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If xy=x+y , what is x in terms of y?
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13 Dec 2017, 00:41
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[GMAT math practice question] If \(xy=\sqrt{x}+\sqrt{y}\) , what is \(\sqrt{x}\) in terms of \(y\)? A. \(1+\sqrt{y}\) B. \(1\sqrt{y}\) C. \(\sqrt{y}1\) D. \(1y\) E. \(1+y\)
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If xy=x+y , what is x in terms of y?
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13 Dec 2017, 04:13
MathRevolution wrote: [GMAT math practice question]
If \(xy=\sqrt{x}+\sqrt{y}\) , what is \(\sqrt{x}\) in terms of \(y\)? A. \(1+\sqrt{y}\) B. \(1\sqrt{y}\) C. \(\sqrt{y}1\) D. \(1y\) E. \(1+y\) Hi.. Since \(\sqrt{x}\) and \(\sqrt{y}\) are there, we can convert x and y too in those terms.. \(xy=\sqrt{x}+\sqrt{y}\)... \(\sqrt{x^2}\sqrt{y^2}=\sqrt{x}+\sqrt{y}\)... \((\sqrt{x}\sqrt{y})(\sqrt{x}+\sqrt{y})=\sqrt{x}+\sqrt{y}\)... \((\sqrt{x}\sqrt{y})(\sqrt{x}+\sqrt{y})(\sqrt{x}+\sqrt{y})=0\)... \((\sqrt{x}\sqrt{y}1)(\sqrt{x}+\sqrt{y})=0\)... .. So either \(\sqrt{x}=\sqrt{y}+1\) or \(\sqrt{x}=\sqrt{y}\) A
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If xy=x+y , what is x in terms of y?
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13 Dec 2017, 09:56
chetan2u wrote: MathRevolution wrote: [GMAT math practice question]
If \(xy=\sqrt{x}+\sqrt{y}\) , what is \(\sqrt{x}\) in terms of \(y\)? A. \(1+\sqrt{y}\) B. \(1\sqrt{y}\) C. \(\sqrt{y}1\) D. \(1y\) E. \(1+y\) Hi.. Since √x and √y are there, we can convert x and y too in those terms.. xy=√x+√y...\(√x^2√y^2=√x+√y\)... (√x√y)(√x+√y)=√x+√y... (√x√y)(√x+√y)(√x+√y)=0.. (√x√y1)(√x+√y)=0 So either √x=1+√y or √x=√y A chetan2u , how did you get from here (√x√y)(√x+√y)(√x+√y)=0 To here? (√x√y1)(√x+√y)=0
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Re: If xy=x+y , what is x in terms of y?
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13 Dec 2017, 17:21
xy= √x+√y Look at equation xy from the perspective as a^2b^2=(a+b)(ab) Here a= √x and b= √y Thus xy=(√x √y)(√x+ √y) √x √y=1=> √x= 1+√y



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Re: If xy=x+y , what is x in terms of y?
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13 Dec 2017, 17:36
genxer123 wrote: chetan2u wrote: MathRevolution wrote: [GMAT math practice question]
If \(xy=\sqrt{x}+\sqrt{y}\) , what is \(\sqrt{x}\) in terms of \(y\)? A. \(1+\sqrt{y}\) B. \(1\sqrt{y}\) C. \(\sqrt{y}1\) D. \(1y\) E. \(1+y\) Hi.. Since √x and √y are there, we can convert x and y too in those terms.. xy=√x+√y...\(√x^2√y^2=√x+√y\)... (√x√y)(√x+√y)=√x+√y... (√x√y)(√x+√y)(√x+√y)=0.. (√x√y1)(√x+√y)=0 So either √x=1+√y or √x=√y A chetan2u , how did you get from here (√x√y)(√x+√y)(√x+√y)=0 To here? (√x√y1)(√x+√y)=0 hi.. (√x√y)(√x+√y)(√x+√y)=0 take out (√x+√y) as it is common \((\sqrt{x}+\sqrt{y})((√x√y1)=0\)
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Re: If xy=x+y , what is x in terms of y?
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13 Dec 2017, 18:45
This is true x−y=√x+√y when x=25; y=16
2516=√25+√16 9=9 So, √X or √25= 5 only A gives 5 as a answer √y+1= √16+1=4+1=5



