Why $(\frac{\partial x}{\partial z})_w$ can be expressed as-$\frac{\frac{\partial (F,G)}{\partial (z,y)}}{\frac{\partial (F,G)}{\partial (x,y)}}$

Question

Consider $F(x,y,z,w)=0$ and $G(x,y,z,w)=0$ . Why $(\frac{\partial x}{\partial z})_w$= -$\frac{\frac{\partial (F,G)}{\partial (z,y)}}{\frac{\partial (F,G)}{\partial (x,y)}}$.Noted that the $(\frac{\partial x}{\partial z})_w$= -$\frac{\frac{\partial (F,G)}{\partial (z,y)}}{\frac{\partial (F,G)}{\partial (x,y)}}$=-$\frac{F_3G_2-F_2G_3}{F_1G_2-F_2G_1}$

Answer 1

The two equations $F(x,y,z,w_0)=0$, $\ G(x,y,z,w_0)=0$ define a curve $\gamma$ in the three-dimensional $(x,y,z)$-plane $w=w_0$ of ${\mathbb R}^4$. This curve should have a parametrization $$\gamma:\quad z\mapsto \bigl(x(z),y(z),z, w_0\bigr)\qquad \qquad (z_0-h<z<z_0+h)\ ,$$ valid in the neighborhood of some "regular" point $(x_0,y_0,z_0,w_0)$. Since $$F\bigl(x(z),y(z),z,w_0\bigr)\equiv0\ ,\qquad G\bigl(x(z),y(z),z,w_0\bigr)\equiv0\ ,$$ taking the derivative with respect to $z$ gives $$F_x x'(z)+F_y y'(z) + F_z=0\ , \qquad G_x x'(z)+G_y y'(z) + G_z=0\ .$$ We now solve for $x'(z)$ and obtain $$x'(z)=-{F_z G_y-G_z F_y \over F_x G_y-G_x F_y}=-{\det\left[{\partial(F,G)\over\partial(z,y)}\right] \over\det\left[{\partial(F,G)\over\partial(x,y)}\right]}\ .$$ In response to a comment by the OP: Here we had $r=3$ equations (incl. $w=w_0$) in $n=4$ variables, so there was one degree of freedom (the $z$ variable). In general you would have $r$ equations in $n$ variables and $d=n-r$ degrees of freedom. If you want explicit formulae you would have to declare beforehand which $d$ variables should be considered as the independent ones. Look at it this way: There is no explicit formula for the solution set of a rank $r$ homogeneous linear system in $n$ unknowns.