I have a linear system $LS$, where each equation contains three variables and the coefficient of each variable is $\pm 1$. For example, I have $x_{a}-x_{b}+x_{c}=p$ ($p$ is the known term).
Suppose that $x_{i}$, due to $LS$, is the linear combination of a set of variables $A_{x_{i}}$, with $x_{i} \notin A_{x_{i}}$, that is $x_{i} = \sum\limits_{x_{j} \in A_{x_{i}}}{k_{j}x_{j}}+p_{i}$ ($p_{i}$ is the known term). Is it possible to find a value $k_{max}$ such that $k_{j} \leq k_{max}$ $\forall k_{j}$?
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$\begingroup$ From $x_a-x_b+x_c=p$ you get $x_a=x_b-x_c+p$ so you can take $k_{\rm max}=1$, right? $\endgroup$– Gerry MyersonCommented Jun 26, 2022 at 12:21
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$\begingroup$ We have a linear system, therefore we have more than only one equation. For example, if we have $x_{a}-x_{b}+x_{c}=1$, $x_{d}-x_{b}+x_{c}=1$ and $-x_{a}-x_{d}+x_{e}= -1$, we have $x_{c}=2x_{a}-1$. $\endgroup$– Mario GiambarioliCommented Jun 26, 2022 at 14:51
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$\begingroup$ Yes, but you also have $x_a=x_b-x_c+1$, $x_b=x_a+x+c-1$, $x_c=x_b-x_a+1$, $x_d=x_b-x_c+1$, and $x_e=x_a+x_d-1$, all with $k_{\rm max}=1$. So, what is it that you really want? $\endgroup$– Gerry MyersonCommented Jun 26, 2022 at 22:48
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$\begingroup$ Suppose that the linear system has $a$ variables and $a$ linear indipendent equations. Thia means that we can find the value of each variable of $LS$. Suppose we solve $LS$ by substitution, what is the max value of the coefficients of the equations that we get by the substitution method soution of $LS$? $\endgroup$– Mario GiambarioliCommented Jun 28, 2022 at 4:23
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$\begingroup$ The maximum value probably depends on how you implement the substitution method. $\endgroup$– Gerry MyersonCommented Jun 28, 2022 at 7:58
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