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I am trying to balance a chemical equation algebraically and have found the resulting system of equations:

$$\begin{align*} 3a+8b+d&=12e+4f+2g\\ a&=e\\ 4a+4c+3d&=4e+3f+g+3h\\ 2b+d&=3e+f\\ c&=h \end{align*}$$ I am able to reduce this to a system of three equations: $$\begin{align*} 8b+d&=9e+4f+2g\\ c+3d&=3f+g\\ 2b+d&=3e+f \end{align*}$$ Now, I recognize that there are more variables than there are equations, and so there is no single solution to this equation. However, I need a solution with reasonably low numbers and only integer values for all the variables ($a,b,c,d,e,f,g,h$ or, by substitution, $b,c,d,e,f,g$). A matrix solution won't work since (in standard $AX=B$ form) the only elements in $B$ are zero and therefore all the variables are zero.

As a side note, the reaction is: $$H_3PO_4 + (NH_4)_2 + MoO_4 + HNO_3 \longrightarrow (NH_4)_3PO_4 + MoO_3 + NH_4NO_3 + H_2O$$ If there is a better way to go about this (besides inspection; I have been trying that for hours) please let me know.

Any ideas?

Martin Sleziak
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public satanic void
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  • You should expect to have an undermined system: the solution shouldn’t be unique since if you e.g. double the quantities of all of species on both sides of the reaction equation, it will still be balanced. – amd Mar 17 '18 at 00:32
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    See https://math.stackexchange.com/q/624/265466 for a few methods. Your question is essentially a duplicate of that one. – amd Mar 17 '18 at 00:35
  • @amWhy I know how to get to the system of equations; the problem is that my equation is so much more complicated than the one in the example listed above that I still do not understand where to go from here. – public satanic void Mar 17 '18 at 01:05
  • You’re wrong when you say that “A matrix solution won’t work because... therefore all the variables are zero.” In fact, the equation $AX=0$ must have nonzero solutions if the reaction equation can be balanced. The real problem that you’re going to run into is that the null space of the matrix for this equation is three-dimensional, so you’re going to have to find some linear combination of the three “canonical” solutions that meets your criteria. – amd Mar 17 '18 at 01:34
  • You have 5 equations and 8 unknowns. So assume that 5 of the unknowns are variables (e.g. $a,b,c,d,e$) and the other 3 are constants. Then solve this equation \begin{align} 3a+8b+d-12e&=4f+2g\ a-e&=0\ 4a+4c+3d-4e&=3f+g+3h\ 2b+d-3e&=f\ c&=h \end{align} as usually to get the solutions $$a=-\frac h3+g+f, ;b=-\frac h3+4\frac g3+3\frac f2 ,; c=h ,; d=-\frac h3+\frac g3+f,; e=-\frac h3+g+f $$ To ensure integer solutions we set $$h=3r,; g=3s ;f=2t$$ and get $$a=-r+3s+2t, ;b=-r+4s+3t ,; c=3r ,; d=-r+s+2t,; e=-r+3s+2t $$ Try small integers for $f,g,h.$ – miracle173 Mar 17 '18 at 06:38

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