Given: $6x+7y \equiv 17 \pmod{42} \tag1$ $21x+5y \equiv 13 \pmod{42} \tag2$
Here's my initial attempt at solving the above system.
$(2) \times 35$: $21x+7y \equiv 35 \pmod{42} \tag3$ $(3)-(1)$: $15x \equiv 18 \pmod{42}$ $5x \equiv 6 \pmod{14}$ $x \equiv 4 \pmod{14}$ $x \equiv 4,18,32 \pmod{42} \tag4$ Substitute $(4)$ into $(2)$: $5y \equiv 13 \pmod{42}$ $y \equiv 11 \pmod{42}$ Hence the solutions in $\mathbb Z_{42}$ are $(4,11), (18,11), (32,11)$. I know this is correctly the solution set because the answers work, and because I've been told the system has 3 solutions.
Then I tried substituting $(4)$ into $(1)$, and also into $(3)$, and each time I got $7y \equiv 35 \pmod{42}$ $7y \equiv 35 \pmod{42}$ $y \equiv 5,11,17,23,29,35,41 \pmod{42}$ Now, I don't understand why substituting $(4)$ into $(1)$ (or $(3)$) instead of into $(2)$ created excess solutions. I would really appreciate it if someone could take a look and explain it to me..thanks!