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

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

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09 Apr 2008, 15:50
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Is |x| < 1?

(1) |x+1| = 2|x-1|
(2) |x-3| ≠ 0
VP
Joined: 10 Jun 2007
Posts: 1459
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Re: DS abs value [#permalink]

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09 Apr 2008, 15:56
young_gun wrote:
Is |x| < 1?

(1) |x+1| = 2|x-1|
(2) |x-3| ≠ 0

C.

(1) |x+1| = 2*|x-1|
x = 3 or x=1/3
INSUFFICIENT

(2) x ≠ 3
INSUFFICIENT

Together, x=1/3
SUFFICIENT
Current Student
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Re: DS abs value [#permalink]

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09 Apr 2008, 15:58
bkk145 wrote:
young_gun wrote:
Is |x| < 1?

(1) |x+1| = 2|x-1|
(2) |x-3| ≠ 0

C.

(1) |x+1| = 2*|x-1|
x = 3 or x=1/3
INSUFFICIENT

(2) x ≠ 3
INSUFFICIENT

Together, x=1/3
SUFFICIENT

bkk, do you think you could show me how in (1) you solved x=3 or x=1/3? I'm not too great with abs values. Thanks!
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Re: DS abs value [#permalink]

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09 Apr 2008, 16:05
C...

1) says x can be 3 or 1/3

3) says x cant be 3..or -3 we dont know..

together..yes..

x=1/3
Current Student
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Re: DS abs value [#permalink]

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09 Apr 2008, 16:08
fresinha12 wrote:
C...

1) says x can be 3 or 1/3

3) says x cant be 3..or -3 we dont know..

together..yes..

x=1/3

can you explain how you got the bold? thx
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Re: DS abs value [#permalink]

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09 Apr 2008, 16:11
1
KUDOS
young_gun wrote:
bkk145 wrote:
young_gun wrote:
Is |x| < 1?

(1) |x+1| = 2|x-1|
(2) |x-3| ≠ 0

C.

(1) |x+1| = 2*|x-1|
x = 3 or x=1/3
INSUFFICIENT

(2) x ≠ 3
INSUFFICIENT

Together, x=1/3
SUFFICIENT

bkk, do you think you could show me how in (1) you solved x=3 or x=1/3? I'm not too great with abs values. Thanks!

Sure, you have
|x+1| = 2*|x-1|
You know that
|x+1| is positive when x>-1 and negative when x<-1
|x-1| is positive when x>1 and negative when x<1
This means that in a number line, the intervals you have are:
x<-1, -1<x<1, x>1

Now, solve the equation according to those intervals:
For x<-1,
-x-1 = 2*(-x+1)
-x-1 = -2x+2
x = 3

For -1<x<1,
x+1 = 2*(-x+1)
x+1 = -2x+2
x = 1/3

For x>1
x+1 = 2*(x-1)
x+1 = 2x-2
x = 3

x equals to 3 or 1/3

I know, all these signs are confusing! I attached a document for you, take a look. It helped me.
Attachments

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Re: DS abs value [#permalink]

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09 Apr 2008, 16:20
young_gun wrote:
Is |x| < 1?
(1) |x+1| = 2|x-1|
(2) |x-3| ≠ 0

1: |x+1| = 2|x-1|
if x is +ve:
x + 1 = 2x - 2
x = 3

if x is -ve:
x + 1 = -2x + 2
x = 1/3

so insufff.

2: if (x-3) is not equal to 0, then x is not 3 but do not know whether x is > than or < 3. NSF

1 & 2: x = 1/3. suff.

C.
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Re: DS abs value [#permalink]

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10 Apr 2008, 23:53
bkk145.

Kudos for the doc

Thanks all for clarifications.
Re: DS abs value   [#permalink] 10 Apr 2008, 23:53
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# Is |x| < 1? (1) |x+1| = 2|x-1| (2) |x-3| 0

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