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# If y = \frac {x^2 - y^2}{x-y} ( x \ne y ), then what is the

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If y = \frac {x^2 - y^2}{x-y} ( x \ne y ), then what is the [#permalink]

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16 Oct 2011, 04:07
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If $$y = \frac {x^2 - y^2}{x-y}$$ ( $$x \ne y$$ ), then what is the value of $$y$$ ?

1. $$x + y = 3$$
2. $$x - y = -3$$

(C) 2008 GMAT Club - m02#25

Hi All-

I'm struggling to see the logic behind the OA in this question. Taking the information from the question stem, we get y= (x+y)(x-y)/ (x-y), therefore y=x+y. Its easy to see that statement 1 is sufficient, but for statement 2, we will still get y=x+y because y=-3(x+y)/-3. We can pick a number of different values for x-y that equal -3 that fit the question stem (x is not equal to y). Can someone with better math skills than myself explain why statement 2 is sufficient? Cheers!

-Ken
[Reveal] Spoiler: OA
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16 Oct 2011, 04:14
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kenman wrote:
If $$y = \frac {x^2 - y^2}{x-y}$$ ( $$x \ne y$$ ), then what is the value of $$y$$ ?

1. $$x + y = 3$$
2. $$x - y = -3$$

(C) 2008 GMAT Club - m02#25

Hi All-

I'm struggling to see the logic behind the OA in this question. Taking the information from the question stem, we get y= (x+y)(x-y)/ (x-y), therefore y=x+y. Its easy to see that statement 1 is sufficient, but for statement 2, we will still get y=x+y because y=-3(x+y)/-3. We can pick a number of different values for x-y that equal -3 that fit the question stem (x is not equal to y). Can someone with better math skills than myself explain why statement 2 is sufficient? Cheers!

-Ken

Mark my words bro, if you follow the way you are solving the questions you will face a difficult time with DS questions.
Always simply the equations till end, else you will complicate the things for you... - GOLDEN RULE.

now when you got y =x+y => x must be 0
now when you will put x=0 in both the statements you will have unique value of Y.
=> both are sufficient.

Now spend at least 15-20 mins on the advice I gave you and try to put in every question. Always simplify and then go ahead.
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16 Oct 2011, 12:44
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16 Oct 2011, 13:13
prateekbhatt wrote:
even i mistakenly chose A...But its little difficult to understand why D?

$${x^2 - y^2}$$ = $$(x+y)*(x-y)$$

The original expression ends up being: $$y = x + y$$ thus $$x=0$$.

From here, both (1) and (2) can be solved:

(1) $$y=3$$
(2) $$-y=-3$$; $$y=3$$
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16 Oct 2011, 14:06
LGOdream wrote:
prateekbhatt wrote:
even i mistakenly chose A...But its little difficult to understand why D?

$${x^2 - y^2}$$ = $$(x+y)*(x-y)$$

The original expression ends up being: $$y = x + y$$ thus $$x=0$$.

From here, both (1) and (2) can be solved:

(1) $$y=3$$
(2) $$-y=-3$$; $$y=3$$

Ohh got it.....I was thinking that if statement 1 is solving the purpose then why to check statement 2 ....i was wrong...

Thanks for solving.....
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22 Oct 2011, 12:13
+1 for D.

The given input is a good indication of what the questions expects (x not equal to y).

Simplifying the given equations we get:

x(x-y) = 0, since x -y = 0 or x is not equal to y, we know that x = 0.

Substituting x=0 in both conditions will give us the value of y. Hence D.
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Re: If y = \frac {x^2 - y^2}{x-y} ( x \ne y ), then what is the [#permalink]

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16 Dec 2011, 07:07
Excellent question kenman. Definitely worth another look. Its easy to miss out on the minor details and make careless mistakes as I did.
D is the right answer. Thanks for the explanation gurpreetsingh.
Kudos to both..
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Re: If y = \frac {x^2 - y^2}{x-y} ( x \ne y ), then what is the [#permalink]

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03 Jan 2012, 22:45
D. From question stem we know

y = (x + y) (x-y) / (x-y)
y = (x + y) , y = ?

(1) x+ y = 3. so y = 3 SUFF.
(2) x -y = -3

x = -3 + y
Subs value of x in rephrased question stem

y = -3 + y + y
y = 3. SUFF.
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Re: If y = \frac {x^2 - y^2}{x-y} ( x \ne y ), then what is the [#permalink]

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06 Feb 2015, 17:14
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Re: If y = \frac {x^2 - y^2}{x-y} ( x \ne y ), then what is the   [#permalink] 06 Feb 2015, 17:14
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