I have:
$A = f_1(B,C,E) \quad \quad (1)$
$B = f_2(A,C,E) \quad \quad (2)$
$C = f_3(A,B,E) \quad \quad (3)$
$D = f_4(B,C,E) \quad \quad (4)$
$E = f_5(A,C,D) \quad \quad (5)$
where the $f()$s are non-linear but tractable functions. My substitution strategy was:
(2) in (1) to get $A = f_6(C,E) \quad \quad \quad \quad \quad \quad (6)$
(1) in (2) to get $B = f_7(C,E) \quad \quad \quad \quad \quad \quad (7)$
(6) and (7) in (3) to get $C = f_8(E) \quad \quad \quad \quad (8)$
(8) in (6) to get $A = f_9(E) \quad \quad \quad \quad \quad \quad \quad (9)$
(8) in (7) to get $B = f_{10}(E) \quad \quad \quad \quad \quad \quad \quad (10)$
(8) and (10) in (4) to get $D = f_{11}(E) \quad \quad \quad (11)$
(8), (9) and (11) in (5) to get $E$ dependent on only known values.
I was then hoping to substitute that clean expression of $E$ into equations (8) through (11) to get clean expressions for the other variables. However, my "clean" expression for $E$ turned out to be $E=\beta E$, where $\beta$ is an ugly but constant term. This is not something I can substitute back like I wanted to.
Without knowing the exact functions, can you diagnose the problem here? What could cause this dead end?
(1), (2) and (3) can indeed be seen as a $3\times3$ system with the parameter $E$, which you can solve for $A,B,C$ in terms of $E$. Then plugging in (4) and (5), you obtain some $E=f(E).$
If you made no substitution mistake and you do reach
$$E=\beta E,$$
where $\beta\ne1$, the solution is $E=0$ (which is not a dead end). If on the other hand $\beta=1$, the system is indeterminate.