Trouble finding all positive solutions on linear diophantine equations with 3 variables I'm having trouble calculating all possible solutions for a system of 2 diophantine equations with 3 variables.
The question is:
Find all the positive solutions in integers of 
$$x  + y + z = 31$$
$$x+2y+3z=41$$
I substituted x in the second equation giving me 
$$x=31-y-z$$
$$31-y-z+2y+3z=41$$
Which gives me the equation:
$$y+2z=10$$
One solution would be $x=2, z=4$. Since I want only positive solutions I have the following inequalities
$$2+2t>0$$ 
and 
$$4-t>0$$
From that:
$$t>1$$ and $$4>t$$
So $4>t>1$. Since I only want integer answers, $t=3$ and $t=2$.
Substituting in the formula I get $$x=23, y=6, z=2$$ and $$x=22, y=8, z=1$$ which are both correct. 
However, in the book there are two other answers, $$x=24, y=4, z=3$$ and $$x=25, y=2, z=3$$
My question is, how do I arrive at those two other answers?

Note that since $y>0$, $2z<y+2z=10$, then $0<z<5$. If $z=1$ we get $y=8$. If $z=2$ we get $y=6$. If $z=3$ we get $y=4$, and if $z=4$ we get $y=2$. Therefore we get the pairs $(y,z)=(8,1), (6,2), (4,3), (2,4)$. Using that $x=31-(y+z)$, you can get the values of $x$. 
