Okay, the first step is to show that I can take the two decoupled equationsI think I see how to show (one way or the other) which it is. I'll doo that in a bit.
\(
Ax = By + C
\;
Dx = Ez + F
\)
This is trivial to do:
\(
\{Ax = By + C\}(D)
\;
\{Dx = Ez + F\}(A)
\)
\(
ADx = BDy + CD
\;
ADx = AEz + AF
\)
yielding
\(
ADx = BDy + CD = AEz + AF
\)
From this it's obvious that if I have the equations in coupled form, I can always decouple them, change one of them, and then recouple them as above. In doing so, I end up making changed to all three expressions (in general) but the changes to two of them cancel out so as to leave the relationship imposed by the corresponding equation unchanged.