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

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

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22 May 2012, 22:05
1
KUDOS
hey kartik,

i also had the same issue initially...then i went thru the link Bunuel has given and it toook me a little time to get it....

so seee.....
the Question states

|x+1| = 2|x-1|
if we take the number line

--------------{-1}----------------{1}---------------

|x+1| if u see will be negative till -1 and after that it will be positive
|x-1| will be negative till 1 and will be positive after that.....

Now we have 3 areas where we need to check

x<-1
-1<=x<=1
x>1

If u check the number line till {-1} both the expression's value will be negative......so when we solve x<-1 we take both of them as Negative

but when we solve -1<=x<=1 the expression |x+1| is positive as its sign changed on x=-1 but the other one |x-1| is still Negative as its sign would change on {1}.
therefore when we solve of this range we take |x+1|= x+1 and 2|x-1| = -2(x-1) = 2(-x+1)

in the third one both are Positive os both the expressions will be positive....

I hope this help you...as it also took me a lot of time to get this concept in my head.........

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

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22 May 2012, 22:30
kartik222 wrote:
Hi Bunuel,

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).

If understand correctly for (A) you are putting '-' in front of both Left/Right Hand Side Equation. Hence coming to "-x-1=2(-x+1)"
Similarly in (C) I guess you are keeping the Left/Right Hand Side Equation both as '+'.

What about (B)?? value of x is between -1 and 1. How did you decide signs' here?
How are you deciding the sign and coming here: x+1=2(-x+1)
I couldn't understand this

I have seen this post: http://gmatclub.com/forum/math-absolute-value-modulus-86462.html as you referred above but ctill couldn't understand it.

I would really appreciate if you can throw a little light on this.

Thanks,

This issue is discussed here: is-x-1-1-x-1-2-x-1-2-x-82478.html#p1088086

Absolute value properties:
When $$x\leq{0}$$ then $$|x|=-x$$, or more generally when $$some \ expression\leq{0}$$ then $$|some \ expression|\leq{-(some \ expression)}$$. For example: $$|-5|=5=-(-5)$$;

When $$x\geq{0}$$ then $$|x|=x$$, or more generally when $$some \ expression\geq{0}$$ then $$|some \ expression|\leq{some \ expression}$$. For example: $$|5|=5$$;

Applying this to $$|x + 1| = 2|x - 1|$$.

If $$-1\leq{x}\leq{1}$$ then $$x+1>0$$ and $$x-1<0$$ so $$|x + 1| =x+1$$ and $$2|x - 1|=-2(x-1)$$ which means that $$|x + 1| = 2|x - 1|$$ becomes: $$x+1=2(-x+1)$$.

Hope it's clear.
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Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]

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30 Aug 2012, 19:12
good explanation sir thanks
a lot learnt new concept in this
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Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]

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31 Aug 2012, 01:36
I'm looking for absolute number problems. I've looked around the forum but i can't find any problem sets.

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

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31 Aug 2012, 01:42
qweert wrote:
I'm looking for absolute number problems. I've looked around the forum but i can't find any problem sets.

Check our Question Banks: viewforumtags.php

PS questions on Absolute Values: search.php?search_id=tag&tag_id=58
DS questions on Absolute Values: search.php?search_id=tag&tag_id=37

Tough inequality and absolute value questions: inequality-and-absolute-value-questions-from-my-collection-86939.html

Theory on absolute values: 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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02 Nov 2012, 01:00
himanshuhpr wrote:
Is |x| < 1?

(1) |x + 1| = 2|x - 1|
(2) |x - 3| > 0

From 1st statement we know that x = 1/3 or x= 3
From 2nd statement we know that that x>3 or x<3

My question is can we combine these 2 results and say (c) is sufficient as x>3 in statement creates an ambiguity and therfore it should be E.

(2) |x - 3| > 0. Absolute value is always more than or equal to zero. Thus this statement just says that $$|x - 3|\neq{0}$$ --> $$x\neq{3}$$.

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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07 Jul 2013, 23:53
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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

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08 Jul 2013, 00:24
1
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yezz wrote:
Is |x| < 1?

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

From St1 we can say that since both LHS and RHS is positive therefore squaring both sides we get

(x+1)^2= 4(x-1)^2
On simplifying we get 3x^2-10x-3= 0 -----> x=3 or x= 1/3

2 ans and hence st 1 is not sufficient. Therefore Option A and D ruled out

St 2 we have |x - 3| ≠ 0 and since Modulus of any no is greater than or equal to 0 (In this case greater than 0) which means x not equal to 3

Not sufficient.

