Solve:

(i) $2-\frac{3}{5}$

(ii) $4+\frac{7}{8}$

(iii) $\frac{3}{5}+\frac{2}{7}$

(iv) $\frac{9}{11}-\frac{4}{15}$

(v) $\frac{7}{10}+\frac{2}{5}+\frac{3}{2}$

(vi) $2\frac{2}{3}+3\frac{1}{2}$

(vii) $8\frac{1}{2}-3\frac{5}{8}$

(i) $2-\frac{3}{5}=\frac{2\times 5}{5}-\frac{3}{5}=\frac{10-3}{5}=\frac{7}{5}$

(ii) $4+\frac{7}{8}=\frac{4\times 8}{8}+\frac{7}{8}=\frac{32+7}{8}=\frac{39}{8}=4\frac{7}{8}$

(iii) $\frac{3}{5}+\frac{2}{7}=\frac{3\times 7}{5\times 7}+\frac{2\times 5}{7\times 5}=\frac{21+10}{35}=\frac{31}{35}$

(iv) $\frac{9}{11}-\frac{4}{15}=\frac{9\times 15}{11\times 15}-\frac{4\times 11}{15\times 11}=\frac{135-44}{165}=\frac{91}{165}$

(v) $\frac{7}{10}+\frac{2}{5}+\frac{3}{2}=\frac{7}{10}+\frac{2\times 2}{5\times 2}+\frac{3\times 5}{2\times 5}=\frac{7+4+15}{10}=\frac{26}{10}=\frac{13}{5}=2\frac{3}{5}$

(vi) $2\frac{2}{3}+3\frac{1}{2}=\frac{8}{3}+\frac{7}{2}=\frac{8\times 2}{3\times 2}+\frac{7\times 3}{2\times 3}=\frac{16+21}{6}=\frac{37}{6}=6\frac{1}{6}$

(vii) $8\frac{1}{2}-3\frac{5}{8}=\frac{17}{2}-\frac{29}{8}=\frac{17\times 4}{2\times 4}-\frac{29}{8}=\frac{68-29}{8}=\frac{39}{8}=4\frac{7}{8}$