Talk:Method of successive substitution
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[edit]The math is wrong. You didn't subtract 3 from both side like you said you should.
Notice 11 is a solution.
Incorrect solution
[edit]The solution presented here is incorrect.
The multiplicative inverse of 4 (mod 12) is not 1. In fact, there is no such inverse.
Two simultaneous congruences x = r (mod a) and n = s (mod b) are only solvable when a = b (mod gcd(a,b)). The solution is unique modulo lcm(a,b). (gcd = greatest common divisor. lcm = lowest common multiple.)
The answer to the example given is x = 11 (mod 12).
multiplicative inverse
[edit]The solution for
4j ≡ 8 (mod 12)
is j ≡ 2 mod 3 (dividing all sides by the GCD(4,8,12)=4 and using the euclidean multiplicative inverse afterwards)
I have corrected the solution on the main page after receiving no comments for a couple of months
I have changed the equations at the start to get a less trivial result. The previous comments don't describe the actual example anymore.
"multiplactive"?
[edit]I have changed "multiplactive" to "multiplicative", since it seems to be an unusual spelling, rather than a technical term. -- 80.168.228.51 09:14, 26 Oct 2004 (UTC)
"relative prime moduli"
[edit]In my last example the moduli are relatively prime, and could be solved easier using the Chinese Remainder Theorem. So I am changing the example again.
Falling back to CRT?
[edit]If the moduli are coprime, the Chinese remainder theorem can be used directly.
Can't we simply adjust one of the simultaneous congruences to make the moduli coprime?
For example: The two simultaneous congruences of the form
x ≡ r1 (mod m1) x ≡ r2 (mod m2)
can be solved directly by first calculating gcd(m1,m2) and replacing e.g.
x ≡ r1 (mod m1)
with
x ≡ r3 (mod m3)
where m3 = m1/gcd(m1,m2) and r3 = r1 mod m3.
Now, the remaining moduli (m2 and m3) are coprime.
problem in method 2?
[edit]I have a problem with the line:
2a + 1 = 3a + 2 (Rewrite the second congruence in terms of its modulus)
Shouldn't this be: 2a + 1 = 3b + 2
since there is no reason for both modili to have the same multiplier. — Preceding unsigned comment added by 193.254.175.5 (talk) 14:15, 4 June 2014 (UTC)