Способ ( метод) подстановки
[m]\left\{\begin {matrix}x=-5-y\\(-5-y)y=4\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}x=-5-y\\-5y-y^2=4\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}x=-5-y\\y^2+5y+4=0\end {matrix}\right.[/m]
D=25-4*4=9
[m]\left\{\begin {matrix}x=-5-y\\y_{1}=\frac{-5-3}{2}\end {matrix}\right.[/m] или[m]\left\{\begin {matrix}x=-5-y\\y_{2}=\frac{-5+3}{2}\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}x=-5-(-4)\\y_{1}=-4\end {matrix}\right.[/m] или[m]\left\{\begin {matrix}x=-5-(-1)\\y_{2}=-1\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}x=-1\\y_{1}=-4\end {matrix}\right.[/m] или[m]\left\{\begin {matrix}x=-4\\y_{2}=-1\end {matrix}\right.[/m]
О т в е т.[b] (-1;-4); (-4:-1)[/b]
б)
[m]\left\{\begin {matrix}x+y=6\\(x-y)(x+y)=12\end {matrix}\right.[/m]
Способ ( метод) подстановки
[m]\left\{\begin {matrix}x+y=6\\(x-y)\cdot 6=12\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}x+y=6\\x-y=2\end {matrix}\right.[/m]
Способ ( метод) сложения
Заменяем одно из уравнений системы суммой двух уравнений
[m]\left\{\begin {matrix}x+y=6\\2x=8\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}4+y=6\\x=4\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}y=2\\x=4\end {matrix}\right.[/m]
О т в е т.[b] (4:2)[/b]
2.
Применим формулу сокращенного умножения к первому уравнению:
[m]\left\{\begin {matrix}(x+y)^2=1\\x^2-xy=6\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}(x+y)=-1\\x^2-xy=6\end {matrix}\right.[/m] или [m]\left\{\begin {matrix}(x+y)=1\\x^2-xy=6\end {matrix}\right.[/m]
Способ ( метод) подстановки
[m]\left\{\begin {matrix}y=-1-x\\x^2-x\cdot (-1-x)=6\end {matrix}\right.[/m] или [m]\left\{\begin {matrix}y=1-x\\x^2-x\cdot (1-x)=6\end {matrix}\right.[/m]
Решаем первую систему:
[m]\left\{\begin {matrix}y=-1-x\\2x^2+x-6=0\end {matrix}\right.[/m]
D=1-4*2*(-6)=49
[m]\left\{\begin {matrix}y=-1-x\\x_{1}=\frac{-1-7}{2}\end {matrix}\right.[/m] или [m]\left\{\begin {matrix}y=-1-x\\x_{1}=\frac{-1+7}{2}\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}y_{1}=-1-(-4)\\x_{1}=-4\end {matrix}\right.[/m] или [m]\left\{\begin {matrix}y_{2}=-1-3\\x_{2}=3\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}y_{1}=3\\x_{1}=-4\end {matrix}\right.[/m] или [m]\left\{\begin {matrix}y_{2}=-4\\x_{2}=3\end {matrix}\right.[/m]
Решаем вторую систему:
[m]\left\{\begin {matrix}y=1-x\\2x^2-x-6=0\end {matrix}\right.[/m]
D=(-1)^2-4*2*(-6)=49
[m]\left\{\begin {matrix}y=1-x\\x_{1}=\frac{1-7}{2}\end {matrix}\right.[/m] или [m]\left\{\begin {matrix}y=1-x\\x_{1}=\frac{1+7}{2}\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}y_{3}=1-(-3)\\x_{3}=-3\end {matrix}\right.[/m] или [m]\left\{\begin {matrix}y_{4}=1-4\\x_{4}=4\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}y_{3}=4\\x_{3}=-3\end {matrix}\right.[/m] или [m]\left\{\begin {matrix}y_{4}=-3\\x_{4}=4\end {matrix}\right.[/m]
О т в е т.[b] (-4;3); (3;-4)(-3;4); (4:-3)[/b]