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!

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*}
