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If x and y are integers and 4xy = x^2*y+4y, what is the [#permalink]
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Updated on: 23 May 2012, 02:15
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If x and y are integers and 4xy = x^2*y+4y, what is the value of xy? (1) yx = 2 (2) x^3 <0 what I did : 4xy = x^2y+4y 4x = x^2+4 x^24x+4 = 0 (x2)^2=0 x = 2 is this wrong?
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Originally posted by kashishh on 23 May 2012, 01:57.
Last edited by Bunuel on 23 May 2012, 02:15, edited 1 time in total.
Edited the OA




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Re: If x and y are integers and 4xy = x^2*y+4y, what is the [#permalink]
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23 May 2012, 02:14
kashishh wrote: If x and y are integers and 4xy = x^2*y+4y, what is the value of xy?
(1) yx = 2 (2) x^3 <0
what I did : 4xy = x^2y+4y 4x = x^2+4 x^24x+4 = 0 (x2)^2=0 x = 2 is this wrong? You cannot divide 4xy=x^2y+4y by y and write 4x=x^2+4, because y can be zero and division be zero is not allowed. Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero. So when you divide by y you assume, with no ground for it, that y does not equal to zero thus exclude a possible solution. If x and y are integers and 4xy = x^2*y+4y, what is the value of xy?\(4xy=x^2*y+4y\) > \(x^2*y4xy+4y=0\) > \(y(x^24x+4)=0\) > \(y(x2)^2=0\) > either \(x=2\) or \(y=0\) (or both). (1) yx = 2 > if \(x=2\) then \(y=4\) and \(xy=8\) but if \(y=0\) then \(x=2\) and \(xy = 0\). Not sufficient. (2) x^3<0 > \(x<0\), so \(x\neq{2}\) which means that \(y\) must equal to zero > \(xy=0\). Sufficient. Answer: B. Hope it's clear.
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Re: If x and y are integers and 4xy = x^2*y+4y, what is the [#permalink]
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25 May 2012, 19:15
thanks bunuel. Its a common error to 'cancel' y on both sides. I fell prey to the same mistake. However, just to clarify, basically when x = 2 then y can take any value and when y=0 x can take any value. In either case xy=0?



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Re: If x and y are integers and 4xy = x^2*y+4y, what is the [#permalink]
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26 May 2012, 02:42



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Re: If x and y are integers and 4xy = x^2*y+4y, what is the [#permalink]
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26 May 2012, 07:09
Bunuel wrote: If x=2 then y can take any value  TRUE. In this case xy=0 only if y=0 but for any other possible value of y xy does not equal to zero (for example consider x=2 and y=1).
Just by looking at this equation y(x^2 4x+4) = 0 either the quadratic term has to equal 0, in which case x=2, or y has to equal to zero. In which scenario will y =1 and x=2 occur?



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Re: If x and y are integers and 4xy = x^2*y+4y, what is the [#permalink]
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28 May 2012, 05:26



