Verified answer

Writing x = 4 + 17n for some integer n, substituting this into the second congruence yields

4 + 17n ≡ 6 (mod 11)

==> 3n ≡ 1 (mod 11)

==> 3n ≡ 12 (mod 11)

==> n ≡ 4 (mod 11)

Writing n = 4 + 11k for some integer k, we have

x = 4 + 17(4 + 11k) = 72 + 187k.

Substituting this into the first congruence,

72 + 187k ≡ 2 (mod 6)

==> 0 + 1k ≡ 2 (mod 6)

==> k ≡ 2 (mod 6).

Writing k = 2 + 6m for some integer m, we have

x = 72 + 187(2 + 6m) = 446 + 1122m.

Hence, x ≡ 446 (mod 1122).

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I hope this helps!