Positive equilibria for a system of eqautions

Question

I have the following system of equations

\begin{align} \frac{dx}{d \tau} &= x \left(1-x-\frac{y}{x+b} \right) \\ \frac{dy}{d \tau} &= cy \left(-1+a\frac{x}{x+b} \right) \end{align}

I am asked to show that if $a<1$, the only nonnegative equilibria are $(0,0), (1,0)$.

So first it is obvious that in order to the equations become $0$ is $(x,y)=(0,0)$

Then I decided $y=0$ and $1-x-\displaystyle \frac{y}{x+b}=0 \Leftrightarrow x=1$ , hence $(x,y)=(1,0)$

In the same way I decided $x=0$ and $-1+a\displaystyle \frac{x}{x+b}=0$, but there is no $y$.

Finally I decided $-1+a\displaystyle \frac{x}{x+b}=0$ and $1-x-\displaystyle \frac{y}{x+b}=0$, and this is difficult.

I can't figure out what to do now. What about the fact that $a<1$?

Can anyone help?

Answer 1

Probably a comment, rather than an answer. But you have neglected that you have $$ \mathbf{y}\left(-1+a\frac{x}{x+b}\right) $$ equated at $x=0$ we have $$ y(-1+0)=-y \implies y = 0 $$

Answer 2

If $x \ne 0$, We have

$$ 1 - a\frac{x}{x+b} = 0 $$ $$ a\frac{x}{x+b} = 1 $$ $$ ax = x + b$$ $$ (a-1)x = b$$ $$ x = \frac{b}{a-1} $$

If $a < 1$ and $b > 0$, $x < 0$ and therefore is not a non-negative solution