Question
Simplify: $\big(3^{-1}+6^{-1}\big)\div\Big(\frac{3}{4}\Big)^{-1}$
✓
Answer
$\big(3^{-1}+6^{-1}\big)\div\Big(\frac{3}{4}\Big)^{-1}$ $=\Big(\frac{1}{3}+\frac{1}{6}\Big)\div\Big(\frac{4}{3}\Big)^1$ $=\bigg(\Big[\frac{1\times2}{3\times2}\Big]+\Big[\frac{1\times1}{6\times1}\Big]\bigg)\div\Big(\frac{4}{3}\Big)$ $=\Big(\frac{2+1}{6}\Big)\div\Big(\frac{4}{3}\Big)$ $=\Big(\frac{3}{6}\Big)\div\Big(\frac{4}{3}\Big)$ $=\Big(\frac{1}{2}\Big)\div\Big(\frac{4}{3}\Big)$ $=\Big(\frac{1}{2}\Big)\times\Big(\frac{3}{4}\Big)$ $=\Big(\frac{3}{8}\Big)$ $\therefore\big(3^{-1}+6^{-1}\big)\div\Big(\frac{3}{4}\Big)^{-1}=\frac{3}{8}$
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