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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 =
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Updated on: 20 Nov 2013, 07:27
Question Stats:
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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 OPEN DISCUSSION OF THIS QUESTION IS HERE: ifxandyarepositiveandx2y2183xythenx140728.html
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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
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Updated on: 23 Jun 2008, 19:13
jimmyjamesdonkey wrote: If x and y are positive and (x^2)(y^2) = 18  3xy, then x^2 = (183y)/y^3 18/y^2 18/(y^2+3y) 9/y^2 Lost the E answer.. the equation can be solved as (xy3)(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.



Director
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Re: MGMAT Question 4
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23 Jun 2008, 19:20
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
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23 Jun 2008, 19:21
how do you get started on this one ??



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Re: MGMAT Question 4
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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 = 183*3/y * y=9..works



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



Director
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Re: MGMAT Question 4
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23 Jun 2008, 23:39
jimmyjamesdonkey wrote: If x and y are positive and (x^2)(y^2) = 18  3xy, then x^2 = (183y)/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
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20 Nov 2013, 07:22
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 =
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20 Nov 2013, 07:28
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. OPEN DISCUSSION OF THIS QUESTION IS HERE: ifxandyarepositiveandx2y2183xythenx140728.html
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Re: If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =
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22 Jan 2014, 13:01
My solution: Say \(X^2Y^2 = (XY)^2 = M^2\) Than \(M^2=183M\) \(M=3=XY\) So \(Y^2X^2=189 ==> X^2=9/Y^2\) hope it helps
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Re: If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =
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21 Oct 2018, 06:21
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Re: If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =
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