Question

Розв"яжіть систему рівнянь
1) {3у-х=4
х(в квадрате)-4ху-2у(в квадрате)=3
2) {у-2х=-3
у(в квадрате)+3ху-3х(в квадрате)=-5

Answer 1

$1)\ \ \left\{\begin{array}{l}3y-x=4\\x^2-4xy-2y^2=3\end{array}\right\ \ \left\{\begin{array}{l}x=3y-4\\(3y-4)^2-4(3y-4)\cdot y-2y^2=3\end{array}\right\\\\\\\left\{\begin{array}{l}x=3y-4\\9y^2-24y+16-12y^2+16y-2y^2-3=0\end{array}\right\ \ \left\{\begin{array}{l}x=3y-4\\5y^2+8y-13=0\end{array}\right$

$\left\{\begin{array}{l}x=3y-4\\y_1=-\dfrac{13}{5}\ ,\ y_2=1\end{array}\right\ \ \left\{\begin{array}{l}\ x_1=-\dfrac{59}{5}\ ,\ x_2=-1\\y_1=-\dfrac{13}{5}\ ,\ y_2=1\end{array}\right$

$Otvet:\ \ \Big(-\dfrac{59}{5}\ ;\ -\dfrac{13}{5}\Big)\ ,\ \Big(-1\ ;\ 1\, \Big)\ .$

$2)\ \ \left\{\begin{array}{l}y-2x=-3\\y^2+3xy-3x^2=-5\end{array}\right\ \ \left\{\begin{array}{l}y=2x-3\\(2x-3)^2+3x(2x-3)-3x^2+5=0\end{array}\right\\\\\\\left\{\begin{array}{l}y=2x-3\\4x^2-12x+9+6x^2-9x-3x^2+5=0\end{array}\right\ \ \left\{\begin{array}{l}y=2x-3\\7x^2-21x+14=0\end{array}\right$

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