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

Khadija Wells 2021-03-06 Answered
Solve the following systems of congruences.
x4(mod 7)
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Expert Answer

Jaylen Fountain
Answered 2021-03-07 Author has 170 answers

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
xa(mod m)
xb(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+xb+.x(mod n) and axbx(mod n).
3) Theorem: Cancellation Law:
If axay(mod n) and (a,n)=1, then x=y(mod n).
Explanation:
Consider 3x+23(mod 8)
By using addition property,
3x+2+(2)3+(2)(mod 8)
3x1(mod 8)
By using multiplication property,
33x=13(mod 8)
 9x3(mod 8)
Since 91(mod 8),
x3(mod 8)
Therefore, the system of congruences is
x4(mod 7)
x3(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+7k3(mod 8)
By using addition property,
4+7k+(4)3+(4)(mod 8)
7k1(mod 8)
Since 17(mod 8),
7k7(mod 8)
Since (7, 8) = I then by using cancellation law,
k1(mod 8)
Thus, x=4+7(1)=11 satisfies the system and x=11(mod 78) or x11(mod 56) gives all solutions to the given system of congruences.

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