One equation, two variables?

We can see it as a system between two lines:

$31x + 29y - 1125 = 0$ and $x = y$ , whose intersection is at the point $P(\frac{75}{4},\frac{75}{4})$

As $\frac{75}{4} = 18,75$ is not an integer, let's valuate the nearest couple of integers $(x = 19, y=18)$ , which sums up to:

$(31\times19) + (29\times18) = 1111$

$1125 - 1111 = 14$ and the difference between the two parameters is $31 - 29 = 2$ , so we can tell that every time we step up x by 1 and step down y by 1 we obtain the result to step up by 2:

$(31\times20) + (29\times17) = 1113$

$(31\times21) + (29\times16) = 1115$

and so on. Hence, being $14 = 2\times7$ we can step up x by 7 and at the same time step down y by 7 obtaining

$x = 19 + 7 = 26, y = 18 - 7 = 11 \Rightarrow (31\times26) + (29\times11) = 806 + 319 = 1125$

31x=1125 mod 29 .. . . And we get residue x= 29k+26

And we put to the first equation

31(29k+26)+29y=1125

And we get y=11-k

And try k=0 and then (x,y) = (26,11)

x+y=26+11= 37

I first added 2y to both sides to get a common factor of x + y on the right side 31x + 29y + 2y =1125 + 2y I then simplified isolating x+y 31x + 31y = 1125 + 2y 31(x+y) = 1125 + 2y x+y = (1125 +2y)/31 I then found that 1125 mod 31 was 9 and thus to make x+y an integer I would need to add 22 to 1125 2y = 22 y = 11 since y turns out to be an integer we can substitute it into the equation to find x and verify its an integer x + 11 = (1125 + 2(11))/31 x = 37 - 11 x = 26 since it indeed is an integer, just add x + y to get the answer (37)

Note: can we just substitute in some values of y in the original equation as that would get the answer faster. (actually did that originally and x must be an integer because x + y comes out to be an integer and y is an integer)