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Math Help - Expression simplification

  1. #1
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    Expression simplification

    Hi

    I have these two expressions

    A+B=D and D=\frac{\text{i}2kA}{\text{i}k-\alpha}

    I need to find B, so I substitute D into the first expression. The problem is I don't know how to get it into the form below

    B=A\frac{(\text{i}k+\alpha)}{(\text{i}k-\alpha)}

    Thanks
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  2. #2
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    Re: Expression simplification

    Quote Originally Posted by bobred View Post
    Hi

    I have these two expressions

    A+B=D and D=\frac{\text{i}2kA}{\text{i}k-\alpha}

    I need to find B, so I substitute D into the first expression. The problem is I don't know how to get it into the form below

    B=A\frac{(\text{i}k+\alpha)}{(\text{i}k-\alpha)}

    Thanks
    You can't. Think of it this way: A + B = D is B = D - A
    Thanks from topsquark
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  3. #3
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    Re: Expression simplification

    B=A\frac{(\text{i}k+\alpha)}{(\text{i}k-\alpha)}

    From this you get  B(ik-\alpha)=A(ik+\alpha)

    The real numbers on the left have to equal the real numbers on the right and the same goes for imaginary numbers
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  4. #4
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    Re: Expression simplification

    Quote Originally Posted by Shakarri View Post
    B=A\frac{(\text{i}k+\alpha)}{(\text{i}k-\alpha)}

    From this you get  B(ik-\alpha)=A(ik+\alpha)

    The real numbers on the left have to equal the real numbers on the right and the same goes for imaginary numbers
    The poster does not say imaginary numbers. When he specifies what his question is about we will be able to handle this properly.
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  5. #5
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    Re: Expression simplification

    Hi

    I find B by hand easy enough, but Maple gives the expression below and I don't know how

    B=A\frac{(\text{i}k+\alpha)}{(\text{i}k-\alpha)}

    The expressions come from boundary values.
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  6. #6
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    Re: Expression simplification

    Oh sorry I missread your post.

    With A+B=D and D=\frac{\text{i}2kA}{\text{i}k-\alpha}

    You get
    A+B=\frac{\text{i}2kA}{\text{i}k-\alpha}


    B=\frac{\text{i}2kA}{\text{i}k-\alpha}-A

    Factorize out A

    B=A\times(\frac{\text{i}2k}{\text{i}k-\alpha}-1)

    Get a common denominator and simplify.
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  7. #7
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    Re: Expression simplification

    Hi, thanks I have got to the point above its just the next step i'm having trouble with.
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