13.378 In a piece of alloy weighing 6 kg, copper is contained. In another piece of alloy weighing 8 kg, copper is contained in a different percentage than in the first piece. From the first piece, a certain part was separated, and from the second - a part twice as heavy as from the first. Each of the separated parts was melted with the remainder of the other piece, after which two new alloys with the same percentage of copper content were obtained. What is the mass of each of the parts separated from the original pieces of alloy?
Solution. Let $x$ be the mass of the part separated from the first alloy piece, $p(q)$ be the concentration of copper in the first (second) alloy. Then $x+(8-2 x)=8-x$ is the mass of the third alloy, $2 x+(6-x)=6+x$ is the mass of the fourth alloy, $p x+(8-2 x) q$ is the amount of copper in the third alloy, $p(6-x)+2 x q$ is the amount of copper in the fourth alloy. According to the problem, the concentrations of copper in the third and fourth alloys are equal, i.e.

$\frac{p x+(8-2 x) q}{8-x}=\frac{p(6-x)+2 x q}{6+x} \Leftrightarrow$

$\Leftrightarrow \frac{(p x-q x)+(8 q-x q)}{8-x}=\frac{(2 x q-2 x p)+(6 p+x p)}{6+x} \Leftrightarrow$

$\Leftrightarrow \frac{x(p-q)}{8-x}+q=\frac{2 x(q-p)}{6+x}+p \Leftrightarrow \frac{(p-q) x}{8-x}+\frac{2 x(p-q)}{6+x}=p-q \Leftrightarrow$

$\Rightarrow \frac{x}{8-x}+\frac{2 x}{6+x}=1 \Rightarrow x=2.4$.

Answer: 2.4 kg.