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Clarification needed on DS problem

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Clarification needed on DS problem

by knight247 » Sat Feb 07, 2015 11:12 pm
If x and y are integers and 4xy=(x^2)(y)+(4y), what is the value of xy?

(1) y-x=2
(2) x^3<0

The OA is [spoiler]B[/spoiler]

My query might be a silly one, so please bear with me. We are given 4xy=(x^2)(y)+(4y) in the problem.

Why can't I divide throughout by y there and get x=2, and then tackle both statements? Why do I need to do the following and then consider various scenarios for y and x?

x^2*y - 4xy+4y=0
y(x^2-4x+4)=0
y(x-2)^2 = 0

Detailed explanations would be appreciated. I apologise if my query is a silly one.
Last edited by knight247 on Wed Apr 01, 2015 4:31 pm, edited 1 time in total.
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by DavidG@VeritasPrep » Sun Feb 08, 2015 3:29 am
Anytime you divide by a variable, you're assuming that the variable in question is not equal to 0. We cannot assume that y is not 0 here. Note that if y were 0, x could be any number.
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by DavidG@VeritasPrep » Sun Feb 08, 2015 3:41 am
An extension of this logic is that initially, we have two scenarios: in the first scenario, y is not zero, in which case, we can divide by y, as you did, and get x^2 - 4x + 4 = 0, which yields x = 2. In the second scenario, y is 0, and x can be anything.

Statement 2 tells us that x is negative. If x is negative, it isn't equal to 2. If x isn't 2, then, according to the logic above, we just proved that y is 0. If y is 0, xy is also 0, irrespective of what x happens to be.
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by Brent@GMATPrepNow » Sun Feb 08, 2015 9:52 am
knight247 wrote: Why can't I divide throughout by y there and get x=2, and then tackle both statements? Why do I need to do the following and then consider various scenarios for y and x?
To show the dangers of dividing by variables, consider a version of a popular "proof" that "proves" 1 = 2:

Let x = y
Multiply both sides by x to get: x² = xy
Subtract y² from both sides to get: x² - y² = xy - y²
Factor both sides: (x + y)(x - y) = y(x - y)
DIVIDE both sides by (x-y) to get: x + y = y
Since we started with the premise that x = y, let's let x = y = 1
When we plug 1 in for x and y we get: 1 + 1 = 1
In other words, 2 = 1

In the bolded step, we divided both sides by x - y. However, since x = y, we were actually dividing both sides by zero, which caused a big problem.

Here's another, more straightforward example:
We know that (0)(1) = (0)(2)
If we divide both sides by 0, do we get 1 = 2? NO.

The same applies with the equation: xy = xz
If we divide both sides by x, do we get y = z? MAYBE, MAYBE NOT
If x = 0, we cannot conclude that y = z
So, if we divide both sides by x, we must consider the possibility that x might equal zero.

Cheers,
Brent
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by Matt@VeritasPrep » Mon Feb 09, 2015 12:29 am
Important: this question is seriously flawed.

Here's an algebraic approach.

x²y + 4y = 4xy
x²y - 4xy + 4y = 0
y * (x² - 4x + 4) = 0
y * (x - 2)² = 0

So y = 0 or (x - 2) = 0, but we don't know which.

S1 tells us that x - y = 2. From the prompt, we know that either y = 0 or x = 2. If y = 0, then x - y = x - 0, so x = 2. If x = 2, then 2 - y = 2, and y = 0. In either case, xy = 0*2 = 0, and Statement 1 is SUFFICIENT, meaning the OA is wrong.

S2 tells us that x³ < 0, so x < 0. Hence x ≠ 2, meaning that (x - 2) = 0 is false and y = 0 is true. This gives us xy = x * 0 = 0, and is SUFFICIENT. However, this value of x contradicts S1 and breaks the problem.

Do NOT trust the source of this question.

EDIT: Just Googled to try to discover the source of this question. It turns out that S1 should read "y - x = 2", so I retract my remark about the source being unreliable. Decided to leave my explanation intact, as it seems useful enough.
Last edited by Matt@VeritasPrep on Mon Feb 09, 2015 12:38 am, edited 1 time in total.
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by Matt@VeritasPrep » Mon Feb 09, 2015 12:37 am
Brent@GMATPrepNow wrote: Let x = y
Multiply both sides by x to get: x² = xy
Subtract y² from both sides to get: x² - y² = xy - y²
Factor both sides: (x + y)(x - y) = y(x - y)
DIVIDE both sides by (x-y) to get: x + y = y
Since we started with the premise that x = y, let's let x = y = 1
When we plug 1 in for x and y we get: 1 + 1 = 1
In other words, 2 = 1
I remember this! As far as I know, it's from Spivak's Calculus circa 1967, though he might have found it elsewhere. Brent, you might enjoy this corollary (due to Anders Kaseorg):
Suppose there are n proofs that 1 = 2. From this we derive that there are n + 1 - 1 = n + 2 - 1 = n + 1 proofs that 1 = 2. Therefore, by induction, there are infinitely many proofs that 1 = 2.
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by Brent@GMATPrepNow » Mon Feb 09, 2015 7:50 am
Sweet!

Cheers,
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