Linearization and the effect of a change

There is the following system of four equations and four endogenous variables $$(K,L,w,q)$$. Assume $$F$$ is a concave function.

$$\partial F(K,L)/\partial K = r + (1-p)$$

$$\partial F(K,L)/\partial L = w$$

$$F(K,L)-rK-wL-q-(1-p)K = 0$$

$$v(w,p)=\bar{v}$$

It seems to linearize the system around the equilibrium values to get $$\frac{\partial K}{\partial p}$$ and $$\frac{\partial L}{\partial p}$$ and it seems that these equations become

$$\frac{\partial K}{\partial p} = \frac{-F_{LL}-F_{KL}\partial w/\partial p}{(F_{KK}F_{LL}-F^2_{KL})}$$

$$\frac{\partial L}{\partial p} = \frac{F_{KK}\partial w/\partial p + F_{KL}}{(F_{KK}F_{LL}-F^2_{KL})}$$

I tried to Taylar expansion but I could not get these equations. How do we get these last equations? These equations are from p.85 Kousky et al. (2006)

Start from the first order conditions: \begin{align*} &F_k = r + (1- p)\\ &F_L = w\\ \end{align*} Now totally differentiate with respect to $$p$$: \begin{align*} &F_{KK} dK + F_{KL} dL = - dp\\ &F_{LK} dK + F_{LL} dL = \frac{\partial w}{\partial p} dp\\ \end{align*} This is equivalent to: \begin{align*} &dK = \frac{- dp - F_{KL} dL}{F_{KK}}\\ &dL = \frac{\frac{\partial w}{\partial p}dp - F_{LK} dK}{F_{LL}} \end{align*} substituting the second into the first gives: \begin{align*} &dK = \frac{(- dp)F_{LL} - F_{KL}(\frac{\partial w}{\partial p} dp - F_{LK} dK)}{F_{KK} F_{LL}},\\ \to &\left(1 - \frac{F_{KL} F_{LK}}{F_{LL} F_{KK}}\right) dK = \frac{-F_{LL}dp- F_{KL}\frac{\partial w}{\partial p} dp}{F_{KK} F_{LL}},\\ \to &dK = \frac{-F_{LL} - F_{KL}\frac{\partial w}{\partial p}}{F_{LL} F_{KK} - (F_{LK})^2} dp \end{align*} This gives: $$\frac{\partial K}{\partial p} = \frac{-F_{LL} - F_{KL}\frac{\partial w}{\partial p}}{F_{LL} F_{KK} - (F_{LK})^2}$$ The expression for $$\frac{\partial L}{\partial p}$$ can be derived in a similar way.