Solving a two variable equation

[mtayab1994]

Homework Statement

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

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?

 

[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

 

[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??

 

[HallsofIvy]

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.

 

[vela]

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??

Yes, that's correct.