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PhoenixFire

  • 2 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.

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  1. Psymon
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    and yet theyre basically reverse graphs O.o

  5. PhoenixFire
    • 2 years ago
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    Ahhh this is why For all \(x \in (0 \rightarrow \infty)\)

  6. Psymon
    • 2 years ago
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    So is that pretty much all thats needed then? Or do you still have to go through the same process?

  7. PhoenixFire
    • 2 years ago
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    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
    • 2 years ago
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    Lol, hey, glad you got it xDD

  9. PhoenixFire
    • 2 years ago
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    Note to self: Pay attention to all the information in the question.

  10. Psymon
    • 2 years ago
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    Lol. We still do silly mistakes in our work like that x_x Good luck ^_^

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