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Math Help - This problem is ridiculous

  1. #1
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    This problem is ridiculous

    If,

    ab/(a+b) = 2
    ac/(a+c)=5
    bc/(b+c)=4

    then find a+b+c.

    I'd really appreciate a step by step solution.
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  2. #2
    Member dokrbb's Avatar
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    Re: This problem is ridiculous

    Quote Originally Posted by RuyHayabusa View Post
    If,

    ab/(a+b) = 2
    ac/(a+c)=5
    bc/(b+c)=4

    then find a+b+c.

    I'd really appreciate a step by step solution.
    Hi there,

    We would solve for one variable and after that replace it in the original equation in order to have only one variable equation:

    You can solve for a or for b, I will show you the example for both:

    \frac{ab}{a+b} = 2 \\ ab = 2a + 2b \\ a = \frac{2a}{b} + 2 \\ a-2= \frac{2a}{b} \\ \frac{2a}{(a-2)} = b \\ a*\frac{2a}{(a-2)} = 2a + 2 \frac{2a}{(a-2)}

    Now, solve for a, replace the value of a in the initial equation and solve for b, then try for the other examples,

    dokrbb

    hint - you will have a =1 and b = -2 , check them in the first equation
    Last edited by dokrbb; April 18th 2013 at 06:06 AM. Reason: the hint added ;)
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  3. #3
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    Re: This problem is ridiculous

    Thanks a lot!
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  4. #4
    MHF Contributor ebaines's Avatar
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    Re: This problem is ridiculous

    Quote Originally Posted by dokrbb View Post
    a*\frac{2a}{(a-2)} = 2a + 2 \frac{2a}{(a-2)}

    Now, solve for a, replace the value of a in the initial equation and solve for b, then try for the other examples,

    ...
    hint - you will have a =1 and b = -2 , check them in the first equation
    This is not correct. The problem is that this equaton simply yields  2a^2 = 2a^2 + 4a - 4a, so it does not yield a solution for a.

    Here's how to do it. Once you have  a = \frac {2b}{b-2}, use the same process on the equation  \frac {bc}{b+c} = 4 to get  c = \frac {4b} {b-4}. Now you have equatons for a and c in terms of b; sub them into  \frac {ac}{a+c} = 5 and solve for b. Then you can sub the solution back into the other two equation to get values for a and c.

    Despite the suggestion in the previous post  a \ne 1 and b \ne -2.
    Last edited by ebaines; April 18th 2013 at 07:19 AM.
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  5. #5
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    Re: This problem is ridiculous

    If you're having trouble following ebaines' post, here's the process to figuring out a, b, and c.

    1. Isolate a in equation 1
    2. sub it into equation 2
    3. Isolate isolate b in equation 2
    4. sub it into equation 3
    5. equation 3 will only have the variable c. Isolating c will yield its value.
    6. sub in your value for c into equation 2 to find the value of b
    7. sub in your value for b into equation 1 to find the value of a
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