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 \(5<7(x+3)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-7x<16\).
Then, dvision by the coefficient of \(x\)gives \(x > -{{16}\over{7}}\).
So we have the following system of inequalities: \(x> -3\,\wedge\; x > -{{16}\over{7}}\)
and this simplifies to \(x\gt-{{16}\over{7}}\).

Suppose \(x+3<0\), i.e. \(x< -3\). Then we get \(5>7(x+3)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-7x>16\).
Then, division by the coefficient of \(x\) gives \(x < -{{16}\over{7}}\).
So we have the following system of inequalities: \(x< -3\,\wedge\; x < -{{16}\over{7}}\)
and this simplifies to \(x\lt -3\).

The solution of the original inequality is \(x\lt -3\;\vee\;x\gt-{{16}\over{7}}\).