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# If xyz 0, is x (y + z) 0?

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If xyz 0, is x (y + z) 0? [#permalink]

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30 Aug 2010, 10:19
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If xyz ≠ 0, is x (y + z) ≥ 0?

(1) │y + z│ = │y│ + │z│
(2) │x + y│ = │x│ + │y│

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-xyz-0-is-x-y-z-107551.html
[Reveal] Spoiler: OA
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Re: If xyz 0, is x (y + z) 0? [#permalink]

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30 Aug 2010, 10:34
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KUDOS
xyz != 0,
xy+xz>= 0 ?

i). |y+z| = |y|+|z|
=> y,z both > 0 or y,z both < 0
So, xy+xz can be either -ve or +ve depending on sign/value of x.
Insufficient.

ii). |x+y| = |x|+|y|
=> x,y both > 0 or x,y both < 0
Again, xy+xz can be either +ve or -ve depending on z's value/sign.
Insufficient.

Combining both of them :
Either , x,y and z all >0
Or, x, y and z all < 0

In both the cases: xy + xz = +ve + (+ve) = +ve
Hence, "C".
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Re: If xyz 0, is x (y + z) 0? [#permalink]

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31 Aug 2010, 10:20
Has to be C, same explanation as Anshu
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Re: If xyz 0, is x (y + z) 0? [#permalink]

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26 Apr 2015, 14:10
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Re: If xyz 0, is x (y + z) 0? [#permalink]

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27 Apr 2015, 01:35
If xyz ≠ 0, is x (y + z) >= 0?

xyz ≠ 0 means that neither of unknowns is 0.

(1) |y + z| = |y| + |z| --> either both $$y$$ and $$z$$ are positive or both are negative, because if they have opposite signs then $$|y+z|$$ will be less than $$|y|+|z|$$ (|-3+1|<|-3|+1). Not sufficient, as no info about $$x$$.

(2) |x + y| = |x| + |y| --> the same here: either both $$x$$ and $$y$$ are positive or both are negative. Not sufficient, as no info about $$z$$.

(1)+(2) Either all three are positive or all three are negative --> but in both cases the product will be positive: $$x(y+z)=positive*(positive+positive)=positive>0$$ and $$x(y+z)=negative*(negative+negative)=negative*negative=positive>0$$. Sufficient.

Answer: C.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-xyz-0-is-x-y-z-107551.html
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Re: If xyz 0, is x (y + z) 0?   [#permalink] 27 Apr 2015, 01:35
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# If xyz 0, is x (y + z) 0?

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