2. In a sports store, over two days, thirteen pairs of sneakers, two sports suits, and one T-shirt were sold, with the same amount of money earned on the first day as on the second day (from the sale of the aforementioned items). One pair of sneakers is cheaper than a sports suit and more expensive than a T-shirt by the same amount. How many pairs of sneakers and suits were sold in one day with a T-shirt?
Answer: 8 pairs of sneakers and no sports suits.

Solution. Let in one day $x$ suits and $y$ pairs of sneakers were sold with a T-shirt. Then in the other day, $(2-x)$ suits and $(13-y)$ pairs of sneakers were sold.

Let $c$ be the price of one pair of sneakers, and $s$ be the price difference.

Then, from the condition of the problem, it follows that

$$
\begin{aligned}
& x(c+s)+y c+(c-s)=(2-x)(c+s)+(13-y) c \\
&(14-2 x-2 y) c=(2 x-3) s .
\end{aligned}
$$

The number $x$ can take the values 0, 1, or 2. Let's consider each of them in turn

1) $x=0 \Rightarrow(2 y-14) c=3 s$. Since $0<s<c$, then $0<(2 y-14) c<3 c \Rightarrow$

$$
0<(2 y-14)<3 \Rightarrow y=8-\text { unit integer }
$$

2) $x=1 \Rightarrow(2 y-12) c=s$. Since $0<s<c$, then $0<(2 y-12) c<$ c $\Rightarrow$

$$
0<(2 y-12)<1 \Rightarrow y-\text { no integers. }
$$

3) $x=2 \Rightarrow(10-2 y) c=s$. Since $0<s<c$, then $0<(10-2 y) c<$ c $\Rightarrow$

$$
0<(10-2 y)<1 \Rightarrow y \text { - no integers. }
$$

Thus, the situation described in the problem can only occur when $x=0, y=8$.