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

We now distinguish two cases, namely \(8x-2>0\) and \(8x-2<0\).
In both cases we multiply the inequality on both sides by \(8x-2\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(8x-2>0\), i.e. \(x> {{1}\over{4}}\). Then we get \(6<-5(8x-2)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(40x<4\).
Then, dvision by the coefficient of \(x\)gives \(x < {{1}\over{10}}\).
So we have the following system of inequalities: \(x> {{1}\over{4}}\,\wedge\; x < {{1}\over{10}}\)
and this simplifies to \(\text{an empty solution set}\).

Suppose \(8x-2<0\), i.e. \(x< {{1}\over{4}}\). Then we get \(6>-5(8x-2)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(40x>4\).
Then, division by the coefficient of \(x\) gives \(x > {{1}\over{10}}\).
So we have the following system of inequalities: \(x< {{1}\over{4}}\,\wedge\; x > {{1}\over{10}}\)
and this simplifies to \({{1}\over{10}}\lt x\lt {{1}\over{4}}\).

The solution of the original inequality is \({{1}\over{10}}\lt x\lt {{1}\over{4}}\).