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If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =

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Director
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If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =  [#permalink]

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New post Updated on: 20 Nov 2013, 07:27
6
10
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

59% (01:09) correct 41% (01:24) wrong based on 342 sessions

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If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =?

A. (18-3y)/y^3
B. 18/y^2
C. 18/(y^2+3y)
D. 9/y^2
E. 36/y^2

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-and-y-are-positive-and-x-2y-2-18-3xy-then-x-140728.html

Originally posted by jimmyjamesdonkey on 23 Jun 2008, 18:44.
Last edited by Bunuel on 20 Nov 2013, 07:27, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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Re: MGMAT Question 4  [#permalink]

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New post Updated on: 23 Jun 2008, 19:13
1
jimmyjamesdonkey wrote:
If x and y are positive and (x^2)(y^2) = 18 - 3xy, then x^2 =

(18-3y)/y^3

18/y^2

18/(y^2+3y)

9/y^2

Lost the E answer.. :)


the equation can be solved as (xy-3)(xy+6) = 0
Now, xy ne -6 as both are positive
xy = 3 is the only solution
x^2 = 9/y^2

D

Originally posted by Sunny143 on 23 Jun 2008, 19:06.
Last edited by Sunny143 on 23 Jun 2008, 19:13, edited 1 time in total.
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Re: MGMAT Question 4  [#permalink]

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New post 23 Jun 2008, 19:20
1
D is correct, but can anyone tell me why we can't divide by Y^2, since we know y is positive?
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Re: MGMAT Question 4  [#permalink]

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New post 23 Jun 2008, 19:21
how do you get started on this one ??
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Re: MGMAT Question 4  [#permalink]

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New post 23 Jun 2008, 19:23
fresinha12 wrote:
looks like D



i just tried the ans choices..

if x^2=9/y^2

x^2*y^2=9 = 18-3*3/y * y=9..works
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Re: MGMAT Question 4  [#permalink]

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New post 23 Jun 2008, 22:41
Just dividing by y^2 will not give the expression for x^2 purely in terms of y.
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Re: MGMAT Question 4  [#permalink]

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New post 23 Jun 2008, 23:39
1
jimmyjamesdonkey wrote:
If x and y are positive and (x^2)(y^2) = 18 - 3xy, then x^2 =

(18-3y)/y^3

18/y^2

18/(y^2+3y)

9/y^2

Lost the E answer.. :)


(x^2)(y^2) = 18 - 3xy ---> (x^2)(y^2) + 3xy -18 = 0 ----> (xy)^2 + 3xy -18 = 0

(xy - 3)(xy + 6) = 0
xy = 3 or - 6, but since x and y are positive, xy must be 3

substitute xy=3 back into the original equation

(x^2)(y^2) = 18 - 9 = 9 ---> x^2 = 9/(y^2)
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Re: If x and y are positive and (x^2)(y^2) = 18 - 3xy, then x^2  [#permalink]

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New post 20 Nov 2013, 07:22
1
The equation x2y2 = 18 – 3xy is really a quadratic, with the xy as the variable.
x2y2 + 3xy – 18 = 0
(xy + 6)(xy – 3) = 0
xy = 3 or -6
However, we are told that x and y are positive so xy must equal 3.
Therefore, x = 3/y and x2 = 9/y2.
Alternatively, this is a VIC (variable in choice) and can be solved by plugging numbers. If we plug a
value for y and find the corresponding value of x, we can check the answers to see which one
matches the value of x.
Looking at the values 3 and 18 in the equation, a y value of 3 makes sense.
x2(3)2 = 18 – 3(x)(3)
9x2 = 18 – 9x
9x2 – 9x + 18 = 0
x2 – x + 2 = 0
(x+ 2)(x – 1) = 0
x = 1, -2
But since x cannot be negative, x = 1
If we plug y = 3 into each of the answer choices, (C) and (D) both give an x value of 1.

We must now plug another value of y to decide between (C) and (D). Ultimately, only (D) represents
the correct value each time.
The correct answer is D
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Re: If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =  [#permalink]

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New post 20 Nov 2013, 07:28
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1
If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =?
A. (18-3y)/y^3
B. 18/y^2
C. 18/(y^2+3y)
D. 9/y^2
E. 36/y^2

\((xy)^2+3(xy)-18=0\) --> solving for \(xy\) --> \(xy=-6\) (not a valid solution: \(xy\) must be positive as both unknowns are positive) or \(xy=3\) --> so \(xy=3\) --> \(x=\frac{3}{y}\) --> \(x^2=\frac{9}{y^2}\).

Answer: D.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-and-y-are-positive-and-x-2y-2-18-3xy-then-x-140728.html
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Re: If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =  [#permalink]

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New post 22 Jan 2014, 13:01
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My solution:

Say \(X^2Y^2 = (XY)^2 = M^2\)

Than \(M^2=18-3M\)

\(M=3=XY\)

So \(Y^2X^2=18-9 ==> X^2=9/Y^2\)

hope it helps
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Re: If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =  [#permalink]

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Re: If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =   [#permalink] 21 Oct 2018, 06:21
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