Solve k-3/k+2>0

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

Solve for k

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

Solution Steps

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

$$k-3<0 k+2<0$$

The solution satisfying both inequalities is k<-2.

$$k<-2$$

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

$$k+2>0 k-3>0$$

The solution satisfying both inequalities is k>3.

$$k>3$$

The final solution is the union of the obtained solutions.

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