Positive equilibria for a system of eqautions 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? 

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
