1. The problem is to graph the solution regions for each system of linear inequalities from 15 to 24. 2. Each inequality represents a half-plane bounded by a line. The solution to the system is the intersection of these half-planes. 3. To graph each system: - Convert each inequality to the line equation by replacing inequality with equality. - Plot the boundary lines. - Determine which side of each line satisfies the inequality by testing a point (usually the origin if not on the line). - Shade the region that satisfies both inequalities. 4. For example, for problem 15: - Inequalities: $y < 2x + 1$ and $y \leq -x - 4$ - Boundary lines: $y = 2x + 1$ and $y = -x - 4$ - Test point (0,0): - For $y < 2x + 1$: $0 < 0 + 1$ true, so shade below $y=2x+1$ line. - For $y \leq -x - 4$: $0 \leq 0 - 4$ false, so shade opposite side of $y=-x-4$ line. - The solution is the intersection of these shaded regions. 5. Repeat this process for each system (16 to 24). 6. The graphs are sets of points $(x,y)$ satisfying both inequalities, forming polygonal or unbounded regions. 7. Since you requested graphs, here are the Desmos-ready functions for each system's boundary lines: 15: $y=2x+1$, $y=-x-4$ 16: $y=3x-2$, $y=x-2$ 17: $y=-\frac{1}{2}x+1$, $y=x+3$ 18: $y=\frac{1}{3}x$, $y=-4x+1$ 19: $2x+3y=5$, $y=2x-3$ 20: $x+4y=3$, $x-y=2$ 21: $y=0.3x+2$, $y=-0.2x+1$ 22: $y=0.25x-4$, $y=-x-3$ 23: $y=-2x-5$, $4x-y=3$ 24: $-6x+4y=8$, $y=-x-1$ "slug": "linear inequalities", "subject": "algebra", "desmos": { "latex": "y=2x+1,y=-x-4", "features": { "intercepts": true, "extrema": true } }, "q_count": 10