We note first that division by zero is not allowed and that for this reason \(6x-2\) may not be equal to zero and that therefore \(x={{1}\over{3}}\) is not a solution.

We now distinguish two cases, namely \(6x-2>0\) and \(6x-2<0\).
In both cases we multiply the inequality on both sides by \(6x-2\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(6x-2>0\), i.e. \(x> {{1}\over{3}}\). Then we get \(5<-6(6x-2)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(36x<7\).
Then, dvision by the coefficient of \(x\)gives \(x < {{7}\over{36}}\).
So we have the following system of inequalities: \(x> {{1}\over{3}}\,\wedge\; x < {{7}\over{36}}\)
and this simplifies to \(\text{an empty solution set}\).

Suppose \(6x-2<0\), i.e. \(x< {{1}\over{3}}\). Then we get \(5>-6(6x-2)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(36x>7\).
Then, division by the coefficient of \(x\) gives \(x > {{7}\over{36}}\).
So we have the following system of inequalities: \(x< {{1}\over{3}}\,\wedge\; x > {{7}\over{36}}\)
and this simplifies to \({{7}\over{36}}\lt x\lt {{1}\over{3}}\).

The solution of the original inequality is \({{7}\over{36}}\lt x\lt {{1}\over{3}}\).