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PhoenixFire
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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.
 one year ago
 one year ago
PhoenixFire Group Title
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.
 one year ago
 one year ago

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

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

PhoenixFire Group TitleBest ResponseYou've already chosen the best response.0
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 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
 one year ago

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

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

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

PhoenixFire Group TitleBest ResponseYou've already chosen the best response.0
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!
 one year ago

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

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

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