Stars and Bars question with restriction that two variables must be equal How would one proceed in their thought process when  trying to solve what appears to be a standard stars and bars equation, with $x_i\geqslant$ 0, e.g.
$$x_1+x_2+x_3+x_4=24$$
Except, let's say that there is a restriction that two of the variables must be equal. We can write the equation as such:
$$x_1+x_2+2x_3=24$$
At first, I tried substituting $2x_3$ with $y$, and solving for 
$$x_1+x_2+y=24$$
using the standard stars and bars method. However, that would result in over counting, because it does not account for restrictions on $y$, since $y$ has to be even.
Any suggestions on how to solve this sort of problem?
EDIT:
I realize that you can solve smaller numbers using cases, but what would be the approach if you had larger numbers (making it impractical to use cases)?
e.g.
$$x_1+x_2+x_3+x_4+2x_5=350$$
 A: If the two identical variables are equal to $x$, the rest of the sum is $n=24-2x$. You can pick positions of the two identical variables in ${4 \choose 2 }$ different ways. 
Now, you have to split $n$ between the two remaining variables and there are ($n+1=25-2x$) ways to do it in general but you have to be careful. Not all splits are valid.
Take for example $x=0$. You cannot choose (12, 12), (0,24) and (24,0) for the last two variable values because you will end up with two many identical variables. The same is true for $x=1,2,3,4,5,7,8$. In these cases you will have $n+1-3=22-2x$ available pairs for the last two variables.
Take now $x=6$ for example. There is only one pair of values that has to be avoided as values of the last two variables: (6,6). The same is true for $x=9,10,11$ (you can easily check that). In these cases you have exactly $n=24-2x$ choices to split $n$ between the last two variables.  
You can drop $x=12$ because it leads to 12+12+0+0 which is illegal. 
Possible values for $x$ are therefore 0, 1,2, ...,11. For each value of $x$ you can put two identical variables in ${4 \choose 2 }$ different places. So that factor multiplies everything. For every single value of $x$ you have either $(22-2x)$ or $(24-2x)$ways to set the remaining two variables. So the final result is:
$$\binom{4}{2}\times\left(\sum_{x\in\{0,1,2,3,4,5,7,8\}}(22-2x)+\sum_{x\in\{6,9,10,11\}}(24-2x)\right)=$$
$$6\times\left(4\times2+\sum_{x=0}^{11}(22-2x)\right)=$$
$$6\times\left(8+12\times22-2\sum_{x=0}^{11}x\right)=6\times(8+12\times22-2\times66)=840$$
