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Thread: Looks simple but I keep messing things up

  1. #1
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    Looks simple but I keep messing things up

    Any idea where I go wrong? The two equations in the system are:
    (1) $\displaystyle x/y-y/x = 1$

    (2) $\displaystyle 1/x-1/y = 1/2$

    Now trying to simplify:

    (1;2) $\displaystyle (xy)(x/y-y/x) = x^2 - y^2 = xy$

    (2;2) $\displaystyle (xy)(1/x-1/y) = y-x = xy/2 -> /*2 -> 2y-2x = xy$

    Substituting 2;2 into 1;2 $\displaystyle -> x^2 -y^2 = 2y-2x$ ->

    $\displaystyle x^2 +2x = y^2 +2y ->$ then $\displaystyle x=y$??? what??? IF x = y, then the original equation doesn´t hold: $\displaystyle x=y -> x/y-y/x =1 -> x/x -x/x = 1?? 1-1 = 1$???
    Solving this seemingly rediculously easy equationssystems for last 40min,



    Oh the correct answer should be $\displaystyle x = 1+sqrt5 -> y = -3-sqrt5$
    $\displaystyle x =1-sqrt5 -> 7 = -3+sqrt5$
    Last edited by mr fantastic; Jan 13th 2010 at 06:21 PM. Reason: Deleted excessive smilies.
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Henryt999 View Post
    Any idea where I go wrong? The two equations in the system are:
    (1) $\displaystyle x/y-y/x = 1$

    (2) $\displaystyle 1/x-1/y = 1/2$

    Now trying to simplify:

    (1;2) $\displaystyle (xy)(x/y-y/x) = x^2 - y^2 = xy$

    (2;2) $\displaystyle (xy)(1/x-1/y) = y-x = xy/2 -> /*2 -> 2y-2x = xy$

    Substituting 2;2 into 1;2 $\displaystyle -> x^2 -y^2 = 2y-2x$ ->

    $\displaystyle x^2 +2x = y^2 +2y ->$ then $\displaystyle x=y$??? what??? IF x = y, then the original equation doesn´t hold: $\displaystyle x=y -> x/y-y/x =1 -> x/x -x/x = 1?? 1-1 = 1$???
    Solving this seemingly rediculously easy equationssystems for last 40min,
    Hint:

    $\displaystyle \frac{x}
    {y} - \frac{y}
    {x} = 1 \Leftrightarrow \frac{x}
    {y} - {\left( {\frac{x}
    {y}} \right)^{ - 1}} = 1 \Leftrightarrow \left\{ {\frac{x}
    {y} = a} \right\} \Leftrightarrow a - {a^{ - 1}} = 1 \Leftrightarrow$

    $\displaystyle \Leftrightarrow {a^2} - a - 1 = 0 \Leftrightarrow {a_{1,2}} = \frac{{1 \pm \sqrt 5 }}
    {2} \Leftrightarrow \frac{x}
    {y} = \frac{{1 \pm \sqrt 5 }}
    {2} \Leftrightarrow \frac{1}
    {y} = \frac{{1 \pm \sqrt 5 }}
    {{2x}}.$
    Last edited by mr fantastic; Jan 13th 2010 at 06:22 PM. Reason: Edited quote
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  3. #3
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    Thanks

    Do you know where I went wrong?
    Last edited by mr fantastic; Jan 13th 2010 at 06:21 PM.
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  4. #4
    Senior Member DeMath's Avatar
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    From the first equation you have:


    $\displaystyle \frac{x}{y} - \frac{y}{x} = 1 ~ \Leftrightarrow ~ \frac{x}
    {y} - \left(\frac{x}{y}\right)^{-1} = 1 ~ \Leftrightarrow ~ \left\{ {\frac{x}{y} = a} \right\} ~ \Leftrightarrow ~ a - a^{-1} = 1 ~ \Leftrightarrow$

    $\displaystyle \Leftrightarrow ~ a^2 - a - 1 = 0 ~ \Leftrightarrow ~ a_{1,2} = \frac{1 \pm \sqrt 5}{2} ~ \Leftrightarrow ~ \frac{x}{y} = \frac{1 \pm \sqrt 5}
    {2} ~ \Leftrightarrow ~ \frac{1}{y} = \frac{1 \pm \sqrt 5}{{2x}}.$

