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.

\(2+ 5k \equiv 3 (mod\ 8)\)

By using addition property,

\(2+ 5k + (—2) \equiv 3 + (—2)(mod\ 8)\)

\(= 5k = 1 (mod\ 8)\)

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.

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.

\(2+ 5k \equiv 3 (mod\ 8)\)

By using addition property,

\(2+ 5k + (—2) \equiv 3 + (—2)(mod\ 8)\)

\(= 5k = 1 (mod\ 8)\)

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.