Question

2.) $x - 2y \geq 12$
$x + 2y \leq -8$

Explanation:

Step1: Rewrite to slope-intercept form

For $x - 2y \geq 12$:
$-2y \geq -x + 12$
$y \leq \frac{1}{2}x - 6$

For $x + 2y \leq -8$:
$2y \leq -x - 8$
$y \leq -\frac{1}{2}x - 4$

Step2: Find intercepts for first line

For $y = \frac{1}{2}x - 6$:
x-intercept: set $y=0$, $0 = \frac{1}{2}x - 6 \implies x=12$ (point $(12, 0)$)
y-intercept: set $x=0$, $y=-6$ (point $(0, -6)$)
Draw a solid line, shade below it.

Step3: Find intercepts for second line

For $y = -\frac{1}{2}x - 4$:
x-intercept: set $y=0$, $0 = -\frac{1}{2}x - 4 \implies x=-8$ (point $(-8, 0)$)
y-intercept: set $x=0$, $y=-4$ (point $(0, -4)$)
Draw a solid line, shade below it.

Step4: Find intersection point

Solve

$$\begin{cases} y = \frac{1}{2}x - 6 \\ y = -\frac{1}{2}x - 4 \end{cases}$$

Set equal: $\frac{1}{2}x - 6 = -\frac{1}{2}x - 4$
$x - 12 = -x - 8$
$2x = 4 \implies x=2$
Substitute $x=2$: $y = \frac{1}{2}(2)-6 = -5$
Intersection: $(2, -5)$

Step5: Identify solution region

The solution is the overlapping shaded area, below both lines, bounded by the two solid lines and their intersection at $(2, -5)$.

Answer:

1. Graph the solid line $y = \frac{1}{2}x - 6$ (through $(12,0)$ and $(0,-6)$) and shade below it.
2. Graph the solid line $y = -\frac{1}{2}x - 4$ (through $(-8,0)$ and $(0,-4)$) and shade below it.
3. The overlapping shaded area (including the lines) is the solution, with the two lines intersecting at $(2, -5)$.