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!