If xyz no= 0, is x (y + z) >= 0? (1) | y + z |=| y| +| z| : DS Archive
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# If xyz no= 0, is x (y + z) >= 0? (1) | y + z |=| y| +| z|

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If xyz no= 0, is x (y + z) >= 0? (1) | y + z |=| y| +| z| [#permalink]

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16 Oct 2007, 02:19
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If xyz no= 0, is x (y + z) >= 0?
(1) | y + z |=| y| +| z|
(2)| x + y| =| x |+ |y|
SVP
Joined: 01 May 2006
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16 Oct 2007, 02:34
(C) for me

x (y + z) >= 0?
<=> sign(x) = sign(x+z) ?

Stat 1
|y + z|=|y| +|z|
<=> sign(y) = sign(z) = sign(x+z)

Nothing about the sign of x.

INSUFF.

Stat 2
|x + y| =|x|+ |y|
<=> sign(x) = sign(y) = sign(x+y)

Nothing about the sign of y+z.

INSUFF.

Both 1 and 2
We have :
o sign(y) = sign(z) = sign(y+z)
o sign(x) = sign(y) = sign(x+y)

So,
sign(x) = sign(y+z) thus x*(y+z) >= 0

SUFF.
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16 Oct 2007, 03:28
OlgaN wrote:
I am confused with this question, because if we are asked >=, and we can prove >, but we still do not know about =, should we think that we definitely know the answer?
Thus either C or E.

But here, it cannot equal 0 .... As the signs of x, y and z are the same and they are different from 0.

So it is more > than possibly =
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16 Oct 2007, 04:05
OlgaN wrote:
Fig wrote:
OlgaN wrote:
I am confused with this question, because if we are asked >=, and we can prove >, but we still do not know about =, should we think that we definitely know the answer?
Thus either C or E.

But here, it cannot equal 0 .... As the signs of x, y and z are the same and they are different from 0.

So it is more > than possibly =

Agree, thanks. Those brutal DS questions - they make me to think!!!!!

It would not be the last one like this
16 Oct 2007, 04:05
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