To determine which point is a solution to the simultaneous inequalities \( y \leq \frac{1}{4}x + 2 \) and \( y > \frac{1}{2}x + 3 \), we can evaluate each point given:
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Point (-10, -1):
- Check \( -1 \leq \frac{1}{4}(-10) + 2 \): \[ -1 \leq -2.5 + 2 \implies -1 \leq -0.5 \quad \text{(False)} \]
- Check \( -1 > \frac{1}{2}(-10) + 3 \): \[ -1 > -5 + 3 \implies -1 > -2 \quad \text{(True)} \]
Result: does not satisfy the first inequality.
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Point (4, -6):
- Check \( -6 \leq \frac{1}{4}(4) + 2 \): \[ -6 \leq 1 + 2 \implies -6 \leq 3 \quad \text{(True)} \]
- Check \( -6 > \frac{1}{2}(4) + 3 \): \[ -6 > 2 + 3 \implies -6 > 5 \quad \text{(False)} \]
Result: does not satisfy the second inequality.
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Point (5, 5):
- Check \( 5 \leq \frac{1}{4}(5) + 2 \): \[ 5 \leq 1.25 + 2 \implies 5 \leq 3.25 \quad \text{(False)} \]
Result: does not satisfy the first inequality.
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Point (6, 7):
- Check \( 7 \leq \frac{1}{4}(6) + 2 \): \[ 7 \leq 1.5 + 2 \implies 7 \leq 3.5 \quad \text{(False)} \]
- Check \( 7 > \frac{1}{2}(6) + 3 \): \[ 7 > 3 + 3 \implies 7 > 6 \quad \text{(True)} \]
Result: does not satisfy the first inequality.
Based on these evaluations, none of the points satisfy both inequalities simultaneously. Therefore, there is no solution among the options provided.
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