$\left \{ {{x+y=-8} \atop {xy=15}} \right.$
(=) $\left \{ {{x=-8-y} \atop {(-8-y)y=15}} \right.$
(=) $\left \{ {{x=-8-y} \atop {-8y-y^2=15}} \right.$
(=) $\left \{ {{x=-8-y} \atop {y^2 +8y + 15 = 0}} \right.$
(=) $\left \{ {{x=-8-y} \atop {\left[ \begin{array}{l}y=-3\\y=-5\end{array} \right.}} \right.$
(=) $\left \{ {{\left[ \begin{array}{l}x=-5\\x=-3\end{array} \right.} \atop {\left[ \begin{array}{l}y=-3\\y=-5\end{array} \right.}} \right.$
Vậy hệ phương trình có nghiệm: (x;y) = {(-5;-3);(-3;-5)}