Is \(|x| < 1\), means is \(x\) in the range (-1,1)? or is \(-1<x<1\) true?

(1) \(|x + 1| = 2|x - 1|\)
There will be two checkpoints (-1 and 1), thus three ranges to test:
A. \(x<-1\) --> \(-x-1=2(-x+1)\) --> \(x=3\) not good, as \(x<-1\);

B. \(-1\leq{x}\leq{1}\) --> \(x+1=2(-x+1)\) --> \(x=\frac{1}{3}\) is in the range \(-1<x<1\). OK.

C. \(x>1\) --> \(x+1=2(x-1)\) --> \(x=3\) is in the range \(x>1\). OK

So we got TWO values of \(x\), \(\frac{1}{3}\) and \(3\). One (\(\frac{1}{3}\)), is in the range (-1,1) and another (\(3\)), is out of this range. Not sufficient.

(2) \(|x - 3|\neq0\)
Just says that \(x\neq{3}\). But we don't know whether \(x\) is in the range (-1,1) or not.

(1)+(2) From (1): \(x=\frac{1}{3}\) or \(x=3\) and from (2) \(x\neq{3}\) --> means \(x\) can have only one value \(\frac{1}{3}\), which IS in the range (-1,1). Sufficient.

Answer: C.
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Is \(|x| < 1\)?

Is \(|x| < 1\), means is \(x\) in the range (-1,1) or is \(-1<x<1\) true?

(1) \(|x + 1| = 2|x - 1|\)
Two key points: \(x=-1\) and \(x=1\) (key points are the values of x when absolute values equal to zero), thus three ranges to check:
---------{-1}--------{1}---------

A. \(x<-1\) (blue range) --> \(|x + 1| = 2|x - 1|\) becomes: \(-x-1=2(-x+1)\) --> \(x=3\), not OK, as this value is not in the range we are checking (\(x<-1\));
B. \(-1\leq{x}\leq{1}\) (green range) --> \(|x + 1| = 2|x - 1|\) becomes: \(x+1=2(-x+1)\) --> \(x=\frac{1}{3}\). OK, as this value is in the range we are checking (\(-1\leq{x}\leq{1}\));
C. \(x>1\) (red range) --> \(|x + 1| = 2|x - 1|\) becomes: \(x+1=2(x-1)\) --> \(x=3\). OK, as this value is in the range we are checking (\(x>1\)).

So we got TWO values of \(x\) (two solutions): \(\frac{1}{3}\) and \(3\), first is in the range (-1,1) but second is out of the range. Not sufficient.

(2) \(|x - 3|\neq{0}\)
Just says that \(x\neq{3}\). But we don't know whether \(x\) is in the range (-1,1) or not.

(1)+(2) \(x=\frac{1}{3}\) or \(x=3\) AND \(x\neq{3}\) --> means \(x\) can have only value \(\frac{1}{3}\), which is in the range (-1,1). Sufficient.

Answer: C.

Check Walker's post: Absolute value and also to practice in absolute value questions check the link of Inequalities in my signature.

Hope it helps.
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I am not clear about how you derived this.
\(x<-1\) (blue range) --> \(|x + 1| = 2|x - 1|\) becomes: \(-x-1=2(-x+1)\)

Same with other two ranges (green and red)

Appreciate your time.

We have: \(|x + 1| = 2|x - 1|\).

Absolute value properties:
If \(x\geq{0}\), then \(|x|=x\) and if \(x\leq{0}\), then \(|x|=-x\).

For the range \(x<-1\) (blue range) --> \(x+1<0\) so \(|x+1|=-(x+1)\) and \((x - 1)<0\) so \(|x-1|=-(x-1)\) --> \(|x + 1| = 2|x - 1|\) becomes: \(-(x+1)=2(-x+1)\)

The same for other ranges.

Check this for more: math-absolute-value-modulus-86462.html

Hope it helps.
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(1) |x+1| = 2|x-1|
Lets solve this thing
+/-(x+1) = +/-2 (x-1)

Case 1 : x+1 = 2(x-1) .. x=3
Case 2 : x+1 = -2(x-1) .. 3x=1 .. x=1/3
Case 3 : -x-1 = 2(x-1) .. 3x=1 .. x=1/3
Case 4 : -x-1 = -2x+2 .. x=3

So x is either 3 or 1/3. Not enough to answer question

(2) This only implies x cannot be 3. Clearly insufficient

(1+2) x can only be 1/3. Sufficient to answer YES

Answer : (c)
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Hello Bunuel,

Since the second statement says |x-3| is not equal to zero, doesn't this also mean |x-3| > 0, in which case x>3 and x<-3 which is sufficient to answer the question?

Appreciate your help.

Thank you. Bunuel

An absolute value is always more than or equal to zero: \(|anything| \geq 0\). Which means that |x - 3| ≠ 0 just excludes x being 3. So, yes, it does mean |x - 3| > 0. But |x - 3| > 0 does not mean that x > 3 or x < -3. It means x < 3 or x > 3: x can be any number but 3.
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|x|<1 = -1<x<1

Statement 1:

|X+1|=2|X-1|
Squaring on both sides
3x^2-10x+3=0
which gives us 2 solutions (1/3, 3) -- Not Sufficient

statement 2:

|x-3| ≠ 0

which also gives more than 2 solutions

Combined sufficient : Ans - C

Hi shrouded1, thought absolute value give +/- two options to calculate. Therefore not sure why the need for calculating four possibilities here?
Could you help explain please? Thanks

Hi brunel, Noticed that - sign is not according to when x<-1 in B range on RHS, therefore how do you decide which side to calculate for - and which side doesn't in those 3 ranges? Thanks for your time in advanced.