We note first that division by zero is not allowed and that for this reason \(2x+5\) may not be equal to zero and that therefore \(x=-{{5}\over{2}}\) is not a solution.

We now distinguish two cases, namely \(2x+5>0\) and \(2x+5<0\).
In both cases we multiply the inequality on both sides by \(2x+5\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(2x+5>0\), i.e. \(x> -{{5}\over{2}}\). Then we get \(9<-5(2x+5)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(10x<-34\).
Then, dvision by the coefficient of \(x\)gives \(x < -{{17}\over{5}}\).
So we have the following system of inequalities: \(x> -{{5}\over{2}}\,\wedge\; x < -{{17}\over{5}}\)
and this simplifies to \(\text{an empty solution set}\).

Suppose \(2x+5<0\), i.e. \(x< -{{5}\over{2}}\). Then we get \(9>-5(2x+5)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(10x>-34\).
Then, division by the coefficient of \(x\) gives \(x > -{{17}\over{5}}\).
So we have the following system of inequalities: \(x< -{{5}\over{2}}\,\wedge\; x > -{{17}\over{5}}\)
and this simplifies to \(-{{17}\over{5}}\lt x\lt -{{5}\over{2}}\).

The solution of the original inequality is \(-{{17}\over{5}}\lt x\lt -{{5}\over{2}}\).