Question

решите систему уравнений :
{y^2+x=2
{x^2+2y^2=3

Answer 1

$\left \{ {{y^2+x=2} \atop {x^2+2y^2=3}} \right. \\y^2=2-x \\x^2+2(2-x)=3 \\x^2+4-2x-3=0 \\x^2-2x+1=0 \\(x-1)^2=0 \\x-1=0 \\x=1 \\y^2=2-1=1 \\y_1=1 \\y_2=-1$
Ответ: (1;1), (1;-1)

Answer 2

$y^2+x=2 \\ x^2 +2y^2=3 \\ y^2=\frac{3-x^2}{2} \\ \frac{3-x^2}{2}+x=2 \\ 3-x^2+2x=4 \\ x^2-2x+1=0 \\ D=4-4=0 \\ x_{1,2}=\frac{2}{2}=1 \\ y^2=\frac{3-1}{2}=1 \\ y_1=1 \\ y_2=-1$