13.339. It was necessary to take several liters of liquid at a temperature of $a^{\circ}$ and another amount of the same liquid at a temperature of $b^{\circ}$, to obtain a mixture temperature of $c^{\circ}$. However, the second liquid was taken in the amount that was supposed to be taken of the first, and vice versa. What temperature did the mixture end up with?
Solution.

Let $m_{\mathrm{a}}, m_{\mathrm{b}}$ be the volume of the liquid taken at temperatures $a^{\circ}$ and $b^{\circ}$ respectively, $x$ be the temperature of the resulting mixture. According to the condition, we have the system $\left\{\begin{array}{l}m_{a}(a-c)+m_{b}(b-c)=0, \\ m_{b}(a-x)+m_{a}(b-x)=0,\end{array}\right.$ from which $x=a+b-c$.

Answer: $a+b-c$.