Question

Решите систему уравнений :
{у=х+2
{х^2+ 2у=3

Answer 1

$left { {{y=x+2,} atop {x^2+ 2y=3,}} right. left { {{y=x+2,} atop {x^2+ 2(x+2)=3,}} right. \ x^2+2x+1=0, \ (x+1)^2=0, \ x=-1, \ left { {{x=-1,} atop {y=1.}} right.$