Solve the following equation and verify your answer:

Solution:

$\frac{x+3}{x-3}+\frac{x+2}{x-2}=2$

or $\frac{x+3}{x-3}=2-\frac{x+2}{x-2}$

or $\frac{x+3}{x-3}=\frac{2 x-4-x-2}{x-2}$

or $\frac{x+3}{x-3}=\frac{x-6}{x-2}$

or $x^{2}-2 x+3 x-6=x^{2}-3 x-6 x+18$ [After c ross multiplication]

or $\mathrm{x}^{2}-\mathrm{x}^{2}+\mathrm{x}+9 \mathrm{x}=18+6$

or $10 \mathrm{x}=24$

or $\mathrm{x}=\frac{24}{10}$

or $\mathrm{x}=\frac{12}{5}$

Thus, $\mathrm{x}=\frac{12}{5}$ is the solution of the given equation.

Check :

Substituting $\mathrm{x}=\frac{12}{5}$ in the given equation, we get :

L. H.S. $=\frac{\frac{12}{5}+3}{\frac{12}{5}-3}+\frac{\frac{12}{5}+2}{\frac{12}{5}-2}=\frac{12+15}{12-15}+\frac{12+10}{12-10}=\frac{27}{-3}+\frac{22}{2}=\frac{54-66}{-6}=\frac{-12}{-6}=2$

R.H.S. $=2$

$\therefore \mathrm{L} . \mathrm{H} . \mathrm{S} .=\mathrm{R} . \mathrm{H} . \mathrm{S} .$ for $\mathrm{x}=\frac{12}{5}$