Producing methane, reactants and products
The answer is:

$$\ce{2CH3COONa(s) + 2NaOH(s) + Ca(OH)2(s) &lt;=&gt; 2CH4(g) + Na2CO3(s) + CaCO3(s) + H2O}$$ 

You have to get same number of equal atoms in each side of the reaction (this technique is called mass balance). If there were charges, they must be balanced also.

Example

There are $2\times 3 + 2\times 1 + 2\,$ hydrogen atoms in left side, and there are $2\times 4 + 2$ $\ce{H}$ in the right side. So both sides give $10$ hydrogen atoms.

I hope it helps.