6.196. $\left\{\begin{array}{l}x^{2}+y^{2}=34 \\ x+y+x y=23\end{array}\right.$

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6.196. $\left\{\begin{array}{l}x^{2}+y^{2}=34 \\ x+y+x y=23\end{array}\right.$
## Solution.

Rewrite the system of equations as $\left\{\begin{array}{l}(x+y)^{2}-2 x y=34, \\ (x+y)+x y=23\end{array}\right.$ and, by setting $\left\{\begin{array}{l}x+y=u, \\ x y=v,\end{array}\right.$, we get $\left\{\begin{array}{l}u^{2}-2 v=34, \\ u+v=23\end{array} \Rightarrow v=23-u\right.$,

$u^{2}-2(23-u)=34, u^{2}+2 u-80=0$, from which

$u_{1}=-10, u_{2}=8 ; v_{1}=23-10=13, v_{2}=23-8=15$.

This system is equivalent to the combination of two systems of equations:

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0359.jpg?height=123&width=780&top_left_y=740&top_left_x=159)

Answer: $(3 ; 5),(5 ; 3)$.