2B. On Monday, three bananas cost as much as a lemon and an orange together. On Tuesday, the prices of all fruits were reduced by the same amount of money, two oranges cost as much as three bananas and one lemon, and the price of half a lemon was 5 denars.

What was the price of one orange on Monday?
Solution. Let $x, y$, and $z$ be the prices of one banana, one lemon, and one orange, respectively, on Monday. Let the price reduction on Tuesday be $r$ denars. Then from the conditions of the problem we have

$$
\left\{\begin{aligned}
3 x & =y+z \\
2(z-r) & =3(x-r)+y-r \\
\frac{1}{2}(y-r) & =5
\end{aligned}\right.
$$

If we algebraically transform the second equation, we get

$2 z=3 x-2 r+y$.

If we substitute $3 x$ from the first equation, we get $2 z=y+z-2 r+y$, so $z=2 y-2 r=2(y-r)$. But $y-r=10$, from which we get $z=2 \cdot 10=20$ denars.

Therefore, the price of one orange on Monday was 20 denars.