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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?

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. _________________