What is the strategy to find all non-negative tuples $(a,b,c,d)$ with a constraint of $a+b+c+d=9$ and $4a+2b+c-d=31$?

Question

I am trying to solve the following problem

Find the coefficient of $x^{31}$ in the expansion of $(x^4+x^2-\tfrac{x}{12}-\tfrac{1}{3x})^9$.

By using multinomial expansion below,

$$ (x^4+x^2-\tfrac{x}{12}-\tfrac{1}{3x})^9 =\sum_{a+b+c+d=9} {9 \choose {a,b,c,d}}(x^4)^a (x^2)^b \left(-\frac{x}{12}\right)^c \left(-\frac{1}{3x}\right)^d $$

and with some algebraic manipulation, I need to find all non-negative tuples $(a,b,c,d)$ with a constraint of $a+b+c+d=9$ and $4a+2b+c-d=31$.

With Mathematica, I did the following.

ClearAll[data]
data = Flatten[Permutations /@ IntegerPartitions[9, {4}, Range[0, 9]], 1];
Select[data, 4 #[[1]] + 2 #[[2]] + #[[3]] - #[[4]] == 31 &]

The solution is $\left( \begin{array}{cccc} 8 & 0 & 0 & 1 \\ 7 & 1 & 1 & 0 \\ \end{array} \right)$.

Question

What is the strategy to find the tuples easily?

Answer 1

By adding the equations, you get $c=20-(5/2)a-(3/2)b$ and by subtracting them, you get $d=(3/2)a+(1/2)b-11$.
Plot the lines $c=0$ and $d=0$ on the $(a,b)$ plane. They, along with the $a$-axis and $b$-axis, bound the region where all four are nonnegative.
Select gridpoints within the region where $a$ and $b$ are eithwr both even or both odd.

Answer 2

For the conditions $a+b+c+d=9$ and $4a+2b+c-d=31$, consider possible values of $a$ starting at 9 and working to $0$. We see that the only possibilities are $a=8$ or $a=7$.

Case : $a=7$

$a=7,b=1,c=1,d=0$

Case : $a=8$

$a=8,b=0,c=0,d=1$

Answer 3

\begin{cases} a+b+c+d=9\\ 4a+2b+c-d=31 \end{cases}

Trial 1 (Failed)

Choose $a=9$ then the equations get reduced to \begin{cases} b+c+d=0\\ 2b+c-d=-5 \end{cases} No solution for $b$, $c$ and $d$.

Trial 2 (Succeeded)

Choose $a=8$ then the equations get reduced to \begin{cases} b+c+d=1\\ 2b+c-d=-1 \end{cases} The only solution is $b=c=0$ and $d=1$.

Trial 3 (Succeeded)

Choose $a=7$ then the equations get reduced to \begin{cases} b+c+d=2\\ 2b+c-d=3 \end{cases} The only solution is $b=c=1$ and $d=0$.

The following trials always fail. As I don't know how to prove with algebra, let list them all.

Trial 4 (Failed)

Choose $a=6$ then the equations get reduced to \begin{cases} b+c+d=3\\ 2b+c-d=7 \end{cases} No solution.

Trial 5 (Failed)

Choose $a=5$ then the equations get reduced to \begin{cases} b+c+d=4\\ 2b+c-d=11 \end{cases} No solution.

Trial 6 (Failed)

Choose $a=4$ then the equations get reduced to \begin{cases} b+c+d=5\\ 2b+c-d=15 \end{cases} No solution.

Trial 7 (Failed)

Choose $a=3$ then the equations get reduced to \begin{cases} b+c+d=6\\ 2b+c-d=19 \end{cases} No solution.

Trial 8 (Failed)

Choose $a=2$ then the equations get reduced to \begin{cases} b+c+d=7\\ 2b+c-d=23 \end{cases} No solution.

Trial 9 (Failed)

Choose $a=1$ then the equations get reduced to \begin{cases} b+c+d=8\\ 2b+c-d=27 \end{cases} No solution.

Trial 10 (Failed)

Choose $a=0$ then the equations get reduced to \begin{cases} b+c+d=9\\ 2b+c-d=31 \end{cases} No solution.