We note first that division by zero is not allowed and that for this reason \(9x-6\) may not be equal to zero and that therefore \(x={{2}\over{3}}\) is not a solution.

We now distinguish two cases, namely \(9x-6>0\) and \(9x-6<0\).
In both cases we multiply the inequality on both sides by \(9x-6\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(9x-6>0\), i.e. \(x> {{2}\over{3}}\). Then we get \(9<7(9x-6)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-63x<-51\).
Then, dvision by the coefficient of \(x\)gives \(x > {{17}\over{21}}\).
So we have the following system of inequalities: \(x> {{2}\over{3}}\,\wedge\; x > {{17}\over{21}}\)
and this simplifies to \(x\gt{{17}\over{21}}\).

Suppose \(9x-6<0\), i.e. \(x< {{2}\over{3}}\). Then we get \(9>7(9x-6)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-63x>-51\).
Then, division by the coefficient of \(x\) gives \(x < {{17}\over{21}}\).
So we have the following system of inequalities: \(x< {{2}\over{3}}\,\wedge\; x < {{17}\over{21}}\)
and this simplifies to \(x\lt {{2}\over{3}}\).

The solution of the original inequality is \(x\lt {{2}\over{3}}\;\vee\;x\gt{{17}\over{21}}\).