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Math Help - Trick 4 solving 4 n

  1. #1
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    Trick 4 solving 4 n

    In the equation:

    Lambda1=(n+1)Lambda2

    What is the trick 2 get the solution:

    n=Lambda2/(Lambda1-Lambda2)
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  2. #2
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    Quote Originally Posted by Jaxwell View Post
    In the equation:
    Lambda1=(n+1)Lambda2
    What is the trick 2 get the solution:
    n=Lambda2/(Lambda1-Lambda2)
    That is actually wrong.

    \begin{gathered}<br />
  \lambda _1  = \left( {n + 1} \right)\lambda _2  \hfill \\<br />
  \lambda _1  = n\lambda _2  + \lambda _2  \hfill \\<br />
  n\lambda _2  = \lambda _1  - \lambda _2  \hfill \\ <br />
\end{gathered}
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  3. #3
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    Thank You,
    You are of course right.

    I meant: n*Lambda 1 =(n+1)Lambda 2

    with above solution?
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  4. #4
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    Well, you see how things are done from my reply.
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  5. #5
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    I understand Your reply fully, thanks:-)
    But since there is an n on both sides the correct equation:
    nLambda 1=(n+1)Lambda 2
    with solution:
    n=lambda 2/(lavmbda 1-lambda 2) has a different technique which I cant figure out. If You know how I would be very grateful
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  6. #6
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    Do you realize how difficult it is to read your postings?
    Why not learn to post in symbols? You can use LaTeX tags
    [tex] n\lambda_1=(n+1)\lambda_2[/tex] gives  n\lambda_1=(n+1)\lambda_2 .
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  7. #7
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    Thank You, I will try that. The equation:

     n\lambda_1=(n+1)\lambda_2

    has the solution:

     n=\lambda_2/(\lambda_1-\lambda_2)

    If You know the technique, I would be highly grateful
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  8. #8
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    Quote Originally Posted by Jaxwell View Post
    Thank You, I will try that. The equation:

     n\lambda_1=(n+1)\lambda_2

    has the solution:

     n=\lambda_2/(\lambda_1-\lambda_2)

    If You know the technique, I would be highly grateful
    Just change slightly what Plato did for your original equation:

     n\lambda_1 = n \lambda_2 + \lambda_2

    \Rightarrow n (\lambda_1 - \lambda_2) = \lambda_2

    and I think the answer is straight ahead.
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  9. #9
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    Quote Originally Posted by Jaxwell View Post
    Thank You, I will try that. The equation:

     n\lambda_1=(n+1)\lambda_2

    has the solution:

     n=\lambda_2/(\lambda_1-\lambda_2)

    If You know the technique, I would be highly grateful
    It's pretty much basic algebra- you have to get n by itself so first you have to "free" it from the right side by using the "distributive" property: (n+1)\lambda_2= n\lambda_2+ \lambda_2. At that point you have
    n= n\lambda2+ \lambda2
    Still trying to get n by itself, subtract n\lambda from both sides:
    n\lambda_1- n\lambda_2= \lambda_2

    Now use the "distributive" property again:
    n(\lambda_1- \lambda_2)= \lambda_2
    and, finally, divide both sides by (\lambda_1- \lambda_2).

    That's essentially just what Plato did before.
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  10. #10
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    Quote Originally Posted by Jaxwell View Post
    I meant: n*Lambda 1 =(n+1)Lambda 2
    Why don't you make your life a bit easier; use x and y for them Lambdas:
    nx = (n + 1)y
    nx = ny + y
    nx - ny = y
    Take over, Rover...
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