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Math Help - Please help solve these system of equations

  1. #1
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    Please help solve these system of equations

    Y = C + I + G
    C = a + bY

    Solve for Y and C, assuming I, G, a, and b are known.

    I would like someone to show the steps in algebraically solving these equations. Thanks.
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  2. #2
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    \displaystyle C=a+b(C+I+G)=a+bC+bI+bG\Rightarrow C-bC=a+bI+bG

    \displaystyle\Rightarrow C(1-b)=a+bI+bG\Rightarrow C=\frac{a+bI+bG}{1-b}

    Now find Y
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  3. #3
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    the book had the following answer for C.

    C = a + b [a + I + G]/(1-b)
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  4. #4
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    Are you sure it isn't \displaystyle C=\frac{a+b[I+G]}{1-b}\mbox{?}
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  5. #5
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    no, I double checked.
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  6. #6
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    Your book is mistaken unless Y = a + I + G +C and you forgot that a in the first post.
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  7. #7
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    I was just able to solve by putting in the value of C into the equation Y = C - a/b.
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  8. #8
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    Quote Originally Posted by nquadr View Post
    I was just able to solve by putting in the value of C into the equation Y = C - a/b.
    Your Y solution shouldn't contain C.
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  9. #9
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    it doesn't I am using the value of C you calculated.
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  10. #10
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    \displaystyle Y=\frac{a+I+G}{1-b}
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  11. #11
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    Hello, nquadr!

    I'd get rid of that book . . . it's juvenile.
    And, I'm afraid, so are you . . .


    The book had the following answer for C.

    . . C \:=\: a + b\,\dfrac{a + I + G}{1-b}

    What a stupid way to publish an answer!

    Did you simplify it?
    It's obvious that it never occured to them !


    I'll start over . . .

    We have: . \begin{Bmatrix} Y &=& C + I + G & [1] \\ C &=& a + bY & [2] \end{Bmatrix}


    \begin{array}{ccccccc}\text{From [1], we have:} & Y - C &=& I + G \\<br />
\text{From [2]. we have:} & \text{-}bY + C &=& a \end{array}

    Add the equations: . Y - bY \:=\:a + I + G

    . . (1-b)Y \:=\:a + I + G \quad\Rightarrow\quad\boxed{ Y \:=\:\dfrac{a + I + G}{1-b}}


    Substitute into [2]: . C \;=\;a + b\,\dfrac{a+I+G}{1-b} . and they stopped here?!


    Simplify: . \displaystyle C \;=\;\frac{a(1-b) + b(a + I + G)}{1-b} \;=\;\frac{a - ab + ab + bI + bG}{1-b}

    . . Therefore: . \boxed{C \;=\;\frac{a+bI + bG}{1-b}}

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