# 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.

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*}