We note first that division by zero is not allowed and that for this reason \(9x+1\) may not be equal to zero and that therefore \(x=-{{1}\over{9}}\) is not a solution.

We now distinguish two cases, namely \(9x+1>0\) and \(9x+1<0\).
In both cases we multiply the inequality on both sides by \(9x+1\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(9x+1>0\), i.e. \(x> -{{1}\over{9}}\). Then we get \(9<2(9x+1)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-18x<-7\).
Then, dvision by the coefficient of \(x\)gives \(x > {{7}\over{18}}\).
So we have the following system of inequalities: \(x> -{{1}\over{9}}\,\wedge\; x > {{7}\over{18}}\)
and this simplifies to \(x\gt{{7}\over{18}}\).

Suppose \(9x+1<0\), i.e. \(x< -{{1}\over{9}}\). Then we get \(9>2(9x+1)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-18x>-7\).
Then, division by the coefficient of \(x\) gives \(x < {{7}\over{18}}\).
So we have the following system of inequalities: \(x< -{{1}\over{9}}\,\wedge\; x < {{7}\over{18}}\)
and this simplifies to \(x\lt -{{1}\over{9}}\).

The solution of the original inequality is \(x\lt -{{1}\over{9}}\;\vee\;x\gt{{7}\over{18}}\).