Question

ПОМОГИТЕ! СРОЧНО! РЕШИТЬ СИСТЕМУ УРАВНЕНИЙ МЕТОДОМ ПОДСТАНОВКИ!

Answer 1

$\begin{cases}xy-2y-4x=-5\\y-3x=-2\end{cases}= > \ \ \begin{cases}y=3x-2\\x(3x-2)-2(3x-2)-4x=-5\end{cases}= >$

$= > \ \ 3x^2-2x-6x+4-4x+5=0\\{} \ \ \quad\ \ \, 3x^2-12x+9=0\qquad|:3$

$\begin{array}{lcl}x^2-4x+3=0\\D=16-12=4\end{array}\qquad\begin{array}{lcl}x_{1,2}=\dfrac{4\pm2}2=1;\ 3\end{array}$

$y_1=3\cdot1-2=1\\y_2=3\cdot3-2=7$

Ответ:  (1; 1) и (3; 7)

$\begin{cases}y=x-1\\x^2+y^2-4x-2y=-1\end{cases}= > \ \ \begin{cases}y=x-1\\x^2+(x-1)^2-4x-2(x-1)+1=0\end{cases}= >$

$= > \ \ x^2+x^2-2x+1-4x-2x+2+1=0$

$2x^2-8x+4=0\qquad|:2\\\\\begin{array}{lcl}x^2-4x+2=0\\D=16-8=8\end{array}\qquad\begin{array}{lcl}x_{1,2}=\dfrac{4\pm2\sqrt2}2=2\pm\sqrt2\end{array}$

$y_1=2-\sqrt2-1=1-\sqrt2\\y_2=2+\sqrt2-1=1+\sqrt2$

Ответ:  $(2-\sqrt2;\ 1-\sqrt2)\ u \ (2+\sqrt2;\ 1+\sqrt2)$

Answer 2

$\displaystyle 1)\left \{ {{xy-2y-4x=-5} \atop {y-3x=-2}} \right. \\\\\left \{ {{xy-2y-4x=-5} \atop {y=-2+3x}} \right. \\\\x(-2+3x)-2(-2+3x)-4x=-5\\x=1,x=3\\y=-2+3*1,y=-2+3*3\\y=1,y=7\\(x_1,y_1)=(1,1),(x_2,y_2)=(3,7)\\\\2)\left \{ {{y=x-1} \atop {x^2+y^2-4x-2y=-1}} \right.\\ \\x^2+(x-1)^2-4x-2(x-1)=-1\\x=2+\sqrt{2},x=2-\sqrt{2}\\ y=2+\sqrt{2}-1,y=2-\sqrt{2}-1\\ y=1+\sqrt{2},y=1-\sqrt{2}\\ (x_1,y_1)=(2+\sqrt{2},1+\sqrt{2}),(x_2,y_2)=(2-\sqrt{2},1-\sqrt{2})$