{13} + 3 = \frac{56}{52} + 3 = \frac{14}{13} + 3 = \frac{14}{13} + \frac{39}{13} = \frac{53}{13}$$ 6. **Interpretation:** The two lines intersect at the point $$\left(\frac{8}{13}, \frac{53}{13}\right)$$. 7. **Solution region:** The solution to the system is the set of points that satisfy both inequalities simultaneously: - Points on or below the line $y = \frac{7}{4}x + 3$ - Points strictly below the line $y = -\frac{3}{2}x + 5$ This region is the intersection of the half-planes defined by these inequalities. **Final answer:** The solution region is all points $$ (x,y) $$ such that $$y \leq \frac{7}{4}x + 3 \quad \text{and} \quad y < -\frac{3}{2}x + 5.$$