To determine which point is a solution to the system of inequalities, we will evaluate each of the given points against the two inequalities:
1. [tex]\( y \leq \frac{1}{2} x - 3 \)[/tex]
2. [tex]\( y + 2x > 6 \)[/tex]
Let's check each point one by one:
### Point A: [tex]\((7, -8)\)[/tex]
1. Test inequality [tex]\( y \leq \frac{1}{2} x - 3 \)[/tex]:
[tex]\[
-8 \leq \frac{1}{2} \cdot 7 - 3
\][/tex]
Simplify the right side:
[tex]\[
-8 \leq \frac{7}{2} - 3 = \frac{7}{2} - \frac{6}{2} = \frac{1}{2}
\][/tex]
Since [tex]\(-8 \leq \frac{1}{2}\)[/tex], this inequality is satisfied.
2. Test inequality [tex]\( y + 2x > 6 \)[/tex]:
[tex]\[
-8 + 2 \cdot 7 > 6
\][/tex]
Simplify the left side:
[tex]\[
-8 + 14 > 6
\][/tex]
[tex]\[
6 > 6
\][/tex]
This inequality is not satisfied since [tex]\(6\)[/tex] is not greater than [tex]\(6\)[/tex].
Therefore, point A [tex]\((7, -8)\)[/tex] is not a solution.
### Point B: [tex]\((2, -3)\)[/tex]
1. Test inequality [tex]\( y \leq \frac{1}{2} x - 3 \)[/tex]:
[tex]\[
-3 \leq \frac{1}{2} \cdot 2 - 3
\][/tex]
Simplify the right side:
[tex]\[
-3 \leq 1 - 3 = -2
\][/tex]
Since [tex]\(-3 \leq -2\)[/tex], this inequality is satisfied.
2. Test inequality [tex]\( y + 2x > 6 \)[/tex]:
[tex]\[
-3 + 2 \cdot 2 > 6
\][/tex]
Simplify the left side:
[tex]\[
-3 + 4 > 6
\][/tex]
[tex]\[
1 > 6
\][/tex]
This inequality is not satisfied since [tex]\(1\)[/tex] is not greater than [tex]\(6\)[/tex].
Therefore, point B [tex]\((2, -3)\)[/tex] is not a solution.
### Point C: [tex]\((5, -2)\)[/tex]
1. Test inequality [tex]\( y \leq \frac{1}{2} x - 3 \)[/tex]:
[tex]\[
-2 \leq \frac{1}{2} \cdot 5 - 3
\][/tex]
Simplify the right side:
[tex]\[
-2 \leq \frac{5}{2} - 3 = \frac{5}{2} - \frac{6}{2} = -\frac{1}{2}
\][/tex]
Since [tex]\(-2 \leq -\frac{1}{2}\)[/tex], this inequality is satisfied.
2. Test inequality [tex]\( y + 2x > 6 \)[/tex]:
[tex]\[
-2 + 2 \cdot 5 > 6
\][/tex]
Simplify the left side:
[tex]\[
-2 + 10 > 6
\][/tex]
[tex]\[
8 > 6
\][/tex]
This inequality is satisfied.
Therefore, point C [tex]\((5, -2)\)[/tex] is a solution.
### Point D: [tex]\((4, 1)\)[/tex]
1. Test inequality [tex]\( y \leq \frac{1}{2} x - 3 \)[/tex]:
[tex]\[
1 \leq \frac{1}{2} \cdot 4 - 3
\][/tex]
Simplify the right side:
[tex]\[
1 \leq 2 - 3 = -1
\][/tex]
Since [tex]\(1 \leq -1\)[/tex] is not true, this inequality is not satisfied.
Therefore, point D [tex]\((4, 1)\)[/tex] is not a solution.
After evaluating all the points, we find that the only point that satisfies both inequalities is:
[tex]\[
\boxed{(5, -2)}
\][/tex]
