Question

Use the distributivity of multiplication of rational numbers over their addition to simplify:

(i) $\frac{3}{5}\times (\frac{35}{24}+\frac{10}{1})$

(ii)$\frac{-5}{4}\times (\frac{8}{5}+\frac{16}{5})$

(iii)$\frac{2}{7}\times (\frac{7}{16}-\frac{21}{4})$

(iv)$\frac{3}{4}\times (\frac{8}{9}-40)$

Answer 1

(i) $\frac{3}{5}\times (\frac{35}{24}+\frac{10}{1})=\frac{3}{5}\times \frac{35}{24}+\frac{3}{5}\times \frac{10}{1}=\frac{3\times 35}{5\times 24}+\frac{3\times 10}{5\times 1}=\frac{1\times 7}{1\times 8}+\frac{3\times 2}{1\times 1}=\frac{7}{8}+6=\frac{7}{8}+\frac{48}{1}=\frac{7+48}{8}=\frac{55}{8}$

(ii)$\frac{-5}{4}\times (\frac{8}{5}+\frac{16}{5})=\frac{-5}{4}\times \frac{8}{5}+\frac{-5}{4}\times \frac{16}{5}=\frac{-5\times 8}{4\times 5}+\frac{-5\times 16}{4\times 5}=\frac{-1\times 2}{1\times 1}+\frac{-1\times 4}{1\times 1}=-2-4=-6$

(iii)$\frac{2}{7}\times (\frac{7}{16}-\frac{21}{4})=\frac{2}{7}\times \frac{7}{16}-\frac{2}{7}\times \frac{21}{4}=\frac{2\times 7}{7\times 16}-\frac{2\times 21}{7\times 4}=\frac{1\times 1}{1\times 8}-\frac{1\times 3}{1\times 2}=\frac{1}{8}-\frac{3}{2}=\frac{1-12}{8}=\frac{-11}{8}$

(iv)$\frac{3}{4}\times (\frac{8}{9}-40)=\frac{3}{4}\times \frac{8}{9}-\frac{3}{4}\times \frac{40}{1}=\frac{3\times 8}{4\times 9}-\frac{3\times 40}{4\times 1}=\frac{1\times 2}{1\times 3}-\frac{3\times 10}{1\times 1}=\frac{2}{3}-\frac{30}{1}=\frac{2-90}{3}=\frac{-88}{3}$