We note first that division by zero is not allowed and that for this reason \(3x-9\) may not be equal to zero and that therefore \(x=3\) is not a solution.

We now distinguish two cases, namely \(3x-9>0\) and \(3x-9<0\).
In both cases we multiply the inequality on both sides by \(3x-9\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(3x-9>0\), i.e. \(x> 3\). Then we get \(1<-5(3x-9)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(15x<44\).
Then, dvision by the coefficient of \(x\)gives \(x < {{44}\over{15}}\).
So we have the following system of inequalities: \(x> 3\,\wedge\; x < {{44}\over{15}}\)
and this simplifies to \(\text{an empty solution set}\).

Suppose \(3x-9<0\), i.e. \(x< 3\). Then we get \(1>-5(3x-9)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(15x>44\).
Then, division by the coefficient of \(x\) gives \(x > {{44}\over{15}}\).
So we have the following system of inequalities: \(x< 3\,\wedge\; x > {{44}\over{15}}\)
and this simplifies to \({{44}\over{15}}\lt x\lt 3\).

The solution of the original inequality is \({{44}\over{15}}\lt x\lt 3\).