[ Quadratic Equations and Systems of Equations ]

Solve the system of equations: $\left\{\begin{array}{c}x y-t^{2}=9, \\ x^{2}+y^{2}+z^{2}=18\end{array}\right.$.
The first method. $x y=9+t^{2} \geq 9$ and $x^{2}+y^{2}=18-z^{2} \leq 18$. Therefore, $18 \geq x^{2}+y^{2} \geq 2 x y \geq 18$. Hence, all inequality signs must turn into equality signs, that is, $x^{2}+y^{2}=2 x y=18$, so $x=y= \pm 3$. From the first equation of the system and the condition $x y=9$, it follows that $t=0$, and from the second equation and the condition $x^{2}+y^{2}=$ 18, it follows that $z=0$.

The second method. $(x-y)^{2}=x^{2}+y^{2}-2 x y=\left(18-z^{2}\right)-\left(9+t^{2}\right)=-z^{2}-2 t^{2}$.

Since the left side takes only non-negative values, and the right side takes only non-positive values, equality is possible if and only if $x=y$ and $t=z=0$. Substituting this result into any of the equations of the system, we get that $x=y= \pm 3$. Then by substituting the obtained values into the other equation, we verify that they are also its solutions.

## Answer

$(3,3,0,0),(-3,-3,0,0)$