Fix $r$ and $d\equiv r+\binom{r}{2}$ and consider $d$ real numbers $b_1,b_2,..., b_d$.
I have the following system of linear equations: $$ \begin{cases} \text{$(\diamond)$ Differences between the elements $\{b_1,b_h\}$ for $h=2,...,r$ $\hspace{1cm}$ ($r-1$ elements)}\\ b_{r+1}=b_1-b_2\\ b_{r+2}=b_1-b_3\\ ...\\ b_{2r-1}=b_1-b_r\\ -------\\ \text{$(\diamond)$ Differences between the elements $\{b_2,b_h\}$ for $h=3,...,r$ $\hspace{1cm}$ ($r-2$ elements)}\\ b_{2r}=b_2-b_3\\ ...\\ b_{3r-3}=b_2-b_r\\ -------\\ \text{$(\diamond)$ Etc. }\\ ....\\ -------\\ \text{$(\diamond)$ Differences between the elements $\{b_{r-1},b_h\}$ for $h=r$ $\hspace{1cm}$ ($1$ element)} \\ b_d=b_{r-1}-b_r \end{cases} $$
Question: For any $r$, I want to re-write in an EQUIVALENT way such that it
has only relations with the $+$ sign on the RHS of each equation
has only 1 variable on the LHS
has all the variables on the LHS different between each other
has the least possible amount of equations
I'm unable to find a generic expression. Below I report my attempts which should also clarify what I'm trying to do.
My attempts: For $r=2$, the original system is $$ \begin{cases} b_3=b_1-b_2 \end{cases} $$ which simply can be written as $$ \begin{cases} b_1=b_2+b_3 \end{cases} $$ which satisfies my requirements.
For $r=3$, the original system is $$ \begin{cases} b_4=b_1-b_2\\ b_5=b_1-b_3\\ b_6=b_2-b_3\\ \end{cases} \Leftrightarrow \begin{cases} b_1=b_2+b_4\\ b_1=b_3+b_5\\ b_2=b_3+b_6 \end{cases} \Leftrightarrow \begin{cases} b_1=b_3+b_6+b_4\\ b_1=b_3+b_5\\ b_2=b_3+b_6 \end{cases} \Leftrightarrow \begin{cases} b_1=b_2+b_4\\ b_2=b_3+b_6\\ b_5=b_6+b_4\\ \end{cases} $$ which satisfies my requirements.
For $r=4$, the original system is $$ \begin{cases} b_5=b_1-b_2\\ b_6=b_1-b_3\\ b_7=b_1-b_4\\ b_8=b_2-b_3\\ b_9=b_2-b_4\\ b_{10}=b_3-b_4 \end{cases} \Leftrightarrow \begin{cases} b_1=b_2+b_5\\ b_1=b_3+b_6\\ b_1=b_4+b_7\\ b_2=b_3+b_8\\ b_2=b_4+b_9\\ b_3=b_4+b_{10} \end{cases}\Leftrightarrow \begin{cases} b_1=b_2+b_5\\ b_1=b_4+b_{10}+b_6\\ b_1=b_4+b_7\\ b_2=b_4+b_{10}+b_8\\ b_2=b_4+b_9\\ b_3=b_4+b_{10} \end{cases}\Leftrightarrow \begin{cases} b_1=b_2+b_5\\ b_2=b_3+b_8\\ b_3=b_4+b_{10}\\ b_7=b_{10}+b_6\\ b_9=b_{10}+b_8\\ \end{cases} $$ which satisfies my requirements. In principle, as I keep increasing $r$, I should be able to find some path, but I don't see anything!