To find out how far Joey is from home after walking away and then back, we need to perform the following calculations:
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Convert the mixed numbers into improper fractions to make calculations easier.
For \( 21 \frac{3}{4} \): \[ 21 \frac{3}{4} = \frac{21 \times 4 + 3}{4} = \frac{84 + 3}{4} = \frac{87}{4} \]
For \( 15 \frac{2}{3} \): \[ 15 \frac{2}{3} = \frac{15 \times 3 + 2}{3} = \frac{45 + 2}{3} = \frac{47}{3} \]
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Joey walked away from home for \( \frac{87}{4} \) meters and then returned \( \frac{47}{3} \) meters.
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To find out how far Joey is from home, we need to subtract the distance he walked back from the distance he walked away:
\[ \text{Distance from home} = \frac{87}{4} - \frac{47}{3} \]
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To subtract these fractions, we need a common denominator. The least common multiple of 4 and 3 is 12.
Convert \( \frac{87}{4} \) to a fraction with a denominator of 12: \[ \frac{87}{4} \times \frac{3}{3} = \frac{261}{12} \]
Convert \( \frac{47}{3} \) to a fraction with a denominator of 12: \[ \frac{47}{3} \times \frac{4}{4} = \frac{188}{12} \]
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Now perform the subtraction: \[ \frac{261}{12} - \frac{188}{12} = \frac{261 - 188}{12} = \frac{73}{12} \]
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Convert \( \frac{73}{12} \) back into a mixed number: \[ 73 \div 12 = 6 \quad \text{(remainder \( 1 \))} \] Thus, \( \frac{73}{12} = 6 \frac{1}{12} \).
Therefore, Joey is \( 6 \frac{1}{12} \) meters away from home.
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