To find how many more hours Eric rides his bike than his little brother, we will subtract the time his little brother rides from the time Eric rides.
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Convert the mixed numbers to improper fractions:
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Eric rides for \( 4 \frac{1}{3} \) hours. To convert this to an improper fraction: \[ 4 \frac{1}{3} = \frac{4 \cdot 3 + 1}{3} = \frac{12 + 1}{3} = \frac{13}{3} \]
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His little brother rides for \( 1 \frac{3}{4} \) hours. To convert this to an improper fraction: \[ 1 \frac{3}{4} = \frac{1 \cdot 4 + 3}{4} = \frac{4 + 3}{4} = \frac{7}{4} \]
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Find a common denominator to subtract the fractions. The denominators are \( 3 \) and \( 4 \). The least common multiple of \( 3 \) and \( 4 \) is \( 12 \).
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Convert \( \frac{13}{3} \) to a denominator of \( 12 \): \[ \frac{13}{3} = \frac{13 \cdot 4}{3 \cdot 4} = \frac{52}{12} \]
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Convert \( \frac{7}{4} \) to a denominator of \( 12 \): \[ \frac{7}{4} = \frac{7 \cdot 3}{4 \cdot 3} = \frac{21}{12} \]
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Subtract the two fractions: \[ \frac{52}{12} - \frac{21}{12} = \frac{52 - 21}{12} = \frac{31}{12} \]
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Convert the improper fraction back to a mixed number:
- Divide \( 31 \) by \( 12 \): \[ 31 \div 12 = 2 \quad \text{(with a remainder of } 7\text{)} \]
- This result can be expressed as: \[ 2 \frac{7}{12} \]
Thus, Eric rides his bike \( 2 \frac{7}{12} \) hours more than his little brother.
The final answer is: \[ \boxed{2} \quad \text{in the first box, and} \quad \boxed{7} \quad \text{in the second box.} \]