To determine which point is a solution to the simultaneous inequalities \( y < \frac{1}{2}x \) and \( y > \frac{1}{3}x + 3 \), we need to analyze each inequality:
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First Inequality: \( y < \frac{1}{2}x \)
- This inequality represents the region below the line \( y = \frac{1}{2}x \).
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Second Inequality: \( y > \frac{1}{3}x + 3 \)
- This inequality represents the region above the line \( y = \frac{1}{3}x + 3 \).
Next, we will plug in each of the given points to see if they satisfy both inequalities.
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Point (3, 6):
- For \( y < \frac{1}{2}x \): \( 6 < \frac{1}{2} \cdot 3 \) → \( 6 < 1.5 \) (false)
- For \( y > \frac{1}{3}x + 3\): \( 6 > \frac{1}{3} \cdot 3 + 3 \) → \( 6 > 4 \) (true)
This point does not satisfy the first inequality.
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Point (25, 12):
- For \( y < \frac{1}{2}x \): \( 12 < \frac{1}{2} \cdot 25 \) → \( 12 < 12.5 \) (true)
- For \( y > \frac{1}{3}x + 3 \): \( 12 > \frac{1}{3} \cdot 25 + 3 \) → \( 12 > \frac{25}{3} + 3 \) → \( 12 > 8.33 + 3 \) → \( 12 > 11.33 \) (true)
This point satisfies both inequalities.
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Point (2, 2):
- For \( y < \frac{1}{2}x \): \( 2 < \frac{1}{2} \cdot 2 \) → \( 2 < 1 \) (false)
- For \( y > \frac{1}{3}x + 3 \): \( 2 > \frac{1}{3} \cdot 2 + 3 \) → \( 2 > 0.67 + 3 \) → \( 2 > 3.67 \) (false)
This point does not satisfy either inequality.
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Point (10, 5):
- For \( y < \frac{1}{2}x \): \( 5 < \frac{1}{2} \cdot 10 \) → \( 5 < 5 \) (false)
- For \( y > \frac{1}{3}x + 3 \): \( 5 > \frac{1}{3} \cdot 10 + 3 \) → \( 5 > 3.33 + 3 \) → \( 5 > 6.33 \) (false)
This point does not satisfy either inequality.
From this analysis, the only point that satisfies both inequalities is:
(25, 12)
Therefore, the answer is: (25,12)
1 answer