Rewriting an expression without the absolute value symbols

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January 13th 2009, 08:15 PM
tiar

Rewriting an expression without the absolute value symbols

$\frac{|x-y|}{|y-x|}<br />$
x cannot equal y.
I'm stumped on this, I'm being told the answer is 1 but I don't understand how.
If someone could explain this indepth I'd be so grateful! Thank you.
January 13th 2009, 08:37 PM
Last_Singularity

Quote:

Originally Posted by tiar

$\frac{|x-y|}{|y-x|}<br />$
x cannot equal y.
I'm stumped on this, I'm being told the answer is 1 but I don't understand how.
If someone could explain this indepth I'd be so grateful! Thank you.

Let $x-y=z$

Then $\frac{|x-y|}{|y-x|} = \frac{|z|}{|-z|} = \frac{|z|}{|z|} = 1$
January 13th 2009, 08:43 PM
tiar

Ohh, I think I get it now. Thank you! I'm very bad with absolute value...
Edit: Wait, I'm still confused. Where does $\frac{|x-y|}{|-x-y|}$ Come from?
(That is, $\frac{|z|}{|-z|}$ )
January 13th 2009, 10:25 PM
earboth

Quote:

Originally Posted by tiar

Ohh, I think I get it now. Thank you! I'm very bad with absolute value...
Edit: Wait, I'm still confused. Where does $\frac{|x-y|}{|-x-y|}$ Come from?
(That is, $\frac{|z|}{|-z|}$ )

If z = x - y then

-z = (-1)(x - y) = -x + y = y - x
January 14th 2009, 03:53 AM
stapel

Quote:

Originally Posted by tiar

Wait, I'm still confused. Where does $\frac{|x-y|}{|-x-y|}$ Come from?

If you do 5 - 3, you get +2.

If you do 3 - 5, you get -2.

If you reverse a subtraction, you kick a "minus" out front.

Have fun! :D

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