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If x and y are both positive and √x^2 + y^2 = 3x−y, then x/y =

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Math Expert
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V
Joined: 02 Sep 2009
Posts: 58458
If x and y are both positive and √x^2 + y^2 = 3x−y, then x/y =  [#permalink]

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New post 17 Oct 2018, 02:26
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

75% (01:49) correct 25% (01:59) wrong based on 63 sessions

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GMAT Club Legend
GMAT Club Legend
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V
Joined: 12 Sep 2015
Posts: 4007
Location: Canada
Re: If x and y are both positive and √x^2 + y^2 = 3x−y, then x/y =  [#permalink]

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New post 17 Oct 2018, 06:38
Top Contributor
Bunuel wrote:
If x and y are both positive and \(\sqrt{x^2 + y^2} = 3x−y\), then x/y =


A. 0
B. 1/4
C. 1/2
D. 3/4
E. 3/2


GIVEN: \(\sqrt{x^2 + y^2} = 3x−y\)
Square both sides to get: \(x^2 + y^2 = (3x-y)^2\)
aka: \(x^2 + y^2 = (3x-y)(3x-y)\)
Expand: \(x^2 + y^2 = 9x^2 - 6xy + y^2\)
Simplify: \(x^2 = 9x^2 - 6xy\)
Simplify more: \(0 = 8x^2 - 6xy\)
Add 6xy to both sides: \(6xy = 8x^2\)
Divide both sides by x: \(6y = 8x\)
Divide both sides by y: \(6 = \frac{8x}{y}\)
Divide both sides by 8: \(6/8 = \frac{x}{y}\)
Simplify: \(3/4 = \frac{x}{y}\)

Answer: D

Cheers,
Brent
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Senior Manager
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D
Joined: 18 Jun 2018
Posts: 263
Re: If x and y are both positive and √x^2 + y^2 = 3x−y, then x/y =  [#permalink]

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New post 17 Oct 2018, 06:43
Bunuel wrote:
If x and y are both positive and \(\sqrt{x^2 + y^2} = 3x−y\), then x/y =


A. 0
B. 1/4
C. 1/2
D. 3/4
E. 3/2


OA: D

\(\sqrt{x^2 + y^2} = 3x−y\)

Squaring both sides, we get

\(x^2 + y^2 = (3x-y)^2\)
\(x^2 + y^2 =9x^2+y^2-6xy\)
\(8x^2-6xy=0\)
\(2x(4x-3y)=0\)

As\(x\neq{0},\quad 4x-3y\) should be equal to Zero.
\(4x-3y=0\)
\(4x=3y\)
\(\frac{x}{y}=\frac{3}{4}\)
Math Expert
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Joined: 02 Aug 2009
Posts: 7977
If x and y are both positive and √x^2 + y^2 = 3x−y, then x/y =  [#permalink]

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New post 17 Oct 2018, 06:46
Bunuel wrote:
If x and y are both positive and \(\sqrt{x^2 + y^2} = 3x−y\), then x/y =


A. 0
B. 1/4
C. 1/2
D. 3/4
E. 3/2



Two ways..

1) Square both sides.. Since all terms are in X and y, squaring should help
\(\sqrt{x^2 + y^2} = 3x−y........x^2+y^2=9x^2+y^2-6xy......8x^2=6xy..........\frac{x^2}{xy}=\frac{x}{y}\)=\(\frac{3}{4}\)

2) substitute x/y=a.....x=ay
\(\sqrt{(ay)^2 + y^2} = 3ay−y\).......\(\sqrt{a^2 + 1} = 3a-1\)......\(a^2+1=9a^2+1-6a.....8a^2-6a=0......a(8a-6)=0\)
But a is not 0, so 8a-6=0...a=6/8=3/4

3) also substitute answer choices and you can get your answer

D
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If x and y are both positive and √x^2 + y^2 = 3x−y, then x/y =   [#permalink] 17 Oct 2018, 06:46
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