We note first that division by zero is not allowed and that for this reason \(7x-2\) may not be equal to zero and that therefore \(x={{2}\over{7}}\) is not a solution.

We now distinguish two cases, namely \(7x-2>0\) and \(7x-2<0\).
In both cases we multiply the inequality on both sides by \(7x-2\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(7x-2>0\), i.e. \(x> {{2}\over{7}}\). Then we get \(4<-8(7x-2)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(56x<12\).
Then, dvision by the coefficient of \(x\)gives \(x < {{3}\over{14}}\).
So we have the following system of inequalities: \(x> {{2}\over{7}}\,\wedge\; x < {{3}\over{14}}\)
and this simplifies to \(\text{an empty solution set}\).

Suppose \(7x-2<0\), i.e. \(x< {{2}\over{7}}\). Then we get \(4>-8(7x-2)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(56x>12\).
Then, division by the coefficient of \(x\) gives \(x > {{3}\over{14}}\).
So we have the following system of inequalities: \(x< {{2}\over{7}}\,\wedge\; x > {{3}\over{14}}\)
and this simplifies to \({{3}\over{14}}\lt x\lt {{2}\over{7}}\).

The solution of the original inequality is \({{3}\over{14}}\lt x\lt {{2}\over{7}}\).