How to solve $2{x_{1}}+2{x_{2}}+{x_{3}}+{x_{4}}={12}$

Question

How many solutions possible for the equation$$2{x_{1}}+2{x_{2}}+{x_{3}}+{x_{4}}={12}$$ all x are non-negative integer.

I see these links but I don't know how to solve this problem.(I know how to solve ${x_1}+{x_2}+{x_3}+{x_4}=12$)

How many solutions possible for the equation $x_1+x_2+x_3+x_4+x_5=55$ if

Enumerating number of solutions to an equation

Answer 1

Solve for $2a+b=12$. For each one of the $7$ solutions $(a,b)=(k_1,k_2)$, solve for possibilities which count $(k_1+1)\times(k_2+1)$.

Eg. A solution to $2a+b=12$ is $(a,b)=(5,2)$.

So, $x_1+x_2=5$ while $x_3+x_4=2$. Within themselves;

$(x_1,x_2)=(0,5)$,

$(x_1,x_2)=(1,4)$,

...

$(x_1,x_2)=(5,0)$

while

$(x_3,x_4)=(0,2)$,

$(x_3,x_4)=(1,1)$,

$(x_3,x_4)=(2,0)$

adding up to $(5+1)\times(2+1)=18$ solutions for $(a,b)=(5,2)$.

Answer 2

Solve $x_{1}+x_{2}+x_{3}+x_{4}=12-x_{1}-x_{2}$ for all natural $x_{1},x_{2}$ s.t $x_{1}+x_{2}\leq6$ s.t and sum the results.