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

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.

Hope it helps.
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It's about expanding the absolute value in different ranges. Discussed in Walker's post on absolute values: math-absolute-value-modulus-86462.html

If you're more comfortable with "squaring" method (note that it's not always applicable) then you can apply it to the first statement too (and not only to the stem).

Is \(|x| < 1\)?

Is \(|x| < 1\) --> is \(-1<x<1\)?

(1) \(|x + 1| = 2|x - 1|\) --> square both sides: \((x+1)^2=4(x-1)^2\) --> \(x^2+2x+1=4x^2-8x+4\) --> \(3x^2-10x+3=0\) --> \(x=\frac{1}{3}\) or \(x=3\) --> 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.

Hope it's clear.
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Answer is C - you beat me to it walker

For statement 1, you need to actually break it down to 2 separate statements to account for both scenarios:

Option #1 ->
x + 1 = 2(x-1)
x + 1 = 2x -2
x = 2x -3
3 + x = 2x
3 = x {or also Option #2}

Option #2 -> x + 1 = -2(x-1)
x + 1 = -2x +2
x = -2x + 1
3x = 1
x = 1/3

Statement (1) is insufficient because |x| could be 3 or 1/3. So the answer is sometimes yes, sometimes no which means insufficient.

Statement (2) is insufficient because we are told that |x -3| ≠ 0. This means that as long as x ≠ 3, then we're ok. There are far too many options so we certainly have a sometimes yes, and sometimes no.

But together, from Statement (1) we have either 3 or 1/3 and from statement (2) we have everything BUT 3. So the only overlap we have for possible values to make both statements true is 1/3. So the answer is Yes, |x| IS less than 1 and we have enough info to answer the question.

|x|<1 means x e (-1,1)

1)|x+1|=2|x-1|
There are 2 key points (x=-1 and x=1) where one of the expressions under modules passes 0 and change its sign.
x<-1: -(x+1)=-2(x-1) --> -x-1=-2x+2 ---> x = 3. But x=3 does not satisfy x<-1 condition.
-1<=x<1: x+1 = -2(x-1) ---> x+1= -2x+2 --> x=1/3. It satisfies the condition.
x>=1: x+1=2x-2 ---> x=3. It satisfies the condition.
Insufficient

2) |x - 3| ≠ 0
x≠3
Insufficient

1&2) Only x=1/3 satisfies both statements. Sufficient.

Jade3, check this post, please: math-absolute-value-modulus-86462.html
If you find it useful/useless, let me know
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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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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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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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On to your question.
Is |x| < 1?
Yes, essentially this just asks you whether x lies between -1 and 1. |x| represents the distance from 0 which will be less than 1 if x lies within -1 < x < 1.

Statement 1: |x+1| = 2|x-1|
Draw the number line. This equation says that distance from -1 is twice the distance from 1. At which point is the distance from -1, twice the distance from 1? If you split the distance (2 units) between -1 and 1 into 3 parts, 1 part away from 1 and 2 parts away from -1 will be the point 1/3. The diagram below will show you this situation. Similarly the distance between -1 and 1 is 2 so if you go 2 units further to the right of 1, you get the point 3 which will be 4 units away from -1 i.e. twice the distance from 1.

x could be 1/3 or 3. Hence x may or may not lie between -1 and 1. Not sufficient.

Statement 2: |x-3| does not equal 0
This statement just says that x is not 3. It is not sufficient alone.

Both together, we get that x must be either 3 or 1/3 and it is not 3. Then x must be 1/3 and must lie between -1 and 1. Sufficient.
Answer (C)
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+1 C

However, I used using other methods: a) Squaring both sides of the equation because |x| = \((x^2)\)^(1/2), b) Combining different scenarios between these two possibilities: (x+1)>0, (x+1)<0, (x-1)>0, (x-1)<0. Therefore, I get four possible scenarios.

Bunuel, could you explain the logic behing your method to solve this problem?
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Hi Bunuel ,
I also used the squaring method....
In what all situations is the squaring method not applicable ???
Can you please explain why ?

Sure, for example it's not always applicable for inequalities.

A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
\(2<4\) --> we can square both sides and write: \(2^2<4^2\);
\(0\leq{x}<{y}\) --> we can square both sides and write: \(x^2<y^2\);

But if either of side is negative then raising to even power doesn't always work.
For example: \(1>-2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we can not square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are non-negative.

B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).
For example:
\(-2<-1\) --> we can raise both sides to third power and write: \(-2^3=-8<-1=-1^3\) or \(-5<1\) --> \((-5)^3=-125<1=1^3\);
\(x<y\) --> we can raise both sides to third power and write: \(x^3<y^3\).

So for our question we can not square x/|x|< x as we don't know the sign of either of side.

Hope it helps.
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In DS questions, the idea is to break down the question stem to a situation that can be dealt with easily. In this question, the question stem may appear a bit too short for your liking. However, you can still break it down to a simpler level.

|x|<1 is satisfied by the range -1<x<1. So, in other words, the question is asking us to find out if -1<x<1. This is what I meant about breaking down the stem. Now that you have done this, you know what to look for in the statements.

Also remember that |x| = √(\(x^2\) ).

Therefore, in statement I, we can say, |x+1| = √(〖\((x+1)〗^2\)) and |x-1| = √(\(〖(x-1)〗^2\) )

So, √(〖\((x+1)〗^2\) ) = 2 √(\(〖(x-1)〗^2\) ). On squaring both sides and simplifying, we get,

3\(x^2\) – 10 x + 3 = 0. Solving this quadratic equation, we get x = 3 or x = 1/3.

This is true. The distance of the number 3 from -1 is indeed double that of its distance from 1. Similarly, the distance of the number 1/3 from -1 is double that of its distance from 1. So, it actually ties in with the definition of |x-a| on which statement I is framed.

However, all this does not still give us a unique value of x. Statement I is insufficient.
Answer options A and D can be eliminated. Possible answer options are B, C or E.

Statement II alone says x is not equal to 3. This is clearly insufficient to determine if -1<x<1. Answer option B can be eliminated.

Combining both statements, we know that x has to be equal to 1/3 since it cannot be equal to 3. The combination of statements is sufficient. Answer option E can be eliminated.

The correct answer option is C.

Hope this helps!

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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Hi Bunuel i have come across same question but in second statement it |x-3|>0

I am bit confused as when we say |x|>3 we mean -3<x<3

I don't know what to make out of |x-3|>0

Can I just deduce that x>3

Posted from my mobile device

Here is that question: https://gmatclub.com/forum/is-x-1-1-x-1 ... fl=similar

|x| > 3 means that x > 3 or x < -3. Think about it this way: |x| > 3 means that the distance between x and 0 is more than 3, so x must be either less than -3 or more than 3.

Now about |x - 3| > 0. The absolute value is always more than or equal to 0. So, |x - 3| > 0 just means that |x - 3| ≠ 0 or x ≠ 3. You can use the distance concept here too: |x - 3| > 0 means that the distance between x and 3 is more than 0, so x can be any number but 3 (for any number but 3 the distance between x and 3 is more that 0).

Hope it helps.
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