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

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Manager
Joined: 05 May 2005
Posts: 77

Kudos [?]: 57 [0], given: 0

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

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08 Feb 2007, 21:24
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Difficulty:

25% (medium)

Question Stats:

77% (00:36) correct 23% (00:59) wrong based on 53 sessions

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

(A) x = 0 and z = 0
(B) x = 1 and y = 1
(C) y = 1 and z = 0
(D) x = 1 or y = 0
(E) x = 1 or z = 0
[Reveal] Spoiler: OA

Last edited by Bunuel on 03 Dec 2012, 02:41, edited 1 time in total.
Renamed the topic and edited the question.

Kudos [?]: 57 [0], given: 0

Senior Manager
Joined: 04 Jan 2006
Posts: 276

Kudos [?]: 43 [0], given: 0

Re: Factoring problem: x,y,z [#permalink]

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08 Feb 2007, 21:33
above720 wrote:
How would you solve?

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

a) x=0 and z=0
b) x=1 and y=1
c) y=1 and z=0
d) x=1 or y=0
e) x=1 or z=0

xy + z = xy + xz

z = xz

case 1: z not= 0, x = z/z = 1

case 2: z = 0, 0 = x0 = 0

Combining 2 cases: x = 1 or z =0

E. is the answer.

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Math Expert
Joined: 02 Sep 2009
Posts: 41886

Kudos [?]: 128668 [1], given: 12181

Re: Factoring problem: x,y,z [#permalink]

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13 Mar 2011, 13:25
1
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Expert's post
dc123 wrote:
why is ans A not a choice?

If xy + z = x(y + z), which of the following must be true?
(A) x = 0 and z = 0
(B) x = 1 and y = 1
(C) y = 1 and z = 0
(D) x = 1 or y = 0
(E) x = 1 or z = 0

$$xy+z=x(y+z)$$ --> $$xy+z=xy+xz$$ --> $$xy$$ cancels out --> $$xz-z=0$$ --> $$z(x-1)=0$$ --> either $$z=0$$ (in this case $$x$$ can take ANY value) OR $$x=1$$ (in this case $$z$$ can take ANY value).

To elaborate more:
As the expression with $$y$$ cancels out, we can say that equation $$xy+z=x(y+z)$$ does not depend on value of $$y$$. Which means that $$y$$ can take ANY value. So all answer choices which specify the exact value of $$y$$ are wrong.

Next: "AND" in answer choices means that BOTH values must be true, but "OR" in answer choices means that EITHER value must be true.

Hope it helps.
_________________

Kudos [?]: 128668 [1], given: 12181

Director
Joined: 01 Feb 2011
Posts: 726

Kudos [?]: 143 [0], given: 42

Re: Factoring problem: x,y,z [#permalink]

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13 Mar 2011, 18:50
i guess you are substituting each options,

Option A holds good, but thats not true in all the cases. Then it wont be a "must be true"

example when x =1 , expression becomes xy+z = x(y+z) => y+z = y+z

So the condition i have chosen x=1 holds good (that is not A).

In short just because it satisfies the expression in one example doesnt mean that its the must.

by solving the expression , you should get z=0 or x=1 , which is a must. Answer is E.

Hope it helps.

dc123 wrote:
why is ans A not a choice?

Kudos [?]: 143 [0], given: 42

Re: Factoring problem: x,y,z   [#permalink] 13 Mar 2011, 18:50
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# If xy + z = x(y + z), which of the following must be true?

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