We note first that division by zero is not allowed and that for this reason \(6x+5\) may not be equal to zero and that therefore \(x=-{{5}\over{6}}\) is not a solution.

We now distinguish two cases, namely \(6x+5>0\) and \(6x+5<0\).
In both cases we multiply the inequality on both sides by \(6x+5\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(6x+5>0\), i.e. \(x> -{{5}\over{6}}\). Then we get \(5<-5(6x+5)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(30x<-30\).
Then, dvision by the coefficient of \(x\)gives \(x < -1\).
So we have the following system of inequalities: \(x> -{{5}\over{6}}\,\wedge\; x < -1\)
and this simplifies to \(\text{an empty solution set}\).

Suppose \(6x+5<0\), i.e. \(x< -{{5}\over{6}}\). Then we get \(5>-5(6x+5)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(30x>-30\).
Then, division by the coefficient of \(x\) gives \(x > -1\).
So we have the following system of inequalities: \(x< -{{5}\over{6}}\,\wedge\; x > -1\)
and this simplifies to \(-1\lt x\lt -{{5}\over{6}}\).

The solution of the original inequality is \(-1\lt x\lt -{{5}\over{6}}\).