Proving the Equation (y-2)/(2-y)=-1 for All Real Numbers y ≠ 2

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This is not a very difficult question, by any means, and I see that the answer does equal -1, but I do not see the mathematical steps that go on to prove that (y-2)/(2-y)=-1 .. Can someone show me the steps that go through that? Thanks in advance.
 
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Dude...

1) (y-2)/(2-y)=-1
2) (y-2)=-1(-y+2)
3) (y-2)=(y-2)

Wait a bit and I'll post the same in TeX.
 
Better way.

[tex]\frac{y-2}{2-y}=\frac{(y-2)(y+2)}{(2-y)(2+y)}=\frac{y^2-4}{4-y^2}=-1[/tex]
 
lol, ok sorry. That was posted slightly incorrectly the first time. Basically, the textbook says "simplify: (y-2)/(2-y) " and the back of the book gives the final answer to be -1, so I can't do the left side right side stuff
 
ahh, yes, I see your second post now . That does seem to be better. Thanks
 
so basically what I do in your second post is I multiply the bottom and top by the conjugate of the denominator?
 
mike_302 said:
so basically what I do in your second post is I multiply the bottom and top by the conjugate of the denominator?
Yes. (I had to look up what's a conjugate to answer your question.:shy:)

In the future, could you please post your homework/assignment questions in the homhttps://www.physicsforums.com/forumdisplay.php?f=152"um?


This forum is more about theoretical discussions.
 
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Wow, I'm stupid...

[tex]\frac{y-2}{2-y}=\frac{-1(2-y)}{2-y} =-1[/tex]

This is what you get for not doing any maths for almost a year.
 
OH! Jeeeze. I see, lol, Thanks. and, yes. Normally I post in the math homework forums but I think I just clicked on the wrong link and got here. I didn't notice that, sorry.
 
ivanetski said:
This forum is more about theoretical discussions.

I thought it was about general math.
 
(y-2)/(2-y) = -1 is an equation

(y-2)/(2-y) is not an equation; it's an expression.
 
mike_302 said:
and the back of the book gives the final answer to be -1, so I can't do the left side right side stuff

It's -1 for all real numbers y not equal to 2. :wink: