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

$\frac{7 \mathrm{x}-1}{4}-\frac{1}{3}\left(2 \mathrm{x}-\frac{1-\mathrm{x}}{2}\right)=\frac{10}{3}$

or $\frac{7 \mathrm{x}-1}{4}-\frac{2 \mathrm{x}}{3}+\frac{1-\mathrm{x}}{6}=\frac{10}{3}$

or $\frac{21 \mathrm{x}-3-8 \mathrm{x}+2-2 \mathrm{x}}{12}=\frac{10}{3}$

or $11 \mathrm{x}-1=40[$ Multiply ing both sides by 12$]$

or $11 \mathrm{x}=40+1$

or $\mathrm{x}=\frac{41}{11}$

Thus, $\mathrm{x}=\frac{41}{11}$ is the solution of the given equation.

Check:

Substituting $\mathrm{x}=\frac{41}{11}$ in the given equation, we get:

L. H.S. $=\frac{7 \times \frac{41}{11}-1}{4}-\frac{1}{3}\left(2 \times \frac{41}{11}-\frac{1-\frac{41}{11}}{2}\right)=\frac{276}{44}-\frac{82}{33}+\frac{-30}{66}=\frac{10}{3}$

R.H.S. $=\frac{10}{3}$

$\therefore$ L. H.S. $=$ R.H. S. for $\mathrm{x}=\frac{41}{11}$