We note first that division by zero is not allowed and that for this reason \(6x+2\) may not be equal to zero and that therefore \(x=-{{1}\over{3}}\) is not a solution.

We now distinguish two cases, namely \(6x+2>0\) and \(6x+2<0\).
In both cases we multiply the inequality on both sides by \(6x+2\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(6x+2>0\), i.e. \(x> -{{1}\over{3}}\). Then we get \(8<3(6x+2)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-18x<-2\).
Then, dvision by the coefficient of \(x\)gives \(x > {{1}\over{9}}\).
So we have the following system of inequalities: \(x> -{{1}\over{3}}\,\wedge\; x > {{1}\over{9}}\)
and this simplifies to \(x\gt{{1}\over{9}}\).

Suppose \(6x+2<0\), i.e. \(x< -{{1}\over{3}}\). Then we get \(8>3(6x+2)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-18x>-2\).
Then, division by the coefficient of \(x\) gives \(x < {{1}\over{9}}\).
So we have the following system of inequalities: \(x< -{{1}\over{3}}\,\wedge\; x < {{1}\over{9}}\)
and this simplifies to \(x\lt -{{1}\over{3}}\).

The solution of the original inequality is \(x\lt -{{1}\over{3}}\;\vee\;x\gt{{1}\over{9}}\).