Question

решите систему уравнений методом подстановки​

Answer 1

Ответ:

$1) x_{1}=-3; y_{1}=6; x_{2}=2;y_{2}=1;\\2)x_{1}=2; y_{1}=-1; x_{2}=-3; y_{2}=-6$

Объяснение:

1)$\left \{ {{x^2-y=3} \atop {x+y=3}} \right.\left \{ {{x^2-y=3} \atop {x=3-y}} \right.\left \{ {{(3-y)^2-y=3} \atop {x=3-y}} \right.\left \{ {{9-6y+y^2-y=3} \atop {x=3-y}} \right.\left \{ {{y^2-7y+6=0} \atop {x=3-y}} \right.\\\\ y^2-7y+6=0\\\\D=b^2-4ac=49-24=25=5^2\\\\y_{1}=\frac{7+5}{2}=6\\\\y_{2}=\frac{7-5}{2}=1\\\\\left \{ {{y_{1}=6} \atop {x_{1}=-3}} \right. \left \{ {{y_{2}=1} \atop {x_{2}=2}} \right.$

2)$\left \{ {{x^2+y=3} \atop {y-x+3=0}} \right. \left \{ {{x^2+y=3} \atop {x=3+y}} \right.\left \{ {{(3+y)^2+y=3} \atop {x=3+y}} \right.\left \{ {{9+6y+y^2+y=3} \atop {x=3+y}} \right.\left \{ {{y^2+7y+6=0} \atop {x=3+y}} \right.\\\\ y^2+7y+6=0\\\\D=b^2-4ac=49-24=25=5^2\\\\y_{1}=\frac{-7+5}{2}=-1\\\\y_{2}=\frac{-7-5}{2}=-6\\\\\left \{ {{y_{1}=-1} \atop {x_{1}=2}} \right. \left \{ {{y_{2}=-6} \atop {x_{2}=-3}} \right.$