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