Question 300927
{{{Y=C+I+G}}}
{{{C=a+bY}}}
Let's substitute to simplify the knowns.
{{{d=I+G}}}
So then, 
1.{{{Y=C+d}}}
2.{{{C=bY+a}}}
You're solving for C and Y.
You can do a direct substitution using eq. 1 in eq. 2 and solving for C.
{{{C=bY+a}}}
{{{C=b(C+d)+a}}}
{{{C=bC+bd+a}}}
{{{C-bC=bd+a}}}
{{{C(1-b)=bd+a}}}
{{{C=(bd+a)/(1-b)}}}
Then go back and solve for Y using eq. 1,
{{{Y=C+d}}}
{{{Y=(bd+a)/(1-b)+d}}}
{{{Y=(bd+a)/(1-b)+d(1-b)/(1-b)}}}
{{{Y=(bd+a+d(1-b))/(1-b)}}}
{{{Y=(bd+a+d-bd)/(1-b)}}}
{{{Y=(a+d)/(1-b)}}}
Now we can go back and re-substitute,

{{{highlight_green(C=(b(I+G)+a)/(1-b))}}}
{{{highlight_green(Y=(a+I+G)/(1-b))}}}