We note first that division by zero is not allowed and that for this reason \(6x-1\) may not be equal to zero and that therefore \(x={{1}\over{6}}\) is not a solution.

We now distinguish two cases, namely \(6x-1>0\) and \(6x-1<0\).
In both cases we multiply the inequality on both sides by \(6x-1\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(6x-1>0\), i.e. \(x> {{1}\over{6}}\). Then we get \(5<-(6x-1)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(6x<-4\).
Then, dvision by the coefficient of \(x\)gives \(x < -{{2}\over{3}}\).
So we have the following system of inequalities: \(x> {{1}\over{6}}\,\wedge\; x < -{{2}\over{3}}\)
and this simplifies to \(\text{an empty solution set}\).

Suppose \(6x-1<0\), i.e. \(x< {{1}\over{6}}\). Then we get \(5>-(6x-1)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(6x>-4\).
Then, division by the coefficient of \(x\) gives \(x > -{{2}\over{3}}\).
So we have the following system of inequalities: \(x< {{1}\over{6}}\,\wedge\; x > -{{2}\over{3}}\)
and this simplifies to \(-{{2}\over{3}}\lt x\lt {{1}\over{6}}\).

The solution of the original inequality is \(-{{2}\over{3}}\lt x\lt {{1}\over{6}}\).