Solve the following

$\frac{4 x+7}{9-3 x}=\frac{1}{4}$

$\Rightarrow 4(4 x+7)=1(9-3 x) \quad$ (by cross multiplication)

$\Rightarrow 16 x+28=9-3 x$

$\Rightarrow 16 x+3 x=9-28$

$\Rightarrow 19 x=-19$

$\Rightarrow x=\frac{-19}{19}=-1$

$\therefore x=-1$