Determine the real numbers $x, y$ such that $x+y=1$ and $x^{3}+y^{3}=19$.

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On a $s_{1}^{3}=x^{3}+3 x^{2} y+3 x y^{2}+y^{3}=x^{3}+y^{3}+3 s_{1} s_{2}$, so $x^{3}+y^{3}=s_{1}^{3}-3 s_{1} s_{2}$. Thus, $s_{1}=1$ and $s_{2}=-6$, so $x$ and $y$ are roots of the polynomial $X^{2}-X-6$, which has 3 and -2 as roots. Therefore, $(x, y)=(3,-2)$ or $(x, y)=(-2,3)$.