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

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

Ответ:

student130

08.10.2020 13:12

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

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

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