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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
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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17 Oct 2018, 02:26
00:00

Difficulty:

35% (medium)

Question Stats:

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

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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

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

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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: $$3/4 = \frac{x}{y}$$

Cheers,
Brent
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Senior Manager
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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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
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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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

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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