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

We now distinguish two cases, namely \(3x-1>0\) and \(3x-1<0\).
In both cases we multiply the inequality on both sides by \(3x-1\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(3x-1>0\), i.e. \(x> {{1}\over{3}}\). Then we get \(3<7(3x-1)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-21x<-10\).
Then, dvision by the coefficient of \(x\)gives \(x > {{10}\over{21}}\).
So we have the following system of inequalities: \(x> {{1}\over{3}}\,\wedge\; x > {{10}\over{21}}\)
and this simplifies to \(x\gt{{10}\over{21}}\).

Suppose \(3x-1<0\), i.e. \(x< {{1}\over{3}}\). Then we get \(3>7(3x-1)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-21x>-10\).
Then, division by the coefficient of \(x\) gives \(x < {{10}\over{21}}\).
So we have the following system of inequalities: \(x< {{1}\over{3}}\,\wedge\; x < {{10}\over{21}}\)
and this simplifies to \(x\lt {{1}\over{3}}\).

The solution of the original inequality is \(x\lt {{1}\over{3}}\;\vee\;x\gt{{10}\over{21}}\).