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    (Original post by Gregorius)
    If you feed this mess to FullSimplify in Mathematica it spits out
    Just noticed that the result falls out of this expression, no need for Mathematica...
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    (Original post by Gregorius)
    I get the required solution out (with a bit of help from Mathematica). If

    \displaystyle \phi = \frac{A - x}{A + x}

    then

    \displaystyle x = A \frac{1 - \phi}{1 + \phi}

    If you set  \phi to be that mess on the RHS of the original equation, you get

    \displaystyle \frac{1 - \phi}{1 + \phi} = \frac{(A+y)(A+z) - (A-y)(A-z)}{(A+y)(A+z) + (A-y)(A-z)}

    If you feed this mess to FullSimplify in Mathematica it spits out

    \displaystyle A \frac{(y+z)}{A^2+yz}

    from which the result follows.
    thanks for that.

    but how did you go from the first eq to the second eq in your post?

    And the last eq, if you divide both sides by A, how do you get yz/A^2?
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    (Original post by Et Tu, Brute?)
    thanks for that.

    but how did you go from the first eq to the second eq in your post?

    And the last eq, if you divide both sides by A, how do you get yz/A^2?
     \displaystyle \phi = \frac{A - x}{A + x} \implies \phi A + \phi x = A - x \implies x + \phi x = A - \phi A \implies x (1+\phi) = A(1-\phi)

    etc etc.

    Then

    \displaystyle x = A \frac{1 - \phi}{1 + \phi} = A^2 \frac{(y+z)}{A^2+yz}

    Now divide top and bottom by A^2.
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    (Original post by Gregorius)
     \displaystyle \phi = \frac{A - x}{A + x} \implies \phi A + \phi x = A - x \implies x + \phi x = A - \phi A \implies x (1+\phi) = A(1-\phi)

    etc etc.

    Then

    \displaystyle x = A \frac{1 - \phi}{1 + \phi} = A^2 \frac{(y+z)}{A^2+yz}

    Now divide top and bottom by A^2.
    finally I can see it!

    Thanks, massive help from everyone here, appreciated.
 
 
 
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