# FAQ: Congruences: Lesson - Solving Linear Congruence Problems

This community-built FAQ covers the “Solving Linear Congruence Problems” exercise from the lesson “Congruences: Lesson”.

Paths and Courses
This exercise can be found in the following Codecademy content:

## FAQs on the exercise Solving Linear Congruence Problems

There are currently no frequently asked questions associated with this exercise – that’s where you come in! You can contribute to this section by offering your own questions, answers, or clarifications on this exercise. Ask or answer a question by clicking reply () below.

If you’ve had an “aha” moment about the concepts, formatting, syntax, or anything else with this exercise, consider sharing those insights! Teaching others and answering their questions is one of the best ways to learn and stay sharp.

## Join the Discussion. Help a fellow learner on their journey.

You can also find further discussion and get answers to your questions over in #get-help.

Agree with a comment or answer? Like () to up-vote the contribution!

Need broader help or resources? Head to #get-help and #community:tips-and-resources. If you are wanting feedback or inspiration for a project, check out #project.

Looking for motivation to keep learning? Join our wider discussions in #community

Found a bug? Report it online, or post in #community:Codecademy-Bug-Reporting

Have a question about your account or billing? Reach out to our customer support team!

None of the above? Find out where to ask other questions here!

I cant reach the solution with the provided material at all. Makes no sense to me

So 2 and 7 are congruent mod 5 because
2 % 5 = 2 and 7 % 5 = 2

details

so 2, 7, 12, 17, … would be in the same congruence class mod 5

and if I wanted to solve
3x + 2 ≡ 1 (mod 5)
like I’d solve a linear equation

I’d subtract 2 from both sides of the ≡
3x ≡ -1 (mod 5)
but I don’t like negative numbers,
so I can add 5 (or a multiple of 5) so that the stuff would be the same mod 5
3x ≡ 4 (mod 5)
I want to divide both sides of the ≡ by 3,
but 4 can’t be divided by 3 evenly [so that I have an integer]
so I can keep adding 5 until I get something I can divide by 3 on the right of the ≡
3x ≡ 9 (mod 5)
I can divide 9 by 3 evenly
so I divide both sides of the ≡ by 3
x ≡ 3 (mod 5)
so the solution for x is 3 or anything else that’s 3 (mod 5)