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If xy ≠ 0 and x^2y^2 − xy = 6, which of the following could be y in te

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If xy ≠ 0 and x^2y^2 − xy = 6, which of the following could be y in te  [#permalink]

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New post Updated on: 30 Jul 2019, 01:28
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If xy ≠ 0 and \(x^2y^2 − xy = 6\), 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 sperumba on 18 Jan 2006, 20:15.
Last edited by Bunuel on 30 Jul 2019, 01:28, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
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If xy ≠ 0 and x^2y^2 − xy = 6, which of the following could be y in te  [#permalink]

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New post 18 Oct 2015, 13:18
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Re: If xy ≠ 0 and x^2y^2 − xy = 6, which of the following could be y in te  [#permalink]

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New post 18 Oct 2015, 23:04
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Bunuel wrote:
If xy ≠ 0 and x^2y^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

Kudos for a correct solution.


\(x^2y^2 − xy = 6\)

taking xy = a, we have
\(a^2 − a= 6\)
=> \(a^2 − a − 6 = 0\)

solving we get a = 3 or -2

if a = xy = 3, we have y = 3/x
if a = xy = -2, we have y = -2/x


Answer choice E

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Re: If xy not equal 0 and x^2*y^2 -xy = 6, which of the following could be  [#permalink]

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New post 28 Aug 2010, 02:54
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take \(xy= a\)

\(a^2 - a- 6=0\)

\((a-3)*(a+2) = 0\)

=> either \(a = 3\) or \(a = -2\) => either \(xy =3\) or \(xy = -2\)

Hence E
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Re: If xy not equal 0 and x^2*y^2 -xy = 6, which of the following could be  [#permalink]

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New post 26 Sep 2010, 09:26
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udaymathapati wrote:
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


Backsolving..
\(I) y= 1/(2x) ==> xy = 1/2\)
substituting
\(1/4 - 1/2 <> 6..\)
\(LHS <> RHS\)

\(II) y = - 2 / x\)
\(xy = -2 ..\)
subtituting in the equation.
LHS = RHS

\(III) y = 3/x ...
xy = 3\)
substituting in the equation.
LHS= RHS


Hence E
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Re: If xy ≠ 0 and x^2y^2 − xy = 6, which of the following could be y in te  [#permalink]

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New post 18 Oct 2015, 13:24
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Bunuel wrote:
If xy ≠ 0 and x^2y^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

Kudos for a correct solution.


x^2y^2 − xy = 6 => (xy-3)(xy+2)=0

=> xy=3 or xy =-2

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

Because the question is COULD BE TRUE, meaning that there exists the value of x: y=1/(2x) and x^2y^2 − xy = 6

=> 1/2x = -2/x or =3/x

=> 1=-4 or 1=6: WRONG

=> only II and III are TRUE

=> Ans: E
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Re: If xy ≠ 0 and x^2y^2 − xy = 6, which of the following could be y in te  [#permalink]

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New post 03 May 2016, 05:36
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Bunuel wrote:
If xy ≠ 0 and x^2y^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

Kudos for a correct solution.


We are given the equation, (x^2)(y^2) – xy = 6, and, although it may not be obvious, the equation is a quadratic-format equation. Thus, our first step is to set the equation equal to zero. We then factor the left side and, finally, solve for xy.

(x^2)(y^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 x^2y^2 − xy = 6, which of the following could be y in te  [#permalink]

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New post 03 May 2016, 05:56
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Bunuel wrote:
If xy ≠ 0 and x^2y^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

Kudos for a correct solution.


Hi,
its a 700 level Q, BUT is literally begging you to make it a simpler Q by using substitution..
we have an equation - \(x^2y^2 − xy = 6\)
we have tto find OUT of 3, which all will be values of y in terms of x..

two ways-


1) factorize and then find answers
2) substitution, as we have our answers right in front of us and any value that fits in IS the answer..


[b]we can actually eliminate I at the first glance as it is giving us a 2 in denominator and hence LHS will become some fraction, but lets still try..[/b]

EQ - \(x^2y^2 − xy = 6\)
I. 1/(2x)......\(x^2\frac{1}{2x}^2 − x\frac{1}{2x} = 6\) ..... \(\frac{1}{4}-\frac{1}{2} = 6\)... NO
II. -2/x .......\(x^2\frac{-2}{x}^2 − x\frac{-2}{x} = 6\) ..... \(4-(-2) = 6\)..YES
III. 3/x .......\(x^2\frac{3}{x}^2 − x\frac{3}{x} = 6\) ..... \(9-3 = 6\)..YES

so II and III are correct
E
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Re: If xy ≠ 0 and x^2y^2 − xy = 6, which of the following could be y in te  [#permalink]

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New post 17 Oct 2016, 04:41
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Bunuel wrote:
If xy ≠ 0 and \(x^2y^2 − xy = 6\), 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

Kudos for a correct solution.


\(x^2y^2 − xy = 6\)

i.e. \(xy(xy − 1) = 6\) [Please observe that xy and (xy-1) are two consecutive number]
But 3*2 = 6
also, (-2)*(-3) = 6

therefore, xy = 3 or xy = -2

i.e. , y = 3/x or y=-2/x

Answer: option E
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Re: If xy ≠ 0 and x^2y^2 − xy = 6, which of the following could be y in te  [#permalink]

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New post 01 Feb 2018, 06:55
Bunuel wrote:
If xy ≠ 0 and \(x^2y^2 − xy = 6\), 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

Kudos for a correct solution.


Why is it not (xy-2)(xy+3)?
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Re: If xy ≠ 0 and x^2y^2 − xy = 6, which of the following could be y in te  [#permalink]

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New post 01 Feb 2018, 07:07
michaelarbeid wrote:
Bunuel wrote:
If xy ≠ 0 and \(x^2y^2 − xy = 6\), 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

Kudos for a correct solution.


Why is it not (xy-2)(xy+3)?



I don't know if the highlighted part is a typo error but here your expression doesn't match with with the given expressions in question

I am assuming that you meant to type \((xy+2)(xy-3)\) which becomes \(x^2*y^2 - xy = 6\)

And if you meant \((xy+2)(xy-3)\) then you are absolutely correct because this also leads us to the correct result

\(xy = -2 or +3\)

Hope this helps!!!
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Re: If xy ≠ 0 and x^2y^2 − xy = 6, which of the following could be y in te  [#permalink]

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New post 01 Feb 2018, 13:25
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 x^2y^2 − xy = 6, which of the following could be y in te  [#permalink]

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New post 15 Sep 2019, 19:16
sperumba wrote:
If xy ≠ 0 and \(x^2y^2 − xy = 6\), 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


Solving the equation
xy = 3 or -2
y = 3/x or -2/x

IMO E

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Re: Maths: Y in terms of X ?  [#permalink]

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Re: Maths: Y in terms of X ?   [#permalink] 30 Sep 2019, 03:23
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