Solve x+8/x-6>0

$$\frac { x + 8 } { x - 6 } > 0$$

Solve for x

$$x\in \left(-\infty,-8\right)\cup \left(6,\infty\right)$$

Solution Steps

For the quotient to be positive, x+8 and x-6 have to be both negative or both positive. Consider the case when x+8 and x-6 are both negative.

$$x+8<0 x-6<0$$

The solution satisfying both inequalities is x<-8.

$$x<-8$$

Consider the case when x+8 and x-6 are both positive.

$$x-6>0 x+8>0$$

The solution satisfying both inequalities is x>6.

$$x>6$$

The final solution is the union of the obtained solutions.

$$x<-8\text{; }x>6$$