We note first that division by zero is not allowed and that for this reason \(5x-7\) may not be equal to zero and that therefore \(x={{7}\over{5}}\) is not a solution.

We now distinguish two cases, namely \(5x-7>0\) and \(5x-7<0\).
In both cases we multiply the inequality on both sides by \(5x-7\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(5x-7>0\), i.e. \(x> {{7}\over{5}}\). Then we get \(1<4(5x-7)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-20x<-29\).
Then, dvision by the coefficient of \(x\)gives \(x > {{29}\over{20}}\).
So we have the following system of inequalities: \(x> {{7}\over{5}}\,\wedge\; x > {{29}\over{20}}\)
and this simplifies to \(x\gt{{29}\over{20}}\).

Suppose \(5x-7<0\), i.e. \(x< {{7}\over{5}}\). Then we get \(1>4(5x-7)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(-20x>-29\).
Then, division by the coefficient of \(x\) gives \(x < {{29}\over{20}}\).
So we have the following system of inequalities: \(x< {{7}\over{5}}\,\wedge\; x < {{29}\over{20}}\)
and this simplifies to \(x\lt {{7}\over{5}}\).

The solution of the original inequality is \(x\lt {{7}\over{5}}\;\vee\;x\gt{{29}\over{20}}\).