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

We now distinguish two cases, namely \(2x-1>0\) and \(2x-1<0\).
In both cases we multiply the inequality on both sides by \(2x-1\) because we then get a linear inequality, for which we know there is a solution method.

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

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

The solution of the original inequality is \({{1}\over{10}}\lt x\lt {{1}\over{2}}\).