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Re: If x and y are integers and 4xy = x^2*y+4y, what is the [#permalink]
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31 May 2012, 20:55
Made silly mistake.... I thought using stmt 1  xy always comes to be 0...
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Re: If x and y are integers and 4xy = x^2*y+4y, what is the [#permalink]
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14 Sep 2013, 03:21
Bunuel wrote: kashishh wrote: If x and y are integers and 4xy = x^2*y+4y, what is the value of xy?
(1) yx = 2 (2) x^3 <0
what I did : 4xy = x^2y+4y 4x = x^2+4 x^24x+4 = 0 (x2)^2=0 x = 2 is this wrong? You cannot divide 4xy=x^2y+4y by y and write 4x=x^2+4, because y can be zero and division be zero is not allowed. Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero. So when you divide by y you assume, with no ground for it, that y does not equal to zero thus exclude a possible solution. If x and y are integers and 4xy = x^2*y+4y, what is the value of xy?\(4xy=x^2*y+4y\) > \(x^2*y4xy+4y=0\) > \(y(x^24x+4)=0\) > \(y(x2)^2=0\) > either \(x=2\) or \(y=0\) (or both). (1) yx = 2 > if \(x=2\) then \(y=4\) and \(xy=8\) but if \(y=0\) then \(x=2\) and \(xy = 0\). Not sufficient. (2) x^3<0 > \(x<0\), so \(x\neq{2}\) which means that \(y\) must equal to zero > \(xy=0\). Sufficient. Answer: B. Hope it's clear. instead of taking LHS to RHS , I brought RHS to LHS and got 4xyx^2y  4y = 0 now y (4xx^2 4 )= 0 y (x^2 +4x  4)= 0 y (x+2)^2= 0 either y = 0 or x = 2 now obviously x=2 does not satisfy 4xy = x^2*y+4y, now I was expecting the same answer both ways. Now am I doing something wrong,or is it just one of those cases where the equation works a certain way and not the other way by the way factorization of x^2 +4x4 =0 is (x+2)^2 right? if x^2 has a coefficient we have to multiply the constant (4) with the coefficient of x^2 do we not.
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Re: If x and y are integers and 4xy = x^2*y+4y, what is the [#permalink]
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14 Sep 2013, 03:29
stne wrote: Bunuel wrote: kashishh wrote: If x and y are integers and 4xy = x^2*y+4y, what is the value of xy?
(1) yx = 2 (2) x^3 <0
what I did : 4xy = x^2y+4y 4x = x^2+4 x^24x+4 = 0 (x2)^2=0 x = 2 is this wrong? You cannot divide 4xy=x^2y+4y by y and write 4x=x^2+4, because y can be zero and division be zero is not allowed. Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero. So when you divide by y you assume, with no ground for it, that y does not equal to zero thus exclude a possible solution. If x and y are integers and 4xy = x^2*y+4y, what is the value of xy?\(4xy=x^2*y+4y\) > \(x^2*y4xy+4y=0\) > \(y(x^24x+4)=0\) > \(y(x2)^2=0\) > either \(x=2\) or \(y=0\) (or both). (1) yx = 2 > if \(x=2\) then \(y=4\) and \(xy=8\) but if \(y=0\) then \(x=2\) and \(xy = 0\). Not sufficient. (2) x^3<0 > \(x<0\), so \(x\neq{2}\) which means that \(y\) must equal to zero > \(xy=0\). Sufficient. Answer: B. Hope it's clear. instead of taking LHS to RHS , I brought RHS to LHS and got 4xyx^2y  4y = 0 now y (4xx^2 4 )= 0 y (x^2 +4x  4)= 0 y (x+2)^2= 0 either y = 0 or x = 2 now obviously x=2 does not satisfy 4xy = x^2*y+4y, now I was expecting the same answer both ways. Now am I doing something wrong,or is it just one of those cases where the equation works a certain way and not the other way by the way factorization of x^2 +4x4 =0 is (x+2)^2 right? if x^2 has a coefficient we have to multiply the constant (4) with the coefficient of x^2 do we not. You could expand \((x+2)^2\) to check your reasoning: \((x+2)^2=x^2+2x+4\), while \(x^2 +4x4=(x^24x+4)=(x2)^2\). Thus, \(y(4xx^2 4 )=0\) can be written as \(y(x2)^2=0\) > \(x=2\) or \(y=0\). Hope it's clear.
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Re: If x and y are integers and 4xy = x^2*y+4y, what is the [#permalink]
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Updated on: 14 Sep 2013, 04:07
Bunuel wrote: stne wrote: instead of taking LHS to RHS , I brought RHS to LHS and got 4xyx^2y  4y = 0 now y (4xx^2 4 )= 0 y (x^2 +4x  4)= 0 y (x+2)^2= 0 either y = 0 or x = 2 now obviously x=2 does not satisfy 4xy = x^2*y+4y, now I was expecting the same answer both ways. Now am I doing something wrong,or is it just one of those cases where the equation works a certain way and not the other way
by the way factorization of x^2 +4x4 =0 is (x+2)^2 right? if x^2 has a coefficient we have to multiply the constant (4) with the coefficient of x^2 do we not.
You could expand \((x+2)^2\) to check your reasoning: \((x+2)^2=x^2+2x+4\), while \(x^2 +4x4=(x^24x+4)=(x2)^2\). Thus, \(y(4xx^2 4 )=0\) can be written as \(y(x2)^2=0\) > \(x=2\) or \(y=0\). Hope it's clear. I think I am still missing something, can you please show me how to factorize x^2+4x4= 0 without taking out the negative sign, answer should be same right? Thank you
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Originally posted by stne on 14 Sep 2013, 03:59.
Last edited by stne on 14 Sep 2013, 04:07, edited 1 time in total.



