To subtract the mixed numbers \(15 \frac{2}{3}\) and \(21 \frac{3}{4}\), we can follow these steps:
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First, convert both mixed numbers to improper fractions.
For \(15 \frac{2}{3}\): \[ 15 \frac{2}{3} = \left(15 \times 3 + 2\right) / 3 = \frac{45 + 2}{3} = \frac{47}{3} \]
For \(21 \frac{3}{4}\): \[ 21 \frac{3}{4} = \left(21 \times 4 + 3\right) / 4 = \frac{84 + 3}{4} = \frac{87}{4} \]
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Now we need to subtract the two improper fractions: \[ 15 \frac{2}{3} - 21 \frac{3}{4} = \frac{47}{3} - \frac{87}{4} \]
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To subtract these fractions, we need a common denominator. The least common multiple of 3 and 4 is 12.
Convert both fractions to have the denominator of 12:
For \(\frac{47}{3}\): \[ \frac{47}{3} = \frac{47 \times 4}{3 \times 4} = \frac{188}{12} \]
For \(\frac{87}{4}\): \[ \frac{87}{4} = \frac{87 \times 3}{4 \times 3} = \frac{261}{12} \]
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Now we can subtract: \[ \frac{188}{12} - \frac{261}{12} = \frac{188 - 261}{12} = \frac{-73}{12} \]
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The result is \(-\frac{73}{12}\). We can express this as a mixed number:
\(-\frac{73}{12}\) can be written as \(-6 \frac{1}{12}\) because \(73 \div 12 = 6\) with a remainder of \(1\).
Therefore, the final answer is: \[ -6 \frac{1}{12} \]