Find the fixed points of the two systems What are the fixed points of the two non linear systems below?
\begin{align}
x(t)& = x(3-x-2y)\\
y(t)& = y(2-x-y)
\end{align}
I know that $(0,0)$, $(0,2)$, $(3,0)$, $(1,1)$ are the fixed points. However, I don't understand some of the steps to obtain this. Any help will be appreciated.
 A: Unless I missed something in the question, the fixed points are given by solutions of $x = x(3-x-2y)$, $y=y(2-x-y)$.
If $x=0$, the first equation is trivially satisfied, and the second gives $y = y(2-y)$, or in other words $y(y-1) = 0$. Hence $(0,0)$ and $(0,1)$ are solutions.
If $y=0$, the second equation is trivially satisfied, and the first gives $x=x(3-x)$ or in other words $x(2-x) = 0$. Hence $(0,0)$ and $(2,0)$ are solutions.
Now suppose $x\neq 0$ and $y\neq 0$. Divide the equations by $x,y$ respectively to get $1 = 3-x-2y$, $1 = 2-x-y$, or equivalently $x+2y-2=0$ and $1-x-y=0$. Adding the first equation to two times the second gives $x=0$ which cannot be, hence there are no solutions with $x\neq 0$ and $y\neq 0$.
Hence the only solutions are $(0,0), (0,1),(2,0)$.
Addendum: Following Hans comments below, I will look for stationary points of the original system. This means solving $0 = x(3-x-2y)$, $0=y(2-x-y)$ instead.
The same approach works: If $x=0$, then the first equation is trivially satisfied, and the second becomes $0=y(2-y)$. The solutions are $(0,0), (0,2)$.
If $y=0$, then the second equation is trivially satisfied, and the first becomes $0 = x(3-x)$. The solutions are $(0,0), (3,0)$.
Now suppose $x\neq 0$ and $y\neq 0$. Divide the equations by $x,y$ respectively to get $3-x-2y=0$, $2-x-y = 0$. Subtracting the first equation from the second yields $y=1$ from which it follows that the solution is $(1,1)$.
Hence the solutions are $(0,0), (3,0), (0,2), (1,1)$.
