Which points are solutions to the linear inequality [tex]y \ \textless \ 0.5x + 2[/tex]? Select three options.

A. [tex](-3, -2)[/tex]
B. [tex](-2, 1)[/tex]
C. [tex](-1, -2)[/tex]
D. [tex](-1, 2)[/tex]
E. [tex](4, -2)[/tex]



Answer :

To determine which points lie below the line defined by the inequality [tex]\( y < 0.5x + 2 \)[/tex], we need to evaluate each point and check if they satisfy the inequality.

Let's analyze each point step-by-step:

1. For the point [tex]\((-3, -2)\)[/tex]:
- Substitute [tex]\( x = -3 \)[/tex] into [tex]\( 0.5x + 2 \)[/tex]:
[tex]\[ y = 0.5(-3) + 2 = -1.5 + 2 = 0.5 \][/tex]
- Check if [tex]\(-2 < 0.5\)[/tex]:
[tex]\[ -2 < 0.5 \quad \text{(True)} \][/tex]
So, [tex]\((-3, -2)\)[/tex] satisfies the inequality.

2. For the point [tex]\((-2, 1)\)[/tex]:
- Substitute [tex]\( x = -2 \)[/tex] into [tex]\( 0.5x + 2 \)[/tex]:
[tex]\[ y = 0.5(-2) + 2 = -1 + 2 = 1 \][/tex]
- Check if [tex]\(1 < 1\)[/tex]:
[tex]\[ 1 < 1 \quad \text{(False)} \][/tex]
So, [tex]\((-2, 1)\)[/tex] does not satisfy the inequality.

3. For the point [tex]\((-1, -2)\)[/tex]:
- Substitute [tex]\( x = -1 \)[/tex] into [tex]\( 0.5x + 2 \)[/tex]:
[tex]\[ y = 0.5(-1) + 2 = -0.5 + 2 = 1.5 \][/tex]
- Check if [tex]\(-2 < 1.5\)[/tex]:
[tex]\[ -2 < 1.5 \quad \text{(True)} \][/tex]
So, [tex]\((-1, -2)\)[/tex] satisfies the inequality.

4. For the point [tex]\((-1, 2)\)[/tex]:
- Substitute [tex]\( x = -1 \)[/tex] into [tex]\( 0.5x + 2 \)[/tex]:
[tex]\[ y = 0.5(-1) + 2 = -0.5 + 2 = 1.5 \][/tex]
- Check if [tex]\(2 < 1.5\)[/tex]:
[tex]\[ 2 < 1.5 \quad \text{(False)} \][/tex]
So, [tex]\((-1, 2)\)[/tex] does not satisfy the inequality.

5. For the point [tex]\((4, -2)\)[/tex]:
- Substitute [tex]\( x = 4 \)[/tex] into [tex]\( 0.5x + 2 \)[/tex]:
[tex]\[ y = 0.5(4) + 2 = 2 + 2 = 4 \][/tex]
- Check if [tex]\(-2 < 4\)[/tex]:
[tex]\[ -2 < 4 \quad \text{(True)} \][/tex]
So, [tex]\((4, -2)\)[/tex] satisfies the inequality.

Thus, the three points that satisfy the inequality [tex]\( y < 0.5x + 2 \)[/tex] are:
[tex]\[ (-3, -2), (-1, -2), \text{and } (4, -2) \][/tex]