We note first that division by zero is not allowed and that for this reason \(3x-5\) may not be equal to zero and that therefore \(x={{5}\over{3}}\) is not a solution.

We now distinguish two cases, namely \(3x-5>0\) and \(3x-5<0\).
In both cases we multiply the inequality on both sides by \(3x-5\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(3x-5>0\), i.e. \(x> {{5}\over{3}}\). Then we get \(2<-(3x-5)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(3x<3\).
Then, dvision by the coefficient of \(x\)gives \(x < 1\).
So we have the following system of inequalities: \(x> {{5}\over{3}}\,\wedge\; x < 1\)
and this simplifies to \(\text{an empty solution set}\).

Suppose \(3x-5<0\), i.e. \(x< {{5}\over{3}}\). Then we get \(2>-(3x-5)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(3x>3\).
Then, division by the coefficient of \(x\) gives \(x > 1\).
So we have the following system of inequalities: \(x< {{5}\over{3}}\,\wedge\; x > 1\)
and this simplifies to \(1\lt x\lt {{5}\over{3}}\).

The solution of the original inequality is \(1\lt x\lt {{5}\over{3}}\).