Let's find all the real number pairs $(x, y)$ that satisfy the following system of equations (where $t$ is a real parameter):

$$
x^{2}+t=1, \quad (1) \quad (x+y) t=0, \quad (2) \quad y^{2}+t=1. \quad (3)
$$
If $t=0$, then from (1) $x^{2}=1, x= \pm 1$; from (3) $y^{2}=1, y= \pm 1$.

In this case, $( \pm 1 ; \pm 1)$ is a solution to the system of equations, which can be easily verified by substitution.

If $t \neq 0$, then from (1) $x^{2}=1-t>0$, that is, for $t>1$ the system of equations has no solution.