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If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?
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15 Oct 2012, 04:17
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If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =? A. \(\frac{(183y)}{y^3}\) B. \(\frac{18}{y^2}\) C. \(\frac{18}{(y^2+3y)}\) D. \(\frac{9}{y^2}\) E. \(\frac{36}{y^2}\)
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If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?
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15 Oct 2012, 04:22
Jp27 wrote: If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =?
A. \(\frac{(183y)}{y^3}\)
B. \(\frac{18}{y^2}\)
C. \(\frac{18}{(y^2+3y)}\)
D. \(\frac{9}{y^2}\)
E. \(\frac{36}{y^2}\) \((xy)^2+3(xy)18=0\) > 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. Hope it's clear.
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Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?
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10 Dec 2012, 23:57
\(x^2y^2 + 3xy = 18\) \(xy (xy + 3) = 18\) I thought of 2 positive numbers such as \(n\) and \(n + 3\) whose product is 18 > \(3\) and \(6\) \(xy = 3\) \(x^2 = 9/y^2\) Answer: D
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Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?
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15 Oct 2012, 04:38
Bunuel wrote: Jp27 wrote: If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =?
A. (183y)/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. 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



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Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?
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15 Oct 2012, 04:44
Jp27 wrote: Bunuel wrote: Jp27 wrote: If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =?
A. (183y)/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. 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=29PS algebra questions: search.php?search_id=tag&tag_id=50Hope it helps.
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Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?
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18 Dec 2012, 21:45
\(x^2y^2+3xy18=0\) \((xy)^2+3xy18=0\) \((xy  3)(xy + 6)=0\) \(xy = 3 & xy = 6\) Since x and y are positive, we choose xy = 3. \(xy = 3\) \(x = \frac{3}{y}\) \(x^2 = \frac{9}{y^2}\) Answer: D
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Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?
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16 Dec 2013, 19:21
Bunuel wrote: Jp27 wrote: If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =?
A. (183y)/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. 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 ?
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11 Jun 2015, 03:21
I don't understand why you can not divide by y^2 here? Pls help



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Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?
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11 Jun 2015, 03:36
noTh1ng wrote: I don't understand why you can not divide by y^2 here? Pls help If you divide x^2 * y^2 = 18 – 3xy by y^2 you get x^2 = 18/y^2  3/y = (183y)/y^2. We don't have (183y)/y^2 among answer choices, thus we need other approach.
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Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?
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11 Jun 2015, 04:34
Bunuel wrote: noTh1ng wrote: I don't understand why you can not divide by y^2 here? Pls help If you divide x^2 * y^2 = 18 – 3xy by y^2 you get x^2 = 18/y^2  3/y = (183y)/y^2. We don't have (183y)/y^2 among answer choices, thus we need other approach. I sometimes don't get the GMAC.. just because it's not among the answer choices doesnt make it wrong imho



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Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?
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11 Jun 2015, 21:45
noTh1ng wrote: Bunuel wrote: noTh1ng wrote: I don't understand why you can not divide by y^2 here? Pls help If you divide x^2 * y^2 = 18 – 3xy by y^2 you get x^2 = 18/y^2  3/y = (183y)/y^2. We don't have (183y)/y^2 among answer choices, thus we need other approach. I sometimes don't get the GMAC.. just because it's not among the answer choices doesnt make it wrong imho If an expression is not in the options, it doesn't make the expression wrong. You might just need to manipulate it further. Here, dividing by y^2 doesn't work: \(x^2 * y^2 = 18 – 3xy\) When you divide by y^2, you get \(x^2 = 18/y^2 – 3x/y\) How do you separate the x and y since you need to write x^2 in terms of y only? You will need to divide by xy and then solve for it. By the way, GMAC is absolutely reasoning based. If something confuses you, especially in Quant, it means you have missed a point.
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Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?
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06 Mar 2018, 13:33
Hi All, This question has a "quirky" design element to it  it takes a bit longer than normal to solve AND you might have to TEST VALUES TWICE to get to the correct answer. You could TEST Y = 3, then X = 1. Then, the answer to the question is also 1 (1² = 1). Plugging Y = 3 into the answer choices gets us 2 answers that match: Answer C = 18/(3² + 9) = 18/18 = 1 Answer D = 9/3² = 9/9 = 1 None of the other answers matches, so we're down to 2 choices. At this point, you can either guess one of the 2 or go back and TEST a new value. If we use Y = 2 in the original equation, then we have 4X² = 18  6X 4X² +6X  18 = 0 2X² +3X  9 = 0 Factoring this down might seem a little strange, but it DOES lead to a solution… (X + 3)(2X  3) = 0 X = 3 or +3/2 Since we're told that X and Y are both POSITIVE, X = 3/2 So now we plug Y = 2 into the answers and we're looking for (3/2)² = 9/4 Answer C: 18/10 = 9/5 Answer D: 9/4 Final Answer: GMAT assassins aren't born, they're made, Rich
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If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?
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07 Apr 2019, 10:15
@ Bunuel wrote: Jp27 wrote: If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =?
A. \(\frac{(183y)}{y^3}\)
B. \(\frac{18}{y^2}\)
C. \(\frac{18}{(y^2+3y)}\)
D. \(\frac{9}{y^2}\)
E. \(\frac{36}{y^2}\) \((xy)^2+3(xy)18=0\) > 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. Hope it's clear. IT's a really basic thing I want to ask you, but I must before my exam, I feel... How do you equate the 2 equations here.. like can you tell me, how is (xy3) (xy+6) derived .. I understand that once we open the brackets, we get the above equation, also that 1 entity must be negative and 1 be positive as we have +3 and 18. But, I ask, how 6 and 3.. is there any other way I am missing ? or is this the right way only.. I welcome any sort of help that will make this crystal clear for me



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If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?
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08 Apr 2019, 07:29
Call \(xy = t\) and solve the equation. You'll get: \(t^2 +3t 18 = 0\) \((t + 6) (t  3) = 0\). \(6  3 = 3\); \(6 * 3 = 18\) \(xy = 6\) \(xy = 3\) Since you consider only positive values, \(x = \frac{3}{y}\), so \(x^2 = \frac{9}{y^2}\). Hope it's clear.
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If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?
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11 Apr 2019, 04:03
"Plug in the value" is a good approach for this one. Take x = 3 and y = 1. Plug in the values and we'll get 9 as the answer. Moreover, from the options only one option D satisfies that. On the more regular approach, I really like the method of taking xy as one variable T and then soving the quadratic equation thus formed.
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If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ?
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