# Thread: Please help solve these system of equations

1. ## 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.

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

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

Now find Y

3. the book had the following answer for C.

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

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

5. no, I double checked.

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

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

8. 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.

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

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

11. 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.

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

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

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

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

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

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

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

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