A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
solve the following system.
\[2x\equiv 3 (mod 5)\]
\[4x\equiv 2 (mod 6)\]
\[3x\equiv 2 (mod 7)\]
anonymous
 one year ago
solve the following system. \[2x\equiv 3 (mod 5)\] \[4x\equiv 2 (mod 6)\] \[3x\equiv 2 (mod 7)\]

This Question is Closed

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1familiar with chinese remainder theorem ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have done an assignment but it only had something like \[x\equiv 3 (mod 5)\] \[x\equiv 2 (mod 6)\] \[x\equiv 2 (mod 7)\]. But am not sure how to use it to solve the problem. @ganeshie8

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1solve each of the linear congruence to get a system with simpler congruences, for example : \(2x\equiv 3 \pmod 5\) simplifies to \(x\equiv 4\pmod{5}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 Please show me a step on how you got the 4

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1I have just guessed it. Plugin x=4 and observe that it satisfies the congruence \[2*4\equiv 8\equiv 3\pmod{5}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 Thanks. I will try doing that and then apply the Chinese remainder theorem

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1For the second congruence, \(4x\equiv 2 \pmod 6\), dividing \(2\) through out gives \[2x\equiv 1\pmod{3}\] then it is easy to eyeball the solution : \[x\equiv 2\pmod{3}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[3x\equiv 2 (mod 7)\] simplifues to \[x\equiv 4 (mod 7)\] Does that look correct? @ganeshie8

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Probably \[x\equiv 10 (mod 7)\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\(x\equiv 4\pmod{7}\) is same as \(x\equiv 3 \pmod{7}\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1so the given system, after transforming into simple linear congruences is \[x\equiv 4\pmod{5}\\x\equiv 2\pmod{3}\\x\equiv 3\pmod{7}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got it. Thanks alot @ganeshie8
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.