We note first that division by zero is not allowed and that for this reason \(5x-3\) may not be equal to zero and that therefore \(x={{3}\over{5}}\) is not a solution.

We now distinguish two cases, namely \(5x-3>0\) and \(5x-3<0\).
In both cases we multiply the inequality on both sides by \(5x-3\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(5x-3>0\), i.e. \(x> {{3}\over{5}}\). Then we get \(3<5(5x-3)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-25x<-18\).
Then, dvision by the coefficient of \(x\)gives \(x > {{18}\over{25}}\).
So we have the following system of inequalities: \(x> {{3}\over{5}}\,\wedge\; x > {{18}\over{25}}\)
and this simplifies to \(x\gt{{18}\over{25}}\).

Suppose \(5x-3<0\), i.e. \(x< {{3}\over{5}}\). Then we get \(3>5(5x-3)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-25x>-18\).
Then, division by the coefficient of \(x\) gives \(x < {{18}\over{25}}\).
So we have the following system of inequalities: \(x< {{3}\over{5}}\,\wedge\; x < {{18}\over{25}}\)
and this simplifies to \(x\lt {{3}\over{5}}\).

The solution of the original inequality is \(x\lt {{3}\over{5}}\;\vee\;x\gt{{18}\over{25}}\).