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Math Help - Algebraic Rearrangement

  1. #1
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    Algebraic Rearrangement

    Hi, I'm new to this site and I hope you guys can help. I'm not actually pre university but an engineer, yet I'm having some problems rearranging an equation and was hoping someone could help. The equation is as follow:

    ((A/B)-1)*(((A/C)^2)-1)=2*(A/C)*((D*E)/F)

    and the equation needs rearranging to find A. Can anyone help?

    Thanks
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Algebraic Rearrangement

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  3. #3
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    Re: Algebraic Rearrangement

    Quote Originally Posted by maloneb View Post
    Hi, I'm new to this site and I hope you guys can help. I'm not actually pre university but an engineer, yet I'm having some problems rearranging an equation and was hoping someone could help. The equation is as follow:

    ((A/B)-1)*(((A/C)^2)-1)=2*(A/C)*((D*E)/F)

    and the equation needs rearranging to find A. Can anyone help?

    Thanks
    I start first by distrubuting the exponents to get:

    (B/A)*(C^2/A^2) = (2ADE)/(CF)

    Next, combine like bases and simplify:

    (BC^2)/(A^3) = (2ADE)/(CF)

    Multiply both sides by A^3 to get:

    BC^2 = (2A^4DE)/(CF)

    Multiply both sides by (CF) to get:

    BC^3F = 2A^4DE

    Divide both sides by 2DE to get:

    (BC^3F)/(2DE) = A^4

    Take the fourth root of each side to get:

    A = fourth_root[(BC^3F)/(2DE)]
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  4. #4
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    Re: Algebraic Rearrangement

    Quote Originally Posted by rdtedm View Post
    I start first by distrubuting the exponents to get:

    (B/A)*(C^2/A^2) = (2ADE)/(CF)

    Next, combine like bases and simplify:

    (BC^2)/(A^3) = (2ADE)/(CF)

    Multiply both sides by A^3 to get:

    BC^2 = (2A^4DE)/(CF)

    Multiply both sides by (CF) to get:

    BC^3F = 2A^4DE

    Divide both sides by 2DE to get:

    (BC^3F)/(2DE) = A^4

    Take the fourth root of each side to get:

    A = fourth_root[(BC^3F)/(2DE)]
    I assumed here, that "-1" meant the inverse.. if its actually minus one, then obviously this is incorrect.
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  5. #5
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    Re: Algebraic Rearrangement

    Quote Originally Posted by rdtedm View Post
    I assumed here, that "-1" meant the inverse.. if its actually minus one, then obviously this is incorrect.
    No, -1 is not inverse, it is actually -1
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  6. #6
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    Clarksville, ARk
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    Re: Algebraic Rearrangement

    It's a cubic equation in A. Have fun !!!
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