To determine which point satisfies the given system of inequalities:
[tex]\[
\begin{cases}
y \leq \frac{1}{2} x - 3 \\
y + 2x > 6
\end{cases}
\][/tex]
we need to check each of the given points against both inequalities. Let's analyze each point in turn:
1. Point (7, -8)
Check the first inequality:
[tex]\[ -8 \leq \frac{1}{2} \cdot 7 - 3 \][/tex]
[tex]\[ -8 \leq 3.5 - 3 \][/tex]
[tex]\[ -8 \leq 0.5 \quad \text{(True)} \][/tex]
Check the second inequality:
[tex]\[ -8 + 2 \cdot 7 > 6 \][/tex]
[tex]\[ -8 + 14 > 6 \][/tex]
[tex]\[ 6 > 6 \quad \text{(False)} \][/tex]
Since both inequalities must be satisfied and the second one is false, point (7, -8) is not a solution.
2. Point (2, -3)
Check the first inequality:
[tex]\[ -3 \leq \frac{1}{2} \cdot 2 - 3 \][/tex]
[tex]\[ -3 \leq 1 - 3 \][/tex]
[tex]\[ -3 \leq -2 \quad \text{(True)} \][/tex]
Check the second inequality:
[tex]\[ -3 + 2 \cdot 2 > 6 \][/tex]
[tex]\[ -3 + 4 > 6 \][/tex]
[tex]\[ 1 > 6 \quad \text{(False)} \][/tex]
Since both inequalities must be satisfied and the second one is false, point (2, -3) is not a solution.
3. Point (5, -2)
Check the first inequality:
[tex]\[ -2 \leq \frac{1}{2} \cdot 5 - 3 \][/tex]
[tex]\[ -2 \leq 2.5 - 3 \][/tex]
[tex]\[ -2 \leq -0.5 \quad \text{(True)} \][/tex]
Check the second inequality:
[tex]\[ -2 + 2 \cdot 5 > 6 \][/tex]
[tex]\[ -2 + 10 > 6 \][/tex]
[tex]\[ 8 > 6 \quad \text{(True)} \][/tex]
Since both inequalities are satisfied, point (5, -2) is a solution to the system.
4. Point (4, 1)
Check the first inequality:
[tex]\[ 1 \leq \frac{1}{2} \cdot 4 - 3 \][/tex]
[tex]\[ 1 \leq 2 - 3 \][/tex]
[tex]\[ 1 \leq -1 \quad \text{(False)} \][/tex]
Check the second inequality:
[tex]\[ 1 + 2 \cdot 4 > 6 \][/tex]
[tex]\[ 1 + 8 > 6 \][/tex]
[tex]\[ 9 > 6 \quad \text{(True)} \][/tex]
Since both inequalities must be satisfied and the first one is false, point (4, 1) is not a solution.
After checking all the points, we find that the only point that satisfies both inequalities is:
[tex]\[ \boxed{3} \][/tex]
