Question

решить систему уравнений 2x+y=7 x^2-y=1

Answer 1

$left { {{2x+y=7} atop {x^2-y=1}} right. Longleftrightarrow left { {{y=7-2x} atop {x^2-(7-2x)=1}} right. Longleftrightarrow left { {{y=7-2x} atop {x^2+2x-8=0}} right. \\ x^2+2x-8=0\ D=4+32=36; sqrt D=6\\ x_{1/2}= frac{-2pm6}{2}\\ x_1= frac{-2-6}{2}=- frac{8}{2}=-4\\ x_2= frac{-2+6}{2}= frac{4}{2}=2\\\ y_1=7-2cdot(-4)=7+8=15\ y_2=7-2cdot(2)=7-4=3$


Ответ: $(-4;15)(2;3)$