Solve. \( 4 \frac{6}{7}+2 \frac{1}{6}=\square \)
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To solve \( 4 \frac{6}{7}+2 \frac{1}{6} \), we first convert the mixed numbers to improper fractions. \( 4 \frac{6}{7} \) becomes \( \frac{4 \times 7 + 6}{7} = \frac{28 + 6}{7} = \frac{34}{7} \). \( 2 \frac{1}{6} \) becomes \( \frac{2 \times 6 + 1}{6} = \frac{12 + 1}{6} = \frac{13}{6} \). Next, we need a common denominator to add these fractions. The least common multiple of 7 and 6 is 42. Now we convert both fractions: \( \frac{34}{7} = \frac{34 \times 6}{7 \times 6} = \frac{204}{42} \), \( \frac{13}{6} = \frac{13 \times 7}{6 \times 7} = \frac{91}{42} \). Now we can add: \( \frac{204}{42} + \frac{91}{42} = \frac{204 + 91}{42} = \frac{295}{42} \). To convert \( \frac{295}{42} \) back to a mixed number, we divide 295 by 42, which gives us 7 as the whole number part (since \( 42 \times 7 = 294 \)) and a remainder of 1. Thus, \( \frac{295}{42} = 7 \frac{1}{42} \). So, \( 4 \frac{6}{7}+2 \frac{1}{6} = 7 \frac{1}{42} \). Final answer: \( 7 \frac{1}{42} \)