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

Question

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

v12345 · Accepted Answer

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

Bunuel · Answer

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.

avigutman · Answer

Video solution from Quant Reasoning: Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1 https://www.youtube.com/watch?v=tzTiumMNOS4

gurpreetsingh · Answer

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

g4gmat · Answer

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

camlan1990 · Answer

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

ScottTargetTestPrep · Answer

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

chetan2u · Answer

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..

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..

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

GMATinsight · Answer

x^2y^2 − xy = 6

i.e.  xy(xy − 1) = 6     
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

EMPOWERgmatRichC · Answer

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:  E

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

bbb123 · Answer

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

EMPOWERgmatRichC · Answer

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

GMATinsight · Answer

Answer: Option E

Video solution by  GMATinsight

https://www.youtube.com/watch?v=Q6J6-0ilDhk

woohoo921 · Answer

avigutman 
Thank you for this clear explanation!

I have a follow-up question. When I was originally solving this equation, I simplified it down to

xy(xy-1)=6
then set xy=6 and xy-1=6 --> xy=7 
so then y=6/x and y=7/x
Is this method incorrect because in order to solve a quadratic equation, you need have the variable expression equal to zero?

Thank you again for all of your time and help.

avigutman · Answer

Your inference was based on the following, erroneous reasoning:  If the product of two numbers is 6, then one of those numbers must be 6. 
In your defence, that reasoning would be correct if the product were zero:  If the product of two numbers is 0, then one of those numbers must be 0. 
Do you see why this only works for zero,  woohoo921 ?

woohoo921 · Answer

Thank you avigutman for your quick response! To clarify, what if it was set equal to a number different than 6 such as 7.... xy(xy-1)=7 Can you still not simplify and set xy=0 and xy-1=7 to solve for xy? In other words, do you need to have the right side of the equation equal to 0? Thank you again

avigutman · Answer

I'm going to respond to your question with a question,  woohoo921 , because I'd rather teach you how to fish than give you a fish.
Say you know that xy(xy-1)=7.
In other words, the product of two numbers (xy and (xy - 1)) is 7.
Is it reasonable to infer that one of those numbers is 7? Why, or why not?

Side note: if we were told that the two numbers were positive integers, we could infer that one of them is 1 and the other is 7 (because 7 is a prime number). And, since we're on the subject, if you are told that the product of  two different positive integers  is the square of a prime number (p^2), you can infer that the two numbers are 1 and p^2. 
How am I coming up with these "rules?" I'm just using quantitative  reasoning .

If you like, come join me at one of my live AMA sessions on Zoom, where we can all practice this kind of reasoning together. Your first AMA is completely free of charge, using the free 7-day trial at  quantreasoning.com

kingbucky · Answer

If  xy 
eq 0  and  (xy)^2 - xy = 6  which of the following could be  y  in terms of  x ?

I.  \frac{1}{2x}  
 II.  -\frac{2}{x}  
 III.  \frac{3}{x}

(xy)^2 - xy = 6 \Rightarrow (xy)((xy) - 1) = 6 ; notice the equation is quadratic with variable  (xy) , hence be on lookout for 2 solutions. Luckily see,  3 \cdot 2 = 6 , one possible value of  xy = 3 . Let's see  -2 \cdot -3 = 6 , we get next possible  xy = -2 .

Therefore, option II and III are right. 
 However, for a safe measure, let's check option I too, by plugging back into the equation. We should make sure that we hadn't jumped over some important step in the way carelessly.

y = \frac{1}{2x} \Rightarrow xy = \frac{1}{2} , but plugging back to  (xy)((xy) - 1) , gives  -\frac{1}{4} . Hence now, we are sure I is not the right answer.

Moreover, since the options did not have "All of the above" choice, we could have skipped checking I as well.

Hence, the right answer should be II & III, which is option E.

ablatt4 · Answer

Questions like this can be confusing because you substitute a variable for another variable which feels wrong.

If we let x^2 y^2 and xy each be one value we can say that xy= z

Therefore z^2 - z -6=0

(z-3)(z-2)=0

z=3,-2

xy= -6

x= -2 OR 3

y= -2 OR 3 (whichever x isn't)

plug both values into each and solve

(E)

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

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From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards