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Source: — Data Sufficiency |
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by shankar.ashwin » Mon Oct 10, 2011 8:50 am
(x+2) = +/- 2(x-2)

Case 1
x+2 = 2x-4
x=6.

Case 2

x+2=-2x+4
3x=2
x=2/3.

Edited..
Last edited by shankar.ashwin on Mon Oct 10, 2011 11:13 am, edited 2 times in total.
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by n@resh » Mon Oct 10, 2011 10:20 am
Neha Mittal wrote:|x+2| = 2|x-2| find range of x


I do not understand why x = 6 is an invalid solution
X = 2/3 and 6 for two different cases.

so range will be : [2/3, 6)
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by n@resh » Mon Oct 10, 2011 10:20 am
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by Whitney Garner » Mon Oct 10, 2011 2:44 pm
n@resh wrote:
Neha Mittal wrote:|x+2| = 2|x-2| find range of x


I do not understand why x = 6 is an invalid solution
X = 2/3 and 6 for two different cases.

so range will be : [2/3, 6)
Just as a side-note, there is no "range" here as [2/3,6) would mean that values of X from (and including) 2/3 all the way up to 6 would satisfy this expression, and that is not the case (just check x=1 for example).

The range would simply be the values [2/3] and [6].

:)
Whit
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