To determine which point is a solution to the simultaneous inequalities \( y < \frac{1}{2}x \) and \( y > \frac{1}{3}x + 3 \), we will test each of the provided points:
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Point (3, 6):
- \( y < \frac{1}{2}x \): \( 6 < \frac{1}{2} \cdot 3 = 1.5 \) ➔ False
- \( y > \frac{1}{3}x + 3 \): \( 6 > \frac{1}{3} \cdot 3 + 3 = 4 \) ➔ True
- Overall: False since the first inequality is not satisfied.
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Point (10, 5):
- \( y < \frac{1}{2}x \): \( 5 < \frac{1}{2} \cdot 10 = 5 \) ➔ False
- \( y > \frac{1}{3}x + 3 \): \( 5 > \frac{1}{3} \cdot 10 + 3 \approx 6.33 \) ➔ False
- Overall: False since both inequalities are not satisfied.
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Point (25, 12):
- \( y < \frac{1}{2}x \): \( 12 < \frac{1}{2} \cdot 25 = 12.5 \) ➔ True
- \( y > \frac{1}{3}x + 3 \): \( 12 > \frac{1}{3} \cdot 25 + 3 \approx 11.33 \) ➔ True
- Overall: True since both inequalities are satisfied.
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Point (2, 2):
- \( y < \frac{1}{2}x \): \( 2 < \frac{1}{2} \cdot 2 = 1 \) ➔ False
- \( y > \frac{1}{3}x + 3 \): \( 2 > \frac{1}{3} \cdot 2 + 3 \approx 4 \) ➔ False
- Overall: False since both inequalities are not satisfied.
From this analysis, the only point that satisfies both inequalities is (25, 12).