Yes, the "steady" in "steady state" means that it does not change. "Steady State" solutions for n and c are constant solutions so the derivatives are 0. The equations become:
$\frac{n}{n+1}- \beta nc=0$ and
$\alpha- \mu c= 0$.
Since the second equation involves only c, no n, I would solve it first:
$c= \frac{\alpha}{\mu}$.
Now put that into the first equation:
$\frac{n}{n+1}- \frac{\alpha \beta}{\mu}n= 0$
$\frac{n}{n+1}= \frac{\alpha \beta}{\mu}n$
$n= \frac{\alpha\beta}{\mu}n(n+ 1)$
Obviously n= 0 is a solution to that so one steady state solution is $c= \frac{\alpha}{\mu}$, $n= 0$.
If n is not 0 we can divide by it to get $1= \frac{\alpha\beta}{\mu}(n+ 1)$.
Then $n+ 1= \frac{\mu}{\alpha\beta}$,
$n= \frac{\mu}{\alpha\beta}- 1= \frac{\mu- \alpha\beta}{\alpha\beta}$.
So if n is not 0 then $c= \frac{\alpha}{\mu}$, $n= \frac{\mu}{\alpha\beta}- 1$ is a steady state solution.
But notice the "$\frac{\mu}{\alpha\beta}- 1$". If $\frac{\mu}{\alpha\beta}= 1$ then we are back to the "n= 0" solution, above. If $\frac{\mu}{\alpha\beta}> 1$ the steady state solution for n is positive, if $\frac{\mu}{\alpha\beta}< 1$ the steady state solution for n is negative.
Was this the entire question or was there some application in which n being positive or negative would be important?
