What number should be subtracted from the sum of 2 \(\frac{1}{6}\) and 2\(\frac{7}{12}\) to give 3\(\frac{1}{4}\)?

Explanation

The sum of 2 \(\frac{1}{6}\) and 2\(\frac{7}{12}\)

= \(\frac{13}{6}\) + \(\frac{31}{12}\)

=  \(\frac{13 \times 2 + 31}{12}\) 

= \(\frac{26 + 31}{12}\)

= \(\frac{57}{12}\)

What should be subtracted from \(\frac{57}{12}\) to give 3\(\frac{1}{4}\)

\(\frac{57}{12}\) - y =  3\(\frac{1}{4}\)

: y = \(\frac{57}{12}\) - 3\(\frac{1}{4}\) = \(\frac{57}{12}\) - \(\frac{13}{4}\)

y = \(\frac{57 - 3 \times 13}{12}\) =  \(\frac{57 - 39}{12}\) 

y =  \(\frac{18}{12}\)

y =  \(\frac{3}{2}\) or 1\(\frac{1}{2}\)