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If xy+z=x(y+z), which of the following must be true? 1)

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Director
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If xy+z=x(y+z), which of the following must be true? 1) [#permalink] New post 04 Jul 2006, 02:10
00:00
A
B
C
D
E

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(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 1 sessions
If xy+z=x(y+z), which of the following must be true?

1) x=0, z=0
2) x=1, y=1
3) y=1, z=0
4) x=1, y=0
5) x=1, z=0

OA will follow.
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Re: PS: xy+z=x(y+z) [#permalink] New post 04 Jul 2006, 02:55
M8 wrote:
If xy+z=x(y+z), which of the following must be true?

1) x=0, z=0
2) x=1, y=1
3) y=1, z=0
4) x=1, y=0
5) x=1, z=0

OA will follow.


4 ?

Am I wrong or does all the values satisfy?
Manager
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 [#permalink] New post 04 Jul 2006, 04:56
xy+z=x(y+z) => xy+z=xy+xz => xz - z = 0

=> z(x-1) = 0 => z=0, x=1

Hence 5
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 [#permalink] New post 04 Jul 2006, 05:02
xy + z = xy + xz

xy 's cancel out

z = xz
xz - z = 0
z(x-1) = 0

z = 0, x = 1

(5)
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Re: PS: xy+z=x(y+z) [#permalink] New post 04 Jul 2006, 07:33
M8 wrote:
If xy+z=x(y+z), which of the following must be true?

1) x=0, z=0
2) x=1, y=1
3) y=1, z=0
4) x=1, y=0
5) x=1, z=0

OA will follow.


xy + z = x (y + z)
xy + z = xy + xz
z = xz
xz - z = 0
z (x - 1) = 0
z = 0, x = 1

arrived at E.

however any value for x and y with z = 0 satisfy the eq xy + z = x (y + z). therefore, the question is not properly structured. 8-)
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 [#permalink] New post 04 Jul 2006, 09:12
I don't think this question was written accurately.

You can deduce the equation to:

xy+ z = xy + xz --> thus y cancels out and its value doesn't matter

z = xz --> thus it only matters that x must equal 1, if that is true z can have any value
Director
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 [#permalink] New post 04 Jul 2006, 23:49
OA is 'E'.
But I was really stuck between A and E guys. What do you think about 'A'?
This is GMATPrep question.
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 [#permalink] New post 13 Jul 2006, 22:22
z=xz -> x = 1
z can be anything
let's try x=1, y=1, z=1
xy+z = x(y+z)
(1)(1) + 1 = 1 (1+1)
2=2 .... z doesn't have to equal 0

This question is poorly written!
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 [#permalink] New post 13 Jul 2006, 23:34
xy+z = x(y+z)
xy+z = xy+xz
z-xz = 0
z(1-x) = 0

z = 0, x = 1
  [#permalink] 13 Jul 2006, 23:34
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