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

$5\left(\frac{7 x+5}{3}\right)-\frac{23}{3}=13-\frac{4 x-2}{3}$

or $\frac{35 x+25}{3}+\frac{4 x-2}{3}=13+\frac{23}{3}$

or $\frac{35 x+25+4 x-2}{3}=\frac{39+23}{3}$

or $39 \mathrm{x}+23=62[$ Multiply ing both sides by 3$]$ or $39 \mathrm{x}=62-23$

or $\mathrm{x}=\frac{39}{39}$

or $\mathrm{x}=1$

Thus, $x=1$ is the solution of the given equation. Check:

Substituting $\mathrm{x}=1$ in the given equation, we get:

L.H.S. $=5\left(\frac{7 \times 1+5}{3}\right)-\frac{23}{3}=\frac{60}{3}-\frac{23}{3}=\frac{37}{3}$

R.H.S. $=13-\frac{4 \times 1-2}{3}=\frac{39-2}{3}=\frac{37}{3}$

$\therefore \mathrm{L} . \mathrm{H} . \mathrm{S} .=\mathrm{R} . \mathrm{H} . \mathrm{S} .$ for $\mathrm{x}=1$