Solve the following system of equations, where $t$ is a real parameter:

$$
\left.\begin{array}{r}
x+y+z=t \\
x+(t+1) y+z=0 \\
x+y-(t+1) z=2 t
\end{array}\right\}
$$
Solution. Express $(x+z)$ from (1) and substitute the obtained expression into (2).

$$
\begin{gathered}
x+z=t-y \\
t-y+(t+1) y=0
\end{gathered}
$$

From here, $t-y+ty+y=0$, which simplifies to $ty=-t$. If $t \neq 0$, then $y=-1$. Substituting $y=-1$ into $(1')$, we get $x=t+1-z$. Substitute this into (3):

$$
t+1-z-1-(t+1)z=2t
$$

If $t \neq -2$, then $z=\frac{-t}{t+2}$, and

$$
x=t+1+\frac{t}{t+2}=\frac{t^2+4t+2}{t+2}
$$

Since we have performed identical transformations throughout, the obtained values are indeed solutions to the system of equations.

If $t=0$, then the system of equations becomes:

$$
\begin{aligned}
& x+y+z=0 \\
& x+y+z=0 \\
& x+y-z=0
\end{aligned}
$$

In this case, $x=-y$ and $z=0$. The system of equations has infinitely many solutions.

If $t=-2$, then

$\left(1''\right)$

$\left(2''\right)$

$\left(3''\right)$

$$
\begin{aligned}
& x+y+z=-2 \\
& x-y+z=0 \\
& x+y+z=-4
\end{aligned}
$$

In this case, the system of equations has no solution, as equations $\left(1''\right)$ and $\left(3''\right)$ contradict each other.