Question 62776: help please
solve the rational inequality (x-2)/(x+3) is less than or equal to 0
is it: (-3,2] (-oo,-3) (-oo,-3)U{2,oo) or [2,oo)
i am very confused
Found 3 solutions by jai_kos, ikleyn, n2:
Answer by jai_kos(139)   (Show Source):
You can put this solution on YOUR website! Given (x-2)/(x+3) is less than or equal to 0
(x-2) /(x-3) < = 0
multiply the above equation by (x -3), we get
(x-2) < = 0
x -2 < =0
Add 2 on both sides, we get
x < = 2
Therefore this in the interval form is given by
(-infinty ,2]
Answer by ikleyn(53712)  (Show Source):
You can put this solution on YOUR website! .
help please
solve the rational inequality (x-2)/(x+3) is less than or equal to 0
is it: (-3,2] (-oo,-3) (-oo,-3)U{2,oo) or [2,oo)
i am very confused
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The solution in the post by @jai_kos in incorrect.
It is incorrect methodologically and gives incorrect answer.
See my correct solution below.
They want you solve this inequality
 <= 0. (1)
The left side rational function can be non-positive if and only if
EITHER the numerator is non-positive and denominator is positive
x - 2 <= 0 and x + 3 > 0 (2)
OR the numerator is non-negative and denominator is negative
x - 2 >= 0 and x + 3 < 0. (3)
In case (2), x <= 2 and x > -3 simultaneously, or
-3 < x <= 2.
In case (3), x >= 2 and x < -3 simultaneously, which has no solutions.
Thus the final solution to the given inequality is this set of real numbers -3 < x <= 2,
or, in the interval notation, the set (-3,2].
Solved.
The error made by @jai_kos is that when he multiplies both sides of the original inequality by (x+3),
he misses the case when (x+3) is negative, which requires different treatment.
This error, which jai_kos makes solving the problem, is a typical error, which beginners make
when trying to solve such inequalities,
until the more experienced teachers/tutors will explain their error and will show a right way solving.
Answer by n2(72) (Show Source):
You can put this solution on YOUR website! .
solve the rational inequality (x-2)/(x+3) <= 0
~~~~~~~~~~~~~~~~~~~~~~~~
They want you solve this inequality
<= 0. (1)
The left side rational function can be non-positive if and only if
EITHER the numerator is non-positive and denominator is positive
x - 2 <= 0 and x + 3 > 0 (2)
OR the numerator is non-negative and denominator is negative
x - 2 >= 0 and x + 3 < 0. (3)
In case (2), x <= 2 and x > -3 simultaneously, or
-3 < x <= 2.
In case (3), x >= 2 and x < -3 simultaneously, which has no solutions.
Thus the final solution to the given inequality is this set of real numbers -3 < x <= 2,
or, in the interval notation, the set (-3,2].
Solved.
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