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If y=x+1/x and x!=0, is xy>0?
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If \(y=\frac{x+1}{x}\) and \(x\neq{}0\), is \(xy>0\) ? (1) \(x^2+2x+1>0\) (2) \(y\neq{0}\) My own question, as always any feedback is appreciated Click here for the OE
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Originally posted by Zarrolou on 16 May 2013, 11:24.
Last edited by Bunuel on 12 Jul 2014, 12:43, edited 3 times in total.
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Re: If y=x+1/x and x!=0, is xy>0?
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16 May 2013, 11:47
Zarrolou wrote: If \(y=\frac{x+1}{x}\) and \(x\neq{}0\), is \(xy>0\) ?
A)\(x^2+2x+1>0\)
B)\(y\neq{0}\)
My own question, as always any feedback is appreciated Kudos to the first correct solution(s)! I think Answer is D question can be written as x+1 = xy so, to prove that xy> 0 we essentially need to prove that x1 > 0 STAT1 x^2 + 2x + 1 > 0 means, (x+1) ^ 2 >0 taking square root on both the sides does not change the inequality so, we have x+1 > 0 which means that xy > 0 So, SUFFICIENT STAT2 y != 0 we know that x+1 = xy => xy is non negative. so, it is either 0 or positive since we know that both x and y ar enot equal to zero so, xy > 0 so, SUFFICIENT Hence, Answer will be D
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Re: If y=x+1/x and x!=0, is xy>0?
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Updated on: 16 May 2013, 11:59
E it is..................... For getting xy>0 we need to prove that both x and y are either greater than 0 I.e positive or both of them are negative......
Xy = x+1
Once negative: Xy= x1 Xy+x=1 X=1/y+1
Once positive: similarly we will get x= 1/y1
We are getting x once positive and negative ......so can't say xy >0 or not....
II X^2+2x+1>0 Gives us (x+1)^2>0 Gives us x>1 and from question stem we know that x is not equal to 0  hence x will be positive. But we still don't know their value/ sign of y. Hence this statement is also not sufficient.
Hope I'm correct.....
Originally posted by nikhilsehgal on 16 May 2013, 11:41.
Last edited by nikhilsehgal on 16 May 2013, 11:59, edited 1 time in total.



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Re: If y=x+1/x and x!=0, is xy>0?
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16 May 2013, 12:23
Agreed I made the mistake of multiplying y by x and having to test four different cases: yx = x+1 yx = x+1 Then I had to test cases for X>0 and X<0 (i.e. four total cases) and each one turned out to hold true. Your way makes more sense and is quicker to boot. for 2.) xy = x+1 so xy could be 0 or greater than zero. We know that X isn't zero from the stem and we know that Y isn't zero from #2. Therefore x/y must be greater or less than zero yielding a positive or negative result. However, because xy = the absolute value of x+1 xy can only be greater than or equil to zero and because both x and y are not zero, product zy MUST be greater than zero. Good work! nktdotgupta wrote: Zarrolou wrote: If \(y=\frac{x+1}{x}\) and \(x\neq{}0\), is \(xy>0\) ?
A)\(x^2+2x+1>0\)
B)\(y\neq{0}\)
My own question, as always any feedback is appreciated Kudos to the first correct solution(s)! I think Answer is D question can be written as x+1 = xy so, to prove that xy> 0 we essentially need to prove that x1 > 0 STAT1 x^2 + 2x + 1 > 0 means, (x+1) ^ 2 >0 taking square root on both the sides does not change the inequality so, we have x+1 > 0 which means that xy > 0 So, SUFFICIENT STAT2 y != 0 we know that x+1 = xy => xy is non negative. so, it is either 0 or positive since we know that both x and y ar enot equal to zero so, xy > 0 so, SUFFICIENT Hence, Answer will be D



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Re: If y=x+1/x and x!=0, is xy>0?
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16 May 2013, 13:23
Zarrolou wrote: If \(y=\frac{x+1}{x}\) and \(x\neq{}0\), is \(xy>0\) ?
A)\(x^2+2x+1>0\)
B)\(y\neq{0}\)
My own question, as always any feedback is appreciated Kudos to the first correct solution(s)! The answer sure does appear to be [D] Statement 1: \(x^2+2x+1>0\) Hence \((x+1)^2\) > 0 Now since the above is a squared term, the value will always be positive except for x = 1. Since the statement doesnt include 0 under the range we can assume that x is not equal to 1. Putting the same values into our main equation i.e. xy = x+1 we, can be sure that x+1 is not equal to zero at any point. Hence the statement would be sufficient, since if we prove x+1 is not equal to zero, then its always greater than 0! Hence xy = x+1 > 0 Statement 2. Now if y is not equal to zero, we can assume x+1 is never equal to zero! Hence we prove the fact by the same above method, that xy = x+1 > 0 The idea in the above question, is to establish that at no point x = 1. At only that value y = 0 and xy = 0. Otherwise for all other possible values x+1 > 0. As always wonderful question Zarrolou! Hope my procedure is correct! Regards, Arpan
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Re: If y=x+1/x and x!=0, is xy>0?
