Data Sufficiency

Is |x| < 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| &#8800; 0

(C)

[spoiler]First I did the equation X+1 = 2x – 2.

I found that X = 3

Then I did the negation so –(X+1) = 2x – 2.

X = 1/3

Therefore – I have two values, one greater than 1 one less than 1 so I can eliminate A/D

I then looked at statement two. The statement obviously does not help us.

Eliminate B.

CE remaining

From statement one we discovered X is either 3 or 1/3. With statement 2, I can eliminate 3 and therefore can definitively answer NO.

Therefore C.
[/spoiler]

Any other thoughts on other methods?

Don't you have to take into account the negation on the other side of (1)?

doclkk wrote:Is |x| < 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| &#8800; 0

(C)

[spoiler]First I did the equation X+1 = 2x – 2.

I found that X = 3

Then I did the negation so –(X+1) = 2x – 2.

X = 1/3

Therefore – I have two values, one greater than 1 one less than 1 so I can eliminate A/D

I then looked at statement two. The statement obviously does not help us.

Eliminate B.

CE remaining

From statement one we discovered X is either 3 or 1/3. With statement 2, I can eliminate 3 and therefore can definitively answer NO.

Therefore C.
[/spoiler]

Any other thoughts on other methods?

C because
Is |x| < 1?
we are asked whether -1<x<1
(1) |x + 1| = 2|x - 1|
means distance from X to -1 is 2 times the distance from x to 1
you can imagine two possible value. one between -1 and 1. and the other on the right side of 1
Not suff.
(2) |x - 3| &#8800; 0
X not equal to 3
but can be anything.
Not suff

Combined
if x not equal to 3. it must be the value between -1 and 1
SUFF

Good solution ToHellAndBack..
I had a longer solution using the critical point approach and arrived at C.

In this approach: stmt 1 gave me 2 solutions 3 or 1/3.
stmt said x != 3.

Combining x = 1/3
hence ans is C.

IMO C

doclkk wrote:Is |x| < 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| &#8800; 0

Any other thoughts on other methods?

Asking whether -1<x<1

1. for x<-1 -(x+1) = -2(x-1) or x+1 = 2x-2 or x = 3
-1<x<1 => x+1 = -2(x-1) or x+1 = -2x+2 or x=1/3
x>1 x+1 = 2x-2 or x =3
so sometime x =3 sometime 1/3 nothing can be said with surety

2. x!= 3

If we combine 1&2 we get x=1/3 sufficient.