To subtract the mixed numbers \(2 \frac{3}{4}\) and \(1 \frac{1}{6}\), follow these steps:
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Convert mixed numbers to improper fractions:
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For \(2 \frac{3}{4}\): \[ 2 \frac{3}{4} = \left(2 \times 4 + 3\right) / 4 = \frac{8 + 3}{4} = \frac{11}{4} \]
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For \(1 \frac{1}{6}\): \[ 1 \frac{1}{6} = \left(1 \times 6 + 1\right) / 6 = \frac{6 + 1}{6} = \frac{7}{6} \]
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Now, subtract the two improper fractions: \[ \frac{11}{4} - \frac{7}{6} \ \text{To perform the subtraction, we need a common denominator.} \]
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The least common multiple of 4 and 6 is 12.
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Convert \(\frac{11}{4}\) to have a denominator of 12: \[ \frac{11}{4} = \frac{11 \times 3}{4 \times 3} = \frac{33}{12} \]
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Convert \(\frac{7}{6}\) to have a denominator of 12: \[ \frac{7}{6} = \frac{7 \times 2}{6 \times 2} = \frac{14}{12} \]
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Now perform the subtraction: \[ \frac{33}{12} - \frac{14}{12} = \frac{33 - 14}{12} = \frac{19}{12} \]
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Convert the improper fraction \(\frac{19}{12}\) back to a mixed number:
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Divide 19 by 12 to find the whole number part: \[ 19 \div 12 = 1 \quad \text{(with a remainder of 7)} \]
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This gives us: \[ 1 \quad \text{(whole part)} + \frac{7}{12} \quad \text{(fractional part)} = 1 \frac{7}{12} \]
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Final answer: \[ 1 \frac{7}{12} \]
So, in the format you asked for, the answer is \([1] [7]/[12]\).
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