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16 May 2013, 13:46
Good job WholeLottaLove, nktdotgupta, arpanpatnaik ! Official explanationIf \(y=\frac{x+1}{x}\) and \(x\neq{}0\), is \(xy>0\) ?Rewrite the question as (multiply by x) \(xy=x+1\) The \(abs\) is \(\geq{0}\), so basically we have to check if \(xy=x+1\neq{0}\) A)\(x^2+2x+1>0\)\((x+1)^2>0\) So \(x\neq{1}\) and \(x+1>0\) This is what we are looking for. Sufficient B)\(y\neq{0}\)\(y=\frac{x+1}{x}\neq{0}\) so \(x+1\neq{0}\) Sufficient The correct answer is DHope everything is clear
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Re: If y=x+1/x and x!=0, is xy>0?
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16 May 2013, 15:41
@Zarrolou,Regret for not responding within time.If \(y=\frac{x+1}{x}\) and \(x\neq{}0\), is \(xy>0\) ?
A)\(x^2+2x+1>0\)
B)\(y\neq{0}\)First lets simplify the inequality \(\frac{x+1}{x}\) –y = 0 > \(\frac{x+1xy}{x}\) = 0 > we know \(x\neq{}0\), then x+1xy must be zero. Hence x+1  xy = 0 > x+1 = xy We are asked whether xy>0 > Whether x+1 > 0 ? > We know the expression within modules can either be zero or greater than zero. For xy to be greater than zero x+1 has to be greater than zero. x+1 will be zero only when x=1 and for any other value of x, x+1 will always be greater than zero So the question can be rephrased as whether \(x\neq{1}\)Statement 1) \(x^2\) + 2x + 1 > 0This is an quadratic inequality. Rule : For any quadratic inequation \(ax^2\) + bx + c > 0, if \(b^2\) – 4ac = 0 and a > 0 then the inequality holds true outside the interval of rootsIn our case \(b^2\) – 4ac = 4 – 4 = 0 and a > 0 so \(x^2\) + 2X + 1 > 0 will hold true for all values beyond the Root(s) of equation (Towards any direction  Positive or Negative) \(x^2\) + 2x + 1 = 0 > x(x+1) +1(x + 1) = 0 > (x+1)(x+1) = 0 > x=Root = 1 So \(x^2\) + 2x + 1 > 0 will hold true for any of x except for 1 This reveals that \(x\neq{1}\) and xy>0 > SufficientStatement 2) \(y\neq{0}\)From the question stem we know \(x\neq{}0\) As per Statement 2, \(y\neq{0}\)>That means both X and Y are nonzero. x+1 = xy xy can be either Positive or Negative x+1 can be Zero or Positive Combining both these inferences we can conclude that XY must be Positive. SufficientAnswer = D
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Re: If y=x+1/x and x!=0, is xy>0?
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24 May 2014, 11:22
Lets make things simpler here. If y=x+1/x, Is xy>0? Now since x+1 is always non negative then, xy>0 always except when x+1=0> x=1. Thus, we need to prove that x is not equal to 1. Statement 1. We have that (x+1)^2>0 Therefore, we have that x+1>0 x+1>0> x>1 or x+1<0, x<1. Therefore, in both cases x is different from 1. Sufficient Statement 2 If y is different from zero it means that x is also different from 1, since only the numerator can give 0. Remember x can't be zero in this case. Sufficient Answer: D Hope this clarifies Cheers! J



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Re: If y=x+1/x and x!=0, is xy>0?
