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Thread: Basic algebra question

  1. #1
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    Basic algebra question

    let say if we have
    x(a)+x(b)+x(c) and a+b+c=d
    we can write it as=x(d)

    but what if we have
    x(a)+y(b)+z(c) and a+b+c=d
    how can we write it then?
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  2. #2
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    Re: Basic algebra question

    if all we are given is $a + b + c = d$

    then there is no simplification of

    $a x + b y + c z$
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  3. #3
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    Re: Basic algebra question

    Thank for our reply
    I think I rewrite my question

    (a + b ) + (c + d ) + (e + f ) = S ------------>Equation 1

    alpha(a + b ) + beta(c + d ) + gemma(e + f )=??? ..............>equation 2


    Is there any way to rewrite this equation semiller to the first one
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  4. #4
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    Re: Basic algebra question

    Quote Originally Posted by gevni View Post
    Thank for our reply
    I think I rewrite my question
    (a + b ) + (c + d ) + (e + f ) = S ------------>Equation 1

    alpha(a + b ) + beta(c + d ) + gemma(e + f )=??? ..............>equation 2
    I think that you are confusing function notation and simple multiplication.
    $\alpha(a + b )$ is function notation meaning that alpha is a function with argument $a+b$.
    As such $\alpha(a + b )\ne\alpha(a )+\alpha(b )$
    If that is not what you mean then what do you mean?
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  5. #5
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    Re: Basic algebra question

    Quote Originally Posted by gevni View Post
    Thank for our reply
    I think I rewrite my question

    (a + b ) + (c + d ) + (e + f ) = S ------------>Equation 1

    alpha(a + b ) + beta(c + d ) + gemma(e + f )=??? ..............>equation 2


    Is there any way to rewrite this equation semiller to the first one
    Suppose $\beta = \gamma = 0$. Then you have for equation 2:
    $\alpha (a+b)+0(c+d)+0(e+f) = \alpha (a+b)$
    That has very little relation left to $S$. You can write:
    $(a+b) = S-(c+d)-(e+f) = S-(c+d+e+f)$
    which gives:
    $\alpha (a+b) + \beta (c+d) + \gamma (e+f) = \alpha (S-c-d-e-f) + \beta (c+d) + \gamma (e+f)$, but this is probably not what you are looking for.

    Alternately, you can have:

    $\alpha (a+b) + \beta (c+d) + \gamma (e+f) = \alpha S + (\beta - \alpha)(c+d) + (\gamma - \alpha)(e+f)$
    Last edited by SlipEternal; Apr 12th 2018 at 01:32 PM.
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