Solve each of the following equation and also check your result in each case:

$(3 \mathrm{x}-8)(3 \mathrm{x}+2)-(4 \mathrm{x}-11)(2 \mathrm{x}+1)=(\mathrm{x}-3)(\mathrm{x}+7)$

or $9 \mathrm{x}^{2}+6 \mathrm{x}-24 \mathrm{x}-16-8 \mathrm{x}^{2}-4 \mathrm{x}+22 \mathrm{x}+11=\mathrm{x}^{2}+7 \mathrm{x}-3 \mathrm{x}-21$

or $\mathrm{x}^{2}-5=\mathrm{x}^{2}+4 \mathrm{x}-21$

or $4 \mathrm{x}=-5+21$

or $\mathrm{x}=\frac{16}{4}=4$

Thus, $x=4$ is the solution of the given equation. Check :

Substituting $x=4$ in the given equation, we get:

L. H.S. $=(3 \times 4-8)(3 \times 4+2)-(4 \times 4-11)(2 \times 4+1)=4 \times 14-5 \times 9=11$

R. H.S. $=(4-3)(4+7)=11$

$\therefore$ L.H.S. $=$ R. H.S. for $\mathrm{x}=4$.