3. From two pieces of alloy weighing $6 \mathrm{~kg}$ and $3 \mathrm{~kg}$ with different percentages of copper, pieces of the same weight are cut off. Each of the cut pieces is then fused with the remainder of the other piece. After this fusion, the percentage of copper in both alloys is equal. How much do the cut pieces weigh?
Solution. Let's denote by $A$ the alloy whose weight is $6 \mathrm{~kg}$, and by $B$ the alloy whose weight is $3 \mathrm{~kg}$. Let in $1 \mathrm{~kg}$ of alloy $A$ there be $u \mathrm{~kg}$ of copper, and in $1 \mathrm{~kg}$ of alloy $B$ there be $v \mathrm{~kg}$ of copper. From the conditions of the problem, we have $u \neq v$. Let the weight of the cut pieces from alloys $A$ and $B$ be $x \mathrm{~kg}$. After combining the cut pieces with the remaining parts of the other pieces, in $1 \mathrm{~kg}$ of alloy $A$ there will be $\frac{(6-x) u+x v}{6} \mathrm{~kg}$ of copper, and in $1 \mathrm{~kg}$ of alloy $B$ there will be $\frac{(3-x) v+x u}{3} \mathrm{~kg}$ of copper. Since the percentage of copper in both alloys after mixing is the same, we have

$$
\frac{(6-x) u+x v}{6}=\frac{(3-x) v+x u}{3}
$$

From the last equation, we get $(u-v)(9 x-18)=0$, from which we have $x=2$, since $u \neq v$.