We note first that division by zero is not allowed and that for this reason \(8x+3\) may not be equal to zero and that therefore \(x=-{{3}\over{8}}\) is not a solution.

We now distinguish two cases, namely \(8x+3>0\) and \(8x+3<0\).
In both cases we multiply the inequality on both sides by \(8x+3\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(8x+3>0\), i.e. \(x> -{{3}\over{8}}\). Then we get \(6<4(8x+3)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-32x<6\).
Then, dvision by the coefficient of \(x\)gives \(x > -{{3}\over{16}}\).
So we have the following system of inequalities: \(x> -{{3}\over{8}}\,\wedge\; x > -{{3}\over{16}}\)
and this simplifies to \(x\gt-{{3}\over{16}}\).

Suppose \(8x+3<0\), i.e. \(x< -{{3}\over{8}}\). Then we get \(6>4(8x+3)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-32x>6\).
Then, division by the coefficient of \(x\) gives \(x < -{{3}\over{16}}\).
So we have the following system of inequalities: \(x< -{{3}\over{8}}\,\wedge\; x < -{{3}\over{16}}\)
and this simplifies to \(x\lt -{{3}\over{8}}\).

The solution of the original inequality is \(x\lt -{{3}\over{8}}\;\vee\;x\gt-{{3}\over{16}}\).