Question

1. $y > \frac{1}{3}x + 1$

$y < \frac{4}{3}x - 2$

1. $y \geq \frac{5}{2}x - 3$

$y \leq \frac{1}{2}x + 1$

1. $y < -4x - 3$

$y \geq 2x + 3$

1. $y > -3$

$y \leq -\frac{4}{3}x + 1$

Explanation:

For Problem 11: $y>\frac{1}{3}x+1$, $y<\frac{4}{3}x-2$

Step1: Graph boundary $y=\frac{1}{3}x+1$

Use dashed line (since $>$), slope $\frac{1}{3}$, y-intercept $(0,1)$.

Step2: Shade above boundary

Region satisfies $y>\frac{1}{3}x+1$.

Step3: Graph boundary $y=\frac{4}{3}x-2$

Use dashed line (since $<$), slope $\frac{4}{3}$, y-intercept $(0,-2)$.

Step4: Shade below boundary

Region satisfies $y<\frac{4}{3}x-2$.

Step5: Find overlapping shaded area

This is the solution region.

Step1: Graph boundary $y=\frac{5}{2}x-3$

Use solid line (since $\geq$), slope $\frac{5}{2}$, y-intercept $(0,-3)$.

Step2: Shade above boundary

Region satisfies $y\geq\frac{5}{2}x-3$.

Step3: Graph boundary $y=-\frac{1}{2}x+1$

Use solid line (since $\leq$), slope $-\frac{1}{2}$, y-intercept $(0,1)$.

Step4: Shade below boundary

Region satisfies $y\leq-\frac{1}{2}x+1$.

Step5: Find overlapping shaded area

This is the solution region.

Step1: Graph boundary $y=-4x-3$

Use dashed line (since $<$), slope $-4$, y-intercept $(0,-3)$.

Step2: Shade below boundary

Region satisfies $y<-4x-3$.

Step3: Graph boundary $y=2x+3$

Use solid line (since $\geq$), slope $2$, y-intercept $(0,3)$.

Step4: Shade above boundary

Region satisfies $y\geq2x+3$.

Step5: Find overlapping shaded area

This is the solution region.

Answer:

The solution is the overlapping shaded area above $y=\frac{1}{3}x+1$ (dashed) and below $y=\frac{4}{3}x-2$ (dashed).

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For Problem 12: $y\geq\frac{5}{2}x-3$, $y\leq-\frac{1}{2}x+1$