To determine how far Joey is from home, we need to consider the distances he walked in both directions.
- Joey walked east for \( 15 \frac{2}{3} \) meters.
- Then, he walked west for \( 21 \frac{3}{4} \) meters.
We can start by converting both mixed numbers to improper fractions:
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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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For \( 21 \frac{3}{4} \): \[ 21 \frac{3}{4} = \frac{21 \times 4 + 3}{4} = \frac{84 + 3}{4} = \frac{87}{4} \]
Now, since Joey is walking in opposite directions, we need to find the net distance from home by subtracting the distance walked east from the distance walked west:
We convert the fractions to a common denominator to subtract them. The least common multiple of the denominators \( 3 \) and \( 4 \) is \( 12 \).
Convert \( \frac{47}{3} \) to twelfths: \[ \frac{47}{3} = \frac{47 \times 4}{3 \times 4} = \frac{188}{12} \]
Convert \( \frac{87}{4} \) to twelfths: \[ \frac{87}{4} = \frac{87 \times 3}{4 \times 3} = \frac{261}{12} \]
Now we can subtract the distances: \[ \text{Net distance} = \frac{261}{12} - \frac{188}{12} = \frac{261 - 188}{12} = \frac{73}{12} \]
Next, we convert \( \frac{73}{12} \) to a mixed number: \[ 73 \div 12 = 6 \quad \text{(with a remainder of 1)} \] Thus, \[ \frac{73}{12} = 6 \frac{1}{12} \]
Therefore, Joey was \( 6 \frac{1}{12} \) meters from home.
Joey was \( 6 \frac{1}{12} \) meters from home.