Question

Use the distributivity of multiplication of rational numbers over their addition to simplify:
(i) $\frac{3}{5}\times \left(\frac{35}{24}+\frac{10}{1}\right)$

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

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

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

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\frac{3}{5}\times (\frac{35}{24}+\frac{10}{1})=\frac{3}{5}\times \frac{35}{24}+\frac{3}{5}\times \frac{10}{1}=\frac{7}{8}+\frac{6}{1}=\frac{7+48}{8}=\frac{55}{8} \\ \left(\mathrm{ii}\right)\frac{-5}{4}\times (\frac{8}{5}+\frac{16}{5})=\frac{-5}{4}\times \frac{8}{5}+\frac{-5}{4}\times \frac{16}{5}=\frac{-2}{1}+\frac{-4}{1}=-6 \\ \left(\mathrm{iii}\right)\frac{2}{7}\times (\frac{7}{16}-\frac{21}{4})=\frac{2}{7}\times \frac{7}{16}-\frac{2}{7}\times \frac{21}{4}=\frac{1}{8}-\frac{3}{2}=\frac{1-12}{8}=\frac{-11}{8} \\ \left(\mathrm{iv}\right)\frac{3}{4}\times (\frac{8}{9}-40)=\frac{3}{4}\times \frac{8}{9}-\frac{3}{4}\times 40=\frac{2}{3}-30=\frac{2-90}{3}=\frac{-88}{3} \end{gathered}$$