Solve x+6/3-x<0

$$\frac { x + 6 } { 3 - x } < 0$$

Solve for x

$$x\in \left(-\infty,-6\right)\cup \left(3,\infty\right)$$

Solution Steps

For the quotient to be negative, x+6 and 3-x have to be of the opposite signs. Consider the case when x+6 is positive and 3-x is negative.

$$x+6>0 3-x<0$$

The solution satisfying both inequalities is x>3.

$$x>3$$

Consider the case when 3-x is positive and x+6 is negative.

$$3-x>0 x+6<0$$

The solution satisfying both inequalities is x<-6.

$$x<-6$$

The final solution is the union of the obtained solutions.

$$x>3\text{; }x<-6$$