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Mathematics 17 Online
OpenStudy (anonymous):

5x=3(mod7)

OpenStudy (anonymous):

x = 3/5

myininaya (myininaya):

5x-3=7k where k is some integer x=2 is a solution

OpenStudy (jamesj):

No agdgd: this is a problem in integers

OpenStudy (anonymous):

x=3/5

OpenStudy (anonymous):

oh I see.

OpenStudy (anonymous):

http://en.wikipedia.org/wiki/Modular_arithmetic I get it.

myininaya (myininaya):

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

myininaya (myininaya):

so we could say x=2+7i for any integer i>=1

OpenStudy (anonymous):

thank you.

myininaya (myininaya):

i think there might be a shorter process but i can't remember

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