Question:

$\frac{9 x-7}{3 x+5}=\frac{3 x-4}{x+6}$

Solution:

$\frac{9 x-7}{3 x+5}=\frac{3 x-4}{x+6}$

or $9 \mathrm{x}^{2}-7 \mathrm{x}+54 \mathrm{x}-42=9 \mathrm{x}^{2}-12 \mathrm{x}+15 \mathrm{x}-20$ [After $c$ ross multiplication]

or $9 x^{2}-9 x^{2}+47 x-3 x=-20+42$

or $44 \mathrm{x}=22$

or $\mathrm{x}=\frac{22}{44}$

or $\mathrm{x}=\frac{1}{2}$

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

Check:

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

L. H.S. $=\frac{9\left(\frac{1}{2}\right)-7}{3\left(\frac{1}{2}\right)+5}=\frac{9-14}{3+10}=\frac{-5}{13}$

R. H.S. $=\frac{3\left(\frac{1}{2}\right)-4}{\frac{1}{2}+6}=\frac{3-8}{1+12}=\frac{-5}{13}$

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