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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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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

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[Reveal] Spoiler: OA

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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, 12:24
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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 18 Oct 2015, 12:37
I believe the question should be structured differently
otherwise it is very ambiguous
is it x^2 * y^2 or

x to the power of 2y^2

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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, 14:22
x^2y^2 − xy = 6

x^2y^2 − xy -6=0
by solving this equation, we get
xy =3 and xy=-2

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

so it satisfies II and III

answer is 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, 15:34
We have xy = 3 or xy =-2 . So y=3/x or -2/x. Answer is 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, 20:44
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.


xy(xy-1) = 6

Possible solutions (2,3) (-2,-3) (1,6) (-1,-6) (3,2) (-3,-2) (-6,-1)

1) 1/2x = y so xy = 1/2 but we know that xy has to be integer - (see solutions above)
2) -2/x =y so xy = -2 (xy-1 = -3) Yes we can get this.
3) 3/x = y so xy =3 (yes we can get this one too)

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 18 Oct 2015, 22:04
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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 ≠ 0 and x^2y^2 − xy = 6, which of the following could be y in te [#permalink]

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New post 02 May 2016, 14:46
this is a (x+1)(x+2) type question on mild steroids

x^2y^2 - xy = 6
x^2y^2 - xy - 6 = 0
(xy - 3)(xy + 2) = 0 (because -3*2 = -6 and because -3xy + 2xy = -xy)
xy = 3, xy = -2
y = 3/x , y = -2/x

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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, 04:36
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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, 04:56
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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, 01:25
Question looks confusing at first sight. Kindly restructure the Question as : (x^2) * (y^2) - xy = 6.

Thanks.

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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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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, 03:41
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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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Re: If xy ≠ 0 and x^2y^2 − xy = 6, which of the following could be y in te   [#permalink] 16 Dec 2017, 04:32
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