5. (CZS) Solve the following system of equations:
$$
\begin{array}{l}
\left|a_{1}-a_{2}\right| x_{2}+\left|a_{1}-a_{3}\right| x_{3}+\left|a_{1}-a_{4}\right| x_{4}=1, \\
\left|a_{2}-a_{1}\right| x_{1}+\left|a_{2}-a_{3}\right| x_{3}+\left|a_{2}-a_{4}\right| x_{4}=1, \\
\left|a_{3}-a_{1}\right| x_{1}+\left|a_{3}-a_{2}\right| x_{2}+\left|a_{3}-a_{4}\right| x_{4}=1, \\
\left|a_{4}-a_{1}\right| x_{1}+\left|a_{4}-a_{2}\right| x_{2}+\left|a_{4}-a_{3}\right| x_{3}=1,
\end{array}
$$
where $\alpha_{1}, \alpha_{2}, a_{3}$, and $\alpha_{4}$ are mutually distinct real numbers.
5. We define $L_{1}=\left|a_{1}-a_{2}\right| x_{2}+\left|a_{1}-a_{3}\right| x_{3}+\left|a_{1}-a_{4}\right| x_{4}$ and analogously $L_{2}$, $L_{3}$, and $L_{4}$. Let us assume w.l.o.g. that $a_{1}<a_{2}<a_{3}<a_{4}$. In that case,
$$
\begin{aligned}
2\left|a_{1}-a_{2}\right|\left|a_{2}-a_{3}\right| x_{2} & =\left|a_{3}-a_{2}\right| L_{1}-\left|a_{1}-a_{3}\right| L_{2}+\left|a_{1}-a_{2}\right| L_{3} \\
& =\left|a_{3}-a_{2}\right|-\left|a_{1}-a_{3}\right|+\left|a_{1}-a_{2}\right|=0, \\
2\left|a_{2}-a_{3}\right|\left|a_{3}-a_{4}\right| x_{3} & =\left|a_{4}-a_{3}\right| L_{2}-\left|a_{2}-a_{4}\right| L_{3}+\left|a_{2}-a_{3}\right| L_{4} \\
& =\left|a_{4}-a_{3}\right|-\left|a_{2}-a_{4}\right|+\left|a_{2}-a_{3}\right|=0 .
\end{aligned}
$$

Hence it follows that $x_{2}=x_{3}=0$ and consequently $x_{1}=x_{4}=1 /\left|\alpha_{1}-a_{4}\right|$. This solution set indeed satisfies the starting equations. It is easy to generalize this result to any ordering of $a_{1}, a_{2}, a_{3}, a_{4}$.