How can I balance this reaction using the algebraic method?
Now all you have to do is variable substitution, which is kinda tedious since it's such a large equation. Remember that solutions are defined up to multiplying everything by a constant factor, so we might as well pick 


$a = 2$ (this guess is inspired by the equation for $\ce{Fe}$)
$\ce{Fe}{:~} e = 1$
$\ce{C}{:~} h = 12$
$\ce{N}{:~} g = 12$
Because $\ce{Mn}{:~} b = f$, we can rewrite $\ce{O}$ as $4c = 4d + 12e + 3g + 2h + i$
Subtract 4 times $\ce{S}$, and we get $0 = -4f + 3g + 2h + i$, so $4f = 60 + i$ (call this $\ce{O}'$)
Now rewrite everything in terms of $b$:


$\ce{K}{:~} 8 + b = d$, so $d = b + 8$
$\ce{Mn}{:~} f = b$
$\ce{O}'{:~} 4b = 60 + i$, so $i = 4b - 60$
$\ce{S}{:~} c = (b + 8) + 3 + b = 2b + 11$
$\ce{H}{:~} 2(2b + 11) = b + 8 + 12 + 2(4b - 60)$, so $4b + 22 = 9b - 100$, so $5b = 122$ and $b = 122/5$. This is a bit unfortunate since later on we have to multiply all variables by 5.

We now have $a=2$, $b=122/5$, $c=299/5$, $d=162/5$, $e=1$, $f=122/5$, $g=12$, $h=12$ and $i=188/5$.
By multiplying everything by 5 we get the final solution:


$a = 10$
$b = 122$
$c = 299$
$d = 162$
$e = 5$
$f = 122$
$g = 60$
$h = 60$
$i = 188$



You can check that these figures will make the equation balanced.