Question

$m= \frac{4-1}{4-3}$ $b=6 (0,6)$
$x + 4y \geq -32$
$x - y < -3$

Explanation:

Step1: Rewrite first inequality to slope-intercept

$$7x + 4y \geq -32 \implies y \geq -\frac{7}{4}x - 8$$

Step2: Rewrite second inequality to slope-intercept

$$x - y < -3 \implies y > x + 3$$

Step3: Graph first boundary line

Plot $y = -\frac{7}{4}x - 8$: y-intercept at $(0, -8)$, x-intercept at $(-\frac{32}{7}, 0) \approx (-4.57, 0)$. Use a solid line (due to $\geq$), shade above the line.

Step4: Graph second boundary line

Plot $y = x + 3$: y-intercept at $(0, 3)$, x-intercept at $(-3, 0)$. Use a dashed line (due to $<$), shade above the line.

Step5: Identify overlapping shaded region

The solution is the area shaded by both inequalities.

Answer:

The solution is the overlapping region above the solid line $y = -\frac{7}{4}x - 8$ and above the dashed line $y = x + 3$ on the coordinate grid.

• Solid line points: $(0, -8)$, $(-4, -1)$, $(4, -15)$
• Dashed line points: $(0, 3)$, $(2, 5)$, $(-2, 1)$