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aboslute value inequalities

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aboslute value inequalities [#permalink] New post 23 Oct 2005, 09:07
|3x-2|<=|2x-5|

Hi,
Can someone let me know if there's another approach to solve this problem besides squaring both sides to remove the absolute value inequalities first?

Thanks.
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 [#permalink] New post 23 Oct 2005, 09:38
If there is nothing mentioned about x, for x=0, this is false. For -ve integer values of x, it is true and for positive values it is true.
Can you please post the whole question?
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 [#permalink] New post 23 Oct 2005, 10:02
gsr wrote:
If there is nothing mentioned about x, for x=0, this is false. For -ve integer values of x, it is true and for positive values it is true.
Can you please post the whole question?


This is the original question:
Find the solution set for the following inequalities |3x-2|<=|2x-5|

The first step of the explanatory answer for this question from 4gmat ebook is squaring both sides to get rid of the absolute value sign.
(3x-2)^2<=(2x-5)^2

However, I read once that
Quote:
"Another way to eliminate an absolute value is to square both sides of the equation. Taking the absolute value makes things non-negative, and squaring makes things non-negative. So, if you square something, you no longer need to take its absolute value. However, be careful when squaring both sides of an equation as this can lead to extraneous solutions."
from http://www.richland.edu/james/lecture/m116/prerequisites.html

Thus, I would like to know if there's an approach other than squaring both sides.

Let me know if you want to see the OE.
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 [#permalink] New post 24 Oct 2005, 09:41
Is the answer the following?

x <= 7/5
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 [#permalink] New post 24 Oct 2005, 09:57
For the record, I came up with x <= 7/5 by using the squaring method. Frankly, I think it's easier than trying to pick numbers to come up with an answer. For this problem, squaring is the much easier method.

|3x-2| <= |2x-5|
(3x-2)^2 <= (2X-5)^2
9x^2 - 12x + 4 <= 4x^2 - 20x + 25
5x^2 + 8x -21 <= 0
(5x-7)(x+3) <= 0
x <= -3 or 5x <= 7
x <= -3 or x <= 7/5

Because we know that x is <= either -3 or 7/5, take the larger number, 7/5.
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 [#permalink] New post 08 Nov 2005, 17:36
If you dont want to square both sides, then here is the other way, which is usually shorter, however, I found that it is lengthier/error prone for me for this question:

for x < 2/3 and x < 5/2 => x < 2/3
x >= 3

for x < 2/3 and x > 5/2
x >= 3/5

for x > 2/3 and x < 5/2 => 2/3 < x < 5/2
x <= 7/5

for x > 2/3 and x > 5/2 => x > 5/2
x <= -3

The only solution that is valid for range is x < 7/5 for 2/3 < x < 5/2
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  [#permalink] 08 Nov 2005, 17:36
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