Let's determine the coefficients $a, b, c$, knowing that the following system of equations has no solution, but if we replace $x-1$ with $|x-1|$ in equation (2), the pair $x=3/4, y=5/8$ satisfies the new system of equations.

$$
\begin{aligned}
& a(x-1)+2 y=1 \\
& b(x-1)+c y=3
\end{aligned}
$$
The second condition gives two equations for the coefficients. If $x=3 / 4$, then $x-1=-1 / 4,|x-1|=1 / 4$, and thus

$$
\frac{5-a}{4}=1, \quad \frac{2 b+5 c}{8}=3
$$

which means $a=1, b=12-(5 / 2) c$. Substituting these into (1) and (2) results in the system of equations

$$
(x-1)+2 y=1, \quad\{12-(5 / 2) c)\}(x-1)+c y=3
$$

which has no solution. If we keep the first equation and substitute $(x-1)$ from the first into the second, we get an equivalent system of equations, so the

$$
\begin{gathered}
(x-1)=1-2 y, \\
(6 c-24) y=(5 / 2) c-9
\end{gathered}
$$

system of equations has no solution. This only follows if $6 c-24=0, c=4$, because then the right-hand side of $\left(2^{\prime}\right)$ is 1, not 0. In this case, $b=2$.

The conditions of the problem are therefore satisfied when $a=1, b=2, c=4$.

László Baranyai (Győr, Révai M. Gymn., 1st year)

Endre Boros (Budapest, I. István Gymn., 1st year)