I have a system of linear equations:
\begin{align*} t_0+t_1-t_2+t_3-t_4+t_5 = a \\ -t_0+t_1+t_2+t_3-t_4+t_5 = b \\ -t_0 + t_1 - t_2 + t_3 + t_4 + t_5 = c \end{align*}
I want to find the range of possible values $(a,b,c) \in \mathbb{R}^3$ can take with the condition that $t_0,t_1,t_2,t_3,t_4,t_5 \geq 0$.
I have tried inverting the system \begin{align*} t_2 &= \frac{1}{2}(b-a)+t_0\\ t_4 &= \frac{1}{2}(c-a) + t_0\\ t_5 &= \frac{1}{2}(b+c)+t_0-t_1-t_3 \end{align*} and fiddling around with inequalities haphazardly, but I'm not sure if I'm getting the full range, and I also have 14 other systems of equations (like this but with different signs) that I also want to examine.
I'm not sure about the other systems you have, but one thing pops out of the equations you have for $t_2$, $t_4$, and $t_5$: They all have a ${}+t_0$ term.
It appears that you can get any triple $(a,b,c)\in\mathbb{R}^3$. To do this, choose $a,b,c\in\mathbb{R}$, and choose $t_1$ and $t_3$ arbitrarily (but nonnegative). Now choose $t_0$ so that $$t_0 \ge \frac{a-b}{2}, \quad t_0 \ge \frac{a-c}{2}, \quad t_0 \ge t_1 + t_3 - \frac{b+c}{2}, \quad t_0 \ge 0.$$ Now choose $t_2,t_4,t_5$ according to the actual equations. The inequalities above guarantee that all the $t$'s are nonnegative.