问题
设由方程组 $\left\{\begin{array}{l}F(x, y, z)=0 \\ G(x, y, z)=0\end{array}\right.$ 确定的隐函数为 $y$ $=$ $y(x)$ 与 $z$ $=$ $z(x)$, 则 $\frac{\mathrm{d} y}{\mathrm{~d} x}$ 和 $\frac{\mathrm{d} z}{\mathrm{~d} x}$ 可以通过解以下哪个线性方程组求出？选项
[A]. $\left\{\begin{array}{l} F_{z}^{\prime}+F_{x}^{\prime} \frac{\mathrm{d} y}{\mathrm{~d} x}+F_{y}^{\prime} \frac{\mathrm{d} z}{\mathrm{~d} x}=0 \\ G_{z}^{\prime}+G_{x}^{\prime} \frac{\mathrm{d} y}{\mathrm{~d} x}+G_{y}^{\prime} \frac{\mathrm{d} z}{\mathrm{~d} x}=0 \end{array}\right.$[B]. $\left\{\begin{array}{l} F_{x}^{\prime}+F_{y}^{\prime} \frac{\mathrm{d} z}{\mathrm{~d} x}+F_{x}^{\prime} \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \\ G_{x}^{\prime}+G_{y}^{\prime} \frac{\mathrm{d} z}{\mathrm{~d} x}+G_{z}^{\prime} \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \end{array}\right.$
[C]. $\left\{\begin{array}{l} F_{x}^{\prime}+F_{y}^{\prime} \frac{\mathrm{d} y}{\mathrm{~d} x}+F_{z}^{\prime} \frac{\mathrm{d} z}{\mathrm{~d} x}=0 \\ G_{x}^{\prime}+G_{y}^{\prime} \frac{\mathrm{d} y}{\mathrm{~d} x}+G_{z}^{\prime} \frac{\mathrm{d} z}{\mathrm{~d} x}=0 \end{array}\right.$
[D]. $\left\{\begin{array}{l} F_{x}+F_{y}^{\prime} \frac{\mathrm{d} y}{\mathrm{~d} x}+F_{x}^{\prime} \frac{\mathrm{d} z}{\mathrm{~d} x}=0 \\ G_{x}+G_{y}^{\prime} \frac{\mathrm{d} y}{\mathrm{~d} x}+G_{z}^{\prime} \frac{\mathrm{d} z}{\mathrm{~d} x}=0 \end{array}\right.$
$\left\{\begin{array}{l} F_{x}^{\prime}+F_{y}^{\prime} \frac{\mathrm{d} y}{\mathrm{~d} x}+F_{z}^{\prime} \frac{\mathrm{d} z}{\mathrm{~d} x}=0 \\ G_{x}^{\prime}+G_{y}^{\prime} \frac{\mathrm{d} y}{\mathrm{~d} x}+G_{z}^{\prime} \frac{\mathrm{d} z}{\mathrm{~d} x}=0 \end{array}\right.$