Помогите пожалуйста :)Задание во вложениях

$\left \{ {{x^2-5xy+y^2=-47} \atop {xy=21}} \right. \\\ \left \{ {{x^2-5\cdot21+y^2=-47} \atop {xy=21}} \right. \\\ \left \{ {{x^2+y^2=58} \atop {xy=21}} \right. \\\ x= \frac{21}{y} \\\ (\frac{21}{y} )^2+y^2=58 \\\ \frac{441}{y^2} +y^2=58 \\\ 441+y^4=78y^2 \\\ a^2-58a+441=0 \\\ D_1=841-441=400 \\\ a_1=49 \\\ a_2=9$
$\\\ y^2=49 \\\ y_1=7 \\\ y_2=-7 \\\ y^2=9 \\\ y_3=3 \\\ y_4=-3 \\\ x_1= \frac{21}{7}=3 \\\ x_2= \frac{21}{-7}=-3 \\\ x_3= \frac{21}{3}=7 \\\ x_4= \frac{21}{3}=-7$
$|x+y|+|x-y|=|3+7|+|3-7|=10+4=14 \\\ |x+y|+|x-y|=|-3-7|+|-3+7|=10+4=14 \\\ |x+y|+|x-y|=|7+3|+|7-3|=10+4=14 \\\ |x+y|+|x-y|=|-7-3|+-7+3|=10+4=14$
Ответ: 14