Solve x-3/x+2>0

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

Solve for x

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

Solution Steps

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

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

The solution satisfying both inequalities is x<-2.

$$x<-2$$

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

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

The solution satisfying both inequalities is x>3.

$$x>3$$

The final solution is the union of the obtained solutions.

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