6.080. $\left\{\begin{array}{l}u^{2}+u v=15, \\ v^{2}+u v=10\end{array}\right.$
## Solution.

Rewrite the system as $\left\{\begin{array}{l}u(u+v)=15, \\ v(u+v)=10\end{array}\right.$ and divide the first equation by the second:

$$
\frac{u(u+v)}{v(u+v)}=\frac{15}{10} \Leftrightarrow \frac{u}{v}=\frac{3}{2} \Leftrightarrow u=\frac{3 v}{2}
$$

Substituting $u=\frac{3 v}{2}$ into the second equation of the system, we get $v^{2}+\frac{3 v^{2}}{2}=10, \frac{5 v^{2}}{2}=10, v^{2}=4$, from which $v_{1}=-2, v_{2}=2$. Then $u_{1}=-3$, $u_{2}=3$.

Answer: $(-3 ;-2)(3 ; 2)$.