Question:
$\frac{9-3 y}{1-9 y}=\frac{8}{5}$
Solution:
Given, $\frac{9-3 y}{1-9 y}=\frac{8}{5}$
$\Rightarrow$ $5(9-3 y)=8(1-9 y)$ [by cross-multiplication]
$\Rightarrow \quad 45-15 y=8-72 y$
$\Rightarrow \quad 72 y-15 y=8-45 \quad$ [transposing $-72 y$ to LHS and 45 to RHS]
$\Rightarrow \quad 57 y=-37$
$\Rightarrow$ $\frac{57 y}{57}=\frac{-37}{57}$ [dividing both sides by 57 ]
$\therefore$ $y=\frac{-37}{57}$