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Math Help - Rational Equations

  1. #1
    TWN
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    Rational Equations

    Please walk me through how to solve these:

    1. [w/(w-1)] + w = [(4w-3)/(w-1)]

    2. {(4n^2)/[(n^2)-9]} - [2n/(n+3)] = 3/(n-3)

    3. [4/(z-2)] - [(z+6)/(z+1)] = 1


    Thanks in advance =)
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  2. #2
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    Quote Originally Posted by TWN View Post
    Please walk me through how to solve these:

    1. [w/(w-1)] + w = [(4w-3)/(w-1)]

    2. {(4n^2)/[(n^2)-9]} - [2n/(n+3)] = 3/(n-3)

    3. [4/(z-2)] - [(z+6)/(z+1)] = 1


    Thanks in advance =)

    1.

    \frac{w}{(w-1)} + w = \frac{(4w-3)}{(w-1)}

    \frac{w(w-1)}{(w-1)} + w = \frac{(4w-3)(w-1)}{(w-1)}

    w + w = (4w-3)

    2w = 4w-3

    3 = 2w

    w = \frac{2}{3}



    2.

    \frac{4n^2}{n^2-9} - \frac{2n}{(n+3)} = \frac{3}{(n-3)}

    \frac{4n^2}{(n+3)(n-3)} - \frac{2n(n-3)}{(n+3)(n-3)} = \frac{3}{(n-3)}

    \frac{4n^2-2n(n-3)}{(n+3)(n-3)} = \frac{3}{(n-3)}

    \frac{4n^2-2n^2+6n}{(n+3)(n-3)} = \frac{3}{(n-3)}

    \frac{2n^2+6n}{(n+3)(n-3)} = \frac{3}{(n-3)}

    \frac{2n(n+3)}{(n+3)(n-3)} = \frac{3}{(n-3)}

    \frac{2n}{(n-3)} = \frac{3}{(n-3)}

    \frac{2n(n-3)}{(n-3)} = \frac{3(n-3)}{(n-3)}

    2n = 3

    n = \frac{2}{3}
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  3. #3
    TWN
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    Thank you for the quick reply!
    First of all, if 2x=3, the solution would be 3/2, not 2/3. This applies to both solutions you provided.
    I tried plugging both 2/3 and 3/2 into both original equations and neither worked. Something seems to be incorrect.
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