Solve the following systems of congruences. 2xequiv 5(mod 3) 3x+2equiv 3(mod 8) 5x+4equiv 5(mod 7)

remolatg 2021-02-18 Answered
Solve the following systems of congruences.
2x5(mod 3)
3x+23(mod 8)
5x+45(mod 7)
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Expert Answer

Aamina Herring
Answered 2021-02-19 Author has 85 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 xy(mod n).
Explanation:
Consider, 2x5(mod 3)
Since 52(mod 3),
2x2(mod 3)
By using multiplication property,
22x22(mod 3)
4x4(mod 3)
Since 41(mod 3),
x=1(mod 3)
Now, consider 5x+45(mod 7)
By using addition property,
5x+4+(4)5+(4)(mod 7)
5x=1(mod 7)
By using multiplication property,
53x13(mod 7)
15x3(mod 7)
Since 151(mod 7),
x3(mod 7)
Therefore, the system of congruences is
x1(mod 3)
x3(mod 7)
Since 3 and 7 are relatively prime then (3,7)=1.
From the first congruence x=1+3k for some integer k and substitute this expression for x into the second congruence.
1+3k=3(mod 7)
By using addition property,
1+3k+(1)3+(1)(mod 7)
3k2(mod 7)
By using multiplication property,
33k23(mod 7)
9k6(mod 7)
Since 92(mod 7)
2k6(mod 7)
Since (2,7)=1 then by using cancellation law,
k(mod 7)
Thus, x=1+3(3)=10 satisfies the system and x10(mod 37) or x(mod 21) gives all solutions to the given systems of congruences.
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