Solve the following equation and verify your answer:

$\frac{x^{2}-(x+1)(x+2)}{5 x+1}=6$

or $\frac{x^{2}-x^{2}-2 x-x-2}{5 x+1}=6$

or $\frac{-3 x-2}{5 x+1}=6$

or $30 \mathrm{x}+6=-3 \mathrm{x}-2[$ After c ross multipl ication $]$

or $30 \mathrm{x}+3 \mathrm{x}=-2-6$

or $33 \mathrm{x}=-8$ or $\mathrm{x}=\frac{-8}{33}$

Thus, $\mathrm{x}=\frac{-8}{33}$ is the solution of the given equation.

Check :

Substituting $\mathrm{x}=\frac{-8}{33}$ in the given equation, we get :

L.H.S. $=\frac{\left(\frac{-8}{33}\right)^{2}-\left(\frac{-8}{33}+1\right)\left(\frac{-8}{33}+2\right)}{5\left(\frac{8}{33}\right)+1}=\frac{\frac{64}{1089}-\frac{25}{33} \times \frac{58}{33}}{\frac{-40}{33}+1}=\frac{\frac{64}{1089}-\frac{1450}{1089}}{\frac{-7}{33}}=\frac{\frac{-1386}{1089}}{\frac{-7}{33}}=\frac{42}{7}=$ R. H. S. $=6$

$\therefore$ L.H.S. $=$ R.H.S. for $\mathrm{x}=\frac{-8}{33}$