Question

solve the following system of inequalities graphically on the set of axes below. state the coordinates of a point in the solution set.
$y \leq -x + 2$
$y \geq \frac{1}{2}x - 4$

Explanation:

Step1: Graph $y=-x+2$ (solid line)

Find intercepts:

• When $x=0$, $y=2$ (y-intercept: $(0,2)$)
• When $y=0$, $0=-x+2 \implies x=2$ (x-intercept: $(2,0)$)

Shade below the line (since $y \leq -x+2$).

Step2: Graph $y=\frac{1}{2}x-4$ (solid line)

Find intercepts:

• When $x=0$, $y=-4$ (y-intercept: $(0,-4)$)
• When $y=0$, $0=\frac{1}{2}x-4 \implies x=8$ (x-intercept: $(8,0)$)

Shade above the line (since $y \geq \frac{1}{2}x-4$).

Step3: Find intersection of shaded regions

The solution is the overlapping shaded area. To confirm the intersection point of the two lines, solve:

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

Set equal: $-x + 2 = \frac{1}{2}x - 4$

$$ 2 + 4 = \frac{1}{2}x + x \\ 6 = \frac{3}{2}x \\ x = 4 $$

Substitute $x=4$ into $y=-x+2$: $y = -4 + 2 = -2$
Intersection point: $(4,-2)$

Step4: Pick a point in the solution

Choose a point in the overlapping shaded area, e.g., $(0,0)$. Verify:

• $0 \leq -0 + 2$: $0 \leq 2$ (true)
• $0 \geq \frac{1}{2}(0) - 4$: $0 \geq -4$ (true)

Answer:

A point in the solution set is $\boldsymbol{(0,0)}$ (other valid points include $(4,-2)$, $(2,-1)$, etc.)