\( \displaystyle2{1 \over 3} = { 2 \times 3 + 1 \over 3}= {7 \over 3}\) ;
\( \displaystyle4{2 \over 5} = { 4 \times 5 + 2 \over 5}= {{22} \over 5}\) ;
\( \displaystyle3{1 \over 4} = { 3 \times 4 + 1 \over 4} ={{13} \over 4}\) ;
\( \displaystyle9{5 \over 7} = { 9 \times 7 + 5 \over 3}= {{68} \over 7}\) ;
\( \displaystyle10{3 \over {10}} = {10 \times 10 + 3 \over {10} }={{103} \over {10}}\).
