4. If $\left\{\begin{array}{l}x^{2}+x y+y=14, \\ y^{2}+x y+x=28,\end{array}\right.$ find $x+y=$ $\qquad$
(2001, TI Cup National Junior High School Mathematics Competition)
Let $x=a+b, y=a-b$.
Transform the original system of equations to
$$
\begin{array}{l}
\left\{\begin{array}{l}
2 a^{2}+2 a b+a-b=14, \\
2 a^{2}-2 a b+a+b=28
\end{array}\right. \\
\Rightarrow a=3 \text { or }-\frac{7}{2} .
\end{array}
$$

From $x+y=2 a$, we get $x+y=6$ or -7.