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# xyz <> 0, is x (y + z) = 0? (1) y + z = y + z (2) x +

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VP
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xyz <> 0, is x (y + z) = 0? (1) y + z = y + z (2) x + [#permalink]

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21 Jan 2005, 09:18
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xyz <> 0, is x (y + z) = 0?

(1) Â¦ y + z Â¦ = Â¦ y Â¦ + Â¦ z Â¦

(2) Â¦ x + y Â¦ = Â¦ x Â¦ + Â¦ y Â¦
Director
Joined: 31 Aug 2004
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21 Jan 2005, 09:52
I pick A, then we know that y <> -z
Current Student
Joined: 28 Dec 2004
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21 Jan 2005, 10:08
A it is...

twixit's explanation is good...we know, x,y,z are not equal to zero...all we need to check for is, if y+z is not equal to zero...statement I doest that.
VP
Joined: 18 Nov 2004
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21 Jan 2005, 10:40
"A" it is....same explanation as above.....y and z has to be of same sign for the statement 1 to be true and they can't be 0.
Director
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21 Jan 2005, 15:01
from solving the stem we get y = -z

A will be suff

B insuff
Director
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21 Jan 2005, 18:31
Pardon me for being really slow to this, but do you guys mind breaking this down further by substituting with actual numbers please? Please ! Thanks
SVP
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21 Jan 2005, 22:15
(I) means y and z have same sign, so that the absolute value of their sum is equal to the sum of their absolute values. In other words y+z can't be zero unless both are zero, but the question has xyz<>0. Therefore we would be able to determine that x(y+z)<>0.

This is why I is sufficient.

(II) By the same reason we get x+y<>0, but we don't know if y+z=0, so not sufficient to tell if x(y+z)=0.
Senior Manager
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23 Jan 2005, 17:53
hi Cn somebody tell me what does <> sign means??? I am lost.

Thanks
Saurabh Malpani
23 Jan 2005, 17:53
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# xyz <> 0, is x (y + z) = 0? (1) y + z = y + z (2) x +

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