Which point is a solution to the linear inequality [tex]\( y \ \textless \ -\frac{1}{2}x + 2 \)[/tex]?

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



Answer :

Let's analyze each point given to determine if it satisfies the linear inequality [tex]\( y < -\frac{1}{2}x + 2 \)[/tex].

### Point (2, 3)
1. Substitute [tex]\( x = 2 \)[/tex] into the inequality:
[tex]\[ y < -\frac{1}{2}(2) + 2 \][/tex]
2. Simplify the right side:
[tex]\[ y < -1 + 2 \][/tex]
[tex]\[ y < 1 \][/tex]
3. Now, compare [tex]\( y \)[/tex] with 1:
[tex]\[ 3 < 1 \][/tex]
This is not true. Hence, the point [tex]\((2, 3)\)[/tex] does not satisfy the inequality.

### Point (2, 1)
1. Substitute [tex]\( x = 2 \)[/tex] into the inequality:
[tex]\[ y < -\frac{1}{2}(2) + 2 \][/tex]
2. Simplify the right side:
[tex]\[ y < -1 + 2 \][/tex]
[tex]\[ y < 1 \][/tex]
3. Now, compare [tex]\( y \)[/tex] with 1:
[tex]\[ 1 < 1 \][/tex]
This is not true since 1 is equal to 1. Hence, the point [tex]\((2, 1)\)[/tex] does not satisfy the inequality.

### Point (3, -2)
1. Substitute [tex]\( x = 3 \)[/tex] into the inequality:
[tex]\[ y < -\frac{1}{2}(3) + 2 \][/tex]
2. Simplify the right side:
[tex]\[ y < -\frac{3}{2} + 2 \][/tex]
[tex]\[ y < -1.5 + 2 \][/tex]
[tex]\[ y < 0.5 \][/tex]
3. Now, compare [tex]\( y \)[/tex] with 0.5:
[tex]\[ -2 < 0.5 \][/tex]
This is true. Therefore, the point [tex]\((3, -2)\)[/tex] satisfies the inequality.

### Point (-1, 3)
1. Substitute [tex]\( x = -1 \)[/tex] into the inequality:
[tex]\[ y < -\frac{1}{2}(-1) + 2 \][/tex]
2. Simplify the right side:
[tex]\[ y < \frac{1}{2} + 2 \][/tex]
[tex]\[ y < 2.5 \][/tex]
3. Now, compare [tex]\( y \)[/tex] with 2.5:
[tex]\[ 3 < 2.5 \][/tex]
This is not true. Hence, the point [tex]\((-1, 3)\)[/tex] does not satisfy the inequality.

After evaluating all the given points, we find that the only point that satisfies the linear inequality [tex]\( y < -\frac{1}{2}x + 2 \)[/tex] is:

[tex]\((3, -2)\)[/tex].