Question

Simple algebra that I apparently cannot do... Can anyone help here? Simplify the following:
\[{x-\frac{1}{x}}={y-\frac{1}{y}}\]

It simplifies to \(x=y\) but I just can't get it to do that.

Answer 1

Alright, I think I got it....
So I'm going to multiply through by x and y on both sides to get:
\[y(x ^{2}-1)=x(y ^{2}-1)\] I'll distribute x and y now:
\[x ^{2}y-y = y ^{2}x-x\]Ill move the lone x and the lone y to opposite sides:
\[x ^{2}y+x=y ^{2}x+y\]Now I'll factor out an x on the left and a y on the right:
\[x(xy+1)=y(xy+1)\]Divide both sides by (xy+1) and x = y

Answer 2

Yes! That works!
But now I have another problem...
\[{x+\frac{1}{x}}={y+\frac{1}{y}}\]
This is supposed to result in two values for x... but I apply the same method and you still get \(x=y\)
Sure you can just \[x(xy-1)-y(xy-1)=>(x-y)(xy-1)=0\] Therefore, \(x=y\) and \(x=\frac{1}{y}\)

So now why could I not have done that with the first one?

Answer 3

I guess I should tell you what I'm actually trying to do...
I'm supposed to prove that
\[f(x)=\frac{1}{2}(x+\frac{1}{x})\] is NOT 1-to-1
and then
\[f(x)=\frac{1}{2}(x-\frac{1}{x})\] IS 1-to-1

to do this I'm using
\(f(x)=f(y) => x=y\) will prove it is 1-to-1

Answer 4

and yet theyre basically reverse graphs O.o

Answer 5

Ahhh this is why
For all \(x \in (0 \rightarrow \infty)\)

Answer 6

So is that pretty much all thats needed then? Or do you still have to go through the same process?

Answer 7

and for one it would give \(x=y\) and \(x=-\frac{1}{y}\) which is not in the domain and therefore ignored. Success! Thank you!

Answer 8

Lol, hey, glad you got it xDD

Answer 9

Note to self: Pay attention to all the information in the question.

Answer 10

Lol. We still do silly mistakes in our work like that x_x Good luck ^_^