Предмет: Алгебра, автор: taeana38

Помогите решить (ДАЮ 35 БАЛЛОВ)

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Ответы

Автор ответа: NNNLLL54

$\left\{\begin{array}{l}x+2y=3\\x^2-3xy=7\end{array}\right\ \ \left\{\begin{array}{l}x=3-2y\\(3-2y)^2-3(3-2y)\, y=7\end{array}\right\ \ \left\{\begin{array}{l}x=3-2y\\9-12y+4y^2-9y+6y^2=7\end{array}\right$

$\left\{\begin{array}{l}x=3-2y\\10y^2-21y+2=0\end{array}\right\ \ \left\{\begin{array}{l}x=3-2y\\y_1=0,1\ ,\ y_2=2\end{array}\right\ \ \left\{\begin{array}{l}x_1=2,8\ ,\ x_2=-1\\y_1=0,1\ ,\ y_2=2\end{array}\right\\\\\\Otvet:\ \ (2,8\ ;\ 0,1)\ ,\ (-1\ ;\ 2)\ .$

$2)\ \ \left\{\begin{array}{l}\dfrac{x}{3}+y=1\\y^2-xy=7\end{array}\right\ \ \left\{\begin{array}{l}y=1-\dfrac{x}{3}\\\Big(1-\dfrac{x}{3}\Big)^2-x\cdot \Big(1-\dfrac{x}{3}\Big)-7=0\end{array}\right\\\\\\\left\{\begin{array}{l}y=1-\dfrac{x}{3}\\1-\dfrac{2x}{3}+\dfrac{x^2}{9}-x+\dfrac{x^2}{3}-7=0\end{array}\right\ \ \left\{\begin{array}{l}y=1-\dfrac{x}{3}\\\dfrac{4}{9}\, x^2-\dfrac{5}{3}\, x-6=0\end{array}\right$

$\left\{\begin{array}{l}y=1-\dfrac{x}{3}\\4x^2-15x-54=0\end{array}\right\ \ \left\{\begin{array}{l}y=1-\dfrac{x}{3}\\x_1=-2,25\ ,\ x_2=6\end{array}\right\ \ \left\{\begin{array}{l}y_1=1,75\ ,\ y_2=-1\\x_1=-2,25\ ,\ x_2=6\end{array}\right\\\\\\Otvet:\ \ (-2,25\ ;\ 1,75\, )\ ,\ (\ 6\ ;-1\, )\ .$