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08 Sep 2016, 03:56
BrushMyQuant, arpanpatnaik, Zarrolou, Narenn & jlgdrAre you guys sure that Answer is D? because your own Solutions are Contradicting your answers. Question: Is xy > 0 Given: \(x\neq{}0\) \(y=\frac{x+1}{x}\) x+1 will always be NonNegative but it can STILL BE 0. Similarly, y CAN ALSO BE 0.Statement 1: \(x^2+2x+1>0\) => \((x+1)^2 > 0\) => x > 1 There is no constraint on y.So, Case 1: If y = 1, x > 0; xy > 0 (x & y must have same sign because x+1 will always be NonNegative.) Case 2: y = 1, x = 0.5; xy > 0 (x & y must have same sign because x+1 will always be NonNegative.) Case 3: y = 0, x > 1; xy = 0 i.e. xy !> 0Statement 1: InsufficientStatement 2: \(y\neq{0}\) Case 1: If y = 1, x > 0; xy > 0 (x & y must have same sign because x+1 will always be NonNegative.) Case 2: y = 1, x = 0.5; xy > 0 (x & y must have same sign because x+1 will always be NonNegative.) Case 3: y = 100, x = 1; xy = x+1 = 0 i.e. y = 0 (because if x = 1 and xy = 0, y = 0). But this case violates the condition given in Statement 2 that \(y\neq{0}\). Therefore, \(x\neq{1}\) and as given \(y\neq{0}\) Hence, xy > 0 (Always) Statement 2: SufficientAnswer: B
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Re: If y=x+1/x and x!=0, is xy>0?
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08 Sep 2016, 04:11
umg wrote: BrushMyQuant, arpanpatnaik, Zarrolou, Narenn & jlgdrAre you guys sure that Answer is D? because your own Solutions are Contradicting your answers. Question: Is xy > 0 Given: \(x\neq{}0\) \(y=\frac{x+1}{x}\) x+1 will always be NonNegative but it can STILL BE 0. Similarly, y CAN ALSO BE 0.Statement 1: \(x^2+2x+1>0\) => \((x+1)^2 > 0\) => x > 1 There is no constraint on y.So, Case 1: If y = 1, x > 0; xy > 0 (x & y must have same sign because x+1 will always be NonNegative.) Case 2: y = 1, x = 0.5; xy > 0 (x & y must have same sign because x+1 will always be NonNegative.) Case 3: y = 0, x > 1; xy = 0 i.e. xy !> 0Statement 1: InsufficientStatement 2: \(y\neq{0}\) Case 1: If y = 1, x > 0; xy > 0 (x & y must have same sign because x+1 will always be NonNegative.) Case 2: y = 1, x = 0.5; xy > 0 (x & y must have same sign because x+1 will always be NonNegative.) Case 3: y = 100, x = 1; xy = x+1 = 0 i.e. y = 0 (because if x = 1 and xy = 0, y = 0). But this case violates the condition given in Statement 2 that \(y\neq{0}\). Therefore, \(x\neq{1}\) and as given \(y\neq{0}\) Hence, xy > 0 (Always) Statement 2: SufficientAnswer: B Notice that x^2+2x+1>0 is true for all value of x except when x = 1. So, for (1) \(x\neq 1\) and if \(x\neq 1\), then \(y\neq 0\). Your third case is not possible.
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If y=x+1/x and x!=0, is xy>0?
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08 Sep 2016, 04:51
Bunuel wrote: umg wrote: BrushMyQuant, arpanpatnaik, Zarrolou, Narenn & jlgdrAre you guys sure that Answer is D? because your own Solutions are Contradicting your answers. Question: Is xy > 0 Given: \(x\neq{}0\) \(y=\frac{x+1}{x}\) x+1 will always be NonNegative but it can STILL BE 0. Similarly, y CAN ALSO BE 0.Statement 1: \(x^2+2x+1>0\) => \((x+1)^2 > 0\) => x > 1 There is no constraint on y.So, Case 1: If y = 1, x > 0; xy > 0 (x & y must have same sign because x+1 will always be NonNegative.) Case 2: y = 1, x = 0.5; xy > 0 (x & y must have same sign because x+1 will always be NonNegative.) Case 3: y = 0, x > 1; xy = 0 i.e. xy !> 0Statement 1: InsufficientStatement 2: \(y\neq{0}\) Case 1: If y = 1, x > 0; xy > 0 (x & y must have same sign because x+1 will always be NonNegative.) Case 2: y = 1, x = 0.5; xy > 0 (x & y must have same sign because x+1 will always be NonNegative.) Case 3: y = 100, x = 1; xy = x+1 = 0 i.e. y = 0 (because if x = 1 and xy = 0, y = 0). But this case violates the condition given in Statement 2 that \(y\neq{0}\). Therefore, \(x\neq{1}\) and as given \(y\neq{0}\) Hence, xy > 0 (Always) Statement 2: SufficientAnswer: B Notice that x^2+2x+1>0 is true for all value of x except when x = 1. So, for (1) \(x\neq 1\) and if \(x\neq 1\), then \(y\neq 0\). Your third case is not possible. Wow! This question is crazier than I anticipated. I see it now. I used that logic in Statement 2 but not in Statement 1. Thanks for pointing it out.
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