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If xy ≠ 0 and x2y2 – xy = 6, which of the following could be

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If xy ≠ 0 and x2y2 – xy = 6, which of the following could be  [#permalink]

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New post Updated on: 10 Feb 2014, 05:08
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If xy ≠ 0 and x^2*y^2 - xy = 6, which of the following could be y in terms of x?

I. 1/(2x)
II. – 2/x
III. 3/x

A. I only
B. II only
C. I and II
D. I and III
E. II and III

Originally posted by naaga on 25 Feb 2011, 08:23.
Last edited by Bunuel on 10 Feb 2014, 05:08, edited 3 times in total.
Renamed the topic, edited the question and the OA.
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If xy ≠ 0 and x2y2 – xy = 6, which of the following could be  [#permalink]

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New post 25 Feb 2011, 09:11
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naaga wrote:
If xy ≠ 0 and x2y2 – xy = 6, which of the following could be y in terms of x?

I. 1/(2x)
II. – 2/x
III. 3/x


A. I only
B. II only
C. I and II
D. I and III
E. II and III


If xy ≠ 0 and x^2*y^2 - xy = 6, which of the following could be y in terms of x?

I. 1/(2x)
II. - 2/x
III. 3/x

\(x^2*y^2-xy=6\) --> \((xy)^2-xy-6=0\) --> \((xy-3)(xy+2)=0\) --> either \(xy-3=0\) and in this case \(y=\frac{3}{x}\) or \(xy+2=0\) and in this case \(y=-\frac{2}{x}\).

Answer: E.

naaga, please format the questions properly.

P.S. This is your third question for today with incorrect OA.
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If xy≠0 and x^2*y^2-xy=6, which of the following could be y  [#permalink]

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New post 16 Jan 2012, 12:46
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4
If xy≠0 and x^2*y^2-xy=6, which of the following could be y in terms of x?

I. 1/(2x)
II. - 2/x
III. 3/x

\(x^2*y^2-xy=6\) --> \((xy)^2-xy-6=0\) --> factor quadratics for xy: \((xy-3)(xy+2)=0\) (or just solve quadratics for xy) --> either \(xy-3=0\) and in this case \(y=\frac{3}{x}\) or \(xy+2=0\) and in this case \(y=-\frac{2}{x}\).

Answer: E.

Factoring Quadratics: http://www.purplemath.com/modules/factquad.htm

Solving Quadratic Equations: http://www.purplemath.com/modules/solvquad.htm

Hope it helps.
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Re: If xy ≠ 0 and x2y2 – xy = 6, which of the following could be  [#permalink]

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New post 07 Jul 2015, 22:37
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Hi All,

This question is quirky in that it tests you on math rules and patterns that you probably know, but in ways that you're not used to thinking about...

We're told that neither X nor Y are equal to 0. We're also told that (X^2)(Y^2) - XY = 6. We're asked which of the following COULD be the value of Y in terms of X...

The first interesting thing about this question is the use of the word COULD....that word implies that there's MORE THAN ONE possible solution.
The second interesting thing is that the 'term' (XY) can be factored out of the 'left side' of the equation. Normally, you look to factor our a single variable or number, but here, it's the product of two variables that you can factor out. Doing so gives us...

XY(XY - 1) = 6

While this looks complicated, there's an easy pattern here:

(number)(number - 1) = 6

Can you think of 2 numbers, that differ by 1, that you can multiply to get 6?

You should be thinking 2 and 3... because (3)(3-1) = 6

So XY = 3 is a possible solution. In this case, Y = 3/X. The wording of the prompt makes me think that there should be another solution though, so is there ANOTHER pair of numbers, that differ by 1, that you can multiply together to get 6? Hint: the numbers do NOT have to be positive....

How about -2 and -3....

(-2)(-2-1) = 6

So XY = -2 is another possible solution. In this case, Y = -2/X

There's only one answer that includes both of those solutions...

Final Answer:

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Re: If xy ≠ 0 and x2y2 – xy = 6, which of the following could be  [#permalink]

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New post 08 Jul 2015, 21:57
1
I tried this in the following way:

Since the values given are of "y", i plugged in these values in the equation for "y".

Starting with 3rd option first which is easier to plug in, x^2(3/x)^2 - x(3/x) = 6 satisfy the equation.

Plugging in y = -2/x in the equation, x^2(-2/x)^2 - x(-2/x) = 6 also satisfy the equation.

plugging in y = 1/(2x) doesn't satisfy the equation, Hence E.

Is my approach correct?
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Re: If xy ≠ 0 and x2y2 – xy = 6, which of the following could be  [#permalink]

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New post 09 Jul 2015, 18:43
1
Hi ash99,

YES, your method for dealing with this question is just as effective as any of the others (it's essentially a variation on TEST THE ANSWERS). Based on how the the 5 answer choices are designed, once you proved that the third and second Roman Numerals were potential solutions, then you could have stopped working (since none of the answers included all 3 Roman Numerals).

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Re: If xy≠0 and x^2*y^2-xy=6, which of the following could be y  [#permalink]

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New post 21 Feb 2017, 10:27
1
Baten80 wrote:
If xy not equal 0 and x^2*y^2 -xy = 6, which of the following could be y in terms of x?

I. 1/2x
II. -2/x
III. 3/x

A. I only
B. II only
C. I and II
D. I and III
E. II and III


We are given the equation x^2y^2 – xy = 6, and although it may not be obvious, the equation is a quadratic-format equation. Thus, our first step is to set one side of the equation to zero. We then factor the other side as a quadratic. .

x^2y^2 – xy – 6 = 0

(xy – 3)(xy + 2) = 0

xy – 3 = 0 or xy + 2 = 0

xy = 3 or xy = -2

Since we need y in terms of x, we can isolate y in both of our equations.

y = 3/x or y = -2/x

Thus, the expressions in II and III are correct.

Answer: E
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Re: If xy ≠ 0 and x2y2 – xy = 6, which of the following could be  [#permalink]

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