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Math Help - Mathematical teaser equation

  1. #1
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    Mathematical teaser equation

    Hello everyone!

    I hope everyone has had a relaxing summer.

    Anyway, I have ventured again into a realm of math that is far beyond my abilities, in other words I am trying to solve some mathematical problems from the book Challenging Mathematical Teasers by J.A.H. Hunter

    The word problem I am currently solving leads to the following two equations

     <br />
x^2-y^2=z^3<br />

    and
    \frac{x}{y}=\frac{y}{z}

    by substiuting we will have x^2-xz=z^3

    This is about as far as I got, but then the solutions says something that I can't quite understand.

    This equation is fully satisfied by: x-z=mz, and x= \frac{z^2}{m}, where m is any rational number.




    Could someone pleade exmplain the above thing to me? Does the author of this book use some systematic approach to solving these kind of equations with 3 unknowns or is it just the mathematician's creativity?


















    All responses are appreciated.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Coach View Post
    Hello everyone!

    I hope everyone has had a relaxing summer.

    Anyway, I have ventured again into a realm of math that is far beyond my abilities, in other words I am trying to solve some mathematical problems from the book Challenging Mathematical Teasers by J.A.H. Hunter

    The word problem I am currently solving leads to the following two equations

     <br />
x^2-y^2=z^3<br />

    and
    \frac{x}{y}=\frac{y}{z}

    by substiuting we will have x^2-xz=z^3

    This is about as far as I got, but then the solutions says something that I can't quite understand.

    This equation is fully satisfied by: x-z=mz, and x= \frac{z^2}{m}, where m is any rational numberd.
    This is just a change of variable, what ever the value of x and z we can write x-z=mz for some m, then because x(x-z)=z^3 we have of necessity that x=z^2/m. The advantage is now that by equating the two expressions for x you have a quadratic relating m and z, so you can solve this for m in terms of z, and so the solutions can be written in tems of a single parameter z.

    (note you don't need to assume that m is rational, at least I can't see why its necessary unless we are looking for integer or rational solutions)

    RonL
    Last edited by CaptainBlack; August 22nd 2008 at 02:38 AM.
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  3. #3
    Newbie
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    Coach you've got the equatin
    x^2 - xz = z^3
    or

    x(x-z) = z^3
    x-z = z^3/x
    now put x= z^2/m

    x-z = zm...
    so you got it.
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  4. #4
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    May i ask a question about "Depreciation"?
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Lucy.Gray View Post
    May i ask a question about "Depreciation"?
    of course! you are more than welcome to... in a new thread
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