Trouble finding all positive solutions on linear diophantine equations with 3 variables

Question

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?

Answer 1

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 $x$.

Answer 2

iF $y+2z=10\implies y=10-2z\implies0<z<5$