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Director
Joined: 14 Oct 2003
Posts: 601
Location: On Vacation at My Crawford, Texas Ranch
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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
_________________
"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
Last edited by Titleist on 23 Oct 2005, 10:55, edited 2 times in total.
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SVP
Joined: 05 Apr 2005
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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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SVP
Joined: 24 Sep 2005
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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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SVP
Joined: 05 Apr 2005
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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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Director
Joined: 21 Aug 2005
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Use x=x and x=-x, both arrive at a solution that is not an integer and >1. So C
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Director
Joined: 14 Oct 2003
Posts: 601
Location: On Vacation at My Crawford, Texas Ranch
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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. 
_________________
"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
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