Question

Solve each of the following equation and also check your result in each case:
$\frac{(1-2x)}{7}-\frac{(2-3x)}{8}=\frac{3}{2}+\frac{x}{4}$

Answer 1

$$\begin{gathered} \frac{1-2\mathrm{x}}{7}-\frac{2-3\mathrm{x}}{8}=\frac{3}{2}+\frac{\mathrm{x}}{4} \\ \mathrm{or}\frac{1-2\mathrm{x}}{7}=\frac{3}{2}+\frac{\mathrm{x}}{4}+\frac{2-3\mathrm{x}}{8} \\ \mathrm{or}\frac{1-2\mathrm{x}}{7}=\frac{12+2\mathrm{x}+2-3\mathrm{x}}{8} \\ \mathrm{or}\frac{1-2\mathrm{x}}{7}=\frac{14-x}{8} \\ \mathrm{or}8-16\mathrm{x}=98-7x \\ \mathrm{or}-16\mathrm{x}+7\mathrm{x}=98-8 \\ \mathrm{or}\mathrm{x}=\frac{-90}{9} \\ =-10 \\ \mathrm{Check}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{1-2\times \left(-10\right)}{7}-\frac{2-3\times \left(-10\right)}{8}=\frac{1+20}{7}-\frac{2+30}{8}=3-4=-1 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{3}{2}+\frac{-10}{4}=\frac{3}{2}+\frac{-5}{2}=\frac{3-5}{2}=-1 \end{gathered}$$

$\therefore$ L.H.S. = R.H.S. for x = $-$10