Question:
Evaluate $\left\{\left(\frac{4}{3}\right)^{-1}-\left(\frac{1}{4}\right)^{-1}\right\}^{-1}$
Solution:
$\left\{\left(\frac{4}{3}\right)^{-1}-\left(\frac{1}{4}\right)^{-1}\right\}^{-1}=\left\{\left(\frac{3}{4}\right)^{1}-\left(\frac{4}{1}\right)^{1}\right\}^{-1}=\left\{\left(\frac{3}{4}\right)-\left(\frac{4}{1}\right)\right\}^{-1}$
The L.C.M. of 4 and 1 is 4 .
$\therefore\left\{\left(\frac{3 \times 1}{4 \times 1}\right)-\left(\frac{4 \times 4}{1 \times 4}\right)\right\}^{-1}$
$=\left\{\frac{3}{4}-\frac{16}{4}\right\}^{-1}=\left\{\frac{3-16}{4}\right\}^{-1}=\left\{\frac{-13}{4}\right\}^{-1}=\left\{\frac{4}{-13}\right\}^{1}=\frac{4}{-13}$
$=\frac{4 \times-1}{-13 \times-1}=\frac{-4}{13}$