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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.

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.

Check Walker's post: Absolute value and also to practice in absolute value questions check the link of Inequalities in my signature.

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: \(-x-1=2(-x+1)\)

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)

Appreciate your time.

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: \(-(x+1)=2(-x+1)\)

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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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.

Question --> Is |x|<1 ? Is -1<x<1?

Statement 1 Square both sides You end up with quadratic 3x^2-10x+30 Factorize (3x-1)(x-3) x= 3 or x=1/3. Two answers, therefore Insuff

Statement 2 x cannot be 3

Statement 1 and 2 together Then it has to be x 1/3 and it is within the range so answer is YES

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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