Solve the following equation and verify your answer:
$\frac{y-(7-8 y)}{9 y-(3+4 y)}=\frac{2}{3}$
$\frac{\mathrm{y}-(7-8 \mathrm{y})}{9 \mathrm{y}-(3+4 \mathrm{y})}=\frac{2}{3}$
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 $\mathrm{y}=\frac{15}{17}$
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