How come the same reaction can be balanced in two distinct ways? I faced this reaction which can be balanced in $2$ distinct ways which are not multiples of each other.

$$\begin{align}\ce{6XeF4 + 12H2O &amp;-&gt; 2XeO3 + 24HF + 4Xe + 3O2}\tag{1}\\[0.4em]
\ce{4XeF4 + 8H2O &amp;-&gt; 2XeO3 + 16HF + 2Xe + O2}\tag{2}\end{align}$$

The first reaction is as it is given in the books and the second reaction was balanced by me.

I sense that only a mechanism can resolve this. I found a similar question on googling, but the answer is unsatisfactory. (It's Yahoo Answers. What can you expect?)

I also found two independent reactions here which on adding produce equation $(2)$. I am confused as to what exactly happens.
As per Nicolau's suggestion, : 

$$\{(1-2)\times\color\red{4}\}-2$$

gives 

$$2\ce{XeO3}\to2\ce{Xe}+\ce{3O2}$$

Thus, I conjecture that the second and third reaction differ by the fact of decomposition of $\ce{XeO3}$

$2$ can be converted into $1$ by multiplying by $\color \red{\frac 3 2}$ and decomposing one of the $\ce{XeO3} $ molecules.

Wikipedia also says that $\ce{XeO3}$ is unstable.