Algebra for economics

[Simonsky]

Hi there,

I'm having a problem with an equation in economics.

It goes like this: C= C(subscript 0) +cY (where Y =[C(subscript 0) +I=G]/(1-c)

So with substitution it goes:

C = C(0) + cY = C(0) + c[C(0) + I + G]/(1-c) = [C(0) +c(I+G)]/(1-c)

It's the last step I don't get where the C(0) leaves the bracket and the whole thing is then divided by (1-c)

Any hints appreciated!

 

[Dr.Peterson]

You are (with a typo fixed) given this:

[MATH]C = C_0 + cY[/MATH]​

[MATH]Y = \frac{C_0+I+G}{1-c}[/MATH]​

Here's the substitution, with an extra step or two shown, introducing a common denominator:

[MATH]C = C_0 + c\left[\frac{C_0+I+G}{1-c}\right][/MATH]​

[MATH] = \frac{C_0(1-c)}{1-c} + \left[\frac{c(C_0+I+G)}{1-c}\right][/MATH]​

[MATH] = \frac{C_0-cC_0 + cC_0+cI+cG}{1-c}[/MATH]​

[MATH] = \frac{C_0+ cI + cG}{1-c}[/MATH]​

[MATH] = \frac{C_0+ c(I + G)}{1-c}[/MATH]​

The C₀ is pulled inside and recombined.

 

[Denis]

Your equations as per DrP:
[MATH]C = C_0 + cY[/MATH][MATH]Y = \frac{C_0+I+G}{1-c}[/MATH]
Your variables make my head spin...
I'd change them, as example:
a = b + cd
d = (b + e + f) / (1 - c)

Makes the solving process less wieldy.

 

[Simonsky]

Many thanks Dr. P and Denis - all quite 'simple' when someone who knows their stuff shows you

The problem is that economics equations look unwieldy as Denis points out.