Solve 5+x/3-x<0

$$\frac { 5 + x } { 3 - x } < 0$$

Solve for x

$$x\in \left(-\infty,-5\right)\cup \left(3,\infty\right)$$

Solution Steps

For the quotient to be negative, x+5 and 3-x have to be of the opposite signs. Consider the case when x+5 is positive and 3-x is negative.

$$x+5>0 3-x<0$$

The solution satisfying both inequalities is x>3.

$$x>3$$

Consider the case when 3-x is positive and x+5 is negative.

$$3-x>0 x+5<0$$

The solution satisfying both inequalities is x<-5.

$$x<-5$$

The final solution is the union of the obtained solutions.

$$x>3\text{; }x<-5$$