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Help with the most disgusting 'solve for x' question I have been given Watch

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\displaystyle \frac{A-x}{A+x} = \frac{A-y}{A+y} \frac{A-z}{A+z}





\[\textrm{The answer is:}\]

 

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

    ffs didn't mean to post. was trying to edit my CORRECT latex code, but the code here is messed up, some help on the fractions would be sweet.
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    (Original post by Et Tu, Brute?)
    Srs


    \displaystyle \frac{A-x}{A+x} = \frac{A-y}{A+y} \frac{A-z}{A+z}

    The answer is:

    x = \frac{}{}
    Assuming that what I've written is what you meant, then let's call everyting on the RHS k:

    \displaysyle \frac{A-x}{A+x} = k \Rightarrow A - x = k(A+x) \Rightarrow A - x = kA + kx

    Now collecting the x terms: \displaystyle A - kA = kx + x.

    Factorise, simplify, isolate x and back-substitute k.
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    (Original post by Zacken)
    Assuming that what I've written is what you meant, then let's call everyting on the RHS k:

    \displaysyle \frac{A-x}{A+x} = k \Rightarrow A - x = k(A+x) \Rightarrow A - x = kA + kx

    Now collecting the x terms: \displaystyle A - kA = kx + x.

    Factorise, simplify, isolate x and back-substitute k.
    thanks, that's what I meant, what is the code you used?
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    (Original post by Et Tu, Brute?)
    thanks, that's what I meant, what is the code you used?
    No problem. You should be able to see the code in my reply, but that's of no import right now. Do you understand the method I'm showcasing? Can you follow it?
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    (Original post by Zacken)
    ... showcasing ...
    very nice!
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    (Original post by Zacken)
    Assuming that what I've written is what you meant, then let's call everyting on the RHS k:

    \displaysyle \frac{A-x}{A+x} = k \Rightarrow A - x = k(A+x) \Rightarrow A - x = kA + kx

    Now collecting the x terms: \displaystyle A - kA = kx + x.

    Factorise, simplify, isolate x and back-substitute k.
    (Original post by Zacken)
    No problem. You should be able to see the code in my reply, but that's of no import right now. Do you understand the method I'm showcasing? Can you follow it?
    you edited in my post, so I can't actually see that part, only the part in your own post which is like:

    \frac{A-x}{A+x}

    which is what I have in the op

    Yeah makes sense, though it doesn't look like that will end up like the answer I have, which brings it back to the latex code, how do I do it? The answer is much messier than the starting expression.
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    (Original post by Et Tu, Brute?)
    you edited in my post, so I can't actually see that part, only the part in your own post which is like:

    \frac{A-x}{A+x}

    which is what I have in the op

    Yeah makes sense, though it doesn't look like that will end up like the answer I have, which brings it back to the latex code, how do I do it? The answer is much messier than the starting expression.
    Just click on the equation and it'll bring up the code.

    You'll need to clean up your answer. It'll come down to simplifying \frac{1-k}{1+k}.
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    (Original post by Zacken)
    Just click on the equation and it'll bring up the code.

    You'll need to clean up your answer. It'll come down to simplifying \frac{1-k}{1+k}.
    op is edited.

    Now I have that out of the way I'll try it out again, but can't see it bringing me to the answer from \frac{1-k}{1+k}, maybe I'm just being pessimistic, its been a long day.
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    Assuming that what I've written is what you meant, then let's call everyting on the RHS k:

    \displaysyle \frac{A-x}{A+x} = k \Rightarrow A - x = k(A+x) \Rightarrow A - x = kA + kx

    Now collecting the x terms: \displaystyle A - kA = kx + x.

    Factorise, simplify, isolate x and back-substitute k.
    (Original post by Et Tu, Brute?)
    op is edited.

    Now I have that out of the way I'll try it out again, but can't see it bringing me to the answer from \frac{1-k}{1+k}, maybe I'm just being pessimistic, its been a long day.
    Can you understand the method in my post above? It simplifies to A(1-k) = x(k+1) \Rightarrow x = \frac{A(1-k)}{1+k}
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    (Original post by Zacken)
    Can you understand the method in my post above? It simplifies to A(1-k) = x(k+1) \Rightarrow x = \frac{A(1-k)}{1+k}
    yeah makes sense,working on it now. Will update
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    (Original post by Zacken)
    Can you understand the method in my post above? It simplifies to A(1-k) = x(k+1) \Rightarrow x = \frac{A(1-k)}{1+k}
    I end up with a complete abomination of an expression
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    (Original post by Et Tu, Brute?)
    I end up with a complete abomination of an expression
    You'll need to look for common factors and cancel them out.
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    (Original post by Zacken)
    You'll need to look for common factors and cancel them out.
    have you worked this through to the answer in the op?

    I am more than happy to spend all night at this, however right not I am not convinced I will end up with the answer in the op, which is killing my motivation to actually do this tedious elementary maths
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    (Original post by Et Tu, Brute?)
    have you worked this through to the answer in the op?

    I am more than happy to spend all night at this, however right not I am not convinced I will end up with the answer in the op, which is killing my motivation to actually do this tedious elementary maths
    Nopes. I can see off the bat that a few things cancel, but I'm not convinced of the given answer myself.
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    (Original post by Zacken)
    Nopes. I can see off the bat that a few things cancel, but I'm not convinced of the given answer myself.
    ****ing knew it.

    This is why I hate physics.
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    (Original post by Et Tu, Brute?)
    ****ing knew it.

    This is why I hate physics.
    I'm not saying it's incorrect, just that I haven't verified it myself. I'll do that in a bit and tell you.
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    (Original post by Zacken)
    I'm not saying it's incorrect, just that I haven't verified it myself. I'll do that in a bit and tell you.
    wolfram alpha'd it. Equations don't equal each other according to that, so I wouldn't waste your time. There is probably some substitution to be made somewhere.
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    (Original post by Et Tu, Brute?)
    wolfram alpha'd it. Equations don't equal each other according to that, so I wouldn't waste your time. There is probably some substitution to be made somewhere.
    Oh well, you can still solve it and get a correct answer that isn't the given one.
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    (Original post by Zacken)
    Oh well, you can still solve it and get a correct answer that isn't the given one.
    well it seemed odd to me that we were asked to rearrange an equation in a physics modules tbh, it makes sense now that it isn't a simple but tedious maths exercise. It most likely involves the inverse Lorentz boost, where exactly it comes in is a different matter altogether. But I suspect that solving that equation for x and taking it no further will yield little if any marks sadly.
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    (Original post by Et Tu, Brute?)
    wolfram alpha'd it. Equations don't equal each other according to that, so I wouldn't waste your time. There is probably some substitution to be made somewhere.
    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.
 
 
 
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