9.1. Let the numbers $x, y, u, v$ be distinct and satisfy the relation $\frac{x+u}{x+v}=\frac{y+v}{y+u}$. Find all possible values of the sum $x+y+u+v$.
Answer. $x+y+u+v=0$.

Solution. Let's bring the difference between the left and right parts of the expression from the condition, equal to zero, to a common denominator and factor the numerator: $\frac{x+u}{x+v}-\frac{y+v}{y+u}=\frac{x u+y u+u^{2}-y v-x v-v^{2}}{(x+v)(y+u)}=\frac{(x+y)(u-v)+(u+v)(u-v)}{(x+v)(y+u)}=\frac{(x+y+u+v)(u-v)}{}$.

Therefore, the numerator of the fraction is equal to 0, and the difference $u-v$ is not equal to 0 by the condition, so the first bracket, which is the sum $x+y+u+v$, must be equal to 0.

Grading criteria. The solution does not explicitly mention that the difference $u-v$ is not equal to 0: minus 2 points.