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

We now distinguish two cases, namely \(x+3>0\) and \(x+3<0\).
In both cases we multiply the inequality on both sides by \(x+3\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(x+3>0\), i.e. \(x> -3\). Then we get \(2<x+3\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-x<1\).
Then, dvision by the coefficient of \(x\)gives \(x > -1\).
So we have the following system of inequalities: \(x> -3\,\wedge\; x > -1\)
and this simplifies to \(x\gt-1\).

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

The solution of the original inequality is \(x\lt -3\;\vee\;x\gt-1\).