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Absolut Valu [#permalink] New post 23 Oct 2005, 06:34
If X>0 and 2|9 + X| = 15|X| - 4|2 - X|, then X could be?

I. An Odd Integer

II. An Even Integer

III. Greater than 1


a) I only
b) II only
c) III only
d) I and II
e) I, II, and III
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Last edited by Titleist on 23 Oct 2005, 09:55, edited 2 times in total.
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Re: Absolut Valu [#permalink] New post 23 Oct 2005, 07:32
C. since x is positive, we can open the modulas as under:
2(9 + X) = 15x - 4(2 - X)
18+2x=15x-8+4x
17x = 26
x=28/17>1
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Re: Absolut Valu [#permalink] New post 23 Oct 2005, 07:38
HIMALAYA wrote:
C. since x is positive, we can open the modulas as under:
2(9 + X) = 15x - 4(2 - X)
18+2x=15x-8+4x
17x = 26
x=28/17>1


This is only correct when 0<x<=2
we have to consider two cases
0<x<=2, |2-X|= 2-X
x>2, |2-x|= x-2
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Re: Absolut Valu [#permalink] New post 23 Oct 2005, 08:12
laxieqv wrote:
HIMALAYA wrote:
C. since x is positive, we can open the modulas as under:
2(9 + X) = 15x - 4(2 - X)
18+2x=15x-8+4x
17x = 26
x=28/17>1


This is only correct when 0<x<=2
we have to consider two cases
0<x<=2, |2-X|= 2-X
x>2, |2-x|= x-2


good point.

2(9 + X) = 15x - 4(2 - X)
18+2x=15x-4l2-xl
18 = 13x-4l2-xl
13x= 18+4l2-xl

now here non of integer (even or odd) values for x makes the equation equal... so only x>1 works. hope so...........
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 [#permalink] New post 23 Oct 2005, 09:47
Use x=x and x=-x, both arrive at a solution that is not an integer and >1. So C
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 [#permalink] New post 23 Oct 2005, 09:56
gsr wrote:
Use x=x and x=-x, both arrive at a solution that is not an integer and >1. So C


You guys are way too smart. :wink:
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"Wow! Brazil is big." —George W. Bush, after being shown a map of Brazil by Brazilian president Luiz Inacio Lula da Silva, Brasilia, Brazil, Nov. 6, 2005

http://www.nytimes.com/2005/11/21/inter ... prexy.html

  [#permalink] 23 Oct 2005, 09:56
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