Question

решите систему уравнений: 3х+у+4=0 х2-у2=2

Answer 1

$displaystyle left { {{3x+y+4=0} atop {x^2-y^2=2}} right.\\ left { {{y=-4-3x} atop {x^2-(-4-3x)^2=2}} right.\\ left { {{y=-4-3x} atop {x^2-16-24x-9x^2=2}} right.\\-8x^2-24x-18=0\\-2(4x^2+12x+9)=0\\D=144-144=0\\x=- frac{12}{8}=-1.5\\y=-4-3(-1.5)=0.5$