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


\(x^2*y^2-xy=6\);

\((xy)^2-xy-6=0\);

\((xy-3)(xy+2)=0\);

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.
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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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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

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

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:

GMAT assassins aren't born, they're made,
Rich
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If anyone is uncomfortable solving this question using equation x2y2−xy=6.
It can be solved by plug in values mentioned: