Solve x/2x-3>0

$$\frac { x } { 2 x - 3 } > 0$$

Solve for x

$$x\in \left(-\infty,0\right)\cup \left(\frac{3}{2},\infty\right)$$

Solution Steps

For the quotient to be positive, x and 2x-3 have to be both negative or both positive. Consider the case when x and 2x-3 are both negative.

$$x<0 2x-3<0$$

The solution satisfying both inequalities is x<0.

$$x<0$$

Consider the case when x and 2x-3 are both positive.

$$2x-3>0 x>0$$

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

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

The final solution is the union of the obtained solutions.

$$x<0\text{; }x>\frac{3}{2}$$