How do I calculate the degree of dissociation in equilibrium? In my textbook, for calculating the percentage dissociation of $\ce{HF}$ for the given equation:
$$\ce{HF + H2O &lt;=&gt; H3O+ + F-}$$

The solution is:

Initial Concentrations
$$[\ce{HF}] = 0.08~\mathrm{M}, \: \ce{[H3O+]} = 0, \:\ce{[F- ]}= 0$$
Equilibrium concentrations
$$[\ce{HF}] = 0.08~\mathrm{M} - x, \: \ce{[H3O+]} = x, \:\ce{[F- ]}= x$$

I am not able to undestand why $x$ is subtracted from 0.08 and not $cx$ [ where $x$ is the degree of dissociation] 

I tried solving the same problem taking $cx$ but not able to get the solution, can anybody explain the difference to me? I have tried asking a similar question earlier too but it is really hard for me to get my head around this concept.
Let us first define the terms needed here.   


Degree of dissociation (DOD)
Degree of dissociation is the fraction of a mole of the reactant that underwent dissociation. It is represented by $\alpha$.
$$\alpha = \frac{\text{amount of substance of the reactant dissociated}}{\text{amount of substance of the reactant present initially}}$$  
Number of moles dissociated
It is defined as the product of the initial concentration of the reactant and the degree of dissociation


Now suppose you have a reaction like this
$$\ce{A-&gt;B + C}$$

The initial state of A is always the concentration of A (should be given in the question) while initial moles of B and C are zero (if anything else is not specified).
The final state of A is always defined as $\text{(amount of substance initially present) - (amount of substance dissociated)}$ while for B and C it is just $\text{amount of substance of A dissociated)}$.

Writing our equation again,
\begin{array}{lcccc}
 &amp; \ce{A &amp;-&gt; &amp;B &amp;+ &amp;C}\\
\text{Initial amount of substance}&amp; a &amp;&amp; 0 &amp;&amp; 0\\
\text{Final amount of substance}&amp;   a - a\cdot(\mathrm{DOD}) &amp;&amp; a\cdot(\mathrm{DOD}) &amp;&amp; a\cdot(\mathrm{DOD})\\
\end{array}