Question

Solve each of the following equation and also check your result in each case:
$\frac{2}{3x}-\frac{3}{2x}=\frac{1}{12}$

Answer 1

$$\begin{gathered} \frac{2}{3\mathrm{x}}-\frac{3}{2\mathrm{x}}=\frac{1}{12} \\ \mathrm{or}\frac{4-9}{6\mathrm{x}}=\frac{1}{12} \\ \mathrm{or}\frac{-5}{6\mathrm{x}}=\frac{1}{12} \\ \mathrm{or}6\mathrm{x}=-60 \\ \mathrm{or}\mathrm{x}=\frac{-60}{6} \\ \mathrm{or}\mathrm{x}=-10 \\ \mathrm{Thus},\mathrm{x}=-10\mathrm{is}\mathrm{the}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting}\mathrm{x}=-10\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{equation},\mathrm{we}\mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{2}{3\times (-10)}-\frac{3}{2\times (-10)}=\frac{2}{-30}-\frac{3}{-20}=\frac{-4+9}{60}=\frac{5}{60}=\frac{1}{12} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{1}{12} \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{for}\mathrm{x}=-10. \end{gathered}$$