To solve the expression "2 wholes and 1 third minus 1 whole and 2 fifths," we can break it down step-by-step.
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Convert to improper fractions:
- \(2\) wholes and \(1\) third can be written as: \[ 2 + \frac{1}{3} = \frac{6}{3} + \frac{1}{3} = \frac{7}{3} \]
- \(1\) whole and \(2\) fifths can be written as: \[ 1 + \frac{2}{5} = \frac{5}{5} + \frac{2}{5} = \frac{7}{5} \]
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Subtract the fractions: \[ \frac{7}{3} - \frac{7}{5} \]
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Find a common denominator:
The least common multiple of \(3\) and \(5\) is \(15\).Convert both fractions:
- \(\frac{7}{3} = \frac{7 \times 5}{3 \times 5} = \frac{35}{15}\)
- \(\frac{7}{5} = \frac{7 \times 3}{5 \times 3} = \frac{21}{15}\)
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Perform the subtraction: \[ \frac{35}{15} - \frac{21}{15} = \frac{35 - 21}{15} = \frac{14}{15} \]
So, \(2\) wholes and \(1\) third minus \(1\) whole and \(2\) fifths equals \(\frac{14}{15}\).