Question

Решите систему:

3х*2+у=4
2х*2-у=1

Answer 1

$left { {{3x^2+y=4} atop {2x^2-y=1}} right. left { {{3x^2+y=4} atop {y=-1+2x^2}} right. \ 3x^2-1+2x^2=4 \ x=1 \ x=-1 \ y=-1+2*1^2 = textgreater y=1 \ y=-1+2*(-1)^2 = textgreater y=1 \ (x_{1},y_{1})=(1,1) \ (x_{2},y_{2})=(-1,1)$