# How does one derive relative demand?

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

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:

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