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(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

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.. ?

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. _________________

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.

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.

Answer: C.

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?

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.

Answer: C.

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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