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 \(4<-(3x+9)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(3x<-13\).
Then, dvision by the coefficient of \(x\)gives \(x < -{{13}\over{3}}\).
So we have the following system of inequalities: \(x> -3\,\wedge\; x < -{{13}\over{3}}\)
and this simplifies to \(\text{an empty solution set}\).

Suppose \(3x+9<0\), i.e. \(x< -3\). Then we get \(4>-(3x+9)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(3x>-13\).
Then, division by the coefficient of \(x\) gives \(x > -{{13}\over{3}}\).
So we have the following system of inequalities: \(x< -3\,\wedge\; x > -{{13}\over{3}}\)
and this simplifies to \(-{{13}\over{3}}\lt x\lt -3\).

The solution of the original inequality is \(-{{13}\over{3}}\lt x\lt -3\).