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Math Help - Algerbra

  1. #1
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    Algerbra

    2/3 = 5n-81/{([n-1]/4)^2 -10} What is n?
    All help greatly appreciated
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  2. #2
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    Quote Originally Posted by ollieman View Post
    2/3 = 5n-81/{([n-1]/4)^2 -10} What is n?
    All help greatly appreciated
    <br />
\frac{2}{3} = \frac{5n-81}{(\frac{(n-1)}{4})^2 -10}


    <br />
\frac{2}{3} = \frac{16(5n-81)}{(n-1)^2 -160}

    <br />
2(~(n-1)^2-160~) = 48(5n-81)

    <br />
 (~(n)^2-159 -2n~) = 24(5n-81)


    Now use formula for roots of quadratic equation
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  3. #3
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    Quote Originally Posted by ollieman View Post
    2/3 = 5n-81/{([n-1]/4)^2 -10} What is n?
    All help greatly appreciated
    Your formatting is ambiguous; I think you mean the following:

    . . . . . \frac{2}{3}\, =\, \frac{5n\, -\, 81}{\left(\frac{n\, -\, 1}{4}\right)^2\, -\, 10}

    If so, then a good first step in solving this rational equation is probably to multiply through, to get rid of the denominators:

    . . . . . 2\left[\left(\frac{n\, -\, 1}{4}\right)^2\, -\, 10\right]\, =\, 3(5n\, -\, 81)

    . . . . . 2\left[\frac{n^2\, -\, 2n\, +\, 1}{16}\, -\, 10\right]\, =\, 15n\, -\, 243

    . . . . . \frac{n^2\, -\, 2n\, +\, 1}{8}\, -\, 20\, =\, 15n\, -\, 243

    . . . . . n^2\, -\, 2n\, +\, 1\, -\, 160\, =\, 120n\, -\, 1944

    Then collect all the terms on one side of the "equals" sign:

    . . . . . n^2\, -\, 122n\, +\, 1785\, =\, 0

    Then apply the Quadratic Formula to solve for "n". Remember to check your solutions in the original equation, making sure that they don't cause any "division by zero" problems.
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