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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.
Is \(|x| < 1\), means is \(x\) in the range (-1,1)? or is \(-1<x<1\) true?
(1) \(|x + 1| = 2|x - 1|\) There will be two checkpoints (-1 and 1), thus three ranges to test: A. \(x<-1\) --> \(-x-1=2(-x+1)\) --> \(x=3\) not good, as \(x<-1\);
B. \(-1\leq{x}\leq{1}\) --> \(x+1=2(-x+1)\) --> \(x=\frac{1}{3}\) is in the range \(-1<x<1\). OK.
C. \(x>1\) --> \(x+1=2(x-1)\) --> \(x=3\) is in the range \(x>1\). OK
So we got TWO values of \(x\), \(\frac{1}{3}\) and \(3\). One (\(\frac{1}{3}\)), is in the range (-1,1) and another (\(3\)), is out of this range. Not sufficient.
(2) \(|x - 3|\neq0\) Just says that \(x\neq{3}\). But we don't know whether \(x\) is in the range (-1,1) or not.
(1)+(2) From (1): \(x=\frac{1}{3}\) or \(x=3\) and from (2) \(x\neq{3}\) --> means \(x\) can have only one value \(\frac{1}{3}\), which IS in the range (-1,1). Sufficient.
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: \(-(x+1)=2(-x+1)\)
Here is my approach here we need to see whether x lies in the zone => (-1,1) statement 1 is a double modulus equality => Remember whenever you encounter a double modulus equality (not inequality ) we can have only two cases => one where both of the signs will be +ve and the other when one is positive and the other is negative(take any one positive and the other negative)
as per this logic => x=3 and x=1/3 => 3 does not satisfy our bound but 1/3 does hence insufficient statement 2 is insufficient too as x can be 0 or 50 combining them we get x≠3 so x=1/3 => which satisfies our bound => C is sufficient
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Is \(|x| < 1\), means is \(x\) in the range (-1,1)? or is \(-1<x<1\) true?
(1) \(|x + 1| = 2|x - 1|\) There will be two checkpoints (-1 and 1), thus three ranges to test: A. \(x<-1\) --> \(-x-1=2(-x+1)\) --> \(x=3\) not good, as \(x<-1\);
B. \(-1\leq{x}\leq{1}\) --> \(x+1=2(-x+1)\) --> \(x=\frac{1}{3}\) is in the range \(-1<x<1\). OK.
C. \(x>1\) --> \(x+1=2(x-1)\) --> \(x=3\) is in the range \(x>1\). OK
So we got TWO values of \(x\), \(\frac{1}{3}\) and \(3\). One (\(\frac{1}{3}\)), is in the range (-1,1) and another (\(3\)), is out of this range. Not sufficient.
(2) \(|x - 3|\neq0\) Just says that \(x\neq{3}\). But we don't know whether \(x\) is in the range (-1,1) or not.
(1)+(2) From (1): \(x=\frac{1}{3}\) or \(x=3\) and from (2) \(x\neq{3}\) --> means \(x\) can have only one value \(\frac{1}{3}\), which IS in the range (-1,1). Sufficient.
For the 2nd statement, since it is modulus, don't we need to consider this as well: -(x-3)!=0 so x!=-3 ? So there will two not equals, x!=3 or x!=-3. Please let me know whether I am right in this logic.
For the 2nd statement, since it is modulus, don't we need to consider this as well: -(x-3)!=0 so x!=-3? So there will two not equals, x!=3 or x!=-3. Please let me know whether I am right in this logic.
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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57% (01:44) correct 
