Question

1. which point would not be a solution to the system of linear inequalities shown below?

$y < 2x + 2$
$y \leq -\frac{5}{2}x - 5$
a. $(-4, -8)$
b. $(-8, -1)$
c. $(-2, -4)$
d. $(-2, -8)$

Explanation:

Step1: Test point A $(-4,-8)$

1. Substitute into $y<2x+2$:

$-8 < 2(-4)+2 = -8+2=-6$, this is true.

1. Substitute into $y\leq -\frac{5}{2}x -5$:

$-8 \leq -\frac{5}{2}(-4)-5 = 10-5=5$, this is true.

Step2: Test point B $(-8,-1)$

1. Substitute into $y<2x+2$:

$-1 < 2(-8)+2 = -16+2=-14$, this is false.

1. Substitute into $y\leq -\frac{5}{2}x -5$:

$-1 \leq -\frac{5}{2}(-8)-5 = 20-5=15$, this is true.
Since it fails the first inequality, this point is not a solution.

Step3: Verify remaining points

1. Point C $(-2,-4)$:

$-4 < 2(-2)+2 = -2$, true.
$-4 \leq -\frac{5}{2}(-2)-5 =5-5=0$, true.

1. Point D $(-2,-8)$:

$-8 < 2(-2)+2 = -2$, true.
$-8 \leq -\frac{5}{2}(-2)-5 =0$, true.

Answer:

B. $(-8,-1)$