Generating functions - difficulty with question 
Find the number of non-negative integers solutions to the equation $$x_1+x_2+x_3+x_4=12$$
  when $x_1=2x_2+2$ and $x_3 \le x_4$.

My try:
Iv'e substituted $x_1$, thus the equation is $3x_2+x_3+x_4=10$. Second, we may define $x_4=x_3+y$ where $y \ge 0$ and the final equation is $3x_2+2x_3+y=10$.
Now we can write $$g(x)=\sum_{k=0}^{\infty} \left(x^{3k} \right) \sum_{i=0}^{\infty} \left(x^{2i} \right) \sum_{j=0}^{\infty} \left(x^j \right)=\frac{1}{(1-x^3)(1-x^2)(1-x)}$$
I know how to continue, but it's far too complicated and messy... I'm looking for an easier way.
Please help, thank you!
 A: One way to simplify the computation would be to take out the $\frac{1}{1-x^3}$, so that you only have to do partial fractions on $\frac{1}{(1-x^2)(1-x)}$.  Then you get
\begin{align*}
\frac{1}{(1-x^2)(1-x)}
&= \frac{1}{4(1+x)} + \frac{1}{4(1-x)} + \frac{1}{2(1-x)^2} \\
&= \frac{1}{4} \sum_{i=0}^\infty (-x)^i
+ \frac{1}{4} \sum_{i=0}^\infty x^i + \frac{1}{2} \sum_{i=0}^\infty (i+1)x^i \\
&= \sum_{i=0}^\infty \left[ \frac{(-1)^i + 1 + 2(i+1)}{4} x^i\right]
\end{align*}
Then,
$$
g(x) = (1 + x^3 + x^6 + x^9 + \cdots) \sum_{i=0}^\infty \left[ \frac{(-1)^i + 1 + 2(i+1)}{4} x^i\right]
$$
Your answer is the coefficient of $x^{10}$ in $g(x)$, which is, using the above,
\begin{align*}
\quad \frac{(-1)^{10} + 1 + 2(10 + 1)}{4}
&+ \frac{(-1)^{7} + 1 + 2(7 + 1)}{4} \\
+ \frac{(-1)^{4} + 1 + 2(4 + 1)}{4}
&+ \frac{(-1)^{1} + 1 + 2(1 + 1)}{4} \\
&= \frac{24}{4} + \frac{16}{4} + \frac{12}{4} + \frac{4}{4} \\
&= 6 + 4 + 3 + 1 \\
&= 14.
\end{align*}
A: You say you are looking for an easier way - so set x2=0,1,2,3 and you will get x1=2,4,6,8 respectively (x2>3 doesn't work). 
The remainder to 12 (10,7,4,1) is to be split between x3 and x4 such that x3≤x4. This yields 14 solutions in total (6+4+3+1).
Certainly not the mathematically most beautiful, but the easiest way to solve this question.
