Question

Solve each of the following equation and also check your result in each case:
$\frac{7x-1}{4}-\frac{1}{3}\left(2x-\frac{1-x}{2}\right)=\frac{10}{3}$

Answer 1

$$\begin{gathered} \frac{7\mathrm{x}-1}{4}-\frac{1}{3}(2\mathrm{x}-\frac{1-\mathrm{x}}{2})=\frac{10}{3} \\ \mathrm{or}\frac{7\mathrm{x}-1}{4}-\frac{2\mathrm{x}}{3}+\frac{1-\mathrm{x}}{6}=\frac{10}{3} \\ \mathrm{or}\frac{21\mathrm{x}-3-8\mathrm{x}+2-2\mathrm{x}}{12}=\frac{10}{3} \\ \mathrm{or}11\mathrm{x}-1=40[\mathrm{Multiply}\text{ing}\mathrm{both}\mathrm{sides}\mathrm{by}12] \\ \mathrm{or}11\mathrm{x}=40+1 \\ \mathrm{or}\mathrm{x}=\frac{41}{11} \\ \mathrm{Thus},\mathrm{x}=\frac{41}{11}\mathrm{is}\mathrm{the}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{equation}. \\ \mathrm{Check}: \\ \mathrm{Substituting}\mathrm{x}=\frac{41}{11}\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{equation},\mathrm{we}\mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{7\times {\displaystyle \frac{41}{11}}-1}{4}-\frac{1}{3}(2\times \frac{41}{11}-\frac{1-{\displaystyle \frac{41}{11}}}{2})=\frac{276}{44}-\frac{82}{33}+\frac{-30}{66}=\frac{10}{3} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{10}{3} \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{for}\mathrm{x}=\frac{41}{11} \end{gathered}$$