6.191. $\left\{\begin{array}{l}x^{2}+y=y^{2}+x \\ y^{2}+x=6\end{array}\right.$

The system of equations is:
\[
\left\{\begin{array}{l}
x^{2} + y = y^{2} + x \\
y^{2} + x = 6
\end{array}\right.
\]
## Solution.

From the condition we have

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ x ^ { 2 } - y ^ { 2 } - x + y = 0 , } \\
{ y ^ { 2 } + x = 6 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(x-y)(x+y)-(x-y)=0, \\
y^{2}+x=6
\end{array}\right.\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
(x-y)(x+y-1)=0 \\
y^{2}+x=6
\end{array}\right.
\end{aligned}
$$

The obtained system is equivalent to the combination of two systems of equations:

1) $\left\{\begin{array}{l}x-y=0, \\ y^{2}+x=6\end{array}\right.$ or 2$)\left\{\begin{array}{l}x+y-1=0, \\ y^{2}+x=6\end{array}\right.$

Solving them, we find

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0354.jpg?height=222&width=802&top_left_y=850&top_left_x=156)

Answer: $(2 ; 2),(-3 ;-3),\left(\frac{1+\sqrt{21}}{2} ; \frac{1-\sqrt{21}}{2}\right),\left(\frac{1-\sqrt{21}}{2} ; \frac{1+\sqrt{21}}{2}\right)$.