2x 5 (mod 7) 4x 2 (mod 5) 3x 9 (mod 11) solve the above congruences

If you're supposed to solve the congruences one at a time:

1:

First find the inverse of 5 modulo 7. Note that , so .

Then find the inverse of 6 modulo 7. Note that , so 6 is its own inverse, and we have

so the first congruence has solution where .

2:

We have , so . Then

and

so the solution is for .

3:

, so , and

, so and

so for .

If they're to be taken simultaneously, use the Chinese remainder theorem.

Combined:

Let's start with

so that two terms will vanish when considering the remainder of modulo 7, 5, or 11, respectively.

Taken modulo 7, we have

Taken modulo 5, we have

but we want to end up with 3, so we multiply the second term by the inverse of 2 modulo 5, then by 3. Note that , so our new choice of is

and taken modulo 5, we end up with a remainder of 3, as desired.

Taken modulo 11, we have

but we want to end up with 3, so we multiply the third term by the inverse of 2 modulo 11, then by 3. Note that , so becomes

By the CRT, the least positive solution is

so the general solution to the system would be for .