Balancing reactions algebraic method limitations
As per the OP's request, here is a solution. The equation is:

$$\ce{a XeF4 + b H2O  -&gt; c XeO3 + d Xe  + e O2  + f HF  \tag 1}$$

So atom balance yields:

$$4a=1f     \tag 2$$
$$2b=1f     \tag 3$$
$$1a=1c + 1d     \tag 4$$
$$1b = 3c + 2e     \tag 5$$

The equations relate the six unknown coefficients $a$ through $f$. But each coefficient is an integer greater than zero, so I proceed as follows:


Assume $a = 1$, which is the minimum non-zero value it might have. Then, from equation $(4)$, $1 = a = c + d$, which is impossible because either $c$ or $d$ would have to be zero.
Assume $a = 2$. Then $b = 4$ because equations $(2)$ and $(3)$ show that $b = 2a$. But then equation $(5)$ shows that $4 = b = 3c + 2e$. Then if $c = 1, e = 0.5$, which is impossible. Obviously, $c$ cannot equal $2$, because that would give a negative value for $e$.
Assume $a = 3$. Then $b = 6$, because $b = 2a$, as above. Then equation $(5)$ shows that $6 = b = 3c + 2e$. Then if $c = 1, e = 1.5$, which is impossible. If $c = 2$, then $e = 0$, which is impossible. As before, $c$ cannot be greater than $2$, because that would give a negative value for $e$.
Assume $a = 4$. Then $b = 8$ and $f = 2b = 16$. Then equation $(5)$ shows that $8 = b = 3c + 2e$. If $c = 1$, then $2e = 5$, so $e = 2.5$, which is impossible. If $c = 2$, then $2e = 2$, so $e = 1$. Then equation $(4)$ shows that $4 = 2 + d$, so $d = 2$. This is a solution. Lastly, $c$ cannot equal $3$, because that would give a negative value for $e$.


Almost final result: $a = 4, b = 8, c = 2, d = 2, e = 1,$ and $f = 16$. Hence

$$\ce{4 XeF4 + 8 H2O  -&gt; 2 XeO3 + 2 Xe  + 1 O2  + 16 HF  \tag 6}$$

For a = 5, the solution is a = 5, b = 10, c = 2, d = 3, e = 2 and f = 20. This is the same as Oscar Lanzi's example solution. For a = 6, the solution is a = 6, b = 12, c = 2, d = 4, e = 3 and f = 24.

For a = 7, a double solution occurs: a = 7, b = 14, c = 2, d = 5, e = 4 and f = 28 and also a = 7, b = 14, c = 4, d = 3, e = 1 and f = 28.

There are an unlimited number of valid balanced equations, as expected. So the a = 4 solution is simply the first solution in the infinite series and it is the minimum coefficient solution, since the total number of atoms is 11a.

Bonus: Mithoron provided a link to a problem Balancing disproportionation reactions answered by Ivan Neretin years ago. The equation that was to be balanced was:

$$\ce{a XeF2 + b H2O  -&gt; c Xe + d XeO3  + e O2  + f HF  \tag 7}$$

So atom balance yields:

$$1a=1c + 1d     \tag 8$$
$$2a=1f     \tag 9$$
$$2b=1f     \tag {10}$$
$$1b = 3d + 2e     \tag {11}$$

The equations relate the six unknown coefficients $a$ through $f$. But each coefficient is an integer greater than zero, so I first note that equations $(9)$ and $(10)$ show that $a = b$. Then equation $(11)$ shows that $b$ must at least equal $5$. Otherwise either $d = 0$ or $e = 0$, which is impossible.

Now I proceed by assuming $a = b = 5$. Then equation $(8)$ gives rise to the following four possibilities: $c = 1$ and $d = 4$; $c = 2$ and $d = 3$; $c = 3$ and $d = 2$; $c = 4$ and $d = 1$. If $c = 1$ and $d = 4$, then equation $(11)$ yields $5 = 12 + 2e$, which is impossible. Likewise for $c = 2$ and $d = 3$ and also for $c = 3$ and $d = 2$. But if $c = 4$ and $d = 1$, then equation $(11)$ yields $5 = 3 + 2e$, so $e = 1$. This is a solution: $a = 5, b = 5, c = 4, d = 1, e = 1,$ and $f = 10$. Hence the balanced equation is:

$$\ce{5 XeF2 + 5 H2O  -&gt; 4 Xe + 1 XeO3  + 1 O2  + 10 HF   }$$

However, as Ivan Neretin beautifully pointed out in his answer, there is also the simpler redox reaction given by 

$$\ce{2 XeF2 + 2 H2O  -&gt; 2 Xe   + 1 O2  + 4 HF   }$$

So any linear combination of these, with positive integer coefficients, is also a valid balanced equation. For example, just add the last two equations together together:

$$\ce{7 XeF2 + 7 H2O  -&gt; 6 Xe + 1 XeO3  + 2 O2  + 14 HF   }$$