Question

решите системные уравнения:
1.
x^2-2xy=7
x-3y=-2

2.
x^2+xy-y^2=1
x-2y=1

Answer 1

Ответ:

Объяснение:

$\left \{ {{x^{2}-2xy=7 } \atop {x-3y=-2}} \right. \\\left \{ {{(3y-2)^{2}-2y(3y-2)=7 } \atop {x=3y-2}} \right.\\\left \{ {{9y{2}-12y+4-6y^{2}+4y-7=0 \atop {x=3y-2}} \right. \\\left \{ {{3y^{2}-8y-3=0 } \atop {x=3y-2}} \right.$

Решим первое уравнение через дискриминант

$3y^{2} -8y-3=0\\D=(-8)^{2} -4*3*(-3)=64+36=100\\y_{1} =\frac{-(-8)+\sqrt{100} }{2*3}=\frac{8+10}{6} =\frac{18}{6} =3\\y_{2} =\frac{-(-8)-\sqrt{100} }{2*3}=\frac{8-10}{6} =-\frac{2}{6} =-\frac{1}{3}$

Получаем две системы уравнений

$\left \{ {{y_{1} =3} \atop {x_{1} =3*3-2}} \right. \\ \left \{ {{y_{1} =3} \atop {x_{1} =7}} \right. \\ \left \{ {{y_{2} =-\frac{1}{3} } \atop {x_{2} =3*(-\frac{1}{3}) -2}} \right. \\ \left \{ {{y_{2} =-\frac{1}{3} } \atop {x_{2} =-3}} \right.$

Ответ: х1=7, у1=3; х2=-3, у2=-(1/3)

$\left \{ {{x^{2}+xy-y^{2} =1 } \atop {x-2y=1}} \right. \\\\\\\left \{ {{(1+2y)^{2}+y(1+2y)-y^{2} =1 } \atop {x=1+2y}} \right.\\\\\left \{ {{1+4y+4y^{2}+y+2y^{2}-y^{2}-1=0 \atop {x=1+2y}} \right. \\ \\\left \{ {{5y^{2}+5y=0 } \atop {x=1+2y}} \right.\\ \\\left \{ {{y(5y+5)=0 } \atop {x=1+2y}} \right.$

Произведение равно нулю когда один из множителей равно нулю

y(5y+5)0

$y_{1} =0\\5y_{2} +5=0\\5y_{2} =-5\\y_{2} =-1$

Получаем две системы уравнений

$\left \{ {{y_{1} =0} \atop {x_{1} =1+2*0}} \right. \\ \left \{ {{y_{1} =0} \atop {x_{1} =1}} \right. \\ \left \{ {{y_{2} =-1} } \atop {x_{2} =1+2*(-1)}} \right. \\ \left \{ {{y_{2} =-1 } \atop {x_{2} =-1}} \right.$

Ответ: х1=1, у1=0; х2=-1, у2=-1