Solve the following equation and verify your answer:

$\frac{15(2-\mathrm{x})-5(\mathrm{x}+6)}{1-3 \mathrm{x}}=10$

or $\frac{30-15 \mathrm{x}-5 \mathrm{x}-30}{1-3 \mathrm{x}}=10$

or $\frac{-20 \mathrm{x}}{1-3 \mathrm{x}}=10$

or $10-30 \mathrm{x}=-20 \mathrm{x}[$ After c ross multiplication $]$

or $-20 \mathrm{x}+30 \mathrm{x}=10$

or $10 \mathrm{x}=10$

or $\mathrm{x}=1$

Thus, $x=1$ is the solution of the given equation. Check :

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

L.H.S. $=\frac{15(2-1)-5(1+6)}{1-3(1)}=\frac{15-35}{-2}=\frac{-20}{-2}=10$

R. H.S. $=10$

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