Modulus question!?

Note that for the first congruence, we have x ≡ 4 (mod 15) is a solution. This implies that

x = 15a + 4, where a is an integer.

We can substitute this into the second congruence as follows:

2[15a + 4] ≡ 2 (mod 14)

or

30a + 8 ≡ 2 (mod 14)

or

30a ≡ -6 (mod 14)

If you observe carefully, you will see that a ≡ 4 (mod 14) is a solution to the above congruence. This implies that

a = 14b + 4, where b is an integer.

We can substitute 14b + 4 for a in the equation for x as follows:

x = 15[ 14b + 4] + 4

x = 210b + 64

From the above equation, it follows that x ≡ 64 (mod 210).

Hence, the solution to the congruence is

x ≡ 64 (mod 210)