To determine which points are solutions to the linear inequality [tex]\( y < 0.5x + 2 \)[/tex], we need to substitute each point into the inequality and check if it holds true.
Let's evaluate each point one by one:
1. Point [tex]\((-3, -2)\)[/tex]
- Substitute [tex]\( x = -3 \)[/tex] into the right-hand side of the inequality:
[tex]\[
y < 0.5(-3) + 2 = -1.5 + 2 = 0.5
\][/tex]
- Now compare [tex]\( y = -2 \)[/tex] with [tex]\( 0.5 \)[/tex]:
[tex]\[
-2 < 0.5
\][/tex]
This is true.
2. Point [tex]\((-2, 1)\)[/tex]
- Substitute [tex]\( x = -2 \)[/tex] into the right-hand side of the inequality:
[tex]\[
y < 0.5(-2) + 2 = -1 + 2 = 1
\][/tex]
- Now compare [tex]\( y = 1 \)[/tex] with [tex]\( 1 \)[/tex]:
[tex]\[
1 < 1
\][/tex]
This is not true.
3. Point [tex]\((-1, -2)\)[/tex]
- Substitute [tex]\( x = -1 \)[/tex] into the right-hand side of the inequality:
[tex]\[
y < 0.5(-1) + 2 = -0.5 + 2 = 1.5
\][/tex]
- Now compare [tex]\( y = -2 \)[/tex] with [tex]\( 1.5 \)[/tex]:
[tex]\[
-2 < 1.5
\][/tex]
This is true.
4. Point [tex]\((-1, 2)\)[/tex]
- Substitute [tex]\( x = -1 \)[/tex] into the right-hand side of the inequality:
[tex]\[
y < 0.5(-1) + 2 = -0.5 + 2 = 1.5
\][/tex]
- Now compare [tex]\( y = 2 \)[/tex] with [tex]\( 1.5 \)[/tex]:
[tex]\[
2 < 1.5
\][/tex]
This is not true.
5. Point [tex]\((1, -2)\)[/tex]
- Substitute [tex]\( x = 1 \)[/tex] into the right-hand side of the inequality:
[tex]\[
y < 0.5(1) + 2 = 0.5 + 2 = 2.5
\][/tex]
- Now compare [tex]\( y = -2 \)[/tex] with [tex]\( 2.5 \)[/tex]:
[tex]\[
-2 < 2.5
\][/tex]
This is true.
Based on the evaluations, the points that satisfy the inequality [tex]\( y < 0.5x + 2 \)[/tex] are:
- [tex]\((-3, -2)\)[/tex]
- [tex]\((-1, -2)\)[/tex]
- [tex]\((1, -2)\)[/tex]
So, the correct options are:
- [tex]\((-3, -2)\)[/tex]
- [tex]\((-1, -2)\)[/tex]
- [tex]\((1, -2)\)[/tex]
