We note first that division by zero is not allowed and that for this reason \(7x-3\) may not be equal to zero and that therefore \(x={{3}\over{7}}\) is not a solution.

We now distinguish two cases, namely \(7x-3>0\) and \(7x-3<0\).
In both cases we multiply the inequality on both sides by \(7x-3\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(7x-3>0\), i.e. \(x> {{3}\over{7}}\). Then we get \(5<-2(7x-3)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(14x<1\).
Then, dvision by the coefficient of \(x\)gives \(x < {{1}\over{14}}\).
So we have the following system of inequalities: \(x> {{3}\over{7}}\,\wedge\; x < {{1}\over{14}}\)
and this simplifies to \(\text{an empty solution set}\).

Suppose \(7x-3<0\), i.e. \(x< {{3}\over{7}}\). Then we get \(5>-2(7x-3)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(14x>1\).
Then, division by the coefficient of \(x\) gives \(x > {{1}\over{14}}\).
So we have the following system of inequalities: \(x< {{3}\over{7}}\,\wedge\; x > {{1}\over{14}}\)
and this simplifies to \({{1}\over{14}}\lt x\lt {{3}\over{7}}\).

The solution of the original inequality is \({{1}\over{14}}\lt x\lt {{3}\over{7}}\).