Congruence help please?

Question

Describe the set of all possible integer solutions for the following system of congruences:

x ≡ 5 (mod 6)
x ≡ 1 (mod 7)

Answer 1

First notice that if x ≡ 5 (mod 6) and x ≡ 1 (mod 7), then the same is true for 42k + x for any integer k. So we want to find an x between 0 and 41 such that x ≡ 5 (mod 6) and x ≡ 1 (mod 7).

The x's between 0 and 41 for which x ≡ 5 (mod 6) are:
6, 11, 17, 23, 29, 35, 41

Looking at these mod 7, we see that only 29 is congruent to 6 mod 7.

So the integers x such that x ≡ 5 (mod 6) and x ≡ 1 (mod 7), are all integers of the form: 42k + 29

Answer 2

From the given congruences we can write

x = 5+6k for integral k and
x = 1+7j for integral j. Then,

7x = 35 + 42k and
6x = 6 + 42j.

Subtraction yields

x = 29 + 42(k-j) where k-j is an integer.

From this we can say that x is congruent to 29 (mod 42).

Answer 3

the first one is 5, 11, 17, 23 etc so i'd describe that as, 6n-1 where n is integer. similarly the 2nd one is 7n-6 where nis integer.