Question

рншите!!!!!!!!!!!!!!!!!

Answer 1

$\left \{ {{3x-y=-10} \atop {x^2+y=10}} \right. \\ y=10-x^2 \\ \\ 3x-(10-x^2)=-10 \\ 3x-10+x^2+10=0 \\ x^2+3x=0 \\ x(x+3)=0 \\ x_1=0 \\ x+3=0 \\ x_2=-3 \\ \\ y_1=10-0^2=10 \\ y_2=10-(-3)^2=10-9=1$