Combining both equation we see that x=1/3 which is in the Question stem range

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

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08 Jul 2013, 15:27
1
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Is |x| < 1?

(1) |x + 1| = 2|x - 1|
(My first problem was to decide how to solve this. I had recently tackled a problem to the effect of |x-z|=|x-y| where I squared both sides to help simplify. In this case, it might be better to find the check points of the LHS and RHS as we are given a variable and an integer as opposed to all variables like the problem I had to solve before)
|x + 1| = 2|x - 1|
x=-1, x=1
so, x<-1, -1<x<1, x>1
x<-1:
-(x+1) = 2 -(x-1)
-x-1 = 2(-x+1)
-x-1 = -2x+2
x=3 INVALID as 3 does not fall within the range of x<-1

-1<x<1
(x+1) = 2 -(x-1)
x+1 = -2x+2
3x=1
x=1/3 VALID as x falls within the range of -1<x<1

x>1
(x+1)=2(x-1)
x+1=2x-2
-x=-3
x=3 VALID as x falls within the range of x>1

We have two valid values for x and therefore we cannot determine a single, correct answer.
INSUFFICIENT

(2) |x - 3| ≠ 0
Tells us nothing about the sign of x. x could be -10 or +10 and |x - 3| ≠ 0 would be valid.
INSUFFICIENT

1+2) x=1/3 OR x=3 AND |x - 3| ≠ 0

Plug in both x values into the above equation. x=3 fails as it does = 0 when the equation says it does not. Therefore, the only remaining x value is 1/3.
SUFFICIENT

(C)

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

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03 Sep 2013, 05:27
From (1) Taking sqaure of both sides ((X+1)^2)/((X-1)^2 = 4, solving this you get X = 3 or 1/3, which is not sufficient.

(2) says X is not = 3

Hence both together tell us the 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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03 Sep 2013, 05:54
I m really sorry, even after reading all the above replies.. i just could'nt get it..

Could someone point out to what i m missing out here..
As per my understanding $$|x| < 1$$ means $$-1 < x < 1$$ .. which eventually means is x=0 ?
Why do we have to consider the cases of x <-1 etc etc.. ?

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

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03 Sep 2013, 05:58
thinktank wrote:
I m really sorry, even after reading all the above replies.. i just could'nt get it..

Could someone point out to what i m missing out here..
As per my understanding $$|x| < 1$$ means $$-1 < x < 1$$ .. which eventually means is x=0 ?
Why do we have to consider the cases of x <-1 etc etc.. ?

Why does "is $$-1 < x < 1$$?" means "is x=0"? We are not told that x is an integer.
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Re: Is |x| < 1? [#permalink]

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30 Jan 2014, 06:18
Economist wrote:
C

1)
if x>1, solving the equality we get x=3.
if -1<x<1, we get x=1/3.
if x<-1, we get x=3.

Not suff
2)
we get x is not equal to 3.

Combining, x can only be 1/3
OA?

Actually your third range is wrong I believe. If x<-1 then x cannot be 3, it isn't a valid solution but the rest is OK

Cheers
J

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

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05 Jul 2014, 00:21
This is very good question
X<-1
-x-1=2(-x+1)

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

X>1
x+1=2(x-1)

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

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02 Sep 2014, 03:17
Bunuel wrote:
tejal777 wrote:
I am having trouble understanding how |x| < 1 translates to -1<x<1.Am missing something:

We have two case,
a. x>0 eg: x=5, |x|=5 so here |x|>0.Therefore,x < 1
b. when x<0,x=-5,|x| is still 5 how does this become x>-1??

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.

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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20 Oct 2014, 22:57
yezz wrote:
Is |x| < 1?

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

1 - On solving the equation, we get the following values of x - 3, 1/3
2 - There can be infinite values of x, except 3

Combining 1 & 2 = x = 1/3

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

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21 Oct 2014, 03:46
Bunuel wrote:
tejal777 wrote:
I am having trouble understanding how |x| < 1 translates to -1<x<1.Am missing something:

We have two case,
a. x>0 eg: x=5, |x|=5 so here |x|>0.Therefore,x < 1
b. when x<0,x=-5,|x| is still 5 how does this become x>-1??

