Solve the following systems of congruences. x≡4(mod 7)

Question

Solve the following systems of congruences.  x≡4(mod 7)

Jaylen Fountain · Accepted Answer

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 $a + x \equiv b + . x (\mod n) a n d a x \equiv b x (\mod n)$ .
3) Theorem: Cancellation Law:
If $a x \equiv a y (\mod n) a n d (a, n) = 1, t h e n x = y (\mod n)$ .
Explanation:
Consider $3 x + 2 \equiv 3 (\mod 8)$
By using addition property,
$3 x + 2 + (- 2) \equiv 3 + (— 2) (\mod 8)$
$\Rightarrow 3 x \equiv 1 (m o d 8)$
By using multiplication property,
$\Rightarrow 3 * 3 x = 1 * 3 (\mod 8)$
$\Rightarrow 9 x \equiv 3 (\mod 8)$
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 $x = 4 + 7 k$ for some integer k and substitute this expression for x into the second congruence.
$4 + 7 k \equiv 3 (\mod 8)$
By using addition property,
$4 + 7 k + (— 4) \equiv 3 + (— 4) (\mod 8)$
$\Rightarrow 7 k \equiv — 1 (\mod 8)$
Since $— 1 \equiv 7 (\mod 8)$ ,
$\Rightarrow 7 k \equiv 7 (\mod 8)$
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 $x = 11 (\mod 7 * 8) o r x \equiv 11 (\mod 56)$ gives all solutions to the given system of congruences.

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

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