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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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Updated on: 04 Dec 2018, 22:04
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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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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.

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

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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.

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

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

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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)

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

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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)

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

(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

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

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06 Mar 2016, 02:24
1
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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09 Mar 2016, 08:15
1
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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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.

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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09 Jun 2018, 04:02
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.

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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Re: Is |x| < 1 ? (1) |x + 1| = 2|x - 1| (2) |x - 3| ≠ 0 &nbs [#permalink] 04 Dec 2018, 20:07
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