Could someone please help explain how they got to this next step (writing in tabular form) in solving this chemical equation? (I have the worked answer, but I don't understand the first step they did).
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Note that $x_1,x_2,x_3$ are just variables of the reaction $x_1\text C_6\text H_{12}\text O_6\to x_2\text C\text O_2+x_3\text C_2\text H_5\text O\text H$ without loss of generality and for an element $\text X$, $k\text X_n$ becomes $kn$ lots of $\text X$. – TheSimpliFire Nov 19 '18 at 07:47
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1@TheSimpliFire except that as written the LHS of each equation isn't $x_1$ but keeps changing. I think you're right that that's how it should have been written though – postmortes Nov 19 '18 at 07:48
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Much related: Automated Balancing of Chemical Equations – Han de Bruijn Dec 04 '20 at 11:56
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If we are to balance the equation, that means finding positive integers $x_1,x_2,x_3$ such that $$ x_1\mathrm{C_6H_{12}O_6}\to x_2\mathrm{CO_2}+x_3\mathrm{C_2H_5OH} $$ has as many of each atom on the left side as it does on the right side.
Looking at carbon first, the number of carbon atoms on the left side is $6x_1$, while on the right side it's $x_2+2x_3$. These two numbers are supposed to be equal, so we put $=$ between them, and that's the first equation.
The two other equations are done similarly for hydrogen and oxygen. However, there is a typo in your picture: it's supposed to be $x_1$ on the left side of all three equations. That is, after all, how many sugar molecules there are on the left-hand side. Thus we get $$ \begin{array}{lccc} \text{Element} &\text{Left side} &&\text{Right side}\\ \text{Carbon}&6x_1&=&x_2+2x_3\\ \text{Hydrogen} &12x_1&=&6x_3\\ \text{Oxygen} &6x_1&=&2x_2+x_3 \end{array} $$
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And the solution is: $\mathrm{C_6H_{12}O_6}\to 2, \mathrm{CO_2} + 2, \mathrm{C_2H_5OH} $ . – Han de Bruijn Nov 22 '20 at 15:16