Question

Решение систем уравнений второй степени

Answer 1

$\left \{ {{y^2+x=61} \atop {x+y=-11}} \right. \\\\x=-11-y\\\\ y^2+(-11-y)=61\\ y^2-11-y-61=0\\ y^2-y-72=0\\ D=1+288=289; \ \sqrt D= 17\\\\ y_{1/2}= \frac{1\pm17}{2}\\\\ y_1=9\\ y_2=-8\\\\ x_1=-11-9\\ x_1=-20\\\\ x_2=-11-(-8)\\ x_2=-11+8\\ x_2=-3$

Ответ: $(-20;9)(-3;-8)$

Answer 2

Х=61-у^2

61-у^2+у=-11
61-у^2+у+11=0
72-у^2+у=0
у^2-у-72=0
д=1-4×72=289
у1=(1-17)/2=-8
у2=(1+17)/2=9
х1=61-(-8)^2=61-64=-3
х2=61-9^2=-20
(-3;-8)(-20;9)