1. Given real numbers $x, y$ satisfy the system of equations
$$
\left\{\begin{array}{l}
x^{3}+y^{3}=19, \\
x+y=1 .
\end{array}\right.
$$

then $x^{2}+y^{2}=$ $\qquad$
II, 1.13.
From $x^{3}+y^{3}=19$, we get $(x+y)\left[(x+y)^{2}-3 x y\right]=19$. Substituting $x+y=1$ yields $x y=-6$. Therefore, $x^{2}+y^{2}=(x+y)^{2}-2 x y=13$.