Answer:
(i) \[2-\frac{3}{5}=\frac{2}{1}-\frac{3}{5}=\frac{10-3}{5}=\frac{7}{5}=1\frac{2}{5}\] \[\left| \begin{align} & \text{ 5}\overset{1}{\overline{\left){\text{ 7}}\right.}} \\ & \frac{-5}{\underline{2\,\,\,\,\,}} \\ \end{align} \right.\] (ii) \[4+\frac{7}{8}=\frac{4}{1}+\frac{7}{8}=\frac{32+7}{8}=\frac{39}{8}=4\frac{7}{8}\]\[\left| \begin{align} & \text{ 8}\overset{4}{\overline{\left){\text{ 39}}\right.}} \\ & \frac{-32}{\underline{7\,\,\,\,}} \\ \end{align} \right.\] (iii) \[\frac{3}{5}+\frac{2}{7}=\frac{21+10}{35}=\frac{31}{35}\] (iv) \[\frac{9}{11}-\frac{4}{15}=\frac{135-44}{165}=\frac{91}{165}\] (v) \[\frac{7}{10}+\frac{2}{5}+\frac{3}{2}=\frac{7+4+15}{10}=\frac{26}{10}\] \[=\frac{26\div 2}{10\div 2}=\frac{13}{5}=2\frac{3}{5}\] \[\left| \begin{align} & \text{ 5}\overset{2}{\overline{\left){\text{ 13}}\right.}} \\ & \underline{\frac{-10}{3}} \\ \end{align} \right.\] (vi) \[2\frac{2}{3}+3\frac{1}{2}=\frac{(2\times 3)+2}{3}+\frac{(3\times 2)+1}{2}\]\[+\frac{8}{3}+\frac{7}{2}\] \[=\frac{16+21}{6}=\frac{37}{6}=6\frac{1}{6}\] \[\left| \begin{align} & \text{ 6}\overset{6}{\overline{\left){\text{ 37}}\right.}} \\ & \underline{\frac{-36}{1}} \\ \end{align} \right.\] (vii) \[8\frac{1}{2}-3\frac{5}{8}=\frac{(8\times 2)+1}{2}-\frac{(3\times 8)+5}{8}\] \[=\frac{17}{2}-\frac{29}{8}\] \[=\frac{68-29}{8}=\frac{39}{8}=4\frac{7}{8}\] \[\left| \begin{align} & \text{ 8}\overset{4}{\overline{\left){\text{ 39}}\right.}} \\ & \underline{\frac{-32}{7}} \\ \end{align} \right.\]
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