We note first that division by zero is not allowed and that for this reason \(4x+6\) may not be equal to zero and that therefore \(x=-{{3}\over{2}}\) is not a solution.

We now distinguish two cases, namely \(4x+6>0\) and \(4x+6<0\).
In both cases we multiply the inequality on both sides by \(4x+6\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(4x+6>0\), i.e. \(x> -{{3}\over{2}}\). Then we get \(1<-5(4x+6)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(20x<-31\).
Then, dvision by the coefficient of \(x\)gives \(x < -{{31}\over{20}}\).
So we have the following system of inequalities: \(x> -{{3}\over{2}}\,\wedge\; x < -{{31}\over{20}}\)
and this simplifies to \(\text{an empty solution set}\).

Suppose \(4x+6<0\), i.e. \(x< -{{3}\over{2}}\). Then we get \(1>-5(4x+6)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(20x>-31\).
Then, division by the coefficient of \(x\) gives \(x > -{{31}\over{20}}\).
So we have the following system of inequalities: \(x< -{{3}\over{2}}\,\wedge\; x > -{{31}\over{20}}\)
and this simplifies to \(-{{31}\over{20}}\lt x\lt -{{3}\over{2}}\).

The solution of the original inequality is \(-{{31}\over{20}}\lt x\lt -{{3}\over{2}}\).