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y in terms of x

by pdonaldson1990 » Tue May 17, 2016 4:32 pm
Good evening ladies and gentleman,

I am a bit disgruntled about this problem, does anyone have any helpful strategies for solving these questions, especially those that involve factoring equations like these?

If xy cannot equal zero 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

Answer: E
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by OptimusPrep » Tue May 17, 2016 6:08 pm
pdonaldson1990 wrote:Good evening ladies and gentleman,

I am a bit disgruntled about this problem, does anyone have any helpful strategies for solving these questions, especially those that involve factoring equations like these?

If xy cannot equal zero 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

Answer: E
x^2y^2-xy=6
(xy)^2 - xy - 6 = 0
(xy)^2 - 3xy + 2xy - 6 = 0

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

Hence xy = 3 , - 2
y = 3/x or -2/x

Correct Option: E
Last edited by OptimusPrep on Thu May 19, 2016 9:55 am, edited 1 time in total.
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by [email protected] » Tue May 17, 2016 10:57 pm
Hi pdonaldson1990,

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

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by GMATGuruNY » Wed May 18, 2016 2:32 am
pdonaldson1990 wrote: If xy cannot equal zero and x²y² - 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

Answer: E
To eliminate x from the equation and make the algebra easier, let x=1.
Plugging x=1 into x²y² - xy = 6, we get:
(1²)y² - (1)y = 6
y² - y - 6 = 0
(y-3)(y+2) = 0
y=3 or y=-2.

The question stem asks for possible values of y.
Thus, our targets are 3 and -2.

Now plug x=1 into I, II and III to see which yields 3 or -2.

I: (1/2)x = (1/2)1 = 1/2
II: -2/x = -2/1 = -2
III: 3/x = 3/1 = 3

Since II and III work, the correct answer choice must include II and III.

The correct answer is E.
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