Quote:
Originally Posted by MattcAnderson
Find all x such that
5*x+6 is congruent to 0 mod 7 expression 1
From the reading, I notice that the greatest common divisor of 5 and 6 is 1. So the techniques presented here should apply.
We make an augmented T table
X 5*X 5*X mod 7
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0 0 0
1 5 5
2 10 3
3 15 1
4 20 6
5 25 4
6 30 2
From expression 1, conclude that 5*x is congruent to 1 mod 7.

Yes, and only one row in your table has 5x mod 7 = 1...
An alternative approach is to multiply by 3 (because 3x5 mod 7=1):
if \(5x+6\equiv 0\pmod{7}\) then \(3(5x+6)\equiv 0\pmod{7}\)
and so \(x+4\equiv 0\pmod{7}\).
I hope this helps!