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5x=3(mod7)
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x = 3/5
5x-3=7k where k is some integer x=2 is a solution
No agdgd: this is a problem in integers
x=3/5
oh I see.
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if k=1 then 5x-3=7 =>x=2 if k=2 then 5x-3=14 => x=not integer if k=3 then 5x-3=21 => x=not integer if k=4 then 5x-3=28 => x=not integer if k=5 then 5x-3=35 => x=not integer if k=6 then 5x-3=42 => x=9 if k=7 then 5x-3=49 => x=not integer
so we could say x=2+7i for any integer i>=1
thank you.
i think there might be a shorter process but i can't remember
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