Задача 69699 ...
Условие
[2x-y=1
[x² + 2xy+ y² =4.
Решение
[m]\left\{\begin {matrix}2x–y=1\\x^2+ 2xy+ y^2 =4.\end {matrix}\right.[/m] ⇒ [m]\left\{\begin {matrix}2x–y=1\\(x + y)^2 =2^2.\end {matrix}\right.[/m] ⇒
[m]\left\{\begin {matrix}2x–y=1\\x + y =2\end {matrix}\right.[/m] или [m]\left\{\begin {matrix}2x–y=1\\x + y =-2\end {matrix}\right.[/m]
Применяем способ сложения:
[m]\left\{\begin {matrix}y=2x–1\\3x =3\end {matrix}\right.[/m] или [m]\left\{\begin {matrix}y=2x–1\\3x =-1\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}y=2\cdot 1-1\\x =1\end {matrix}\right.[/m] или [m]\left\{\begin {matrix}y=2\cdot (-\frac{1}{3})-1\\x =-\frac{1}{3}\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}y=1\\x =1\end {matrix}\right.[/m] или [m]\left\{\begin {matrix}y=-\frac{5}{3}\\x =-\frac{1}{3}\end {matrix}\right.[/m]