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

We now distinguish two cases, namely \(8x-1>0\) and \(8x-1<0\).
In both cases we multiply the inequality on both sides by \(8x-1\) because we then get a linear inequality, for which we know there is a solution method.

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

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

The solution of the original inequality is \(x\lt {{1}\over{8}}\;\vee\;x\gt{{1}\over{4}}\).