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

We now distinguish two cases, namely \(3x-2>0\) and \(3x-2<0\).
In both cases we multiply the inequality on both sides by \(3x-2\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(3x-2>0\), i.e. \(x> {{2}\over{3}}\). Then we get \(2<-7(3x-2)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(21x<12\).
Then, dvision by the coefficient of \(x\)gives \(x < {{4}\over{7}}\).
So we have the following system of inequalities: \(x> {{2}\over{3}}\,\wedge\; x < {{4}\over{7}}\)
and this simplifies to \(\text{an empty solution set}\).

Suppose \(3x-2<0\), i.e. \(x< {{2}\over{3}}\). Then we get \(2>-7(3x-2)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(21x>12\).
Then, division by the coefficient of \(x\) gives \(x > {{4}\over{7}}\).
So we have the following system of inequalities: \(x< {{2}\over{3}}\,\wedge\; x > {{4}\over{7}}\)
and this simplifies to \({{4}\over{7}}\lt x\lt {{2}\over{3}}\).

The solution of the original inequality is \({{4}\over{7}}\lt x\lt {{2}\over{3}}\).