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Math Help - Rewriting an equation

  1. #1
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    Rewriting an equation

    I wasn't really sure where to put this problem, it is a college course but I'm having issues with the basic algebra. I figured this would be the best spot...


    The textbook (macro econ) gives an equation and wants it in terms of Y, and gives the answer in two steps. I just cant figure out how the author did it.

    Y=c0+c1*Y-c1*T+I+G
    It then rearranges to
    (1-c1)*Y=c0 +I+G-c1*T
    Then it divides both sides by (1-c1) for
    Y=1/(1-c1)*[c0+I+G-c1*T]

    What I don't understand is where he gets the (1-c1) part. He obviously subtracted c1*Y from the right side and added it to the Y on the left hand side of the equation, but I don't know what identity he used to convert Y-c1*Y to (1-c1)*Y. Any help on clearing this up would be greatly appreciated. If any further clarification is necessary please let me know.

    Thanks for your time!

    Clarification: The c1 and c0 are c subscript 1 and c subscript 2 (not that it really makes a difference)
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  2. #2
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    Quote Originally Posted by rhymenoceros View Post
    ...
    What I don't understand is where he gets the (1-c1) part. He obviously subtracted c1*Y from the right side and added it to the Y on the left hand side of the equation, but I don't know what identity he used to convert Y-c1*Y to (1-c1)*Y.

    ...
    If you factor out a value from a sum you must divide each summand by the factor. The results of the division have to be written in parantheses:

    Example:

    3+4+5 = 7\left(\dfrac37+\dfrac47+\dfrac57\right)

    With your problem:

    y-c_1 \cdot y = y\left(\dfrac yy - \dfrac{c_1 \cdot y}y \right)= y \left(1 - c_1\right)

    Maybe I should mention that \dfrac yy = \boxed{1}
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  3. #3
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    Awesome, thanks for the help and your time!
    Last edited by rhymenoceros; October 23rd 2008 at 12:36 PM. Reason: More confusion... Then less....
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