Question

Simplify: $(3^{-1}+6^{-1})÷{\left(\frac{3}{4}\right)}^{-1}.$

Answer 1

$$\begin{gathered} \left(3^{-1}+6^{-1}\right)÷{\left(\frac{3}{4}\right)}^{-1} \\ =\left(\frac{1}{3}+\frac{1}{6}\right)÷{\left(\frac{4}{3}\right)}^1 \\ =\left(\left[{\displaystyle \frac{1\times 2}{3\times 2}}\right]+\left[\frac{1\times 1}{6\times 1}\right]\right)÷\left(\frac{4}{3}\right) \\ =\left(\frac{2+1}{6}\right)÷\left(\frac{4}{3}\right) \\ =\left(\frac{3}{6}\right)÷\left(\frac{4}{3}\right) \\ =\left(\frac{1}{2}\right)÷\left(\frac{4}{3}\right) \\ =\left(\frac{1}{2}\right)\times \left(\frac{3}{4}\right) \\ =\frac{3}{8} \end{gathered}$$

$\therefore \left(3^{-1}+6^{-1}\right)÷{\left(\frac{3}{4}\right)}^{-1}=\frac{3}{8}$