Solve the following equation and verify your answer:

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

or, $12 x^{2}+20 x+21 x+35=12 x^{2}+16 x+6 x+8 \quad$ [Cross multiply $]$

or, $12 \mathrm{x}^{2}-12 \mathrm{x}^{2}+41 \mathrm{x}-22 \mathrm{x}=8-35$

or, $19 \mathrm{x}=-27$

or, $\mathrm{x}=\frac{-27}{19}$

Thus, $\mathrm{x}=\frac{-27}{19}$ is the solution of the given equation

Check :

Substituting $\mathrm{x}=\frac{-27}{19}$ in the given equation, we get:

L. H.S. $=\frac{3\left(\frac{-27}{19}\right)+5}{4\left(\frac{-27}{19}\right)+2}=\frac{-81+95}{-108+38}=\frac{14}{-70}=\frac{-1}{5}$

R. H.S. $=\frac{3\left(\frac{-27}{19}\right)+4}{4\left(\frac{27}{19}\right)+7}=\frac{-81+76}{-108+133}=\frac{-5}{25}=\frac{-1}{5}$

$\therefore$ L. H.S. $=$ R. H.S. for $\mathrm{x}=\frac{-27}{19}$