An equation in two variables? Given 13x + 35y = 2000. How do I find positive integer solutions for this equation (without hit and trial).

My work :- I know I can use Bezout's Theorem to find integer solutions to this equation if I have first solution (x., y.). But I just want positive solutions. To find the first solution I tried using Euclid's algorithm but could not progress further. Please help. Or if you have a smaller way of doing this type of questions feel free to post them as answers. (for all integer solutions or only (+)ve integer solutions).
It's immediately obvious that $x$ has to be a multiple of $5$, if it has any hope of satisfying this. Hence write $x=5z$ and so we get:

$$35y+65z=2000$$

Now you can apply what Alfred suggested because $35+65=100$ which divides $2000$.
$13x+35y=400(70-65)$

$\iff13(x-2000)=35(800-y)$

$\implies \dfrac{13(x-2000)}{35}=800-y $ which is an integer

$\implies35$ divides $13(x-2000)$ hence $x-2000$

$x\equiv2000\pmod{35},x=35t+2000,t&gt;-\dfrac{2000}{35}&gt;-57,$

$13(35t)=35(800-y)\iff y=800-13t&gt;0\iff t&lt;?$