We note first that division by zero is not allowed and that for this reason \(8x+8\) may not be equal to zero and that therefore \(x=-1\) is not a solution.

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

Suppose \(8x+8>0\), i.e. \(x> -1\). Then we get \(9<4(8x+8)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-32x<23\).
Then, dvision by the coefficient of \(x\)gives \(x > -{{23}\over{32}}\).
So we have the following system of inequalities: \(x> -1\,\wedge\; x > -{{23}\over{32}}\)
and this simplifies to \(x\gt-{{23}\over{32}}\).

Suppose \(8x+8<0\), i.e. \(x< -1\). Then we get \(9>4(8x+8)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-32x>23\).
Then, division by the coefficient of \(x\) gives \(x < -{{23}\over{32}}\).
So we have the following system of inequalities: \(x< -1\,\wedge\; x < -{{23}\over{32}}\)
and this simplifies to \(x\lt -1\).

The solution of the original inequality is \(x\lt -1\;\vee\;x\gt-{{23}\over{32}}\).