Question # Solve the following systems of congruences. xequiv 2(mod 5)

Congruence
ANSWERED Solve the following systems of congruences.
$$x\equiv 2(mod\ 5)$$ 2020-10-29
Formula used:
1) Theorem: System of congruences:
Let m and n be relatively prime and a and b integers. There exists an integer x that satisfies the system of congruences
$$x \equiv a(mod\ m)$$
$$x \equiv b(mod\ n)$$
Furthermore, any two solutions x and y are congruent modulo mn.
2) Theorem: Addition and Multiplication Properties:
If $$a \equiv b(mod\ n)$$ and x is any integer, then $$a + x -= b + x(mod\ n)\ and\ ax \equiv bx (mod\ n)$$.
Explanation:
Consider the system of congruences $$x \equiv (mod\ 5)$$
$$x \equiv 3 (mod\ 8)$$
Since 5 and 8 are relatively prime, then $$(5, 8) = 1$$.
Then, by using a theorem, there exists an integer x that satisfies the system of congruences.
From the first congruence $$x = 2 + 5k$$ for some integer k and substitute this expression for x into the second congruence.
$$2+ 5k \equiv 3 (mod\ 8)$$
$$2+ 5k + (—2) \equiv 3 + (—2)(mod\ 8)$$
$$= 5k = 1 (mod\ 8)$$
By using multiplication property,
$$\Rightarrow 2+5k+(-2) \equiv 3+(-2)(mod\ 8)$$
$$\Rightarrow 5k \equiv 1(mod\ 8)$$
By using multiplication property,
$$5*5k -\equiv 5*1(mod\ 8)$$
$$\Rightarrow 25k \equiv 5(mod\ 8)$$
Since $$25 \equiv 1 (mod\ 8)$$,
$$\Rightarrow k \equiv 5 (mod\ 8)$$
Thus, $$x = 2 + 5(5) = 27$$ satisfies the system and $$x \equiv 27 (mod\ 5 * 8)\ or\ x \equiv 27 (mod\ 40)$$ gives all solutions to the given system of congruences.