9.2. Solve the system of equations: $\left\{\begin{array}{c}x^{4}+x^{2} y^{2}+y^{4}=481 \\ x^{2}+x y+y^{2}=37\end{array}\right.$.
Solution:

$$
\begin{gathered}
\left\{\begin{array} { c } 
{ ( x ^ { 2 } + y ^ { 2 } ) ^ { 2 } - ( x y ) ^ { 2 } = 4 8 1 } \\
{ x ^ { 2 } + y ^ { 2 } + x y = 3 7 }
\end{array} \Rightarrow \left\{\begin{array}{c}
\left(x^{2}+y^{2}+x y\right)\left(x^{2}+y^{2}-x y\right)=481 \\
x^{2}+y^{2}+x y=37
\end{array}=>\right.\right. \\
\left\{\begin{array}{c}
x^{2}+y^{2}-x y=\frac{481}{37}=13 \\
x^{2}+y^{2}+x y=37
\end{array}\right.
\end{gathered}
$$

We make the substitutions: $x^{2}+y^{2}=t$ and $x y=n$, then $\left\{\begin{array}{l}t-n=13 \\ t+n=37\end{array}\right.$. Solving the system of linear equations, we get: $t=25$ and $n=12$.

$$
\left\{\begin{array} { c } 
{ x ^ { 2 } + y ^ { 2 } = 2 5 } \\
{ x y = 1 2 }
\end{array} \Rightarrow \left\{\begin{array}{c}
x^{2}+2 x y+y^{2}=25+2 \cdot 12=49 \\
x^{2}-2 x y+y^{2}=25-2 \cdot 12=1
\end{array} \Rightarrow>\left\{\begin{array}{c}
x+y= \pm 7 \\
x-y= \pm 1
\end{array}\right.\right.\right.
$$

Solving the 4 systems of linear equations, we get: ( $4 ; 3$ ), ( $3 ; 4$ ), ( $-3 ;-4$ ) and ( $-4 ;-3$ ).

Answer: $(4 ; 3),(3 ; 4),(-3 ;-4$ and and ( $(-4 ;-3)$.

| Criteria | points |
| :--- | :---: |
| 1. Complete solution of the problem. | 7 |
| 2. Correct solution path, but some solutions are lost. The solution becomes fully correct upon correction. Errors in solving systems of linear equations, with the correct algorithm. | 4 |
| 3. The problem is not solved, but the degree of the initial system is reduced. | 2 |
| 4. All four solutions are found, without extra ones, but it is not proven why there are no others. | 1 |
| 5. Incorrect solution. | 0 |