5 wholes and 2 thirds minus 3 wholes and 1 sixth

1 answer

To perform the subtraction of \(5\) wholes and \(2\) thirds minus \(3\) wholes and \(1\) sixth, we can first convert everything into improper fractions.

  1. Convert the whole numbers and fractions to improper fractions:

    • \(5\) wholes and \(2\) thirds can be converted as follows: \[ 5 = \frac{5 \times 3}{3} = \frac{15}{3} \] So, \(5\) wholes and \(2\) thirds is: \[ 5 \frac{2}{3} = \frac{15}{3} + \frac{2}{3} = \frac{17}{3} \]

    • \(3\) wholes and \(1\) sixth can be converted as follows: \[ 3 = \frac{3 \times 6}{6} = \frac{18}{6} \] So, \(3\) wholes and \(1\) sixth is: \[ 3 \frac{1}{6} = \frac{18}{6} + \frac{1}{6} = \frac{19}{6} \]

  2. Now we need to subtract \(\frac{19}{6}\) from \(\frac{17}{3}\). To do this, we need a common denominator. The least common multiple of \(3\) and \(6\) is \(6\).

  3. Convert \(\frac{17}{3}\) to have a denominator of \(6\): \[ \frac{17}{3} = \frac{17 \times 2}{3 \times 2} = \frac{34}{6} \]

  4. Now perform the subtraction: \[ \frac{34}{6} - \frac{19}{6} = \frac{34 - 19}{6} = \frac{15}{6} \]

  5. Simplify \(\frac{15}{6}\): \[ \frac{15}{6} = \frac{5}{2} \]

Thus, the final result of \(5\) wholes and \(2\) thirds minus \(3\) wholes and \(1\) sixth is: \[ \frac{5}{2} \quad \text{or} \quad 2 \frac{1}{2} \text{ (as a mixed number)} \]