Linear Diophantine using substitution $15x+16y=17$

Question

I'm not sure what part went wrong if someone could point out. I did the question again with Euclidean algorithm and got answers $x=-17+16t$ and $y=17-15t$, so with that in mind, this is what I did

$16y \equiv 17\pmod{15}$
$16y \equiv 32\pmod{15}$

$y \equiv 2\pmod {15}$

Sub back into equation

$15x+16(2+15t)=17$

$x=1-16t$ and $y=2+15t$

since $t$ is an integer

$t=0\; x,y=(1,2)$ and $t=1\; x,y=(-15,17)$ by using $u=a/d$, $v=b/d$

my answer was $(-15+16t,17-15t)$

Answer 1

From $15x = (17-16\times 2)-(16\times 15t)$ you should have $x=-1-16t$

So you are missing a minus sign in your solution.

And since $t$ is an arbitrary integer you can replace by $u=1-t$ and get the same set of solutions

$$(x,y)=(-1-16t,2+15t)=(-17+16u,17-15u)$$