We note first that division by zero is not allowed and that for this reason \(5x-2\) may not be equal to zero and that therefore \(x={{2}\over{5}}\) is not a solution.

We now distinguish two cases, namely \(5x-2>0\) and \(5x-2<0\).
In both cases we multiply the inequality on both sides by \(5x-2\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(5x-2>0\), i.e. \(x> {{2}\over{5}}\). Then we get \(9<-4(5x-2)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(20x<-1\).
Then, dvision by the coefficient of \(x\)gives \(x < -{{1}\over{20}}\).
So we have the following system of inequalities: \(x> {{2}\over{5}}\,\wedge\; x < -{{1}\over{20}}\)
and this simplifies to \(\text{an empty solution set}\).

Suppose \(5x-2<0\), i.e. \(x< {{2}\over{5}}\). Then we get \(9>-4(5x-2)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(20x>-1\).
Then, division by the coefficient of \(x\) gives \(x > -{{1}\over{20}}\).
So we have the following system of inequalities: \(x< {{2}\over{5}}\,\wedge\; x > -{{1}\over{20}}\)
and this simplifies to \(-{{1}\over{20}}\lt x\lt {{2}\over{5}}\).

The solution of the original inequality is \(-{{1}\over{20}}\lt x\lt {{2}\over{5}}\).