We note first that division by zero is not allowed and that for this reason \(8x+1\) may not be equal to zero and that therefore \(x=-{{1}\over{8}}\) is not a solution.

We now distinguish two cases, namely \(8x+1>0\) and \(8x+1<0\).
In both cases we multiply the inequality on both sides by \(8x+1\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(8x+1>0\), i.e. \(x> -{{1}\over{8}}\). Then we get \(6<-6(8x+1)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(48x<-12\).
Then, dvision by the coefficient of \(x\)gives \(x < -{{1}\over{4}}\).
So we have the following system of inequalities: \(x> -{{1}\over{8}}\,\wedge\; x < -{{1}\over{4}}\)
and this simplifies to \(\text{an empty solution set}\).

Suppose \(8x+1<0\), i.e. \(x< -{{1}\over{8}}\). Then we get \(6>-6(8x+1)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(48x>-12\).
Then, division by the coefficient of \(x\) gives \(x > -{{1}\over{4}}\).
So we have the following system of inequalities: \(x< -{{1}\over{8}}\,\wedge\; x > -{{1}\over{4}}\)
and this simplifies to \(-{{1}\over{4}}\lt x\lt -{{1}\over{8}}\).

The solution of the original inequality is \(-{{1}\over{4}}\lt x\lt -{{1}\over{8}}\).