# Solving a two variable equation

1. Apr 26, 2012

### mtayab1994

1. The problem statement, all variables and given/known data

For every (x,y) in Z^2 solve : 6x-5y=1

3. The attempt at a solution

I did 6x=5y+1 and then i said let x=5n+1 so then we get 6(5n+1)=5y+1 and that gives that y=6n+1 so the solution is (5n+1,6n+1). Is this the correct way of solving this?

2. Apr 26, 2012

### ehild

I do not quite follow your way of thinking but the result is correct.

I would say that 6x-5y=5(x-y)+x=1, x=1+5(y-x) and let n=y-x (integer)

Then x=5n+1, and substituting for x into the original equation (as you did), it follows that y= 6n+1.

ehild

3. Apr 26, 2012

### mtayab1994

Yes, that's a clever way that you did. You can also work with congruences .

6x Ξ 1(mod5) and split up the 6x into 5x+x and then you'll be left with x Ξ 1(mod5).

Is that correct as well??

4. Apr 26, 2012

### HallsofIvy

Staff Emeritus
First it is incorrect to say "For every (x, y) in Z2 solve...". It should be simply "For (x, y) in Z2 solve ..."

The simplest way to solve this is to note that 6- 5= 1 so (1, 1) is a solution. Further if we take x= 1- 5k and y= 1+ 6k, 6x+ 5k= 6(1+ 5k)- 5(1+ 6k)= 6+ 30k- 5- 30k= 1 for all k. That is, any (x,y) of the form (1+ 5k, 1+ 6k), for k any integer, which is, of course, the same as your solution, is a correct solution.

5. Apr 27, 2012

### vela

Staff Emeritus
Yes, that's correct.