How to find fixed point I have recurrence equations of
\begin{align*}
x_{n+1} &:= 0.5\cdot (x_{n}^2+2\cdot y_{n}-2\cdot x_{n}\cdot y_{n}-2\cdot y_{n}^2) \\
y_{n+1} &:= -0.8\cdot (x_{n}^2+2\cdot y_{n}-2\cdot x_{n}\cdot y_{n}-2\cdot y_{n}^2)
\end{align*}
How to find $x$ and $y$?
 A: Here is a start. Multiply the first eq. by $.8$ and the second by $.5$ and add them to get the nice relation

$$ y_{n+1} =  - \frac{8}{5} x_{n+1}. $$

Substitute in the first eq. for $y_n$ and solve for $x_n$. I think you can take it from here. 
Added After substituting for $y_n$ in the first equation, assume $\lim_{n\to \infty}x_n =x$ which implies $\lim_{n\to \infty}x_{n+1}=x$ and solve the quadratic equation in $x$. To find $y$ use the above relation.
A: Hint 
As Will Jagy wrote, since you want a fixed point, drop the subscripts and solve. But, first notice that, ratioing the two expressions gives $y=-\frac{8 x}{5}$. So, replace $y$ in the first equation. After simplification, the equation becomes a very simple quadratic equation in $x$ that you will not have any problem to solve. Do not forget that you were already told that $(0,0)$ is alreasy a fixed point; so $x$ should probably be factored out and let you with a simple linear equation in $x$. When you have $x$, compute $y$ according to what we wrote at the beginning.
