Why do two equivalent systems of equations produce a different result?

Question

I've been reading a book about material and energy balance when I faced a confusing problem when solving a system of equations. I came to notice some new concepts (that were already there, but didn't put much attention to) that now I believe are more important than they seem. I think it is a math equation the one I have, it's also a simple one but I'd like to get a clear and correct answer before I draw my own conclusions.

System 1

$$0.09F_8+F_6=r_1-r_2$$ $$0.27F_8=r_2+0.13(1-x)F_1$$ $$0.5F_8=r_1+xF_1$$ $$0.09F_8=-3r_1-r_2+0.65(1-x)F_1$$ $$0.05F_8=-r_1-r_2+0.22(1-x)F_1$$

$F_6$,$F_8$,$F_1$,$x$,$r_1$ and $r_2$ are the variables

Now if I say that $N=xF_1$ I get the System 2

$$0.09F_8+F_6=r_1-r_2$$ $$0.27F_8=r_2+0.13(F_1-N)$$ $$0.5F_8=r_1+N$$ $$0.09F_8=-3r_1-r_2+0.65(F_1-N)$$ $$0.05F_8=-r_1-r_2+0.22(F_1-N)$$

For this system $F_6$,$F_8$,$F_1$,$N$,$r_1$ and $r_2$ are the variables

All the $F_n$ variables are mass stream flow rates, the $r_n$ variables are called reaction rates to account for the mass that "appears" or "dissappears" from a stream flow rate of a certain compound when a chemical reaction takes place within the system and $x$ is the mole fraction of a compound in $F_1$. I mention this because for this systems of equations a negative answer for any variable will not make any physical sense, even for the reaction rates. This systems of equations arise using the principle of conservation of mass

I know there are six unknown variables and only five equations in each system. In order to solve this problem there is something called a basis that my book defines as

As a further consequence of homogeneity of the balance equations, if none of the stream flow rates is assigned a value in the problem statement, then any one of the stream flows can be assigned an arbitrary magnitude for purposes of calculation

And in case it is important what mentions of homogeneity is

A system of equations in which values of a set of variables can be uniformly scaled such that the resulting values continue to satisfy the equations said is said to be homogeneous in those variables. Formally stated, an equation $f(x,y)=0$ two variables x and y are homogeneous in y if, given any solution $(\bar{x},\bar{y})$ any constant times $\bar{y}$ is also a solution

On practice it basically means that you can assume the value of a stream flow rate, that is I can assume that $F_8=100 \text{ mol}.$ so that your system of equations get reduced in unknowns and you are able to solve it. Oftenly when you have to assume stream flow rates the problem asks for a intensive property of the physical system, it can be, at what temperature does the products of whatever-unit will come out if I feed it a stream with such and such composition? to answer such type of questions it doesn't matter how much of a flow goes in or out you will get the correct temperature

So, what is the problem? In system one I assumed $F_1=100 \text{ mol/hr}$ to not deal with a non-linear equation and solved for the rest of variables, that way I got a negative result for $r_2$, that is an obvious red flag. But for System 2 I assumed $F_8=100 \text{ mol/hr}$ and got only positive answers. In my world it didn't had to matter which flow I chose to assume its value I shouldn't get such different results. If I had assumed the value for $N$ on system 2 I should still get positive values for every variable. And to clarify, the equations are not wrong, I have already verified them many times.

When solving for System 1 and the assumption $F_1=100 \text{ mol/hr}$ I got

• $F_6=15.88 \text{ mol/hr}$
• $F_8=45.08 \text{ mol/hr}$
• $F_1=100 \text{ mol/hr}$
• $x=0.03$
• $r_1=19.51\text{ mol/hr}$
• $r_2=-0.43382\text{ mol/hr}$

When solving for System 2 and the assumption $F_8=100 \text{ mol/hr}$ I got

• $F_6=28.24 \text{ mol/hr}$
• $F_8=202.72 \text{ mol/hr}$
• $F_1=100 \text{ mol/hr}$
• $N=10.72 \text{ mol/hr}$
• $r_1=39.28\text{ mol/hr}$
• $r_2=2.04\text{ mol/hr}$

Answer 1

Take these three equations $$0.27F_8=r_2+0.13(F_1-N)$$ $$0.09F_8=-3r_1-r_2+0.65(F_1-N)$$ $$0.05F_8=-r_1-r_2+0.22(F_1-N)$$ and eliminate $r_2$ by adding the first to each of the others: $$0.36F_8+3r_1=0.78(F_1-N)$$ $$0.32F_8+r_1=0.35(F_1-N).$$ Now subtract the first of these from 3 times the second: $$0.6F_8=0.27(F_1-N),$$ or $$20F_8=9(F_1-N).$$ Then $$r_1=0.35(F_1-N) - 0.32F_8=0.206(F_1-N)$$ and $$r_2=0.27F_8-0.13(F_1-N)=-0.0085(F_1-N)$$ so it looks as if not all variables can be positive.

Answer 2

The two systems are mathematically equivalent, but you have solved both of them wrongly, probably because of computational mistakes.

The second system of equations can be reformulated as $$ \underbrace{\pmatrix{0&0&-0.09&1&-1\\ 0.13&-0.13&0.27&0&-1\\ -1&0&0.5&-1&0\\ 0.65&-0.65&0.09&3&1\\ 0.22&-0.22&0.05&1&1}}_A \ \pmatrix{N\\ F_1\\ F_8\\ r_1\\ r_2} =\pmatrix{F_6\\ 0\\ 0\\ 0\\ 0}. $$ It can be verified that $A$ is an invertible matrix. Therefore $$ \pmatrix{N\\ F_1\\ F_8\\ r_1\\ r_2} =A^{-1}\pmatrix{F_6\\ 0\\ 0\\ 0\\ 0} =F_6A^{-1}\pmatrix{1\\ 0\\ 0\\ 0\\ 0} =F_6\pmatrix{0.109195\\ 5.856322\\ 2.586207\\ 1.183908\\ -0.048851}. $$ Hence the only nonnegative solution is $F_6=0$ and $N=F_1=F_8=r_1=r_2=0$. That is, in the original system, the only nonnegative solutions are $F_1=F_6=F_8=r_1=r_2=0$ and $x\ge0$.