Question

Решите систему x+y=3 y↑2-xy=-1

Answer 1

$\left \{ {{x+y=3} \atop { y^{2}-xy =-1}} \right. x=3-y y^{2} -y(3-y)=-1 y^{2} -3y+ y^{2} =-1 2 y^{2} -3y +1=0 D= \sqrt{9-8} = \sqrt{1} =1 y_{1}= \frac{3+1}{4} =1 y_{2} = \frac{3-1}{4} = \frac{1}{2} x_{1} =3-1=2 x_{2} = 3- \frac{1}{2} =2.5$