Solve the following system of equations:

$$
\begin{aligned}
& \frac{x}{b+c}+\frac{y}{c-a}=a+b \\
& \frac{y}{c+a}+\frac{z}{a-b}=b+c \\
& \frac{z}{a+b}+\frac{x}{b-c}=c+a
\end{aligned}
$$
Solve for $y$ from (1), $z$ from (3), substitute the expressions into (2), and collect the terms containing $x$ on one side, and the terms not containing $x$ on the other side:

$$
\begin{gathered}
y=(c-a)(a+b)+\frac{a-c}{b+c} x \\
z=(a+b)(c+a)+\frac{a+b}{c-b} x \\
\left(\frac{a-c}{(b+c)(c+a)}+\frac{a+b}{(c-b)(a-b)}\right) \cdot x= \\
=\frac{(a-c)(a+b)}{c+a}+\frac{(a+b)(c+a)}{b-a}+b+c
\end{gathered}
$$

After bringing to a common denominator and multiplying through:

$$
\frac{2\left(a b c+a^{2} c+c^{2} b+b^{2} a\right)}{\left(c^{2}-b^{2}\right)(c+a)(a-b)} \cdot x=\frac{2\left(a b c+a^{2} c+c^{2} b+b^{2} a\right)}{(c+a)(b-a)}
$$

thus, their quotient, and based on $\left(1^{\prime}\right)$ and $\left(3^{\prime}\right)$, the other two unknowns are:

$$
x=b^{2}-c^{2}, \quad y=(c-a)(a+b+c-b)=c^{2}-a^{2}, \quad z=a^{2}-b^{2}
$$

provided the simplifying factor is non-zero:

$$
K=a b c+a^{2} c+c^{2} b+b^{2} a \neq 0
$$

Of course, we also assumed that all denominators are non-zero - otherwise, at least one of the equations would be meaningless, and thus the system of equations would be meaningless, i.e.,

$$
\begin{gathered}
(a+b)(a-b)(b+c)(b-c)(c+a)(c-a)= \\
\quad=\left(a^{2}-b^{2}\right)\left(b^{2}-c^{2}\right)\left(c^{2}-a^{2}\right) \neq 0
\end{gathered}
$$

If $K=0$, then (4) is satisfied by any number in place of $x$, and thus the system of equations has infinitely many solutions:

$$
\begin{aligned}
& x=t, \quad y-(c-a)(a+b)+\frac{a-c}{b+c} \cdot t \\
& z=(a+b)(c+a)+\frac{a+b}{c-b} \cdot t
\end{aligned}
$$

where $t$ is any number. For example, we get such a case from $K=0$ if we choose $a$ to be 1, $c$ to be -2, and $b$ to be one of the roots of the resulting $b^{2}+2 b-2=0$ equation, $\sqrt{3}-1$.

In every such case, a dependency exists in the system of equations, and after multiplying two equations by suitable numbers and adding them, we get the remaining equation, for example, in the case of the given numerical example, the system is:

$$
\frac{x}{\sqrt{3}-3}-\frac{y}{3}=\sqrt{3}, \quad-y+\frac{z}{2-\sqrt{3}}=\sqrt{3}-3, \quad \frac{x}{1+\sqrt{3}}+\frac{z}{\sqrt{3}}=-1
$$

and here, adding the triple of the first equation and the $(2 \sqrt{3}+3)$-fold of the third equation, we get the second equation.

Eff Lajos (Budapest, Fazekas M. gyak. g. II. o. t.)

Note. Most solutions did not consider the consequences of the $K=0$ case.

Some incorrectly stated that the possibility of $K=0$ is ruled out by assumption (5).

Searching for a numerical example leading to $K=0$ was not part of the task, but it brings one closer to clarity on the issue. However, the mere statement of the condition $K \neq 0$ is sufficient with the knowledge of a first-grader.