{ x-y = 1
{ xy = 6
{ x^2-3y^2 = 1
{ x-2y = 1
[m]\left\{\begin {matrix}(1+2y)^2-3y^2=1\\x=1+2y\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}1+4y+4y^2-3y^2=1\\x=1+2y\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}4y+y^2=0\\x=1+2y\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}y(4+y)=0\\x=1+2y\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}y=0\\x=1+2\cdot 0\end {matrix}\right.[/m] или [m]\left\{\begin {matrix}4+y=0\\x=1+2y\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}y=0\\x=1+2\cdot 0\end {matrix}\right.[/m] или [m]\left\{\begin {matrix}y=-4\\x=1+2\cdot (-4)\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}y_{1}=0\\x_{1}=1\end {matrix}\right.[/m] или [m]\left\{\begin {matrix}y_{2}=-4\\x_{2}=-7\end {matrix}\right.[/m]
О т в е т. (1;0); (-7:-4)
1.
[m]\left\{\begin {matrix}x-y=1\\xy=6\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}x=1+y\\(1+y)y=6\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}x=1+y\\y^2+y-6=0\end {matrix}\right.[/m]
D=1-4*(-6)=25
[m]\left\{\begin {matrix}x_{1}=1+(-3)\\y_{1}=-3\end {matrix}\right.[/m] или [m]\left\{\begin {matrix}x_{1}=1+2\\y_{2}=2\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}x_{1}=-2\\y_{1}=-3\end {matrix}\right.[/m] или [m]\left\{\begin {matrix}x_{1}=3\\y_{2}=2\end {matrix}\right.[/m]
О т в е т. (-2;-3);(3;2)