Single-digit even natural number solutions to the equation $a+b+c+d = 24$ such that $a+b > c+d$ 
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Two algebra questions 

How to approach the below question:

How many single-digit even natural number solutions are there for the equation $a+b+c+d = 24$ such that $a+b > c+d$?


The single-digit even natural numbers are $2$, $4$, $6$, $8$. The sum is $24$, quite big. So the number of possibilities is not large. Almost any careful listing will do the job. But here is a possible systematic approach. 
The average of our numbers must be $6$. If they are all $6$, we violate $a+b\gt c+d$. 
So there is at least one $8$. There can't be three $8$'s, that makes the sum too big. So the number of $8$'s is $1$ or $2$. 
Now listing should be straightforward. Do (i) one $8$ and (ii) two $8$'s.
(i) There is only one $8$, and the average is $6$, so we must have one $8$, two $6$'s, and one $4$. Where can they be? Because $a+b\gt c+d$, one of $a$ or $b$ must be $8$, and the other $6$. And therefore one of $c$ or $d$ is $6$ and the other $4$. List all cases. You should get $4$.
(ii) All yours!
