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Thread: All reals numbers

  1. #1
    Super Member dhiab's Avatar
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    All reals numbers

    Find all reals numbers x , y : x + y = x y
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  2. #2
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    my attempt
    x+y=xy for real number
    (0,0),(2,2),(-1,1/2),(1/2,-1)


    Edit\\
    $\displaystyle x+y=xy $
    $\displaystyle \Rightarrow \frac{1}{x}+\frac{1}{y}=1 \ which\ is\ a\ geometrical\ hyperbola$

    $\displaystyle \Rightarrow \frac{1}{y}=1 - \frac{1}{x} \quad or \quad \frac{1}{x}=1 - \frac{1}{y}$
    $\displaystyle \quad \Rightarrow \quad y=\frac{x}{x-1}\quad or \quad x=\frac{y}{y-1}$
    therefore $\displaystyle x \neq 1\ and\ y \neq 1 $ so,there are many solutions.
    Last edited by ramiee2010; Oct 19th 2009 at 10:25 PM. Reason: correction
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  3. #3
    Super Member dhiab's Avatar
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    Quote Originally Posted by ramiee2010 View Post
    my attempt
    x+y=xy for real number
    (0,0),(2,2),(-1,1/2),(1/2,-1)


    Edit\\
    $\displaystyle x+y=xy $
    $\displaystyle \Rightarrow \frac{1}{x}+\frac{1}{y}=1 \ which\ is\ a\ geometrical\ hyperbola$

    $\displaystyle \Rightarrow \frac{1}{y}=1 - \frac{1}{x} \quad or \quad \frac{1}{x}=1 - \frac{1}{y}$
    $\displaystyle \quad \Rightarrow \quad y=\frac{x}{x-1}\quad or \quad x=\frac{y}{y-1}$
    therefore $\displaystyle x \neq 1\ and\ y \neq 1 $ so,there are many solutions.
    Hello THANK YOU
    I'can write the set of solutions is :

    $\displaystyle S = \left\{ {} \right.\left( {x,\frac{x}{{x - 1}}} \right),\left( {\frac{x}{{x - 1}},x} \right)....\left. {x \in \Re ..\left( {x \ne 1} \right)} \right\}$
    Last edited by dhiab; Oct 19th 2009 at 10:56 PM.
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