We note first that division by zero is not allowed and that for this reason \(4x+8\) may not be equal to zero and that therefore \(x=-2\) is not a solution.

We now distinguish two cases, namely \(4x+8>0\) and \(4x+8<0\).
In both cases we multiply the inequality on both sides by \(4x+8\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(4x+8>0\), i.e. \(x> -2\). Then we get \(7<-9(4x+8)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(36x<-79\).
Then, dvision by the coefficient of \(x\)gives \(x < -{{79}\over{36}}\).
So we have the following system of inequalities: \(x> -2\,\wedge\; x < -{{79}\over{36}}\)
and this simplifies to \(\text{an empty solution set}\).

Suppose \(4x+8<0\), i.e. \(x< -2\). Then we get \(7>-9(4x+8)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(36x>-79\).
Then, division by the coefficient of \(x\) gives \(x > -{{79}\over{36}}\).
So we have the following system of inequalities: \(x< -2\,\wedge\; x > -{{79}\over{36}}\)
and this simplifies to \(-{{79}\over{36}}\lt x\lt -2\).

The solution of the original inequality is \(-{{79}\over{36}}\lt x\lt -2\).