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