6.187. $\left\{\begin{array}{l}(x+1)(y+1)=10 \\ (x+y)(xy+1)=25\end{array}\right.$
## Solution.

Rewrite the system of equations as $\left\{\begin{array}{l}x y+x+y+1=10, \\ (x+y)(x y+1)=25\end{array}\right.$ and introduce

the substitution $\left\{\begin{array}{l}x+y=u, \\ x y=v .\end{array}\right.$

With respect to $u$ and $v$, the system becomes $\left\{\begin{array}{l}v+u=9, \\ u(v+1)=25 .\end{array}\right.$

From here, $v=9-u \Rightarrow u(9-u+1)=25 \Leftrightarrow u^{2}-10 u+25=0,(u-5)^{2}=0$, hence $u=5 ; v=4$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0351.jpg?height=127&width=582&top_left_y=851&top_left_x=113)

Answer: $(4 ; 1),(1 ; 4)$.