11.1. Find all solutions of the system of equations in real numbers:

$$
\left\{\begin{array}{c}
x y+z+t=1 \\
y z+t+x=3 \\
z t+x+y=-1 \\
t x+y+z=1
\end{array}\right.
$$
Answer. $(1,0,-1,2)$.

Solution. Subtract the second equation from the first, the third from the second, the fourth from the third, and the first from the fourth, factorize each difference, and we get: $\quad(x-z)(y-1)=-2,(y-t)(z-1)=4,(z-x)(t-1)=-2,(y-t)(x-1)=0 . \quad$ From the second equality, $y-t \neq 0$, so from the fourth equality, $x=1$. From the first and third equalities, $x-z \neq 0$ and $y-1=-2=1-t$, hence $y=2-t$. Substitute the found expressions into the first and second equations of the original system, we get $z=-1, t=2$, from which $y=0$. Thus, we obtain the unique solution to the system: $(1,0,-1,2)$.

Grading criteria. Guessed the answer with verification: 1 point. Subtractions of equations were performed and the solution was brought to the stage $(x-z)(y-1)=-2, . .: 2$ points. From here, $x=1$ was found: another 1 point.