Lotka Volterra with alternate resource Say we have a variation of the Lotka Volterra system:
$\frac{dx}{dt} = ax-bxy-cx^2$
$\frac{dy}{dt} = ny+mxy-py^2$
where $a,b,c,m,n,p$ are positive constants and $x_0$ and $y_0$ represent the initial population. The values of $a,b,c,m,n,p,x_0,y_0$ don't matter as we can experiment with different ones.
I'm having trouble finding the equilibrium points and their stability. Any help on this is appreciated.
 A: To find the critical points, we want $x' = 0, y' = 0$, so have:
$$ \dfrac{dx}{dt} = ax-bxy-cx^2 = 0 \\ \dfrac{dy}{dt} = ny+mxy-py^2 = 0 $$
We can write this as:
$$x(a-cx -by) = 0 \\ y(n + mx -py) = 0 $$
It is clear that we satisfy these equations with $x = 0, y = 0$.
If we choose $x = 0$ in the first equation, the second gives us:
$$ y = \dfrac{n}{p}$$
If we choose $y = 0$ in the second equation, the first gives us:
$$ x = \dfrac{a}{c}$$
Lastly, we are left analyzing the cases:
$$a-cx -by = 0 \\ n + mx -py = 0 $$
Using your favorite method for a $2x2$ (RREF), this simultaneously gives:
$$ x = \dfrac{a p-b n}{b m+c p}, y = \dfrac{a m+c n}{b m+c p}$$
We are left with four critical points to investigate as:
$$(x, y) = (0, 0), \left(0 , \dfrac{n}{p}\right), \left(\dfrac{a}{c} , 0 \right), \left(\dfrac{a p-b n}{b m+c p}, \dfrac{a m+c n}{b m+c p}\right)$$
It is obvious that some of the constants cannot be zero for obvious reasons, for example, $p \ne 0$ in the second critical point.
You can test these critical points in the original system to verify they simultaneously produce $x ' = 0, y' = 0$. 
Next, start analyzing these points for stability (your turn).
