1. Solve the system:

$$
\left\{\begin{array}{l}
(x+y)(x+y+z)=18 \\
(y+z)(x+y+z)=30 \\
(z+x)(x+y+z)=24
\end{array}\right.
$$
1. Answer: $(1,2,3)$ and $(-1,-2,-3)$

Add all the equations, we get $2(x+y+z)^{2}=72, x+y+z= \pm 6$. Substitute

$$
\left\{\begin{array}{l}
x+y= \pm 3 \\
y+z= \pm 5 \\
z+x= \pm 4
\end{array}\right.
$$

We get two solutions $x= \pm 1, y= \pm 2, z= \pm 3$ (the plus and minus signs occur simultaneously).