## PhoenixFire 3 years ago 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.

1. Psymon

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

2. PhoenixFire

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?

3. PhoenixFire

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

4. Psymon

and yet theyre basically reverse graphs O.o

5. PhoenixFire

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

6. Psymon

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

7. PhoenixFire

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!

8. Psymon

Lol, hey, glad you got it xDD

9. PhoenixFire

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

10. Psymon

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