We note first that division by zero is not allowed and that for this reason \(7x+2\) may not be equal to zero and that therefore \(x=-{{2}\over{7}}\) is not a solution.

We now distinguish two cases, namely \(7x+2>0\) and \(7x+2<0\).
In both cases we multiply the inequality on both sides by \(7x+2\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(7x+2>0\), i.e. \(x> -{{2}\over{7}}\). Then we get \(4<5(7x+2)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-35x<6\).
Then, dvision by the coefficient of \(x\)gives \(x > -{{6}\over{35}}\).
So we have the following system of inequalities: \(x> -{{2}\over{7}}\,\wedge\; x > -{{6}\over{35}}\)
and this simplifies to \(x\gt-{{6}\over{35}}\).

Suppose \(7x+2<0\), i.e. \(x< -{{2}\over{7}}\). Then we get \(4>5(7x+2)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-35x>6\).
Then, division by the coefficient of \(x\) gives \(x < -{{6}\over{35}}\).
So we have the following system of inequalities: \(x< -{{2}\over{7}}\,\wedge\; x < -{{6}\over{35}}\)
and this simplifies to \(x\lt -{{2}\over{7}}\).

The solution of the original inequality is \(x\lt -{{2}\over{7}}\;\vee\;x\gt-{{6}\over{35}}\).