Solve 3-4y/2y-3<0

$$\frac{ 3-4y }{ 2y-3 } < 0$$

Solve for y

$$y\in \left(-\infty,\frac{3}{4}\right)\cup \left(\frac{3}{2},\infty\right)$$

Solution Steps

For the quotient to be negative, 3-4y and 2y-3 have to be of the opposite signs. Consider the case when 3-4y is positive and 2y-3 is negative.

$$3-4y>0 2y-3<0$$

The solution satisfying both inequalities is y<\frac{3}{4}.

$$y<\frac{3}{4}$$

Consider the case when 2y-3 is positive and 3-4y is negative.

$$2y-3>0 3-4y<0$$

The solution satisfying both inequalities is y>\frac{3}{2}.

$$y>\frac{3}{2}$$

The final solution is the union of the obtained solutions.

$$y<\frac{3}{4}\text{; }y>\frac{3}{2}$$