Question

Решите систему уравнений: {(x+2)√y-3=0 {x+y=4

Answer 1

$\left \{ {{(x+2) \sqrt{y-3}=0 } \atop {x+y=4}} \right. ==\ \textgreater \ x=4-y \\ \\ (4-y+2) \sqrt{y-3} =0 \\ (6-y) \sqrt{y-3}=0 \\ \\ 6-y=0 \\ y_1=6 \\ \\ \sqrt{y-3}=0 \\ y-3=0 \\ y_2=3 \\ \\ x+y=4 \\ x+6=4 \\ x_1=-2 \\ \\ x+3=4 \\ x_2=1$