Solve x-2/x-3>0

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

Solve for x

$$x\in \left(-\infty,2\right)\cup \left(3,\infty\right)$$

Solution Steps

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

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

The solution satisfying both inequalities is x<2.

$$x<2$$

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

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

The solution satisfying both inequalities is x>3.

$$x>3$$

The final solution is the union of the obtained solutions.

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