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

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Manager
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If xy+z = x(y+z), which of the following must be true? a. [#permalink]

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31 Mar 2008, 00:36
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If xy+z = x(y+z), which of the following must be true?
a. x=0 & z=0
b. x=1 & y=1
c. y=1 & z=0
d. x=1 or y=0
e. x=1 or z=0

I am missing something in this problem..
Please, help to figure out the point here..

Thank you

Last edited by chica on 31 Mar 2008, 23:41, edited 1 time in total.
Manager
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31 Mar 2008, 00:53
chica wrote:
If xy+z = x(y+z), which of the following must be true?
a. x=0 & z=0
b. x=1 & y=1
c. y=1 & z=0
d. x=1 & y=0
e. x=1 & z=0

I am missing something in this problem..
Please, help to figure out the point here..

Thank you

if xy+z = xy+xz

then either x=1 or z=0
i go with e
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31 Mar 2008, 20:35
chica wrote:
If xy + z = x (y + z), which of the following must be true?
a. x=0 & z=0
b. x=1 & y=1
c. y=1 & z=0
d. x=1 & y=0
e. x=1 & z=0

I am missing something in this problem.. Please, help to figure out the point here..Thank you

I know E is correct but I find A is also a correct one.

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

but how to prove that A is also correct?
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Joined: 28 Mar 2008
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31 Mar 2008, 21:17
GMAT TIGER wrote:
chica wrote:
If xy + z = x (y + z), which of the following must be true?
a. x=0 & z=0
b. x=1 & y=1
c. y=1 & z=0
d. x=1 & y=0
e. x=1 & z=0

I am missing something in this problem.. Please, help to figure out the point here..Thank you

I know E is correct but I find A is also a correct one.

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

but how to prove that A is also correct?

Actually, you can see that xz=z or x=1. so you can eliminate the choices where x is zero first.
Manager
Joined: 19 Dec 2007
Posts: 87
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31 Mar 2008, 23:58
daszero wrote:
GMAT TIGER wrote:
chica wrote:
If xy + z = x (y + z), which of the following must be true?
a. x=0 & z=0
b. x=1 & y=1
c. y=1 & z=0
d. x=1 or y=0
e. x=1 or z=0

I am missing something in this problem.. Please, help to figure out the point here..Thank you

I know E is correct but I find A is also a correct one.

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

but how to prove that A is also correct?

Actually, you can see that xz=z or x=1. so you can eliminate the choices where x is zero first.

I got the question.. thank you
I was thinking in a different way.

Anyway, x should be 1 to met the condition for xz = z, and y can be any for the given equation OR x can be any unless z = 0.
The only choice that points this is E. In A, x not necessarily must be 0 when z is 0 and vice versa.
Re: GMATPrep: PS problem   [#permalink] 31 Mar 2008, 23:58
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