# Solve the nonlinear inequality that has the variable in the denominator. 2/x+3=<1/x-3

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to solve 2/x+3=<1/x-3

2/(x+3)=<1/(x-3)

=> 2/(x+3) - 1/(x-3) =< 0

=> [2(x-3) - (x+3)]/(x - 3)(x + 3) =< 0

=> (2x - 6 - x-3)/(x - 3)(x + 3) =< 0

=> (x-9)/(x - 3)(x + 3) =< 0

This is less than or equal to 0 if one of the terms is less than 0 or all the three terms are less than or equal to 0.

If all the three terms are less than or equal to 0:

• (x-9)=< 0 , (x - 3) =< 0 and (x + 3) =< 0

=> x=< 9 , x=< 3, x=< -3

x=< -3 satisfies all the three conditions.

If one of the terms is less than or equal to 0:

• x=< 9 , x > 3 and x > -3

=> 9 >= x > 3

• x=< 3 , x > 9 and x > -3 , gives no solutions
• x=< -3 , x > 3 and x > 9, gives no solutions.

Therefore the values of x are (-inf , -3] U (3 ; 9]

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

First, we'll move the fraction 1/(x-3) to the left side:

2/(x+3) - 1/(x-3) =< 0

We'll determine the least common denominator:

LCD = (x - 3)(x + 3)

We'll multiply the fractions by LCD:

[2(x-3) - (x+3)]/(x - 3)(x + 3) =< 0

We'll remove the brackets:

(2x - 6 - x-3)/(x - 3)(x + 3) =< 0

(x-9)/(x - 3)(x + 3) =< 0

The values for x are negative if x<-3 and if x belongs to the range (3 ; 9].