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

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

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Is |x| < 1 ?


(1) |x + 1| = 2|x - 1|

(2) |x - 3| ≠ 0


Can anybody suggest a general approach to solve this kind of problems ? I am facing problems to solve these type of questions where equations are containing absolute values together with inequalities... please help.

Originally posted by msand on 31 Dec 2009, 07:24.
Last edited by Bunuel on 04 Dec 2018, 22:04, edited 3 times in total.
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Re: Is |x| < 1 ? (1) |x + 1| = 2|x - 1| (2) |x - 3| ≠ 0  [#permalink]

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New post 12 Feb 2010, 14:52
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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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Re: Is |x| < 1 ? (1) |x + 1| = 2|x - 1| (2) |x - 3| ≠ 0  [#permalink]

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New post 31 Dec 2009, 07:51
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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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Re: Is |x| < 1 ? (1) |x + 1| = 2|x - 1| (2) |x - 3| ≠ 0  [#permalink]

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New post 12 Feb 2010, 03:52
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IMO C

you have done right, but do consider the cases x>1 and x<-1 also
and they will give x=3 and using 2nd equation this isnt possible.

Thus both taken together are sufficient.
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Re: Is |x| < 1 ? (1) |x + 1| = 2|x - 1| (2) |x - 3| ≠ 0  [#permalink]

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New post 08 Jun 2010, 05:16
Thank you Bunuel for your great explanation!

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

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New post 26 Jun 2010, 02:14
Bunuel wrote:
Is \(|x| < 1\)?


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 (\(x>1\)).




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

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New post 26 Jun 2010, 03:13
1
JoyLibs wrote:
Bunuel wrote:
Is \(|x| < 1\)?


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 (\(x>1\)).




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

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New post 23 Oct 2010, 08:09
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tatane90 wrote:
Is |x|< 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| ≠ 0

thanks in advance


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

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New post 06 Mar 2016, 02:24
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Square both sides of statement (1) and we get x=3 or x=1/3.
So statement (1) is insufficient.

statement (2) tells us that X is not equal to 3. Clearly, statement (2) alone is insufficient.

But if we combine statement (1) and statement (2), we get x= 1/3 which is definitely less than 1.

Hence C .
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Re: Is |x| < 1 ? (1) |x + 1| = 2|x - 1| (2) |x - 3| ≠ 0  [#permalink]

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New post 09 Mar 2016, 08:15
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Here is my approach
here we need to see whether x lies in the zone => (-1,1)
statement 1 is a double modulus equality => Remember whenever you encounter a double modulus equality (not inequality ) we can have only two cases => one where both of the signs will be +ve and the other when one is positive and the other is negative(take any one positive and the other negative)

as per this logic => x=3 and x=1/3 => 3 does not satisfy our bound but 1/3 does
hence insufficient
statement 2 is insufficient too as x can be 0 or 50
combining them we get x≠3 so x=1/3 => which satisfies our bound => C is sufficient
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Re: Is |x| < 1 ? (1) |x + 1| = 2|x - 1| (2) |x - 3| ≠ 0  [#permalink]

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New post 09 Jun 2018, 01:07
Bunuel wrote:
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.



Hi Bunuel,

For the 2nd statement, since it is modulus, don't we need to consider this as well: -(x-3)!=0 so x!=-3 ?
So there will two not equals,
x!=3 or x!=-3.
Please let me know whether I am right in this logic.
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Re: Is |x| < 1 ? (1) |x + 1| = 2|x - 1| (2) |x - 3| ≠ 0  [#permalink]

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New post 09 Jun 2018, 04:02
kkrrsshh wrote:
For the 2nd statement, since it is modulus, don't we need to consider this as well: -(x-3)!=0 so x!=-3?
So there will two not equals,
x!=3 or x!=-3.
Please let me know whether I am right in this logic.


Hey kkrrsshh ,

The error is in the highlighted statement above.

-(x-3)!=0

=> - x + 3 !=0

=> -x != - 3

=> x != 3

Does that make sense?
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Re: Is |x| < 1 ? (1) |x + 1| = 2|x - 1| (2) |x - 3| ≠ 0  [#permalink]

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