# The general solution of a $2$-market equilibrium

The question i'm currently grappling with in relation to market equilibrium is:

(a0-b0)+(a1-b1)p1+(a2-b2)p2 = 0
(α0-β0)+(α1-β1)p1+(α2-β2)p2 = 0
ci is used for the shorthand form of (ai-bi) where i is a subscript.
γi is used for the short hand form of (αi-βi) where is a subscript.
c0+c1p1+c2p2=0
γ0+γ1p1+γ2p2=0


where α,β,a and b are all parametric constants. The $0,1,2$ are supposed to be in subscript but i'm not sure how to get that. I'm trying to work through the steps to get an equation in terms of $p1$ but after reaching $p2$ = $-(c0+c1p1)$/$c2$ and subbing that into equation 2, i haven't been able to find a neat form. Any assistance would be much appreciated!

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Are you trying to solve for $p_1, p_2$ in terms of $(a_i-b_i)$ and $(α_i-β_i)$? – user2468 Aug 17 '12 at 2:26

So we have: $$c_1 p_1 + c_2 p_ 2 = -c_0 \tag{1}$$ $$γ_1 p_ 1 + γ_2 p_2 = -γ_0 \tag{2}$$ Multiply $(2)$ by $-c_2/γ_2$ to get: $$\frac{-c_2γ_1}{γ_2} p_ 1 - c_2 p_2 = \frac{c_2γ_0}{γ_2} \tag{3}$$ Now add $(1) + (3)$ to get: $$\Big( c_1 + \frac{-c_2γ_1}{γ_2} \Big) p_1 = -c_0 + \frac{c_2γ_0}{γ_2} \tag{4}$$ Hence $p_1$ is $$p_1 = \frac{ -c_0 + \frac{c_2γ_0}{γ_2} }{ \Big( c_1 + \frac{-c_2γ_1}{γ_2} \Big) } = \frac{-c_0 \gamma_2 + c_2 \gamma_0}{ c_1 \gamma_2 - c_2 \gamma_1} \tag{4}$$ To find $p_2$ you can either substitute $p_1$ in $(1)$ or repeat the above steps with $−c1/γ1$ instead.
Alternatively though if you know about matrices, you can put $(1)$ and $(2)$ in matrix form as: $$\pmatrix{c_1 & c_2 \\ \gamma_1 & \gamma_2} \pmatrix{p_1 \\ p_2} = \pmatrix{-c_0 \\ -\gamma_0}$$ and solve for $$\pmatrix{p_1 \\ p_2}.$$ – user2468 Aug 17 '12 at 2:39