Given two systems of congruences with one common parameter $A$ I need to find all values of $A$ for which both systems have the same solution sets. The systems are: $$x\equiv5\bmod6\\x\equiv A\bmod35$$and$$x\equiv A\bmod10\\x\equiv 14\bmod21$$
I solved it by first finding x in both systems, by CRT and then since the answers were both $\bmod 210$ I equated both solutions with that modulus and solved for $A$.
From the first system I got $x\equiv36A+35\bmod210$, and from the second one $x\equiv21A-70\bmod210$. Equating the $x$'s I got $A\equiv3\bmod6$. Is it correct? And what would I have to do if the moduli weren't equal? Would finding the LCA and multiplying both solutions be correct?
There is an error somewhere. It is simpler to split into prime moduli using CRT:
$\qquad \begin{align} x\equiv\, \color{#c00}1\!\!\!\pmod 2,\ &\ 2\!\!\!\pmod 3,\ A\!\!\!\pmod 5,\ \color{#0a0}A\!\!\!\pmod 7\\ x\equiv \color{#c00}A\!\!\!\pmod 2,\ &\ 2\!\!\!\pmod 3,\ A\!\!\!\pmod 5,\ \ \color{#0a0}0\!\!\!\pmod 7 \end{align}$
By CRT these are equivalent $\iff \color{#c00}A\equiv \color{#c00}1\pmod 2,\ \color{#0a0}A\equiv \color{#0a0}0\pmod 7\,$
Solving we find $\bmod 2\!:\,\ 1\equiv A\equiv 7k\equiv k\,$ so $\, A = 7k = 7(1\!+\!2n) = 7\!+\!14n$