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The first two sets of equations actually don't allow for a generic solution.
Solve[eq1 /. {a[1] -> 0, c[1] -> 0}, b[1]]
{}
Reduce shows a solution given certain conditions:
Reduce[eq1 /. {a[1] -> 0, c[1] -> 0}, b[1]]
d[1] != 0 && x == 1/d[1] && b[1] == 2 d[1]^2 z[1]
Let's assume t...