To determine which point is a solution to the simultaneous inequalities \( y < \frac{1}{2}x \) and \( y > \frac{1}{3}x + 3 \), we can analyze each point given in the options one by one by substituting the coordinates into the inequalities.
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For the point (25, 12):
- Check \( y < \frac{1}{2}x \):
- \( 12 < \frac{1}{2}(25) = 12.5 \) → True
- Check \( y > \frac{1}{3}x + 3 \):
- \( 12 > \frac{1}{3}(25) + 3 \)
- \( 12 > \frac{25}{3} + 3 \)
- \( 12 > 8.33 + 3 \)
- \( 12 > 11.33 \) → True
- This point is a solution.
- Check \( y < \frac{1}{2}x \):
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For the point (2, 2):
- Check \( y < \frac{1}{2}x \):
- \( 2 < \frac{1}{2}(2) = 1 \) → False
- Not a solution.
- Check \( y < \frac{1}{2}x \):
-
For the point (3, 6):
- Check \( y < \frac{1}{2}x \):
- \( 6 < \frac{1}{2}(3) = 1.5 \) → False
- Not a solution.
- Check \( y < \frac{1}{2}x \):
-
For the point (10, 5):
- Check \( y < \frac{1}{2}x \):
- \( 5 < \frac{1}{2}(10) = 5 \) → False (equal, so not less than)
- Not a solution.
- Check \( y < \frac{1}{2}x \):
From the checks above, only the point (25, 12) satisfies both inequalities.
Final Answer: (25, 12)