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Is x < 1 ? (1) x + 1 = 2x  1 (2) x  3 ≠ 0
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Is x < 1 ? (1) x + 1 = 2x  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.
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Originally posted by msand on 31 Dec 2009, 07:24.
Last edited by Bunuel on 04 Dec 2018, 22:04, edited 3 times in total.




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Re: Is x < 1 ? (1) x + 1 = 2x  1 (2) x  3 ≠ 0
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12 Feb 2010, 14:52
Is \(x < 1\), means is \(x\) in the range (1,1)? or is \(1<x<1\) true? (1) \(x + 1 = 2x  1\) There will be two checkpoints (1 and 1), thus three ranges to test: A. \(x<1\) > \(x1=2(x+1)\) > \(x=3\) 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. \(x>1\) > \(x+1=2(x1)\) > \(x=3\) 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. Answer: C.
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Re: Is x < 1 ? (1) x + 1 = 2x  1 (2) x  3 ≠ 0
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31 Dec 2009, 07:51
Is \(x < 1\)? Is \(x < 1\), means is \(x\) in the range (1,1) or is \(1<x<1\) true? (1) \(x + 1 = 2x  1\) 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) > \(x + 1 = 2x  1\) becomes: \(x1=2(x+1)\) > \(x=3\), not OK, as this value is not in the range we are checking (\(x<1\)); B. \(1\leq{x}\leq{1}\) (green range) > \(x + 1 = 2x  1\) 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) > \(x + 1 = 2x  1\) becomes: \(x+1=2(x1)\) > \(x=3\). 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. Hope it helps.
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Re: Is x < 1 ? (1) x + 1 = 2x  1 (2) x  3 ≠ 0
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12 Feb 2010, 03:52
IMO C you have done right, but do consider the cases x>1 and x<1 also and they will give x=3 and using 2nd equation this isnt possible. Thus both taken together are sufficient.
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Re: Is x < 1 ? (1) x + 1 = 2x  1 (2) x  3 ≠ 0
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08 Jun 2010, 05:16
Thank you Bunuel for your great explanation!
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Re: Is x < 1 ? (1) x + 1 = 2x  1 (2) x  3 ≠ 0
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26 Jun 2010, 02:14
Bunuel wrote: Is \(x < 1\)?
A. \(x<1\) (blue range) > \(x + 1 = 2x  1\) becomes: \(x1=2(x+1)\) > \(x=3\), not OK, as this value is not in the range we are checking (\(x<1\)); B. \(1\leq{x}\leq{1}\) (green range) > \(x + 1 = 2x  1\) 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) > \(x + 1 = 2x  1\) becomes: \(x+1=2(x1)\) > \(x=3\). 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) > \(x + 1 = 2x  1\) becomes: \(x1=2(x+1)\) Same with other two ranges (green and red) Appreciate your time.



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Re: Is x < 1 ? (1) x + 1 = 2x  1 (2) x  3 ≠ 0
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26 Jun 2010, 03:13
JoyLibs wrote: Bunuel wrote: Is \(x < 1\)?
A. \(x<1\) (blue range) > \(x + 1 = 2x  1\) becomes: \(x1=2(x+1)\) > \(x=3\), not OK, as this value is not in the range we are checking (\(x<1\)); B. \(1\leq{x}\leq{1}\) (green range) > \(x + 1 = 2x  1\) 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) > \(x + 1 = 2x  1\) becomes: \(x+1=2(x1)\) > \(x=3\). 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) > \(x + 1 = 2x  1\) becomes: \(x1=2(x+1)\) Same with other two ranges (green and red) Appreciate your time. We have: \(x + 1 = 2x  1\). 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 \(x1=(x1)\) > \(x + 1 = 2x  1\) becomes: \((x+1)=2(x+1)\) The same for other ranges. Check this for more: mathabsolutevaluemodulus86462.htmlHope it helps.
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Re: Is x < 1 ? (1) x + 1 = 2x  1 (2) x  3 ≠ 0
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23 Oct 2010, 08:09
tatane90 wrote: Is x< 1? (1) x + 1 = 2x  1 (2) x  3 ≠ 0
thanks in advance (1) x+1 = 2x1 Lets solve this thing +/(x+1) = +/2 (x1) Case 1 : x+1 = 2(x1) .. x=3 Case 2 : x+1 = 2(x1) .. 3x=1 .. x=1/3 Case 3 : x1 = 2(x1) .. 3x=1 .. x=1/3 Case 4 : x1 = 2x+2 .. x=3 So x is either 3 or 1/3. Not enough to answer question (2) This only implies x cannot be 3. Clearly insufficient (1+2) x can only be 1/3. Sufficient to answer YES Answer : (c)
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Re: Is x < 1 ? (1) x + 1 = 2x  1 (2) x  3 ≠ 0
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06 Mar 2016, 02:24
Square both sides of statement (1) and we get x=3 or x=1/3. So statement (1) is insufficient. statement (2) tells us that X is not equal to 3. Clearly, statement (2) alone is insufficient. But if we combine statement (1) and statement (2), we get x= 1/3 which is definitely less than 1. Hence C .
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Re: Is x < 1 ? (1) x + 1 = 2x  1 (2) x  3 ≠ 0
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09 Mar 2016, 08:15
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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Re: Is x < 1 ? (1) x + 1 = 2x  1 (2) x  3 ≠ 0
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09 Jun 2018, 01:07
Bunuel wrote: Is \(x < 1\), means is \(x\) in the range (1,1)? or is \(1<x<1\) true?
(1) \(x + 1 = 2x  1\) There will be two checkpoints (1 and 1), thus three ranges to test: A. \(x<1\) > \(x1=2(x+1)\) > \(x=3\) 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. \(x>1\) > \(x+1=2(x1)\) > \(x=3\) 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.
Answer: C. Hi Bunuel, For the 2nd statement, since it is modulus, don't we need to consider this as well: (x3)!=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.



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Re: Is x < 1 ? (1) x + 1 = 2x  1 (2) x  3 ≠ 0
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09 Jun 2018, 04:02
kkrrsshh wrote: For the 2nd statement, since it is modulus, don't we need to consider this as well: (x3)!=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. Hey kkrrsshh , The error is in the highlighted statement above. (x3)!=0 =>  x + 3 !=0 => x !=  3 => x != 3 Does that make sense?
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Re: Is x < 1 ? (1) x + 1 = 2x  1 (2) x  3 ≠ 0
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