Question

which point would not be a solution to the system of linear inequalities shown below?
$y < -\frac{3}{2}x - 2$
$y > x + 6$
answer attempt 1 out of 2
$(-10, -1)$
$(-12, 3)$
$(-10, 5)$
$(-2, -1)$

Explanation:

Step1: Test point (-10,-1)

First inequality: $-1 < -\frac{3}{2}(-10) - 2 = 15 - 2 = 13$ (True)
Second inequality: $-1 > -10 + 6 = -4$ (True)

Step2: Test point (-12,3)

First inequality: $3 < -\frac{3}{2}(-12) - 2 = 18 - 2 = 16$ (True)
Second inequality: $3 > -12 + 6 = -6$ (True)

Step3: Test point (-10,5)

First inequality: $5 < -\frac{3}{2}(-10) - 2 = 15 - 2 = 13$ (True)
Second inequality: $5 > -10 + 6 = -4$ (True)

Step4: Test point (-2,-1)

First inequality: $-1 < -\frac{3}{2}(-2) - 2 = 3 - 2 = 1$ (True)
Second inequality: $-1 > -2 + 6 = 4$ (False)

Answer:

$\boldsymbol{(-2, -1)}$