Question

Решите систему уравнений:​

Answer 1

Ответ:

$\left \{ {{x^{2} +4y=24} \atop {x+y=3}} \right. \\\left \{ {{x^{2} +4(3-x)=24} \atop {y=3-x}} \right. \\x^{2} +12-4x=24;\\x^{2} -4x-12=0;\\D=4+12=16>0;\\x_{1}=2+4=6;\\x_{2}=2-4=-2;\\\\y_{1}=3-x_{1}=3-6=-3;\\y_{2}=3-x_{2}=3-(-2)=5$

Объяснение: