To solve the system of inequalities and determine which point is a solution, we need to check each given point against both inequalities:
[tex]\[
\begin{cases}
y \leq \frac{1}{2} x - 3 \\
y + 2x > 6
\end{cases}
\][/tex]
Let's check each point one by one.
Option A: (7, -8)
For [tex]\( y \leq \frac{1}{2}x - 3 \)[/tex]:
Substitute [tex]\( x = 7 \)[/tex] and [tex]\( y = -8 \)[/tex]:
[tex]\[
-8 \leq \frac{1}{2}(7) - 3 \implies -8 \leq 3.5 - 3 \implies -8 \leq 0.5 \quad \text{(True)}
\][/tex]
For [tex]\( y + 2x > 6 \)[/tex]:
Substitute [tex]\( x = 7 \)[/tex] and [tex]\( y = -8 \)[/tex]:
[tex]\[
-8 + 2(7) > 6 \implies -8 + 14 > 6 \implies 6 > 6 \quad \text{(False)}
\][/tex]
Since one inequality is false, point (7, -8) is not a solution.
Option B: (2, -3)
For [tex]\( y \leq \frac{1}{2}x - 3 \)[/tex]:
Substitute [tex]\( x = 2 \)[/tex] and [tex]\( y = -3 \)[/tex]:
[tex]\[
-3 \leq \frac{1}{2}(2) - 3 \implies -3 \leq 1 - 3 \implies -3 \leq -2 \quad \text{(True)}
\][/tex]
For [tex]\( y + 2x > 6 \)[/tex]:
Substitute [tex]\( x = 2 \)[/tex] and [tex]\( y = -3 \)[/tex]:
[tex]\[
-3 + 2(2) > 6 \implies -3 + 4 > 6 \implies 1 > 6 \quad \text{(False)}
\][/tex]
Since one inequality is false, point (2, -3) is not a solution.
Option C: (5, -2)
For [tex]\( y \leq \frac{1}{2}x - 3 \)[/tex]:
Substitute [tex]\( x = 5 \)[/tex] and [tex]\( y = -2 \)[/tex]:
[tex]\[
-2 \leq \frac{1}{2}(5) - 3 \implies -2 \leq 2.5 - 3 \implies -2 \leq -0.5 \quad \text{(True)}
\][/tex]
For [tex]\( y + 2x > 6 \)[/tex]:
Substitute [tex]\( x = 5 \)[/tex] and [tex]\( y = -2 \)[/tex]:
[tex]\[
-2 + 2(5) > 6 \implies -2 + 10 > 6 \implies 8 > 6 \quad \text{(True)}
\][/tex]
Since both inequalities are true, point (5, -2) is a solution.
Option D: (4, 1)
For [tex]\( y \leq \frac{1}{2}x - 3 \)[/tex]:
Substitute [tex]\( x = 4 \)[/tex] and [tex]\( y = 1 \)[/tex]:
[tex]\[
1 \leq \frac{1}{2}(4) - 3 \implies 1 \leq 2 - 3 \implies 1 \leq -1 \quad \text{(False)}
\][/tex]
For [tex]\( y + 2x > 6 \)[/tex]:
Substitute [tex]\( x = 4 \)[/tex] and [tex]\( y = 1 \)[/tex]:
[tex]\[
1 + 2(4) > 6 \implies 1 + 8 > 6 \implies 9 > 6 \quad \text{(True)}
\][/tex]
Since one inequality is false, point (4, 1) is not a solution.
The only point that satisfies both inequalities is (5, -2). Therefore, the correct answer is:
C. (5, -2)
