Problem 1. A cup of Coca-Cola, three sandwiches, and seven croissants cost 127 denars. A cup of Coca-Cola, four sandwiches, and ten croissants cost 166 denars. How much together does one cup of Coca-Cola, one sandwich, and one croissant cost?
Solution. Let's denote the price of the Coca-Cola by $c$, the price of the sandwich by $s$, and the price of the bun by $k$. From the problem statement, we have

$$
\begin{gathered}
c+3 s+7 k=127 \\
c+4 s+10 k=166
\end{gathered}
$$

The second equation can be rewritten as

$$
(c+3 s+7 k)+(s+3 k)=166
$$

from which we get

$$
s+3 k=166-127=39
$$

If we add the initial equations, we get

$$
(c+3 s+7 k)+(c+4 s+10 k)=293
$$

which is

$$
2(c+s+k)+5(s+3 k)=293
$$

so from $s+3 k=39$ we have

$$
2(c+s+k)+5 \cdot 39=293, \text{ i.e. } c+s+k=98: 2=49
$$

Thus, one sandwich, one Coca-Cola, and one bun together cost 49 denars.