2
$\begingroup$

Suppose I have two markets, Home and Foreign. Suppose that

$$\frac{p_1}{p_2} = \frac{c_2^F}{c_1^F}$$

$$\frac{p_1}{p_2} = \frac{c_2^H}{c_1^H}$$

Supposedly I am supposed to be able to show

$$\frac{p_1}{p_2} = \frac{c_2^F + c_2^H}{c_1^F+ c_1^H}$$

But by rules of simple algebra, it would seem

$$\frac{p_1}{p_2} = \frac{c_2^Fc_1^H+c_1^Fc_2^H}{2c_1^Fc_1^H}$$

MY QUESTION: What step am I missing to get $$\frac{p_1}{p_2} = \frac{c_2^F + c_2^H}{c_1^F+ c_1^H}$$ ?

$\endgroup$
2
$\begingroup$

Take your two beginning expressions and cross multiply them:

$$p_1 c_1^F = p_2 c_2^F$$

$$p_1 c_1^H = p_2 c_2^H$$

Then subtract each right hand side for both equations. Multiply one of the expressions by $-1$ on each side:

$$p_1 c_1^F - p_2 c_2^F = 0$$

$$-p_1 c_1^H + p_2 c_2^H = 0$$

And set each expression equal to each other:

$$p_1 c_1^F - p_2 c_2^F = -p_1 c_1^H + p_2 c_2^H$$

Put the $p_1$ and $p_2$ terms together:

$$p_1 c_1^F + p_1 c_1^H = p_2 c_2^F + p_2 c_2^H$$

Factor the $p_1$ and $p_2$ terms and rearrange:

$$p_1 (c_1^F + c_1^H) = p_2 (c_2^F + c_2^H) \implies \frac{p_1}{p_2} = \frac{c_2^F + c_2^H}{c_1^F+ c_1^H}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.