6.200. $\left\{\begin{array}{l}(x+y)^{2}+2 x=35-2 y, \\ (x-y)^{2}-2 y=3-2 x\end{array}\right.$
## Solution.

Let's rewrite the system of equations as

$$
\left\{\begin{array}{l}
x^{2}+2 x y+y^{2}+2 x=35-2 y \\
x^{2}-2 x y+y^{2}-2 y=3-2 x
\end{array}\right.
$$

Subtracting the second equation from the first, we get

$4 x y+2 x+2 y=32-2 y+2 x, x=\frac{8-y}{y}$.

Substituting this value of $x$ into the first equation of the system, we obtain

$\left(\frac{8-y}{y}+y\right)^{2}+2 \cdot \frac{8-y}{y}=35-2 y \Rightarrow y^{4}-20 y^{2}+64=0$,

which gives $y_{1}=-2, y_{2}=2, y_{3}=-4, y_{4}=4 ; x_{1}=-5, x_{2}=3, x_{3}=-3, x_{4}=1$.

Answer: $(-5 ;-2),(3 ; 2),(-3 ;-4),(1 ; 4)$.