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

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Senior Manager
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If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ? [#permalink]  15 Oct 2012, 03:17
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68% (02:45) correct 32% (02:50) wrong based on 89 sessions
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
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 27058
Followers: 4184

Kudos [?]: 40429 [3] , given: 5421

Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ? [#permalink]  15 Oct 2012, 03:22
3
KUDOS
Expert's post
Jp27 wrote:
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 $xy=3$ --> $x=\frac{3}{y}$ --> $x^2=\frac{9}{y^2}$.

Hope it's clear.
_________________
Senior Manager
Joined: 22 Dec 2011
Posts: 299
Followers: 3

Kudos [?]: 132 [0], given: 32

Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ? [#permalink]  15 Oct 2012, 03:38
Bunuel wrote:
Jp27 wrote:
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 $xy=3$ --> $x=\frac{3}{y}$ --> $x^2=\frac{9}{y^2}$.

Hope it's clear.

Hi Bunuel - thanks for your swift response.

Where I can get similar question like these for practice?

I have done the roots and equation set in your signature, are there any other sets that I can use for practice? Sincere thanks!

cheers
Math Expert
Joined: 02 Sep 2009
Posts: 27058
Followers: 4184

Kudos [?]: 40429 [1] , given: 5421

Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ? [#permalink]  15 Oct 2012, 03:44
1
KUDOS
Expert's post
Jp27 wrote:
Bunuel wrote:
Jp27 wrote:
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 $xy=3$ --> $x=\frac{3}{y}$ --> $x^2=\frac{9}{y^2}$.

Hope it's clear.

Hi Bunuel - thanks for your swift response.

Where I can get similar question like these for practice?

I have done the roots and equation set in your signature, are there any other sets that I can use for practice? Sincere thanks!

cheers

Check our question banks (viewforumtags.php) for more questions.

DS algebra questions: search.php?search_id=tag&tag_id=29
PS algebra questions: search.php?search_id=tag&tag_id=50

Hope it helps.
_________________
Senior Manager
Joined: 13 Aug 2012
Posts: 464
Concentration: Marketing, Finance
GMAT 1: Q V0
GPA: 3.23
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Kudos [?]: 255 [1] , given: 11

Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ? [#permalink]  10 Dec 2012, 22:57
1
KUDOS
$x^2y^2 + 3xy = 18$
$xy (xy + 3) = 18$

I thought of 2 positive numbers such as $n$ and $n + 3$ whose product is 18 --> $3$ and $6$

$xy = 3$
$x^2 = 9/y^2$

_________________

Impossible is nothing to God.

Senior Manager
Joined: 13 Aug 2012
Posts: 464
Concentration: Marketing, Finance
GMAT 1: Q V0
GPA: 3.23
Followers: 17

Kudos [?]: 255 [0], given: 11

Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ? [#permalink]  18 Dec 2012, 20:45
$x^2y^2+3xy-18=0$
$(xy)^2+3xy-18=0$
$(xy - 3)(xy + 6)=0$
$xy = 3 & xy = -6$ Since x and y are positive, we choose xy = 3.

$xy = 3$
$x = \frac{3}{y}$
$x^2 = \frac{9}{y^2}$

_________________

Impossible is nothing to God.

Intern
Joined: 11 Oct 2013
Posts: 18
Location: United Kingdom
GMAT 1: 490 Q32 V25
GPA: 3.9
WE: Other (Other)
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Kudos [?]: 8 [0], given: 34

Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ? [#permalink]  16 Dec 2013, 18:21
Bunuel wrote:
Jp27 wrote:
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 $xy=3$ --> $x=\frac{3}{y}$ --> $x^2=\frac{9}{y^2}$.

Hope it's clear.

very elegant&simple...if you spot one variable in two
like it
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Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?   [#permalink] 16 Dec 2013, 18:21
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