We note first that division by zero is not allowed and that for this reason \(6x-4\) may not be equal to zero and that therefore \(x={{2}\over{3}}\) is not a solution.

We now distinguish two cases, namely \(6x-4>0\) and \(6x-4<0\).
In both cases we multiply the inequality on both sides by \(6x-4\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(6x-4>0\), i.e. \(x> {{2}\over{3}}\). Then we get \(1<8(6x-4)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-48x<-33\).
Then, dvision by the coefficient of \(x\)gives \(x > {{11}\over{16}}\).
So we have the following system of inequalities: \(x> {{2}\over{3}}\,\wedge\; x > {{11}\over{16}}\)
and this simplifies to \(x\gt{{11}\over{16}}\).

Suppose \(6x-4<0\), i.e. \(x< {{2}\over{3}}\). Then we get \(1>8(6x-4)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-48x>-33\).
Then, division by the coefficient of \(x\) gives \(x < {{11}\over{16}}\).
So we have the following system of inequalities: \(x< {{2}\over{3}}\,\wedge\; x < {{11}\over{16}}\)
and this simplifies to \(x\lt {{2}\over{3}}\).

The solution of the original inequality is \(x\lt {{2}\over{3}}\;\vee\;x\gt{{11}\over{16}}\).