Constructing an equation for the precipitation of a salt
Precipitation is usually a single or double displacement reaction, and refers to the formation of an insoluble, solid salt from soluble ions. For instance, silver bromide, $\ce{AgBr}$, is insoluble in water. However, it is made up of $\ce{Ag+}$ and $\ce{Br-}$ ions, which we can introduce in the form of soluble salts:

$$\ce{LiBr(aq) + AgNO3(aq) -&gt; LiNO3(aq) + AgBr(s)}$$

In an aqueous solution, lithium bromide and silver nitrate underwent a reaction, forming lithium nitrate and silver bromide. In the products, silver bromide is denoted with the state symbol (s), which stands for "solid". This reflects its insolubility in water and hence its tendency to "precipitate".

The net ionic equation is when we only write the ions that participate in the reaction. The ions that do not participate are called spectator ions. For example, in the reaction above, only $\ce{Ag+}$ and $\ce{Br-}$ participated in precipitation. Thus, the net ionic equation for the reaction above is:

$$\ce{Ag+(aq) + Br-(aq) -&gt; AgBr(s)}$$

One way of deriving this net ionic equation is to realise that salts such as $\ce{LiBr}$ and $\ce{AgNO3}$ do not exist as molecules in water, but rather dissociate to give their constituent ions:

$$\begin{align}
\ce{LiBr(s) &amp;-&gt;[dissolution] Li+(aq) + Br-(aq)} \\
\ce{AgNO3(s) &amp;-&gt;[dissolution] Ag+(aq) + NO3-(aq)}
\end{align}$$

However, $\ce{AgBr}$ is insoluble and does not dissociate into ions. If we go back to the first equation and replace all the soluble salts with their constituent ions, we get

$$\ce{\underbrace{Li+(aq) + Br-(aq)}_{from LiBr} + \underbrace{Ag+(aq) + NO3-(aq)}_{from AgNO3} -&gt; \underbrace{Li+(aq) + NO3-(aq)}_{from LiNO3} + \underbrace{AgBr(s)}_{does not dissociate}}$$

The spectator ions which do not take part in the reaction are $\ce{Li+}$ and $\ce{NO3-}$, as they are unchanged in this reaction. Subtracting them from both the left- and right-hand sides, we obtain the net ionic equation:

$$\ce{Ag+(aq) + Br-(aq) -&gt; AgBr(s)}$$

If we take this idea and apply it back to $\ce{CaCO3}$, we first need to come up with soluble sources of the ions $\ce{Ca^2+}$ and $\ce{CO3^2-}$. A possible choice is $\ce{CaCl2}$ and $\ce{Na2CO3}$, although that is hardly the only option.

The overall equation may be written as

$$\ce{CaCl2(aq) + Na2CO3(aq) -&gt; 2NaCl(aq) + CaCO3(s)}$$

and the net ionic equation as

$$\ce{Ca^2+(aq) + CO3^2-(aq) -&gt; CaCO3 (s)}$$

where $\ce{Na+}$ and $\ce{Cl-}$ are the spectator ions.

Finally, here are some web pages you might find helpful:

http://www.occc.edu/KMBailey/Chem1115Tutorials/Net_Ionic_Eqns.htm

http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Reactions_in_Aqueous_Solutions/Precipitation_Reactions