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# Is x=y? 1) (x-y)= (x^2-y^2) 2) X and Y are each greater than

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SVP
Joined: 28 Dec 2005
Posts: 1545

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Is x=y? 1) (x-y)= (x^2-y^2) 2) X and Y are each greater than [#permalink]

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05 Jan 2009, 19:20
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Is x=y?

1) (x-y)= (x^2-y^2)

2) X and Y are each greater than 1

Kudos [?]: 179 [0], given: 2

Intern
Joined: 01 Jan 2009
Posts: 7

Kudos [?]: [0], given: 0

Schools: UCB, UCLA

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05 Jan 2009, 22:13
a. x-y = x^2 - Y^2
= ( x + y) (x -y)

For this to be true either x + y = 1,0 or x - y = 0

b. Tells us that both numbers are greater than 1.

c. Since both numbers are greater than 1 , x + y is not equal to 0.
which leaves x -y = 0

therefore x = y

Option C

Kudos [?]: [0], given: 0

Senior Manager
Joined: 02 Nov 2008
Posts: 276

Kudos [?]: 117 [0], given: 2

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05 Jan 2009, 22:31
pmenon wrote:
Is x=y?

1) (x-y)= (x^2-y^2)

2) X and Y are each greater than 1

Statement 1:
By plugging in numbers you can disprove this. Example:
x = 1
y = 0

1 - 0 = 1^2 - 0^2

Statement 2:
X & Y are greater than 2....real nice but doesn't help us out

Statements 1 & 2:
Prove that X & Y are indeed equal

Kudos [?]: 117 [0], given: 2

Manager
Joined: 09 Jul 2008
Posts: 111

Kudos [?]: 42 [0], given: 1

Location: Dallas, TX
Schools: McCombs 2011

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05 Jan 2009, 22:37
pmenon wrote:
Is x=y?

1) (x-y)= (x^2-y^2)

2) X and Y are each greater than 1

1) INSUFF
can be written as
$$x^2-y^2 - (x-y) = 0$$
$$(x+y)(x-y) - (x-y) = 0$$
$$(x-y) [(x+y)-1] = 0$$
==> $$x=y$$ or $$x+y=1$$

2) INSUFF. Only tells us both are > 1

1 and 2 combined, we can say that x+y=1 not possible because both are >1. Then x=y

Kudos [?]: 42 [0], given: 1

SVP
Joined: 28 Dec 2005
Posts: 1545

Kudos [?]: 179 [0], given: 2

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06 Jan 2009, 06:04
you guys are correct. so here is my question regarding statement 1:

x^2-y^2 = (x-y)(x+y) = x-y

why can i not just divide both sides by x-y to get x+y=1 ? This is what I did, and I answered A because from x+y=1, its clear that x does not equal y.

Kudos [?]: 179 [0], given: 2

Intern
Joined: 30 Sep 2008
Posts: 36

Kudos [?]: 1 [0], given: 1

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06 Jan 2009, 09:44
pmenon wrote:
you guys are correct. so here is my question regarding statement 1:

x^2-y^2 = (x-y)(x+y) = x-y

why can i not just divide both sides by x-y to get x+y=1 ? This is what I did, and I answered A because from x+y=1, its clear that x does not equal y.

you can divide both side but you have to make sure that number is not zero
but this problem x - y has a possible to be zero

Kudos [?]: 1 [0], given: 1

Re: DS: Equations   [#permalink] 06 Jan 2009, 09:44
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