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If xy=x+y , what is x in terms of y?
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13 Dec 2017, 18:48
chetan2u wrote: hi.. (√x√y)(√x+√y)(√x+√y)=0 take out (√x+√y) as it is common \((\sqrt{x}+\sqrt{y})((√x√y1)=0\) chetan2u , thanks. And now I shall see if I can get my brain to unfreeze.
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Re: If xy=x+y , what is x in terms of y?
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15 Dec 2017, 08:27
=> Since \(xy=(\sqrt{x}+ \sqrt{y})(\sqrt{x}\sqrt{y}) and xy=\sqrt{x}+\sqrt{y}, we have \sqrt{x} \sqrt{y} = 1. Thus \sqrt{x} = 1+\sqrt{y}.\) Therefore, the answer is A. Answer: A
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Re: If xy=x+y , what is x in terms of y?
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06 Apr 2018, 08:00
Turkish wrote: This is true x−y=√x+√y when x=25; y=16
2516=√25+√16 9=9 So, √X or √25= 5 only A gives 5 as a answer √y+1= √16+1=4+1=5 How the hell did you think of these numbers.. amazing.. any advise?
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Re: If xy=x+y , what is x in terms of y?
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14 Jan 2019, 08:11
chetan2u Please confirm if my approach is correct xy= √x+√y Squaring both sides X^2y^2= x + y (X+Y)(XY)=X+Y X+Y=(X+Y)/(X+Y) X+Y=1 X=1+Y Square root both sides √x=1 +√y



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Re: If xy=x+y , what is x in terms of y?
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14 Jan 2019, 10:01
Manat wrote: chetan2u Please confirm if my approach is correct xy= √x+√y Squaring both sides X^2y^2= x + y (X+Y)(XY)=X+Y X+Y=(X+Y)/(X+Y) X+Y=1 X=1+Y Square root both sides √x=1 +√y No.. When you square two sides.. xy= √x+√y. => \((xy)^2=(√x+√y)^2.....x^2+y^22xy=x+y+2√(xy)\)
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Re: If xy=x+y , what is x in terms of y?
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14 Jan 2019, 10:14
chetan2u wrote: Manat wrote: chetan2u Please confirm if my approach is correct xy= √x+√y Squaring both sides X^2y^2= x + y (X+Y)(XY)=X+Y X+Y=(X+Y)/(X+Y) X+Y=1 X=1+Y Square root both sides √x=1 +√y No.. When you square two sides.. xy= √x+√y. => \((xy)^2=(√x+√y)^2.....x^2+y^22xy=x+y+2√(xy)\) Right! i understood your solution, Is the below solution correct xy=√x+√y √x^2−√y^2=√x+√y (√x√y)(√x+√y)=√x+√y (√x√y)=(√x+√y)/(√x+√y) √x√y=1 √x=1+√y



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Re: If xy=x+y , what is x in terms of y?
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14 Jan 2019, 10:45
MathRevolution wrote: [GMAT math practice question]
If \(xy=\sqrt{x}+\sqrt{y}\) , what is \(\sqrt{x}\) in terms of \(y\)? A. \(1+\sqrt{y}\) B. \(1\sqrt{y}\) C. \(\sqrt{y}1\) D. \(1y\) E. \(1+y\) Took values and solved \(xy=\sqrt{x}+\sqrt{y}\) \(169=\sqrt{16}+\sqrt{9}\) \(\sqrt{x}\) = \(1+\sqrt{y}\) \(\sqrt{16}\) = \(1+\sqrt{9}\) 4 = 1+3 Answer A
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Re: If xy=x+y , what is x in terms of y?
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14 Jan 2019, 10:48
Manat wrote: chetan2u wrote: Manat wrote: chetan2u Please confirm if my approach is correct xy= √x+√y Squaring both sides X^2y^2= x + y (X+Y)(XY)=X+Y X+Y=(X+Y)/(X+Y) X+Y=1 X=1+Y Square root both sides √x=1 +√y No.. When you square two sides.. xy= √x+√y. => \((xy)^2=(√x+√y)^2.....x^2+y^22xy=x+y+2√(xy)\) Right! i understood your solution, Is the below solution correct xy=√x+√y √x^2−√y^2=√x+√y (√x√y)(√x+√y)=√x+√y (√x√y)=(√x+√y)/(√x+√y) √x√y=1 √x=1+√y Yes that is correct... However a point .. (√x√y)(√x+√y)=√x+√y.. Do not cross multiply as you do not know whether √x=√y=0.. So you have two answers..one as you have found and other could be √x=√y=0... But only one option is given so it is ok
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Re: If xy=x+y , what is x in terms of y?
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13 Sep 2019, 06:22
MathRevolution wrote: [GMAT math practice question]
If \(xy=\sqrt{x}+\sqrt{y}\) , what is \(\sqrt{x}\) in terms of \(y\)? A. \(1+\sqrt{y}\) B. \(1\sqrt{y}\) C. \(\sqrt{y}1\) D. \(1y\) E. \(1+y\) \([1]xy=\sqrt{x}+\sqrt{y}\) \([2]xy=\sqrt{x}^2\sqrt{y}^2…xy=(\sqrt{x}+\sqrt{y})(\sqrt{x}\sqrt{y})\) \([equate:1,2]\sqrt{x}+\sqrt{y}=(\sqrt{x}+\sqrt{y})(\sqrt{x}\sqrt{y})…\sqrt{x}\sqrt{y}=1…\sqrt{x}=1+\sqrt{y}\) Answer (A)




Re: If xy=x+y , what is x in terms of y?
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