We note first that division by zero is not allowed and that for this reason \(8x-4\) may not be equal to zero and that therefore \(x={{1}\over{2}}\) is not a solution.

We now distinguish two cases, namely \(8x-4>0\) and \(8x-4<0\).
In both cases we multiply the inequality on both sides by \(8x-4\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(8x-4>0\), i.e. \(x> {{1}\over{2}}\). Then we get \(4<8(8x-4)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-64x<-36\).
Then, dvision by the coefficient of \(x\)gives \(x > {{9}\over{16}}\).
So we have the following system of inequalities: \(x> {{1}\over{2}}\,\wedge\; x > {{9}\over{16}}\)
and this simplifies to \(x\gt{{9}\over{16}}\).

Suppose \(8x-4<0\), i.e. \(x< {{1}\over{2}}\). Then we get \(4>8(8x-4)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-64x>-36\).
Then, division by the coefficient of \(x\) gives \(x < {{9}\over{16}}\).
So we have the following system of inequalities: \(x< {{1}\over{2}}\,\wedge\; x < {{9}\over{16}}\)
and this simplifies to \(x\lt {{1}\over{2}}\).

The solution of the original inequality is \(x\lt {{1}\over{2}}\;\vee\;x\gt{{9}\over{16}}\).