$$\begin{align*} -abx + bcy + (c^2 - a^2)z &= c, \\ -acx + (b^2 - a^2)y + bcz &= b, \\ ax + cy + bz &= 0. \end{align*}$$
Next, eliminate $z$ from these by first subtracting $c^2-a^2$ times the second from $bc$ times the first, and then subtracting $c$ times the third from the second.
These operations yield the following pair of equations for $x$, $y$:
$$\begin{align*} ca(c^2 - a^2 - b^2)x + (b^2c^2 - (c^2 - a^2)(b^2 - a^2))y &= ba^2, \\ -2acx + (b^2 - a^2 - c^2)y &= b. \end{align*}$$
Equivalently,
$$\begin{align*} a(b^2 + c^2 - a^2)y - c(a^2 + b^2 - c^2)x &= ab, \\ (c^2 + a^2 - b^2)y + 2acx &= -b, \end{align*}$$
which becomes, using the cosine rule,
$$y \cos A - x \cos C = \frac{a}{2ac} \quad \text{and} \quad y \cos B + x = -\frac{b}{2ac}$$
Now eliminate $y$ from these by subtracting $\cos A$ times the second from $\cos B$
times the first, thereby giving an equation for $x$:
$$-(\cos B \cos C + \cos A)x = \frac{a \cos B + b \cos A}{2ac} = \frac{1}{2a}$$
where we have used that $c = a \cos B + b \cos A$. But
$$\cos A = \cos(\pi - B - C) = -\cos(B + C) = -\cos B \cos C + \sin B \sin C$$
and $\sin B \sin C \neq 0$. Thus, we can solve the previous equation for $x$:
$$x = - \frac{abc}{2(ab \sin C)(ac \sin B)} = - \frac{abc}{8(ABC)^2}$$
where $(ABC)$ stands for the area of $ABC$. In the same way it can be shown
that
$$y = \frac{abc \cos B}{8(ABC)^2}, \quad z = \frac{abc \cos C}{8(ABC)^2}, \quad \text{and} \quad w = \frac{abc \cos A}{8(ABC)^2}.$$