Question

Решите систему х^2+4y=8 x+y=2 и 2. х^2-3y=-9 x+y=3

Answer 1

$\left \{ {{x^2+4(2-x)=8} \atop {y=2-x}} \right.$
$x^2+8-4x=8$
$x^2-4x=0$
$x(x-4)=0; x_1=0; x_2=4;y_1=2;y_2=-2$

$\left \{ {{x^2-3(3-x)=-9} \atop {y=(3-x)}} \right.$
$x^2-9+3x=-9$
$x^2+3x=0$
$x(x+3)=0;x_1=0;x_2=-3;y_1=3;y_2=6$