We note first that division by zero is not allowed and that for this reason \(7x+9\) may not be equal to zero and that therefore \(x=-{{9}\over{7}}\) is not a solution.

We now distinguish two cases, namely \(7x+9>0\) and \(7x+9<0\).
In both cases we multiply the inequality on both sides by \(7x+9\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(7x+9>0\), i.e. \(x> -{{9}\over{7}}\). Then we get \(9<6(7x+9)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-42x<45\).
Then, dvision by the coefficient of \(x\)gives \(x > -{{15}\over{14}}\).
So we have the following system of inequalities: \(x> -{{9}\over{7}}\,\wedge\; x > -{{15}\over{14}}\)
and this simplifies to \(x\gt-{{15}\over{14}}\).

Suppose \(7x+9<0\), i.e. \(x< -{{9}\over{7}}\). Then we get \(9>6(7x+9)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-42x>45\).
Then, division by the coefficient of \(x\) gives \(x < -{{15}\over{14}}\).
So we have the following system of inequalities: \(x< -{{9}\over{7}}\,\wedge\; x < -{{15}\over{14}}\)
and this simplifies to \(x\lt -{{9}\over{7}}\).

The solution of the original inequality is \(x\lt -{{9}\over{7}}\;\vee\;x\gt-{{15}\over{14}}\).