# Solve the following systems of congruences. xequiv 4(mod 7)

Question
Congruence
Solve the following systems of congruences.
$$x\equiv 4(mod\ 7)$$

2021-03-07
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 $$a = b(mod\ n)$$ and x is any integer, then $$a + x \equiv b +.x(mod\ n)\ and\ ax \equiv bx (mod\ n)$$.
3) Theorem: Cancellation Law:
If $$ax \equiv ay (mod\ n)\ and\ (a, n) = 1,\ then\ x = y(mod\ n)$$.
Explanation:
Consider $$3x + 2 \equiv 3 (mod\ 8)$$
$$3x +2 + (-2) \equiv 3 + (—2) (mod\ 8)$$
$$\Rightarrow 3x \equiv 1 (mod\ 8)$$
By using multiplication property,
$$\Rightarrow 3*3x=1*3(mod\ 8)$$
$$Rightarrow\ 9x \equiv 3 (mod\ 8)$$
Since $$9 \equiv 1 (mod\ 8)$$,
$$\Rightarrow x \equiv 3(mod\ 8)$$
Therefore, the system of congruences is
$$x \equiv 4 (mod\ 7)$$
$$x \equiv 3(mod\ 8)$$
Since 7 and 8 are relatively prime then $$(7, 8) = 1$$.
Then, by using theorem there exists an integer x that satisfies the system of congruences.
From the first congruence $$x = 4 + 7k$$ for some integer k and substitute this expression for x into the second congruence.
$$4+ 7k \equiv 3 (mod\ 8)$$
$$4+7k + (—4) \equiv 3+ (—4) (mod\ 8)$$
$$\Rightarrow 7k \equiv —1(mod\ 8)$$
Since $$— 1 \equiv 7(mod\ 8)$$,
$$\Rightarrow 7k \equiv 7 (mod\ 8)$$
Since (7, 8) = I then by using cancellation law,
$$\Rightarrow k \equiv 1 (mod\ 8)$$
Thus, $$x = 4 + 7(1) = 11$$ satisfies the system and $$x = 11 (mod\ 7 * 8)\ or\ x \equiv 11 (mod\ 56)$$ gives all solutions to the given system of congruences.

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