Question

решите систему 5у+х^2=4; х+5у+2=0

Answer 1

$\left \{ {{5y+x^{2} =4} \atop {x+5y+2=0}} \right.\\\\-\left \{ {{x^{2}+5y=4 } \atop {x+5y=-2}} \right.\\ -------\\x^{2}-x=6\\x^{2}-x-6=0\\x_{1} =3\\x_{2}=-2\\5y=-x-2\\y=-\frac{x+2}{5} \\y_{1}=-\frac{3+2}{5}=-1\\y_{2}=-\frac{-2+2}{5}=0$

Ответ ; (3 ; - 1) , ( - 2 ; 0)