Writing a balanced chemical equation with linear systems











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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).



Chemical reaction






linear-algebra







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  • Note that $x_1,x_2,x_3$ are just variables of the reaction $x_1text C_6text H_{12}text O_6to x_2text Ctext O_2+x_3text C_2text H_5text Otext H$ without loss of generality and for an element $text X$, $ktext X_n$ becomes $kn$ lots of $text X$.
    – TheSimpliFire
    5 hours ago








  • 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
    5 hours ago

















up vote
0
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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).



Chemical reaction






linear-algebra







share|cite|improve this question









New contributor




Mit34 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • Note that $x_1,x_2,x_3$ are just variables of the reaction $x_1text C_6text H_{12}text O_6to x_2text Ctext O_2+x_3text C_2text H_5text Otext H$ without loss of generality and for an element $text X$, $ktext X_n$ becomes $kn$ lots of $text X$.
    – TheSimpliFire
    5 hours ago








  • 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
    5 hours ago















up vote
0
down vote

favorite









up vote
0
down vote

favorite











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).



Chemical reaction






linear-algebra







share|cite|improve this question









New contributor




Mit34 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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).



Chemical reaction






linear-algebra




linear-algebra






share|cite|improve this question









New contributor




Mit34 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









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Mit34 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 5 hours ago









Arthur

107k7103186




107k7103186






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asked 5 hours ago









Mit34

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Mit34 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • Note that $x_1,x_2,x_3$ are just variables of the reaction $x_1text C_6text H_{12}text O_6to x_2text Ctext O_2+x_3text C_2text H_5text Otext H$ without loss of generality and for an element $text X$, $ktext X_n$ becomes $kn$ lots of $text X$.
    – TheSimpliFire
    5 hours ago








  • 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
    5 hours ago




















  • Note that $x_1,x_2,x_3$ are just variables of the reaction $x_1text C_6text H_{12}text O_6to x_2text Ctext O_2+x_3text C_2text H_5text Otext H$ without loss of generality and for an element $text X$, $ktext X_n$ becomes $kn$ lots of $text X$.
    – TheSimpliFire
    5 hours ago








  • 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
    5 hours ago


















Note that $x_1,x_2,x_3$ are just variables of the reaction $x_1text C_6text H_{12}text O_6to x_2text Ctext O_2+x_3text C_2text H_5text Otext H$ without loss of generality and for an element $text X$, $ktext X_n$ becomes $kn$ lots of $text X$.
– TheSimpliFire
5 hours ago






Note that $x_1,x_2,x_3$ are just variables of the reaction $x_1text C_6text H_{12}text O_6to x_2text Ctext O_2+x_3text C_2text H_5text Otext H$ without loss of generality and for an element $text X$, $ktext X_n$ becomes $kn$ lots of $text X$.
– TheSimpliFire
5 hours ago






1




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
5 hours ago






@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
5 hours ago












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If we are to balance the equation, that means finding positive integers $x_1,x_2,x_3$ such that
$$
x_1C_6H_{12}O_6to x_2CO_2+x_3C_2H_5OH
$$
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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    If we are to balance the equation, that means finding positive integers $x_1,x_2,x_3$ such that
    $$
    x_1C_6H_{12}O_6to x_2CO_2+x_3C_2H_5OH
    $$
    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
    $$
    begin{array}{lccc}
    text{Element} &text{Left side} &&text{Right side}\
    text{Carbon}&6x_1&=&x_2+2x_3\
    text{Hydrogen} &12x_1&=&6x_3\
    text{Oxygen} &6x_1&=&2x_2+x_3
    end{array}
    $$






    share|cite|improve this answer



























      up vote
      1
      down vote













      If we are to balance the equation, that means finding positive integers $x_1,x_2,x_3$ such that
      $$
      x_1C_6H_{12}O_6to x_2CO_2+x_3C_2H_5OH
      $$
      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
      $$
      begin{array}{lccc}
      text{Element} &text{Left side} &&text{Right side}\
      text{Carbon}&6x_1&=&x_2+2x_3\
      text{Hydrogen} &12x_1&=&6x_3\
      text{Oxygen} &6x_1&=&2x_2+x_3
      end{array}
      $$






      share|cite|improve this answer

























        up vote
        1
        down vote










        up vote
        1
        down vote









        If we are to balance the equation, that means finding positive integers $x_1,x_2,x_3$ such that
        $$
        x_1C_6H_{12}O_6to x_2CO_2+x_3C_2H_5OH
        $$
        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
        $$
        begin{array}{lccc}
        text{Element} &text{Left side} &&text{Right side}\
        text{Carbon}&6x_1&=&x_2+2x_3\
        text{Hydrogen} &12x_1&=&6x_3\
        text{Oxygen} &6x_1&=&2x_2+x_3
        end{array}
        $$






        share|cite|improve this answer














        If we are to balance the equation, that means finding positive integers $x_1,x_2,x_3$ such that
        $$
        x_1C_6H_{12}O_6to x_2CO_2+x_3C_2H_5OH
        $$
        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
        $$
        begin{array}{lccc}
        text{Element} &text{Left side} &&text{Right side}\
        text{Carbon}&6x_1&=&x_2+2x_3\
        text{Hydrogen} &12x_1&=&6x_3\
        text{Oxygen} &6x_1&=&2x_2+x_3
        end{array}
        $$







        share|cite|improve this answer














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        edited 5 hours ago

























        answered 5 hours ago









        Arthur

        107k7103186




        107k7103186






















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