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PhoenixFire
 one year 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.
PhoenixFire
 one year 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.

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Psymon
 one year ago
Best ResponseYou've already chosen the best response.1Alright, 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: \[x ^{2}yy = y ^{2}xx\]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

PhoenixFire
 one year ago
Best ResponseYou've already chosen the best response.0Yes! 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 \[x(xy1)y(xy1)=>(xy)(xy1)=0\] Therefore, \(x=y\) and \(x=\frac{1}{y}\) So now why could I not have done that with the first one?

PhoenixFire
 one year ago
Best ResponseYou've already chosen the best response.0I 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 1to1 and then \[f(x)=\frac{1}{2}(x\frac{1}{x})\] IS 1to1 to do this I'm using \(f(x)=f(y) => x=y\) will prove it is 1to1

Psymon
 one year ago
Best ResponseYou've already chosen the best response.1and yet theyre basically reverse graphs O.o

PhoenixFire
 one year ago
Best ResponseYou've already chosen the best response.0Ahhh this is why For all \(x \in (0 \rightarrow \infty)\)

Psymon
 one year ago
Best ResponseYou've already chosen the best response.1So is that pretty much all thats needed then? Or do you still have to go through the same process?

PhoenixFire
 one year ago
Best ResponseYou've already chosen the best response.0and for one it would give \(x=y\) and \(x=\frac{1}{y}\) which is not in the domain and therefore ignored. Success! Thank you!

Psymon
 one year ago
Best ResponseYou've already chosen the best response.1Lol, hey, glad you got it xDD

PhoenixFire
 one year ago
Best ResponseYou've already chosen the best response.0Note to self: Pay attention to all the information in the question.

Psymon
 one year ago
Best ResponseYou've already chosen the best response.1Lol. We still do silly mistakes in our work like that x_x Good luck ^_^
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