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

We now distinguish two cases, namely \(2x+6>0\) and \(2x+6<0\).
In both cases we multiply the inequality on both sides by \(2x+6\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(2x+6>0\), i.e. \(x> -3\). Then we get \(4<2(2x+6)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-4x<8\).
Then, dvision by the coefficient of \(x\)gives \(x > -2\).
So we have the following system of inequalities: \(x> -3\,\wedge\; x > -2\)
and this simplifies to \(x\gt-2\).

Suppose \(2x+6<0\), i.e. \(x< -3\). Then we get \(4>2(2x+6)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-4x>8\).
Then, division by the coefficient of \(x\) gives \(x < -2\).
So we have the following system of inequalities: \(x< -3\,\wedge\; x < -2\)
and this simplifies to \(x\lt -3\).

The solution of the original inequality is \(x\lt -3\;\vee\;x\gt-2\).