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PhoenixFire
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.
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
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?
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
and yet theyre basically reverse graphs O.o
Ahhh this is why For all \(x \in (0 \rightarrow \infty)\)
So is that pretty much all thats needed then? Or do you still have to go through the same process?
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!
Lol, hey, glad you got it xDD
Note to self: Pay attention to all the information in the question.
Lol. We still do silly mistakes in our work like that x_x Good luck ^_^