It is currently 20 Oct 2017, 04:08

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
Senior Manager
Joined: 10 Mar 2008
Posts: 358

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

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

### Show Tags

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
Joined: 12 Jul 2008
Posts: 514

Kudos [?]: 162 [0], given: 0

Schools: Wharton

### Show Tags

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

Kudos [?]: 162 [0], given: 0

Intern
Joined: 18 Sep 2008
Posts: 5

Kudos [?]: 2 [0], given: 0

### Show Tags

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.

Kudos [?]: 2 [0], given: 0

Senior Manager
Joined: 10 Mar 2008
Posts: 358

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

### Show Tags

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...

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

Senior Manager
Joined: 10 Mar 2008
Posts: 358

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

### Show Tags

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

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

Re: Absolute value   [#permalink] 29 Sep 2008, 10:56
Display posts from previous: Sort by