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

Stan Shunpike
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1 Answers1

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Using algebra you get that both expressions are the same: $$(c_2^F+c_2^H)/(c_1^F+c_1^H)=(c_2^Fc_1^H+c_1^Fc_2^H)/(2c_1^Fc_1^H)$$ operating: $$2c_1^Fc_1^Hc_2^F+2c_1^Fc_1^Hc_2^H=c_2^F(c_1^H)^2+c_1^Hc_1^Fc_2^H+c_2^Fc_1^Hc_1^F+(c_1^F)^2c_2^H$$ then: $$c_1^Fc_1^Hc_2^F+c_1^Fc_1^Hc_2^H=c_2^F(c_1^H)^2+(c_1^F)^2c_2^H $$ looking the first two equalities, we get : $$c_1^Fc_2^H=c_2^Fc_1^H$$ replace that in our last equality and you get that both expressions are the same. Reversing the steps deliver you to get the "not so common" equality. It's a little tricky.

thebooort
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