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Number Properties

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Number Properties

by sparkle6 » Wed Sep 21, 2011 7:31 am
If xy does not equal 0 and (x^2)(y^2) - xy = 6, which of the following could be y in terms of x?

1. 1/(2x)
2. -2/x
3. 3/x

A. 1 only
B. 2 only
C. 1 and 2
D. 1 and 3
E. 2 and 3
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by Abhishek009 » Wed Sep 21, 2011 7:36 am
sparkle6 wrote:If xy does not equal 0 and (x^2)(y^2) - xy = 6, which of the following could be y in terms of x?

1. 1/(2x)
2. -2/x
3. 3/x

A. 1 only
B. 2 only
C. 1 and 2
D. 1 and 3
E. 2 and 3
(x^2)(y^2) - xy = 6

x^2y^2 - xy = 6

xy (xy - 1 ) = 6

let xy = a

Now

a ( a - 1 ) = 6

So , a = 3

or xy = 3

Hence Y = 3/x
Abhishek
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by shankar.ashwin » Wed Sep 21, 2011 7:38 am
Substitute values of

y = 1. 1/(2x)
2. -2/x
3. 3/x

in the equation (x^2)(y^2) - xy = 6

Only for (2) and (3) you get 6.
E IMO

You could also do ;

xy(xy-1) = 6

So; xy=3 (or) xy = -2

So y = 3/x (or) y = -2/x
sparkle6 wrote:If xy does not equal 0 and (x^2)(y^2) - xy = 6, which of the following could be y in terms of x?

1. 1/(2x)
2. -2/x
3. 3/x

A. 1 only
B. 2 only
C. 1 and 2
D. 1 and 3
E. 2 and 3
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by gmatclubmember » Wed Sep 21, 2011 7:48 am
Consider x as any constant and solve for y using the Determinant method.
That gives you 2 values of y in terms of x.

Cheers
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by cavaliar » Wed Sep 21, 2011 7:49 am
Scenario 1

xy * (xy-1)= 6

xy = -2
xy-1 = -3

y= -2/x ----->1

Scenario 2

xy *xy-1 = 6

xy = 3
xy-1 = 2

y = 3/x ---->2

Thus answer E
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by GMATGuruNY » Wed Sep 21, 2011 8:03 am
sparkle6 wrote:If xy does not equal 0 and (x^2)(y^2) - xy = 6, which of the following could be y in terms of x?

1. 1/(2x)
2. -2/x
3. 3/x

A. 1 only
B. 2 only
C. 1 and 2
D. 1 and 3
E. 2 and 3
x²y² - xy - 6 = 0.
(xy + 2)(xy - 3) = 0.
Thus, xy = -2 or xy = 3.

If xy = -2, then y = -2/x.
If xy = 3, then y = 3/x.

The correct answer is E.
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