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Re: If x and y are integers and 4xy = x^2*y+4y, what is the [#permalink]
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14 Sep 2013, 04:02
stne wrote: stne wrote: Bunuel wrote: instead of taking LHS to RHS , I brought RHS to LHS and got 4xyx^2y  4y = 0 now y (4xx^2 4 )= 0 y (x^2 +4x  4)= 0 y (x+2)^2= 0 either y = 0 or x = 2 now obviously x=2 does not satisfy 4xy = x^2*y+4y, now I was expecting the same answer both ways. Now am I doing something wrong,or is it just one of those cases where the equation works a certain way and not the other way
by the way factorization of x^2 +4x4 =0 is (x+2)^2 right? if x^2 has a coefficient we have to multiply the constant (4) with the coefficient of x^2 do we not.
You could expand \((x+2)^2\) to check your reasoning: \((x+2)^2=x^2+2x+4\), while \(x^2 +4x4=(x^24x+4)=(x2)^2\). Thus, \(y(4xx^2 4 )=0\) can be written as \(y(x2)^2=0\) > \(x=2\) or \(y=0\). Hope it's clear. I think I am still missing something, can you please show me how to factorize x^2+4x4= 0 without taking out the negative sign, answer should be same right? Thank you You can factor x^2 + 4x  4 only as (x2)^2.
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Re: If x and y are integers and 4xy = x^2*y+4y, what is the [#permalink]
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14 Sep 2013, 05:40
Bunuel wrote: stne wrote: Bunuel wrote: You could expand \((x+2)^2\) to check your reasoning: \((x+2)^2=x^2+2x+4\), while \(x^2 +4x4=(x^24x+4)=(x2)^2\). Thus, \(y(4xx^2 4 )=0\) can be written as \(y(x2)^2=0\) > \(x=2\) or \(y=0\).
Hope it's clear.
I think I am still missing something, can you please show me how to factorize x^2+4x4= 0 without taking out the negative sign, answer should be same right? Thank you You can factor x^2 + 4x  4 only as (x2)^2. Thank you +1, I was under the impression that it was possible to factor x^2 +4x4 as it is and it was not necessary to take out the negative sign. One of those mistakes , waiting to happen.
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Re: If x and y are integers and 4xy = x^2*y+4y, what is the [#permalink]
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03 Jan 2014, 12:14
Hi Bunuel, Thanks for the wonderful response. Please let me know that we should you use it as truth statement for DS question (Algebra) that "Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero". I surely follow it that reduce variable in algebra only when I am sure that they are not equal to 0.
Please do explain one doubt .What you said make logical sense but it contradicts from my previous knowledge that if we have an equation and two expression are equal to each other than we can reduce the common variable such as y from both side and there is no where mention that variable not equal to 0.



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Re: If x and y are integers and 4xy = x^2*y+4y, what is the [#permalink]
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03 Jan 2014, 12:19
vikrantgulia wrote: Hi Bunuel, Thanks for the wonderful response. Please let me know that we should you use it as truth statement for DS question (Algebra) that "Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero". I surely follow it that reduce variable in algebra only when I am sure that they are not equal to 0.
Please do explain one doubt .What you said make logical sense but it contradicts from my previous knowledge that if we have an equation and two expression are equal to each other than we can reduce the common variable such as y from both side and there is no where mention that variable not equal to 0. Yes, the property is true. Consider simple example: xy=x. If you reduce this expression by x you get definite answer that y=1 but if there x=0, then y is not necessary to be 1, it can be any number. Correct solution would be xy=x > x(y1)=0 > x=0 OR y=1. Does this make sense?
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