Question

Решить систему уравнений y^2-2xy=32 и x^2+6xy+9y^2=100.
Жду)

Answer 1

$left { {{y^2-2xy=32} atop {x^2+6xy+9y^2=100}} right. \\ left { {{y^2-2xy=32} atop {(x+3y)^2-10^2=0}} right. \\ left { {{y^2-2xy=32} atop {x^2+6xy+9y^2=100}} right. \\ left { {{y^2-2xy=32} atop {[x+3y+10]*[x+3y-10]=0}} right. \\ left { {{y^2-2xy=32} atop {x+3y+10=0}} right. or left { {{y^2-2xy=32} atop {x+3y-10=0}} right. \\ left { {{y^2-2y(-3y-10)-32=0} atop {x=-3y-10}} right. or left { {{y^2-2y(-3y+10)-32=0} atop {x=-3y+10}} right.$

$left { {{y^2+6y^2+20y-32=0} atop {x=-3y-10}} right. or left { {{y^2+6y^2-20y-32=0} atop {x=-3y+10}} right. \\ left { {{7y^2+20y-32=0} atop {x=-3y-10}} right. or left { {{7y^2-20y-32=0} atop {x=-3y+10}} right. \\ D_1=D_2=20^2-4*7*(-32)=400+896=1296=36^2\\ left { {y=frac{-20pm36}{2*7}=frac{-10pm18}{7}} atop {x=-3y-10}} right. or left { {{y=frac{20pm36}{2*7}=frac{10pm18}{7}} atop {x=-3y+10}} right.$

$left { {y=frac{-20pm36}{2*7}=frac{-10pm18}{7}} atop {x=-3y-10}} right. or left { {{y=frac{20pm36}{2*7}=frac{10pm18}{7}} atop {x=-3y+10}} right. \\ left { {{y=-4 or y=frac{8}{7}} atop {x=-3y-10}} right. or left { {{y=4 or y=-frac{8}{7}} atop {x=-3y+10}} right. \\ (2; -4) or (-frac{94}{7}; frac{8}{7}) or (-2; 4) or (frac{94}{7}; -frac{8}{7})\\ (pm2; mp4) or (pmfrac{94}{7}; mpfrac{8}{7})$