We note first that division by zero is not allowed and that for this reason \(9x+7\) may not be equal to zero and that therefore \(x=-{{7}\over{9}}\) is not a solution.

We now distinguish two cases, namely \(9x+7>0\) and \(9x+7<0\).
In both cases we multiply the inequality on both sides by \(9x+7\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(9x+7>0\), i.e. \(x> -{{7}\over{9}}\). Then we get \(8<-9(9x+7)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(81x<-71\).
Then, dvision by the coefficient of \(x\)gives \(x < -{{71}\over{81}}\).
So we have the following system of inequalities: \(x> -{{7}\over{9}}\,\wedge\; x < -{{71}\over{81}}\)
and this simplifies to \(\text{an empty solution set}\).

Suppose \(9x+7<0\), i.e. \(x< -{{7}\over{9}}\). Then we get \(8>-9(9x+7)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(81x>-71\).
Then, division by the coefficient of \(x\) gives \(x > -{{71}\over{81}}\).
So we have the following system of inequalities: \(x< -{{7}\over{9}}\,\wedge\; x > -{{71}\over{81}}\)
and this simplifies to \(-{{71}\over{81}}\lt x\lt -{{7}\over{9}}\).

The solution of the original inequality is \(-{{71}\over{81}}\lt x\lt -{{7}\over{9}}\).