    So you get

    $\displaystyle \left\{ \begin{gathered}
    \frac{x}
    {y} - \frac{y}
    {x} = 1, \hfill \\
    \frac{1}
    {x} - \frac{1}
    {y} = \frac{1}
    {2}; \hfill \\
    \end{gathered} \right.~\Leftrightarrow{\color{white}...}$$\displaystyle \left[\begin{gathered}\left\{\begin{gathered}\frac{1}{y} = \frac{1 - \sqrt 5}{2x}, \hfill \\ \frac{1}{x} - \frac{1 - \sqrt 5}{2x} = \frac{1}{2}; \hfill \\ \end{gathered} \right. \hfill \\ \left\{\begin{gathered}\frac{1}{y} = \frac{1 + \sqrt 5 }{2x}, \hfill \\ \frac{1}{x} - \frac{1 + \sqrt 5}{2x} = \frac{1}{2}; \hfill \\ \end{gathered}\right. \hfill \\ \end{gathered}\right. ~\Leftrightarrow{\color{white}...}$$\displaystyle \left[ \begin{gathered}\left\{ \begin{gathered}\frac{1}
    {y} = \frac{1}
    {2}\frac{1 - \sqrt 5}
    {{1 + \sqrt 5 }}, \hfill \\
    x = 1 + \sqrt 5 ; \hfill \\
    \end{gathered} \right. \hfill \\
    \left\{ \begin{gathered}
    \frac{1}
    {y} = \frac{1}
    {2}\frac{{1 + \sqrt 5 }}
    {{1 - \sqrt 5 }}, \hfill \\
    x = 1 - \sqrt 5 ; \hfill \\
    \end{gathered}\right. \hfill \\
    \end{gathered}\right.~\Leftrightarrow{\color{white }...}$$\displaystyle \left[\begin{gathered}\left\{ \begin{gathered}x_1 = 1 + \sqrt 5 , \hfill \\y_1 = - 3 - \sqrt 5 . \hfill \\ \end{gathered}\right. \hfill \\ \left\{\begin{gathered}x_2 = 1 - \sqrt 5 , \hfill \\ y_2 = - 3 + \sqrt 5 . \hfill \\ \end{gathered}\right. \hfill \\ \end{gathered}\right.$

    _________________
    $\displaystyle \frac{1}{y} = \frac{1}{2}\frac{1 - \sqrt 5}{1 + \sqrt 5} ~\Leftrightarrow{\color{white}...}$$\displaystyle y = \frac{2\left(1 + \sqrt 5\right)}{1 - \sqrt 5} = \frac{2\left(1 + \sqrt 5\right)^2}{\left(1 - \sqrt 5\right)\left(1 + \sqrt 5\right)} = \frac{2\left(1 + 2\sqrt 5 + 5\right)}{1 - 5} = - \frac{6 + 2\sqrt 5}{2} = -3 - \sqrt 5 .$

    $\displaystyle \frac{1}{y} = \frac{1}{2}\frac{1 + \sqrt 5}{1 - \sqrt 5}~\Leftrightarrow{\color{white}...}$$\displaystyle y = \frac{2\left(1 - \sqrt 5\right)}{1 + \sqrt 5} = \frac{2\left(1 - \sqrt 5\right)^2}{\left(1 + \sqrt 5\right)\left(1 - \sqrt 5\right)} = \frac{2\left(1 - 2\sqrt 5 + 5\right)}{1 - 5} = - \frac{6 - 2\sqrt 5}{2} = -3 + \sqrt 5 .$


    Do you understand this?
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  5. #5
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    Hello, Henryt999!

    This is how I solved it . . .


    Solve the system: .$\displaystyle \begin{array}{cccc}\dfrac{x}{y}-\dfrac{y}{x} &=& 1 & [1] \\ \\[-3mm]
    \dfrac{1}{x}-\dfrac{1}{y} &=& \dfrac{1}{2} & [2]\end{array}$

    $\displaystyle \begin{array}{cccccc}
    \text{Multiply [1] by }xy: & x^2 - y^2 &=& xy & [3] \\ \\
    \text{Multiply [2] by }2xy: & 2y - 2x &=& xy & [4] \end{array}$


    Equate [3] and [4]: .$\displaystyle x^2-y^2 \:=\:2y-2x \quad\Rightarrow\quad x^2 - y^2 + 2x - 2y \:=\:0 $

    Factor: .$\displaystyle (x-y)(x+y) + 2(x-y) \:=\:0$

    Factor: .$\displaystyle (x-y)(x+y+2) \:=\:0$


    We have two equations to solve:

    . . $\displaystyle x - y \:=\:0 \quad\Rightarrow\quad x \:=\:y $ . . . which does not check (extraneous root)

    . . $\displaystyle x+y+2\:=\:0 \quad\Rightarrow\quad y \:=\:-x-2\;\;[5]$


    Substitute into [4]: .$\displaystyle 2(-x-2) - 2x(-x-2) \:=\:x(-x-2) \quad\Rightarrow\quad x^2-2x-4 \:=\:0$

    Quadratic Formula: .$\displaystyle x \;=\;\frac{2\pm\sqrt{20}}{2} \:=\:\frac{2\pm2\sqrt{5}}{2} \quad\Rightarrow\quad\boxed{ x\:=\:1\pm\sqrt{5}}$

    Substitute into [5]: .$\displaystyle y \:=\:-(1\pm\sqrt{5}) - 2 \:=\:-1 \mp\sqrt{5} -2 \quad\Rightarrow\quad\boxed{ y \:=\:-3\mp\sqrt{5}}$

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