Solve the following equation and verify your answer:

$\frac{1-9 y}{19-3 y}=\frac{5}{8}$

or $8-72 \mathrm{y}=95-15 \mathrm{y}[$ After c ross multiplication $]$

or $95-15 \mathrm{y}=8-72 \mathrm{y}$

or $72 \mathrm{y}-15 \mathrm{y}=8-95$

or $57 \mathrm{y}=-87$

or $\mathrm{y}=\frac{-87}{57}$

or $\mathrm{y}=\frac{-29}{19}$

Thus $\mathrm{y}=\frac{-29}{19}$ is the solution of the given equation. Check :

Substituting $\mathrm{y}=\frac{-29}{19}$ in the given equation, we get :

L.H. S. $=\frac{1-9\left(\frac{-29}{19}\right)}{19-3\left(\frac{-29}{19}\right)}=\frac{19+261}{361+87}=\frac{280}{448}=\frac{5}{8}$

R.H. S. $=\frac{5}{8}$

$\therefore$ L.H.S. $=$ R.H. S. for $\mathrm{y}=\frac{-29}{19}$