Let's analyze the problem step by step. We have two unknowns: \(x\) (number of green marbles) and \(y\) (number of yellow marbles). We're given the following conditions:
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"The son of twice the number of green marbles and the number of yellow marbles in a bag is more than 20": This can be interpreted as: \[ 2x + y > 20 \]
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"The total number of green and yellow marbles is less than 15": This translates to: \[ x + y < 15 \]
Now, we will check the given candidate solutions: (10, 5), (6, 7), (8, 6), (3, 12).
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For (10, 5):
- \(2x + y = 2(10) + 5 = 20 + 5 = 25 > 20\) (satisfied)
- \(x + y = 10 + 5 = 15 < 15\) (not satisfied)
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For (6, 7):
- \(2x + y = 2(6) + 7 = 12 + 7 = 19 > 20\) (not satisfied)
- \(x + y = 6 + 7 = 13 < 15\) (satisfied)
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For (8, 6):
- \(2x + y = 2(8) + 6 = 16 + 6 = 22 > 20\) (satisfied)
- \(x + y = 8 + 6 = 14 < 15\) (satisfied)
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For (3, 12):
- \(2x + y = 2(3) + 12 = 6 + 12 = 18 > 20\) (not satisfied)
- \(x + y = 3 + 12 = 15 < 15\) (not satisfied)
After checking the candidate solutions, we find that only the pair \((8, 6)\) satisfies both inequalities:
- \(2x + y > 20\)
- \(x + y < 15\)
Thus, the only valid solution to the system of inequalities is \((8, 6)\).
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