Решите систему уравнений х+у=13 ху=30

Решите задачу:

$\left \{ {{x=13-y} \atop {(13-y)y=30}} \right. \\ \left \{ {{x=13-y} \atop {(13y-y^2=30}} \right. \\ \left \{ {{x=13-y} \atop {y^2-13y+30=0}} \right. \\ D = 13^2-4*30 = 169-120 = 49 = 7^2\\ y = \frac{13 \pm 7}{2} \\\\ \left[\begin{array}{ccc}y = 10\\y = 3\end{array}\right\\\\ \left[\begin{array}{ccc} \left \{ {y = 10 \atop {x=3}} \right. \\\\ \left \{ {{y=3} \atop {x=10}} \right. \end{array}\right$