To solve the problem, we need to first convert the mixed numbers into improper fractions.
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Convert 7 whole and 1 fourth to an improper fraction: \[ 7 \frac{1}{4} = \frac{(7 \times 4) + 1}{4} = \frac{28 + 1}{4} = \frac{29}{4} \]
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Convert 3 whole and 2 thirds to an improper fraction: \[ 3 \frac{2}{3} = \frac{(3 \times 3) + 2}{3} = \frac{9 + 2}{3} = \frac{11}{3} \]
Now, we need to perform the subtraction: \[ \frac{29}{4} - \frac{11}{3} \]
To subtract these fractions, we need a common denominator. The least common multiple of 4 and 3 is 12.
Convert each fraction to have the denominator of 12:
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Convert \(\frac{29}{4}\) to a denominator of 12: \[ \frac{29}{4} = \frac{29 \times 3}{4 \times 3} = \frac{87}{12} \]
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Convert \(\frac{11}{3}\) to a denominator of 12: \[ \frac{11}{3} = \frac{11 \times 4}{3 \times 4} = \frac{44}{12} \]
Now we can subtract: \[ \frac{87}{12} - \frac{44}{12} = \frac{87 - 44}{12} = \frac{43}{12} \]
Finally, we can convert \(\frac{43}{12}\) back into a mixed number: \[ \frac{43}{12} = 3 \frac{7}{12} \quad (\text{since } 43 \div 12 = 3 \text{ R } 7) \]
So, the result of \(7\) wholes and \(1\) fourth minus \(3\) wholes and \(2\) thirds is: \[ 3 \frac{7}{12} \]