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?

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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|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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+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 ?
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Nice explaination Bunuel, Thanks! 1+

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 x<-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) --> x=3.

Similarly for other ranges.

For more check Absolute Value chapter for Math Book: math-absolute-value-modulus-86462.html

Hope it helps.
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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.........

good explanation sir thanks
a lot learnt new concept in this
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