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

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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\(x^2y^2 − xy = 6\)

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

kudos, if you like the post

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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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

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

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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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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\(x^2y^2 − xy = 6\)

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

IMO E

Posted from my mobile device
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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:

why can't
x2y2−xy=6
xy(xy - 1) = 6
so xy = 6 or xy = 7???

Hi bbb123,

To start, you can actually prove that neither of those solutions is correct by 'plugging' it back into the original equation:

IF.... XY = 6, then (XY)(XY - 1) = (6)(5) = 30..... NOT 6.
IF.... XY = 7, then (XY)(XY - 1) = (7)(6) = 42..... NOT 6.

If you want to approach this algebraically, then you will almost certainly have to set one side equal to 0...

(X^2)(Y^2) - XY - 6 = 0

This is solvable (Bunuel walks through the steps in his approach). If you recognize the Number Property involved though, you don't have to use Algebra at all.... Can you think of two CONSECUTIVE INTEGERS that multiply together to equal 6? (Hint: there are actually two pairs that yield that result).

GMAT assassins aren't born, they're made,
Rich
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Hi Rich,

i was looking for the same explanation! i could not understand where i was going wrong.

Just to confirm on my understanding , whenever we are solving an algebraic equation , we need to use the 'zero product property'
ie if we have ax2+bx = z ; we would have to make the equation as ax2+bx-z=0 and only then solve.

Thanks!

Answer: Option E

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Alternatively,

\(x^2*y^2-xy= 6\) -->\( xy (xy-1) = 6\) --> if you look closely, these are two consecutive numbers. xy and xy-1 only two numbers that satisfy the equation are 3 and 2 ( \( xy=3* xy-1=2 = 6\); \(y= 3/x \)) OR ( \(xy = -2\) ,\( xy-1 = - 3;\) \(y=-2/x \))