Is x < 1?
(1) x + 1 = 2x  1
(2) x  3 ≠ 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?
Absolute value fun!
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C becausedoclkk wrote:Is x < 1?
(1) x + 1 = 2x  1
(2) x  3 ≠ 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?
Is x < 1?
we are asked whether 1<x<1
(1) x + 1 = 2x  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 ≠ 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
The powers of two are bloody impolite!!

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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.
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.

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Asking whether 1<x<1doclkk wrote:Is x < 1?
(1) x + 1 = 2x  1
(2) x  3 ≠ 0
Any other thoughts on other methods?
1. for x<1 (x+1) = 2(x1) or x+1 = 2x2 or x = 3
1<x<1 => x+1 = 2(x1) or x+1 = 2x+2 or x=1/3
x>1 x+1 = 2x2 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.
Charged up again to beat the beast