To add the mixed numbers \(-3 \frac{1}{6}\) and \(5 \frac{3}{4}\), we can follow these steps:
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Convert the mixed numbers to improper fractions:
For \(-3 \frac{1}{6}\): \[ -3 \frac{1}{6} = -\left(3 \times 6 + 1\right)/6 = -\left(18 + 1\right)/6 = -\frac{19}{6} \]
For \(5 \frac{3}{4}\): \[ 5 \frac{3}{4} = \left(5 \times 4 + 3\right)/4 = \left(20 + 3\right)/4 = \frac{23}{4} \]
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Find a common denominator for the fractions \(-\frac{19}{6}\) and \(\frac{23}{4}\). The least common multiple of 6 and 4 is 12.
Convert \(-\frac{19}{6}\) to a denominator of 12: \[ -\frac{19}{6} = -\frac{19 \times 2}{6 \times 2} = -\frac{38}{12} \]
Convert \(\frac{23}{4}\) to a denominator of 12: \[ \frac{23}{4} = \frac{23 \times 3}{4 \times 3} = \frac{69}{12} \]
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Add the fractions: \[ -\frac{38}{12} + \frac{69}{12} = \frac{-38 + 69}{12} = \frac{31}{12} \]
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Convert the improper fraction back to a mixed number: \(\frac{31}{12}\) can be written as:
- The whole number part is \(31 \div 12 = 2\) (with a remainder of \(7\)), so we have: \[ \frac{31}{12} = 2 \frac{7}{12} \]
Thus, the final answer is: \[ \boxed{2 \frac{7}{12}} \]
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