We note first that division by zero is not allowed and that for this reason \(6x-7\) may not be equal to zero and that therefore \(x={{7}\over{6}}\) is not a solution.

We now distinguish two cases, namely \(6x-7>0\) and \(6x-7<0\).
In both cases we multiply the inequality on both sides by \(6x-7\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(6x-7>0\), i.e. \(x> {{7}\over{6}}\). Then we get \(7<-9(6x-7)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(54x<56\).
Then, dvision by the coefficient of \(x\)gives \(x < {{28}\over{27}}\).
So we have the following system of inequalities: \(x> {{7}\over{6}}\,\wedge\; x < {{28}\over{27}}\)
and this simplifies to \(\text{an empty solution set}\).

Suppose \(6x-7<0\), i.e. \(x< {{7}\over{6}}\). Then we get \(7>-9(6x-7)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(54x>56\).
Then, division by the coefficient of \(x\) gives \(x > {{28}\over{27}}\).
So we have the following system of inequalities: \(x< {{7}\over{6}}\,\wedge\; x > {{28}\over{27}}\)
and this simplifies to \({{28}\over{27}}\lt x\lt {{7}\over{6}}\).

The solution of the original inequality is \({{28}\over{27}}\lt x\lt {{7}\over{6}}\).