We note first that division by zero is not allowed and that for this reason \(4x-5\) may not be equal to zero and that therefore \(x={{5}\over{4}}\) is not a solution.

We now distinguish two cases, namely \(4x-5>0\) and \(4x-5<0\).
In both cases we multiply the inequality on both sides by \(4x-5\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(4x-5>0\), i.e. \(x> {{5}\over{4}}\). Then we get \(3<-4(4x-5)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(16x<17\).
Then, dvision by the coefficient of \(x\)gives \(x < {{17}\over{16}}\).
So we have the following system of inequalities: \(x> {{5}\over{4}}\,\wedge\; x < {{17}\over{16}}\)
and this simplifies to \(\text{an empty solution set}\).

Suppose \(4x-5<0\), i.e. \(x< {{5}\over{4}}\). Then we get \(3>-4(4x-5)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(16x>17\).
Then, division by the coefficient of \(x\) gives \(x > {{17}\over{16}}\).
So we have the following system of inequalities: \(x< {{5}\over{4}}\,\wedge\; x > {{17}\over{16}}\)
and this simplifies to \({{17}\over{16}}\lt x\lt {{5}\over{4}}\).

The solution of the original inequality is \({{17}\over{16}}\lt x\lt {{5}\over{4}}\).