Analyzing the stability of equilibria There's a model with a condition $r>\mu$:
$$\begin{align}
S'&=r(S+I)-\beta SI-\mu S \\
I'&=\beta SI-(\mu +\alpha)I
\end{align}$$
I can easily see that the equilibria of the second equation are either $I=0$ or $\beta S=\mu +\alpha$. If $I=0$ the equilibria for the first equation is $rS=\mu S$. Since $r>\mu$ it must be the case that $S=0$. My textbook states, "the equilibrium  $(0,0)$ is unstable." I cannot see how though. 
If the equilibrium is unstable the system moves away from the equilibrium. If stable the system moves towards the equilibrium. Since $(0,0)$ is an equilibrium for the system how can we gain any information about its stability since it makes the system equal zero?
 A: Starting with the given system of equations
$S'= r(S+I)-\beta SI-\mu S, \tag{1}$
$I'=\beta SI-(\mu +\alpha)I = I(\beta S - (\mu + \alpha)), \tag{2}$
we see that it indeed does have an equilibrium point at $(S, I) = (0, 0)$.  This means precisely that $S'(0, 0) = I'(0, 0) = 0$, as our OP SOULed_Outt has shown and as may readily be affirmed by direct substitution of $(S, I) = (0, 0)$ into (1)-(2).  Taking $S' = (dS/dt)$, $I' = (dI/dt)$, we see by unicity of solutions that the fact that $S'(0, 0) = I'(0, 0) = 0$ means that, if the system is initialized at $(0, 0)$, then the entire trajectory is given by $S(t) = I(t) = 0$ for all $t$.
The question of the stability of $(0, 0)$, however, addresses the issue of how trajectories initialized near, but not at, $(0, 0)$ behave as $t \to \infty$.  This question may often, but not always, be resolved by examining the linearization of  a given system at the equilibrium in question.  In the present case, we may compute such linearization by calculating the Jacobean matrix of (1)-(2) at $(0, 0)$.  We have, at any point $(S, I)$,
$J(S, I) = \begin{bmatrix} \dfrac{\partial S'}{\partial S} & \dfrac{\partial S'}{\partial I} \\ \dfrac{\partial I'}{\partial S} & \dfrac{\partial I'}{\partial I} \end{bmatrix}; \tag{3}$
we compute the partial derivatives by using (1), (2), finding
$\dfrac{\partial S'}{\partial S} = r - \mu -\beta I, \tag{4}$
$\dfrac{\partial S'}{\partial I} = r - \beta S, \tag{5}$
$\dfrac{\partial I'}{\partial S} = \beta I, \tag{6}$
$\dfrac{\partial I'}{\partial I} = \beta S - \mu - \alpha. \tag{7}$
When these four expressions are evaluated at $(S, I) = (0, 0)$ we find that
$J(0, 0) = \begin{bmatrix} r - \mu & r \\ 0 & -\mu - \alpha \end{bmatrix}. \tag{8}$
Now suppose we examine the eigenvalues of $J(0, 0)$; they satisfy the characteristic polynomial $p_J(\lambda)$, where
$p_J(\lambda) = \det (\begin{bmatrix} r - \mu - \lambda & r \\ 0 & -\mu - \alpha - \lambda \end{bmatrix}) = (r - \mu - \lambda)(-\mu -\alpha - \lambda). \tag{9}$
It is easily read off from (9) that the zeroes of $p_J(\lambda)$ occur at $\lambda_1 = r - \mu$, $\lambda_2 = -\mu - \alpha$.  Since $r > \mu$ by hypothesis, we see that $\lambda_1 > 0$; $\lambda_2$ may be positive, negative or zero depending on the relative values of $\mu$ and $\alpha$.  In the event that $\lambda_2 \ne 0$, then the matrix $J(0, 0)$ is known as hyperbolic, i.e. the eigevalues all have non-zero real parts.  In such a case we may invoke a major result, the so-called stable manifold theorem, which affirms that, in some neighborhood of $(0, 0)$, the solution set of (1), (2) is in fact well-described by the solution set of the system linearized about the equilibrium point $(0, 0)$; a more detailed discussion of this theorem, including linked references, may be found in my answers to this question.  The linearized system is in fact
$\begin{pmatrix} S \\ I \end{pmatrix}' = J(0, 0)\begin{pmatrix} S \\ I \end{pmatrix} = \begin{bmatrix} r - \mu & r \\ 0 & -\mu - \alpha \end{bmatrix}\begin{pmatrix} S \\ I \end{pmatrix}; \tag{10}$
since $\lambda_1 > 0$, we see (intuitively; the stable manifold theorem lends rigor to this intuition) that any trajectory initialized sufficiently close to the line through $(0, 0)$ tangent to an eigenvector corresponding to $\lambda_1$, which is in fact easily seen to be of the form $(k, 0)^T$ for $0 \ne k \in \Bbb R$, will in fact move away from $(0, 0)$ with increasing $t$; thus the equilibrium must be unstable.  When $\lambda_2 = 0$, we encounter a singular case which must be handled by more subtle methods; nevertheless the "eigensolution" to the linearized system $(S(t), I(t)) = e^{\lambda_1 t}(k, 0)$, which corresponds to the initial point $(k, 0)$ at $t = 0$, will still move away from $(0, 0)$ with increasing $t$, at least locally near $(0, 0)$, so the system apparently cannot be stable in this situation.  
The geometrcal interpretation of this analysis is that, for $\lambda_2 >0$ , any point near $(0, 0)$ will be moved away from it by the flow; when $\lambda_2 < 0$, then $(0, 0)$ is a saddle point, and all the integral curves save two resemble hyperbolas, which may flow towards $(0, 0)$ for awhile but will eventually turn and head away.  The remaining two trajectories will approach $(0, 0)$ as $t \to \infty$, but from opposing directions; though do in fact "stabilize" to $(0, 0)$ as $t \to \infty$, they form a special case; the presence of the eigenvalue $\lambda_1 > 0$ implies that $(0, 0)$ is an unstable equilibrium point.
The information regarding the stability of $(0, 0)$ is encoded in the matrix $J(0, 0)$, via its eigenstructure; similar observations apply to any other critical points this system may possess, and indeed to the equilibria of general, nonlinear, autonomous ODEs.  
Hope this helps.  Cheerio,
and as always, 
Fiat Lux!!!
