We note first that division by zero is not allowed and that for this reason \(8x+9\) may not be equal to zero and that therefore \(x=-{{9}\over{8}}\) is not a solution.

We now distinguish two cases, namely \(8x+9>0\) and \(8x+9<0\).
In both cases we multiply the inequality on both sides by \(8x+9\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(8x+9>0\), i.e. \(x> -{{9}\over{8}}\). Then we get \(6<-8(8x+9)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(64x<-78\).
Then, dvision by the coefficient of \(x\)gives \(x < -{{39}\over{32}}\).
So we have the following system of inequalities: \(x> -{{9}\over{8}}\,\wedge\; x < -{{39}\over{32}}\)
and this simplifies to \(\text{an empty solution set}\).

Suppose \(8x+9<0\), i.e. \(x< -{{9}\over{8}}\). Then we get \(6>-8(8x+9)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(64x>-78\).
Then, division by the coefficient of \(x\) gives \(x > -{{39}\over{32}}\).
So we have the following system of inequalities: \(x< -{{9}\over{8}}\,\wedge\; x > -{{39}\over{32}}\)
and this simplifies to \(-{{39}\over{32}}\lt x\lt -{{9}\over{8}}\).

The solution of the original inequality is \(-{{39}\over{32}}\lt x\lt -{{9}\over{8}}\).