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.

Hope it helps.

Hi Bunuel,
I got most of it, just one doubt here
in case of |x|, we either consider x>0 or x<0,
why are we considering values between -1<x<1.
the reason i am asking this is because the question doesnt ask us to do that, right?

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

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21 Oct 2014, 05:25
arnabs wrote:
Bunuel wrote:
tejal777 wrote:
I am having trouble understanding how |x| < 1 translates to -1<x<1.Am missing something:

We have two case,
a. x>0 eg: x=5, |x|=5 so here |x|>0.Therefore,x < 1
b. when x<0,x=-5,|x| is still 5 how does this become x>-1??

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.

Hope it helps.

Hi Bunuel,
I got most of it, just one doubt here
in case of |x|, we either consider x>0 or x<0,
why are we considering values between -1<x<1.
the reason i am asking this is because the question doesnt ask us to do that, right?

It seems that you need to brush up fundamentals.

Theory on Abolute Values: math-absolute-value-modulus-86462.html
Absolute value tips: tips-and-hints-for-specific-quant-topics-with-examples-172096.html#p1381430
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Re: Is |x| < 1 ? (1) |x + 1| = 2|x – 1| (2) |x – 3| > 0 [#permalink]

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13 Jul 2015, 05:51
1
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apple08 wrote:
Need help for a simple solution to get the answer as I m confuse with this question , many thankss
Is |x| < 1 ?

(1) |x + 1| = 2|x – 1|

(2) |x – 3| > 0

Really appreciate it , Many Thanks

Good question.

For absolute values, you need to take into consideration all possible values. For example, if I say |x| <1, then -1<x<1. We get this by looking at 2 cases :

1. When x>= 0, |x|=x ---> x<1. This value of x<1 does satisfy x>=0. So keep this.

2. When x<0 , |x| = -x ----> -x<1 ----> x>-1 . This value of x>-1 (think x= -0.7 or -0.5) still satisfies x<0. So keep this value as well.

Thus, from 1 and 2 , |x| <1 is nothing but a fancy way of saying -1<x<1. You can also visualise this from numberline:

|x| is the distance of x from 0 (as |x| = |x-0|) on the number line. Thus , when I say |x| <1 this means that distance of x from 0 is less than 1 on either side (as we can approach 0 from 2 directions, 1 from positive and 1 from the negative!).

So, the question actually asks is |x| <1 ----> is -1<x<1.

Now lets look at the 2 statements,

Per statement 1, |x + 1| = 2|x – 1|

Now, your points of analysis are x+1 =0 ----> x=-1 and x-1=0---> x=1. Thus you will have 3 regions to look at:

1. x<-1
2. -1<x<1
3. x>1

When you look at the above 3 regions, you will see that case (1) is not going to satisfy while case (2) will give you x=1/3 and case 3 will give you x = 3. So we see that -1<x<1 is true if x=1/3 but is false if x=3. Thus this statement is not sufficient.

Per statement 2, |x-3| >0 ----> this is true for all x as if you substitute x = -1 or -2 or -0.75 or 7 or 10,|x-3| will always greater than 0 as |x| is DISTANCE of x from 0 (as |x| = |x-0|). As such |x-3| is also distance of x from 3 and as distance is always > 0, this is true for all values of x. Note, that we will not take x = 3 as we have been given |x-3| > 0 and not |x-3| >= 0. Thus this statement is not sufficient.

Combining 1 and 2 we get,

x=1/3 or x=3

Now, |x-3| >0 will not be true if x=3 as |3-3| = 0 and NOT >0. Thus only 1 value of x is possible ----> x = 1/3. Hence both the statements together are sufficient. C is the correct answer.

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

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17 Aug 2016, 22:48
Bunuel wrote:
We have two case,
a. x>0 eg: x=5, |x|=5 so here |x|>0.Therefore,x < 1
b. when x<0,x=-5,|x| is still 5 how does this become x>-1??

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.

Hope it helps.

I don't understand the statement 1, B part. (I am bad with inequalities) so if -1<=x<=1 then the equation becomes x+1=2(-x+1) and that gives x = 1/3; but what about the possibility of the equation -(x+1)=2(x-1) ? When do we consider that inequality? I am confused?

PS: I am a new user
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Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0   [#permalink] 17 Aug 2016, 22:48

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