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If x-y=x+y , what is x in terms of y?

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If x-y=x+y , what is x in terms of y?  [#permalink]

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New post 13 Dec 2017, 00:41
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[GMAT math practice question]

If \(x-y=\sqrt{x}+\sqrt{y}\) , what is \(\sqrt{x}\) in terms of \(y\)?

A. \(1+\sqrt{y}\)
B. \(1-\sqrt{y}\)
C. \(\sqrt{y}-1\)
D. \(1-y\)
E. \(1+y\)

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If x-y=x+y , what is x in terms of y?  [#permalink]

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New post 13 Dec 2017, 04:13
3
MathRevolution wrote:
[GMAT math practice question]

If \(x-y=\sqrt{x}+\sqrt{y}\) , what is \(\sqrt{x}\) in terms of \(y\)?

A. \(1+\sqrt{y}\)
B. \(1-\sqrt{y}\)
C. \(\sqrt{y}-1\)
D. \(1-y\)
E. \(1+y\)



Hi..
Since \(\sqrt{x}\) and \(\sqrt{y}\) are there, we can convert x and y too in those terms..

\(x-y=\sqrt{x}+\sqrt{y}\)...
\(\sqrt{x^2}-\sqrt{y^2}=\sqrt{x}+\sqrt{y}\)...
\((\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})=\sqrt{x}+\sqrt{y}\)...
\((\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})-(\sqrt{x}+\sqrt{y})=0\)...
\((\sqrt{x}-\sqrt{y}-1)(\sqrt{x}+\sqrt{y})=0\)...
..
So either \(\sqrt{x}=\sqrt{y}+1\) or \(\sqrt{x}=-\sqrt{y}\)
A
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If x-y=x+y , what is x in terms of y?  [#permalink]

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New post 13 Dec 2017, 09:56
chetan2u wrote:
MathRevolution wrote:
[GMAT math practice question]

If \(x-y=\sqrt{x}+\sqrt{y}\) , what is \(\sqrt{x}\) in terms of \(y\)?

A. \(1+\sqrt{y}\)
B. \(1-\sqrt{y}\)
C. \(\sqrt{y}-1\)
D. \(1-y\)
E. \(1+y\)



Hi..
Since √x and √y are there, we can convert x and y too in those terms..

x-y=√x+√y...\(√x^2-√y^2=√x+√y\)...
(√x-√y)(√x+√y)=√x+√y...
(√x-√y)(√x+√y)-(√x+√y)=0..
(√x-√y-1)(√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-√y-1)(√x+√y)=0
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Re: If x-y=x+y , what is x in terms of y?  [#permalink]

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New post 13 Dec 2017, 17:21
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1
x-y= √x+√y
Look at equation x-y from the perspective as a^2-b^2=(a+b)(a-b)
Here a= √x and b= √y
Thus x-y=(√x- √y)(√x+ √y)
√x- √y=1=> √x= 1+√y
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Re: If x-y=x+y , what is x in terms of y?  [#permalink]

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New post 13 Dec 2017, 17:36
1
genxer123 wrote:
chetan2u wrote:
MathRevolution wrote:
[GMAT math practice question]

If \(x-y=\sqrt{x}+\sqrt{y}\) , what is \(\sqrt{x}\) in terms of \(y\)?

A. \(1+\sqrt{y}\)
B. \(1-\sqrt{y}\)
C. \(\sqrt{y}-1\)
D. \(1-y\)
E. \(1+y\)



Hi..
Since √x and √y are there, we can convert x and y too in those terms..

x-y=√x+√y...\(√x^2-√y^2=√x+√y\)...
(√x-√y)(√x+√y)=√x+√y...
(√x-√y)(√x+√y)-(√x+√y)=0..
(√x-√y-1)(√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-√y-1)(√x+√y)=0


hi..
(√x-√y)(√x+√y)-(√x+√y)=0
take out (√x+√y) as it is common
\((\sqrt{x}+\sqrt{y})((√x-√y-1)=0\)
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Re: If x-y=x+y , what is x in terms of y?  [#permalink]

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New post 13 Dec 2017, 18:45
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This is true x−y=√x+√y when x=25; y=16

25-16=√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 x-y=x+y , what is x in terms of y?  [#permalink]

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New post 13 Dec 2017, 18:48
chetan2u wrote:

hi..
(√x-√y)(√x+√y)-(√x+√y)=0
take out (√x+√y) as it is common
\((\sqrt{x}+\sqrt{y})((√x-√y-1)=0\)

chetan2u , thanks. And now I shall see if I can get my brain to unfreeze.
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Re: If x-y=x+y , what is x in terms of y?  [#permalink]

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New post 15 Dec 2017, 08:27
=>

Since \(x-y=(\sqrt{x}+ \sqrt{y})(\sqrt{x}-\sqrt{y}) and x-y=\sqrt{x}+\sqrt{y}, we have \sqrt{x} -\sqrt{y} = 1.
Thus \sqrt{x} = 1+\sqrt{y}.\)

Therefore, the answer is A.

