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

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

Suppose \(6x+9>0\), i.e. \(x> -{{3}\over{2}}\). Then we get \(1<9(6x+9)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-54x<80\).
Then, dvision by the coefficient of \(x\)gives \(x > -{{40}\over{27}}\).
So we have the following system of inequalities: \(x> -{{3}\over{2}}\,\wedge\; x > -{{40}\over{27}}\)
and this simplifies to \(x\gt-{{40}\over{27}}\).

Suppose \(6x+9<0\), i.e. \(x< -{{3}\over{2}}\). Then we get \(1>9(6x+9)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-54x>80\).
Then, division by the coefficient of \(x\) gives \(x < -{{40}\over{27}}\).
So we have the following system of inequalities: \(x< -{{3}\over{2}}\,\wedge\; x < -{{40}\over{27}}\)
and this simplifies to \(x\lt -{{3}\over{2}}\).

The solution of the original inequality is \(x\lt -{{3}\over{2}}\;\vee\;x\gt-{{40}\over{27}}\).