Given inequality is,
$\frac{6x}{4x-1} <\frac{1}{2} \Rightarrow \frac{12x}{4x-1}<1$
$ \Rightarrow \frac{12x}{4x-1} -1 < 0 \Rightarrow \frac{8x+1}{4x-1} <0$ ...(i)
For above inequality (i), we have two cases either $(I) \: 8x + 1 > 0, 4x - 1 < 0$
or $(II) \: 8x + 1 < 0, 4x - 1 > 0$
From (I), we get $x > \frac{-1}{8} , x < \frac{1}{4} $
$ \therefore $ $ x \in \left(- \frac{1}{8} ,\frac{1}{4} \right) $ ....(ii)
From (II), $x < \frac{-1}{8} , x > \frac{1}{4}$ , both of which can not be true
$ \therefore $ (ii) represent the solution set of given inequality.