How many positive integer solutions are to a system of linear equations? 
Given the linear system of equations:
  $$
\begin{cases}
x_1 + x_2 + x_3 = n\\
x_1 + x_2 + x_3 + x_4 + x_5 = 3n\\
2x_1 +2x_2 +2x_3 +x_4 +x_5 +3x_6 +3x_7 +3x_8 =10n
\end{cases}
$$
  how many solutions are in $\mathbb N\cup\{0\}$?
The solution must not be using sum notation like $\sum y$.

I know how to find the number of solutions to the regular equations like $x_1+x_2+x_3+\dots=n$ but I'm not sure how to do this for a system of equations. I thought of substituting some $x$'s:
$$
x_1 + x_2 + x_3 =3n - (x_4 + x_5)\\
x_4 +x_5=10n-2(x_1 +x_2 +x_3)  -3(x_6 +x_7 +x_8)\\
\implies
x_1 + x_2 + x_3=3n-(10n-2(x_1 +x_2 +x_3)  -3(x_6 +x_7 +x_8))\\
\implies x_1+x_2+x_3+3(x_6+x_7+x_8)=7n \quad *
$$
As far as I understand finding the number of solutions for the system is equivalent to finding the number of solutions to the equation *.
The only next step from here I can think of is using generating functions:
$$
(1+x+x^2+\dots)^3(1+x^3+x^6+x^9+\dots)^3
$$
and we need to find the coefficient of $x^{7n}$.
From the closed form identities we have:
$$
\sum_{k=0}^{\infty} {3-1+k\choose k}x^k\cdot \sum_{i=0}^{\infty} {3-1+i\choose i}x^{3i}
$$
But I have no idea now how to find the coefficient of $7n$ from here and certainly not without using some kind of sum notation.

This is more of a hint rather than a solution but still:
You have $$x_1+x_2+x_3=n.$$ Substituting this into your second equation gives 
$$x_4+x_5=2n.$$
Substituting both in the third gives 
$$x_6+x_7+x_8=2n$$
If I've understood correctly you can solve this new system of equations since each $x_i$ appears in exactly $1$ equation.