Answer: A
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Re: If x-y=x+y , what is x in terms of y?  [#permalink]

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New post 06 Apr 2018, 08:00
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Turkish wrote:
This is true x−y=√x+√y when x=25; y=16

25-16=√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 x-y=x+y , what is x in terms of y?  [#permalink]

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New post 14 Jan 2019, 08:11
chetan2u Please confirm if my approach is correct

x-y= √x+√y

Squaring both sides

X^2-y^2= x + y
(X+Y)(X-Y)=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 x-y=x+y , what is x in terms of y?  [#permalink]

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New post 14 Jan 2019, 10:01
Manat wrote:
chetan2u Please confirm if my approach is correct

x-y= √x+√y

Squaring both sides

X^2-y^2= x + y
(X+Y)(X-Y)=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..
x-y= √x+√y. => \((x-y)^2=(√x+√y)^2.....x^2+y^2-2xy=x+y+2√(xy)\)
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Re: If x-y=x+y , what is x in terms of y?  [#permalink]

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New post 14 Jan 2019, 10:14
chetan2u wrote:
Manat wrote:
chetan2u Please confirm if my approach is correct

x-y= √x+√y

Squaring both sides

X^2-y^2= x + y
(X+Y)(X-Y)=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..
x-y= √x+√y. => \((x-y)^2=(√x+√y)^2.....x^2+y^2-2xy=x+y+2√(xy)\)



Right! i understood your solution,


Is the below solution correct

x-y=√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 x-y=x+y , what is x in terms of y?  [#permalink]

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New post 14 Jan 2019, 10:45
MathRevolution wrote:
[GMAT math practice question]

If \(x-y=\sqrt{x}+\sqrt{y}\) , what is \(\sqrt{x}\) in terms of \(y\)?

A. \(1+\sqrt{y}\)
B. \(1-\sqrt{y}\)
C. \(\sqrt{y}-1\)
D. \(1-y\)
E. \(1+y\)


Took values and solved
\(x-y=\sqrt{x}+\sqrt{y}\)
\(16-9=\sqrt{16}+\sqrt{9}\)

\(\sqrt{x}\) = \(1+\sqrt{y}\)
\(\sqrt{16}\) = \(1+\sqrt{9}\)

4 = 1+3

Answer A
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Re: If x-y=x+y , what is x in terms of y?  [#permalink]

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New post 14 Jan 2019, 10:48
1
Manat wrote:
chetan2u wrote:
Manat wrote:
chetan2u Please confirm if my approach is correct

x-y= √x+√y

Squaring both sides

X^2-y^2= x + y
(X+Y)(X-Y)=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..
x-y= √x+√y. => \((x-y)^2=(√x+√y)^2.....x^2+y^2-2xy=x+y+2√(xy)\)



Right! i understood your solution,


Is the below solution correct

x-y=√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 x-y=x+y , what is x in terms of y?  [#permalink]

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New post 13 Sep 2019, 06:22
MathRevolution wrote:
[GMAT math practice question]

If \(x-y=\sqrt{x}+\sqrt{y}\) , what is \(\sqrt{x}\) in terms of \(y\)?

A. \(1+\sqrt{y}\)
B. \(1-\sqrt{y}\)
C. \(\sqrt{y}-1\)
D. \(1-y\)
E. \(1+y\)


\([1]x-y=\sqrt{x}+\sqrt{y}\)
\([2]x-y=\sqrt{x}^2-\sqrt{y}^2…x-y=(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})\)
\([equate:1,2]\sqrt{x}+\sqrt{y}=(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})…\sqrt{x}-\sqrt{y}=1…\sqrt{x}=1+\sqrt{y}\)

Answer (A)
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Re: If x-y=x+y , what is x in terms of y?   [#permalink] 13 Sep 2019, 06:22
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