We note first that division by zero is not allowed and that for this reason \(8x+2\) may not be equal to zero and that therefore \(x=-{{1}\over{4}}\) is not a solution.

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

Suppose \(8x+2>0\), i.e. \(x> -{{1}\over{4}}\). Then we get \(7<9(8x+2)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-72x<11\).
Then, dvision by the coefficient of \(x\)gives \(x > -{{11}\over{72}}\).
So we have the following system of inequalities: \(x> -{{1}\over{4}}\,\wedge\; x > -{{11}\over{72}}\)
and this simplifies to \(x\gt-{{11}\over{72}}\).

Suppose \(8x+2<0\), i.e. \(x< -{{1}\over{4}}\). Then we get \(7>9(8x+2)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-72x>11\).
Then, division by the coefficient of \(x\) gives \(x < -{{11}\over{72}}\).
So we have the following system of inequalities: \(x< -{{1}\over{4}}\,\wedge\; x < -{{11}\over{72}}\)
and this simplifies to \(x\lt -{{1}\over{4}}\).

The solution of the original inequality is \(x\lt -{{1}\over{4}}\;\vee\;x\gt-{{11}\over{72}}\).