Please help solve these system of equations

**nquadr** · January 8th 2011, 12:28 PM

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.

**dwsmith** · January 8th 2011, 12:30 PM

$\displaystyle C=a+b(C+I+G)=a+bC+bI+bG\Rightarrow C-bC=a+bI+bG$

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

Now find Y

**nquadr** · January 8th 2011, 01:02 PM

the book had the following answer for C.

C = a + b [a + I + G]/(1-b)

**dwsmith** · January 8th 2011, 01:04 PM

Are you sure it isn't $\displaystyle C=\frac{a+b[I+G]}{1-b}\mbox{?}$

**nquadr** · January 8th 2011, 01:07 PM

no, I double checked.

**dwsmith** · January 8th 2011, 01:11 PM

Your book is mistaken unless Y = a + I + G +C and you forgot that a in the first post.

**nquadr** · January 8th 2011, 01:22 PM

I was just able to solve by putting in the value of C into the equation Y = C - a/b.

**dwsmith** · January 8th 2011, 01:25 PM

Originally Posted by nquadr

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.

**nquadr** · January 8th 2011, 01:26 PM

it doesn't I am using the value of C you calculated.

**dwsmith** · January 8th 2011, 01:30 PM

$\displaystyle Y=\frac{a+I+G}{1-b}$

**Soroban** · January 8th 2011, 02:17 PM

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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