Solve the following equation and verify your answer:

$\frac{\mathrm{y}-(7-8 \mathrm{y})}{9 \mathrm{y}-(3+4 \mathrm{y})}=\frac{2}{3}$

or $\frac{9 \mathrm{y}-7}{5 \mathrm{y}-3}=\frac{2}{3}$

or $27 \mathrm{y}-21=10 \mathrm{y}-6$ [After c ross multipl ication]

or $27 \mathrm{y}-10 \mathrm{y}=-6+21$

or $17 \mathrm{y}=15$

or $\mathrm{y}=\frac{15}{17}$

Thus, $\mathrm{y}=\frac{10}{17}$ is the solution of the given equation.

Check :

Substituting $\mathrm{y}=\frac{15}{17}$ in the given equation, we get:

L. H.S. $=\frac{9\left(\frac{15}{17}\right)-7}{5\left(\frac{15}{17}\right)-3}=\frac{135-119}{75-51}=\frac{16}{24}=\frac{2}{3}$

R.H.S. $=\frac{2}{3}$

$\therefore \mathrm{L} . \mathrm{H} . \mathrm{S} .=\mathrm{R} . \mathrm{H} . \mathrm{S} .$ for $\mathrm{y}=\frac{15}{17}$