Question

Решите 2 задание 1 вариант

Answer 1

$\left\{\begin{array}{l}2x-y=1\ ,\\x^2+2xy+y^2=4\end{array}\right\ \ \left\{\begin{array}{l}y=2x-1\ ,\\(x+y)^2=4\end{array}\right\ \ \left\{\begin{array}{l}y=2x-1\ ,\\(x+2x-1)^2=4\end{array}\right\ \ \\\\\\\left\{\begin{array}{l}y=2x-1\ ,\\(3x-1)^2=4\end{array}\right\ \ \left\{\begin{array}{l}y=2x-1\ ,\\3x-1=\pm 2\end{array}\right$

$a)\ \ \left\{\begin{array}{l}y=2x-1\ ,\\3x-1=-2\end{array}\right\ \ \left\{\begin{array}{l}y=2x-1\ ,\\3x=-1\end{array}\right\ \ \left\{\begin{array}{l}y=-\dfrac{2}{3}-1\ ,\\x=-\dfrac{1}{3}\end{array}\right\ \ \left\{\begin{array}{l}y=-1\dfrac{2}{3}\ ,\\x=-\dfrac{1}{3}\end{array}\right$

$b)\ \ \left\{\begin{array}{l}y=2x-1\ ,\\3x-1=2\end{array}\right\ \ \left\{\begin{array}{l}y=2x-1\ ,\\3x=3\end{array}\right\ \ \left\{\begin{array}{l}y=1\ ,\\x=1\end{array}\right\\\\\\Otvet:\ \ \Big(-\dfrac{1}{3}\ ;-1\dfrac{2}{3}\ \Big)\ ,\ (\ 1\ ;\ 1\ )\ .$