Question

Multiply:

(i) $\frac{3}{4}\mathrm{by}\frac{5}{7}$
(ii) $\frac{9}{8}\mathrm{by}\frac{32}{3}$
(iii) $\frac{7}{6}\mathrm{by}24$
(iv) $\frac{-2}{3}\mathrm{by}\frac{6}{7}$
(v) $\frac{-12}{5}\mathrm{by}\frac{10}{-3}$
(vi) $\frac{25}{-9}\mathrm{by}\frac{3}{-10}$
(vii) $\frac{-7}{10}\mathrm{by}\frac{-40}{21}$
(viii) $\frac{-36}{5}\mathrm{by}\frac{20}{-3}$
(ix) $\frac{-13}{15}\mathrm{by}\frac{-25}{26}$

Answer 1

$\begin{array}{l}\\ \left(\mathrm{i}\right)\frac{3}{4}\times \frac{5}{7}=\frac{(3\times 5)}{(4\times 7)}=\frac{15}{28}\\ \left(\mathrm{ii}\right)\frac{{\cancel{9}}^3}{{\cancel{8}}_1}\times \frac{\sout{3}{\sout{2}}^4}{{\sout{3}}_1}=\frac{(3\times 4)}{(1\times 1)}=12\\ \left(\mathrm{iii}\right)\frac{7}{{\sout{6}}_1}\times \frac{\sout{2}{\sout{4}}^4}{1}=7\times 4=28\\ \left(\mathrm{iv}\right)\frac{-2}{{\sout{3}}_1}\times \frac{{\sout{6}}^2}{7}=\frac{(-2\times 2)}{7}=\frac{-4}{7}\\ \text{(v) We need a positive denominator.}\\ \text{∴}\frac{10}{-3}\times \frac{-1}{-1}=\frac{-10}{3}\\ \text{=}\frac{-\sout{1}{\sout{2}}^4}{{\sout{5}}_1}\times \frac{-\sout{1}{\sout{0}}^2}{{\sout{3}}_1}\\ =\left(-4\right)\times \left(-2\right)\\ =8\\ \end{array}$
$\begin{array}{l}\left(\mathrm{vi}\right)\frac{\sout{2}{\sout{5}}^5}{-{\sout{9}}_3}\times \frac{{\sout{3}}^1}{-\sout{1}{\sout{0}}_2}=\frac{5}{3}\times \frac{1}{2}=\frac{5}{6}\\ \left(\mathrm{vii}\right)\frac{-{\sout{7}}^1}{\sout{1}{\sout{0}}_1}\times \frac{-\sout{4}{\sout{0}}^4}{\sout{2}{\sout{1}}_3}=\frac{4}{3}\\ \left(\mathrm{viii}\right)\frac{-\sout{3}{\sout{6}}^{12}}{{\sout{5}}_1}\times \frac{\sout{2}{\sout{0}}^4}{-{\sout{3}}_1}=12\times 4=48\\ \left(\mathrm{ix}\right)\frac{-\sout{1}{\sout{3}}^1}{\sout{1}{\sout{5}}_3}\times \frac{-\sout{2}{\sout{5}}^5}{\sout{2}{\sout{6}}_2}=\frac{-1}{3}\times \frac{-5}{2}=\frac{5}{6}\end{array}$