Solve the following

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

$\Rightarrow \frac{30-15 y-5 y-30}{1-3 y}=10$

$\Rightarrow \frac{-20 y}{1-3 y}=10$

$\Rightarrow 1(-20 y)=10(1-3 y)$     (by cross multiplication)

$\Rightarrow-20 y=10-30 y$

$\Rightarrow-20 y+30 y=10$

$\Rightarrow 10 y=10$

$\Rightarrow y=\frac{10}{10}=1$

$\therefore y=1$