Example 25. Solve the system

$$
\left\{\begin{array}{l}
y^{x^{2}+7 x+12}=1 \\
x+y=6 \\
y>0
\end{array}\right.
$$
Solution. The first equation under the condition $y>0$ is equivalent to the system of equations

$$
y=1, \quad x^{2}+7 x+12=0
$$

i.e., the system of equations

$$
y=1, \quad x=-3, \quad x=-4
$$

Therefore, the given system is equivalent to the combination of three systems:

$$
\left\{\begin{array} { l } 
{ y = 1 , } \\
{ x + y = 6 , }
\end{array} \quad \left\{\begin{array} { l } 
{ y > 0 } \\
{ x = - 3 } \\
{ x + y = 6 , }
\end{array} \quad \left\{\begin{array}{l}
y>0 \\
x=-4 \\
x+y=6
\end{array}\right.\right.\right.
$$

i.e., the combination of

$$
\left\{\begin{array} { l } 
{ x = 5 , } \\
{ y = 1 , }
\end{array} \quad \left\{\begin{array} { l } 
{ x = - 3 , } \\
{ y = 9 , }
\end{array} \quad \left\{\begin{array}{l}
x=-4 \\
y=10
\end{array}\right.\right.\right.
$$

Thus, the set of all solutions to the given system consists of three ordered pairs: $(5 ; 1),(-3 ; 9),(-4 ; 10)$.