Solve the following equation and verify your answer:

$\frac{7 x-2}{5 x-1}=\frac{7 x+3}{5 x+4}$

or $35 \mathrm{x}^{2}+28 \mathrm{x}-10 \mathrm{x}-8=35 \mathrm{x}^{2}+15 \mathrm{x}-7 \mathrm{x}-3[$ After cross multiplication $]$

or $35 x^{2}-35 x^{2}+18 x-8 x=-3+8$

or $10 x=5$

or $x=\frac{5}{10}$ or $x=\frac{1}{2}$

Thus, $x=\frac{1}{2}$ is the solution of the given equation.

Check:

Substituting $x=\frac{1}{2}$ in the given equation, we get:

L. H.S. $=\frac{7\left(\frac{1}{2}\right)-2}{5\left(\frac{1}{2}\right)-1}=\frac{7-4}{5-2}=\frac{3}{3}=1$

R.H.S. $=\frac{7\left(\frac{1}{2}\right)+3}{5\left(\frac{1}{2}\right)+4}=\frac{7+6}{5+8}=\frac{13}{13}=1$

$\therefore \mathrm{L} . \mathrm{H} . \mathrm{S} .=\mathrm{R} . \mathrm{H} . \mathrm{S} .$ for $\mathrm{x}=\frac{1}{2}$