To determine which point is a solution to the system of inequalities:
[tex]\[
\begin{array}{l}
y \leq \frac{1}{2} x - 3 \\
y + 2x > 6
\end{array}
\][/tex]
we will check each point to see if it satisfies both inequalities.
A. [tex]\((7, -8)\)[/tex]
1. Check [tex]\( y \leq \frac{1}{2} x - 3 \)[/tex]:
[tex]\[
-8 \leq \frac{1}{2}(7) - 3 \\
-8 \leq 3.5 - 3 \\
-8 \leq 0.5 \quad \text{(True)}
\][/tex]
2. Check [tex]\( y + 2x > 6 \)[/tex]:
[tex]\[
-8 + 2(7) > 6 \\
-8 + 14 > 6 \\
6 > 6 \quad \text{(False)}
\][/tex]
Since the second inequality is not satisfied, point [tex]\((7, -8)\)[/tex] is not a solution.
B. [tex]\((2, -3)\)[/tex]
1. Check [tex]\( y \leq \frac{1}{2} x - 3 \)[/tex]:
[tex]\[
-3 \leq \frac{1}{2}(2) - 3 \\
-3 \leq 1 - 3 \\
-3 \leq -2 \quad \text{(True)}
\][/tex]
2. Check [tex]\( y + 2x > 6 \)[/tex]:
[tex]\[
-3 + 2(2) > 6 \\
-3 + 4 > 6 \\
1 > 6 \quad \text{(False)}
\][/tex]
Since the second inequality is not satisfied, point [tex]\((2, -3)\)[/tex] is not a solution.
C. [tex]\((5, -2)\)[/tex]
1. Check [tex]\( y \leq \frac{1}{2} x - 3 \)[/tex]:
[tex]\[
-2 \leq \frac{1}{2}(5) - 3 \\
-2 \leq 2.5 - 3 \\
-2 \leq -0.5 \quad \text{(True)}
\][/tex]
2. Check [tex]\( y + 2x > 6 \)[/tex]:
[tex]\[
-2 + 2(5) > 6 \\
-2 + 10 > 6 \\
8 > 6 \quad \text{(True)}
\][/tex]
Since both inequalities are satisfied, point [tex]\((5, -2)\)[/tex] is a solution.
D. [tex]\((4, 1)\)[/tex]
1. Check [tex]\( y \leq \frac{1}{2} x - 3 \)[/tex]:
[tex]\[
1 \leq \frac{1}{2}(4) - 3 \\
1 \leq 2 - 3 \\
1 \leq -1 \quad \text{(False)}
\][/tex]
Since the first inequality is not satisfied, point [tex]\((4, 1)\)[/tex] is not a solution.
Therefore, the correct point that satisfies both inequalities is:
[tex]\[
\boxed{C. (5, -2)}
\][/tex]
