We note first that division by zero is not allowed and that for this reason \(2x+8\) may not be equal to zero and that therefore \(x=-4\) is not a solution.

We now distinguish two cases, namely \(2x+8>0\) and \(2x+8<0\).
In both cases we multiply the inequality on both sides by \(2x+8\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(2x+8>0\), i.e. \(x> -4\). Then we get \(4<7(2x+8)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-14x<52\).
Then, dvision by the coefficient of \(x\)gives \(x > -{{26}\over{7}}\).
So we have the following system of inequalities: \(x> -4\,\wedge\; x > -{{26}\over{7}}\)
and this simplifies to \(x\gt-{{26}\over{7}}\).

Suppose \(2x+8<0\), i.e. \(x< -4\). Then we get \(4>7(2x+8)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-14x>52\).
Then, division by the coefficient of \(x\) gives \(x < -{{26}\over{7}}\).
So we have the following system of inequalities: \(x< -4\,\wedge\; x < -{{26}\over{7}}\)
and this simplifies to \(x\lt -4\).

The solution of the original inequality is \(x\lt -4\;\vee\;x\gt-{{26}\over{7}}\).