Question

which point would be a solution to the system of linear inequalities shown below?
$y \leq -\frac{2}{3}x - 2$ $y \geq -\frac{1}{2}x - 1$
answer attempt 1 out of 2
$\circ$ $(6, -5)$ $\circ$ $(0, -1)$
$\circ$ $(-6, -7)$ $\circ$ $(-12, 5)$

Explanation:

Step1: Test point (6, -5)

First inequality: $-5 \leq -\frac{2}{3}(6) - 2 = -4 - 2 = -6$ → $-5 \leq -6$ is false.
Second inequality: $-5 \geq -\frac{1}{2}(6) - 1 = -3 - 1 = -4$ → $-5 \geq -4$ is false.

Step2: Test point (-6, -7)

First inequality: $-7 \leq -\frac{2}{3}(-6) - 2 = 4 - 2 = 2$ → $-7 \leq 2$ is true.
Second inequality: $-7 \geq -\frac{1}{2}(-6) - 1 = 3 - 1 = 2$ → $-7 \geq 2$ is false.

Step3: Test point (0, -1)

First inequality: $-1 \leq -\frac{2}{3}(0) - 2 = -2$ → $-1 \leq -2$ is false.
Second inequality: $-1 \geq -\frac{1}{2}(0) - 1 = -1$ → $-1 \geq -1$ is true.

Step4: Test point (-12, 5)

First inequality: $5 \leq -\frac{2}{3}(-12) - 2 = 8 - 2 = 6$ → $5 \leq 6$ is true.
Second inequality: $5 \geq -\frac{1}{2}(-12) - 1 = 6 - 1 = 5$ → $5 \geq 5$ is true.

Answer:

D. $(-12, 5)$