To solve the problem \( 7 \frac{2}{4} + 2 \frac{3}{8} \), we first convert the mixed numbers into improper fractions.
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Convert \( 7 \frac{2}{4} \) to an improper fraction: \[ 7 \frac{2}{4} = 7 \times 4 + 2 = 28 + 2 = 30 \quad \text{so} \quad 7 \frac{2}{4} = \frac{30}{4} \]
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Convert \( 2 \frac{3}{8} \) to an improper fraction: \[ 2 \frac{3}{8} = 2 \times 8 + 3 = 16 + 3 = 19 \quad \text{so} \quad 2 \frac{3}{8} = \frac{19}{8} \]
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Now we add the two improper fractions \( \frac{30}{4} \) and \( \frac{19}{8} \). To do this, we need a common denominator. The least common multiple of 4 and 8 is 8.
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Convert \( \frac{30}{4} \) to a fraction with a denominator of 8: \[ \frac{30}{4} = \frac{30 \times 2}{4 \times 2} = \frac{60}{8} \]
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Now add \( \frac{60}{8} \) and \( \frac{19}{8} \): \[ \frac{60}{8} + \frac{19}{8} = \frac{60 + 19}{8} = \frac{79}{8} \]
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Convert \( \frac{79}{8} \) back to a mixed number: \[ 79 \div 8 = 9 \quad \text{with a remainder of} \quad 7 \quad \text{so} \quad \frac{79}{8} = 9 \frac{7}{8} \]
Thus, the final answer is: \[ 9 \frac{7}{8} \]
In the requested format: \[ 9 \quad 7/8 \]