Question

Solve each of the following equation and also check your result in each case:

Answer 1

$$\begin{gathered} \frac{3}{4}\mathrm{x}+4\mathrm{x}=\frac{7}{8}+6\mathrm{x}-6 \\ \mathrm{or}\frac{3}{4}\mathrm{x}-2\mathrm{x}=\frac{7}{8}-6 \\ \mathrm{or}\frac{3\mathrm{x}-8\mathrm{x}}{4}=\frac{7-48}{8} \\ \mathrm{or}\frac{-5\mathrm{x}}{4}=\frac{-41}{8} \\ \mathrm{or}-40\mathrm{x}=-164 \\ \mathrm{or}\mathrm{x}=\frac{-164}{-40}=\frac{41}{10} \\ \mathrm{Check}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{3}{4}\times \frac{41}{10}+4\times \frac{41}{10}=\frac{123}{40}+\frac{164}{10}=\frac{123+656}{40}=\frac{779}{40} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{7}{8}+6\times \frac{41}{10}-6=\frac{7}{8}+\frac{246}{10}-6=\frac{35+984-240}{40}=\frac{779}{40} \end{gathered}$$

$\therefore$ L.H.S. = R.H.S. for x = $\frac{41}{10}$