Diophantine equation with "extra" conditions

Question

My question is how can I solve Diophantine equations with additional restrictions. For example, what about $x+2y+5z=40$ coupled with $x+y+z=20$ where $x,y,z>0$?

Answer 1

The difference between any two integer solutions is an integer multiple of the cross product $(3,-4,1).$ By inspection, I can see that $(0,20,0)$ is a solution. Next $$ (3, 16,1), $$ $$ (6, 12,2) $$ $$ (9, 8,3) $$ $$ (12, 4,4) $$

Answer 2

$x+2y+5z=40$ and $x+y+z=20$ and $x,y,z > 0$

Subtracting you get $y + 4z = 20$, that is $y = 20-4z$

$y>0$ and $z>0$ implies $z \in \{1,2,3,4\}$.

Substituting $y=20-4z$ into $x+y+z=20$, we get $x=3z$.

So $x = 3z, \ y = 20-4z$, and $z \in \{1,2,3,4\}$.