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Is |x|< 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0

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Senior Manager
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Is |x|< 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]

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New post 29 Sep 2008, 09:34
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Is |x|< 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| ≠ 0

Kudos [?]: 376 [0], given: 0

Director
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Joined: 12 Jul 2008
Posts: 514

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Schools: Wharton
Re: Absolute value [#permalink]

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New post 29 Sep 2008, 09:41
vksunder wrote:
Is |x|< 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| ≠ 0


Is -1 < x < 1?

(1) Insufficient

If x > 1:
x+1 = 2(x-1)
x+1 = 2x - 2
x = 3

If x < -1
-(x+1) = -2(x-1)
-x - 1 = -2x + 2
x = 3 --> No solution

If -1 < x < 1
x + 1 = -2(x-1)
x + 1 = -2x + 2
3x = 1
x = 1/3

x = 1/3 or 3

(2) Insufficient
x ≠ 3

(1) and (2) Sufficient
x = 1/3

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Re: Absolute value [#permalink]

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New post 29 Sep 2008, 10:05
Hi,

Dealing with absolutes is a bummer for me.
Could you please explain the reasoning behind the following step :
If -1 < x < 1
x + 1 = -2(x-1)
x + 1 = -2x + 2
3x = 1
x = 1/3

Why is the RHS -2(x-1) ?
Thanks.

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Senior Manager
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Re: Absolute value [#permalink]

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New post 29 Sep 2008, 10:55
dhegdex - I'll try and explain this part: -1 < x < 1

Basically absolute values have 2 solutions. So, in this case |X|<1 has two solutions. One for X > 0 and other one for X < 0

1. x>0: x<1
2. x<0: -(x)<1 which is equal to x>-1

The above solution translates to: -1 < x < 1

Hope this makes sense. I've been struggling with Absolute value problems lately...

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Senior Manager
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Re: Absolute value [#permalink]

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New post 29 Sep 2008, 10:56
Thanks for the wonderful explanation guys!

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Re: Absolute value   [#permalink] 29 Sep 2008, 10:56
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