# Rewriting an expression without the absolute value symbols

• Jan 13th 2009, 08:15 PM
tiar
Rewriting an expression without the absolute value symbols
$\displaystyle \frac{|x-y|}{|y-x|}$
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.
• Jan 13th 2009, 08:37 PM
Last_Singularity
Quote:

Originally Posted by tiar
$\displaystyle \frac{|x-y|}{|y-x|}$
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 $\displaystyle x-y=z$

Then $\displaystyle \frac{|x-y|}{|y-x|} = \frac{|z|}{|-z|} = \frac{|z|}{|z|} = 1$
• Jan 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 $\displaystyle \frac{|x-y|}{|-x-y|}$ Come from?
(That is, $\displaystyle \frac{|z|}{|-z|}$)
• Jan 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 $\displaystyle \frac{|x-y|}{|-x-y|}$ Come from?
(That is, $\displaystyle \frac{|z|}{|-z|}$)

If z = x - y then

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

Originally Posted by tiar
Wait, I'm still confused. Where does $\displaystyle \frac{|x-y|}{|-x-y|}$ 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