1. The problem is to analyze the system of inequalities: $$-5x + 2y \leq 2$$ $$4x + 3y > -9$$ 2. To understand the solution region, we first convert each inequality into an equation to find boundary lines: $$-5x + 2y = 2$$ $$4x + 3y = -9$$ 3. Solve each for $y$ to express in slope-intercept form $y = mx + b$: For the first line: $$-5x + 2y = 2 \implies 2y = 5x + 2 \implies y = \frac{5}{2}x + 1$$ For the second line: $$4x + 3y = -9 \implies 3y = -4x - 9 \implies y = -\frac{4}{3}x - 3$$ 4. The inequalities correspond to regions relative to these lines: - For $$-5x + 2y \leq 2$$, or $$y \leq \frac{5}{2}x + 1$$, the solution is the region on or below the first line. - For $$4x + 3y > -9$$, or $$y > -\frac{4}{3}x - 3$$, the solution is the region above the second line. 5. The solution to the system is the intersection of these two regions. 6. To find the intersection point of the boundary lines, solve the system: $$y = \frac{5}{2}x + 1$$ $$y = -\frac{4}{3}x - 3$$ Set equal: $$\frac{5}{2}x + 1 = -\frac{4}{3}x - 3$$ Multiply both sides by 6 (common denominator) to clear fractions: $$6 \times \left(\frac{5}{2}x + 1\right) = 6 \times \left(-\frac{4}{3}x - 3\right)$$ $$3 \times 5x + 6 = -2 \times 4x - 18$$ $$15x + 6 = -8x - 18$$ Add $8x$ to both sides: $$15x + 8x + 6 = -18$$ $$23x + 6 = -18$$ Subtract 6: $$23x = -24$$ Divide both sides by 23: $$x = \frac{\cancel{23}x}{\cancel{23}} = \frac{-24}{23}$$ Substitute back to find $y$: $$y = \frac{5}{2} \times \left(-\frac{24}{23}\right) + 1 = -\frac{120}{46} + 1 = -\frac{60}{23} + 1 = -\frac{60}{23} + \frac{23}{23} = -\frac{37}{23}$$ 7. The intersection point is $$\left(-\frac{24}{23}, -\frac{37}{23}\right)$$. 8. The solution region is all points satisfying: $$y \leq \frac{5}{2}x + 1$$ $$y > -\frac{4}{3}x - 3$$ which is the area between the two lines, including the first line but excluding the second.