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Math Help - Fractional Algebriac Equation help!!

  1. #1
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    Fractional Algebriac Equation help!!

    Hi, when I try to solve the equation below, I come up with 36, as seen below. I multiply each term by two in order to get rid of the fractions. The correct answer is 72. What am I doing wrong?

    a+b+c+d = 180

    b = a/2
    c=b/2
    d=3c


    a + a/2 + (a/2)/2 + 3[(a/2)/2] = 180
    which becomes...
    a + a/2 + (a/2 * 1/2) + 3((a/2) * (1/2)) = 180
    which becomes...
    2a + a + a + 6a = 360
    which becomes...
    10a=360
    which becomes...
    a=36
    Last edited by Lotrnerd; September 16th 2012 at 01:30 PM.
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  2. #2
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    Re: Fractional Algebriac Equation help!!

    Quote Originally Posted by Lotrnerd View Post
    Hi, when I try to solve the equation below, I come up with 36, as seen below. I multiply each term by to in order to get ride of the fractions. However, I get an answer of 36, but the answer is 72. What am I doing wrong?

    a+b+c+d = 180

    b = a/2
    c=b/2
    d=3c


    a + a/2 + (a/2)/2 + 3[(a/2)/2] = 180
    which becomes...
    a + a/2 + (a/2 * 1/2) + 3((a/2) * (1/2)) = 180
    which becomes...

    a + a/2 + a/4 + 3a/4 = 180

    4a + 2a + a + 3a = 720
    fixed ... now finish it
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  3. #3
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    Re: Fractional Algebriac Equation help!!

    Thanks, so is it a rule that I need to solve within the parentheses, before I decide to multiply each term by a number?
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  4. #4
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    Re: Fractional Algebriac Equation help!!

    Quote Originally Posted by Lotrnerd View Post
    Thanks, so is it a rule that I need to solve within the parentheses, before I decide to multiply each term by a number?
    using the order of operations helps ...
    Thanks from Lotrnerd
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