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absolute value

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24 May 2008, 15:06
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Man, I hate these types of problems and I don't know why. Anyway,

is |x| = y - z ?

(1) x + y = z
(2) x < 0

I did a rephrase of is x = y - z when x is positive AND is x = z - y when x is negative. I'm not even sure if this rephrase is correct and I don't really know where to go from here or how to attack these problems in the future. Any explanation would be greatly appreciated. Thanks in advance.
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24 May 2008, 15:59
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brokerbevo wrote:
Man, I hate these types of problems and I don't know why. Anyway,

is |x| = y - z ?

(1) x + y = z
(2) x < 0

The question is asking is Y > Z

statement 1: we don't know the value of x, so insuff
statement 2: we don't know y or z, insuff

together: if x is negative, then x + y is taking something away from y, and x + y = z, therefore y is greater than z

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25 May 2008, 17:50
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Rather, the question is asking: is Y greater or equal to Z assuming there is an equality between x and y-z (or z-y).
Remember the definition of |x|. |x| is always greater or equal to 0.
If x > 0, then |x| = x.
If x = 0 then |x| = 0.
If x < 0, the |x| = -X. Since x < 0, we have to take -X to make it > 0.
So is |x| = y-z ?
We'll have a definite answer to the question if we have an equality between x and y-z (or z-y) And if we know the sign of x or y-z.
1/ x + y = z
so x = z - y--- Not enough information because we need to know the sign of x or y-z.
2/ x < 0 --- Not enough information because we need a relationship between x and y-z
Together we have both an equality and the sign of x.
so |x| = |z-y| = y-z because x being < 0, so is z-y. z-y < 0 ; therefore |x| = |z- y| = -(z-y) = y-z.
That's all, folks.

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Last edited by TheGMATDoctor on 13 Jul 2010, 12:09, edited 1 time in total.
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27 May 2008, 06:18
Wow, thanks everybody. Great explanations!!
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27 May 2008, 06:26
Great explanations.

To the original poster: Would you mind confirming the official answer?
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27 May 2008, 06:50
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Factorials were someone's attempt to make math look exciting!!!

Re: absolute value   [#permalink] 27 May 2008, 06:50
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