- #1
Korg
- 6
- 0
Homework Statement
Solve the following system of simulataneous congruences:
2X+3Y = 2(mod 5)
X+2Y = 3 (mod 5)
X+2Y = 3 (mod 7)
4X+3Y = 4 (mod 7)
Homework Equations
The Attempt at a Solution
I know how to solve a similar problem if there was just one variable, using the Chinese remainder theorem but I'm not quite sure how to adapt it here. I know the CRT is applicable since 5 and 7 are coprime.
I started trying to solve it by using the two X+2Y congruences and got that X+2Y = 3 (mod 35) . but I'm not sure if this even helps to solve the whole system.
Can anyone help at all?