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 $a+b > c+d$?


Here is one approach you can use:
The even single-digit (presumably in base 10) natural numbers are precisely 2, 4, 6, 8.
Thus, if $y$ and $z$ are such numbers, then $4\leq y+z\leq 16$, and of course $y+z$ is even. 
Break the problem up as follows: find the number of solutions $w$, $x$ to $w+x=24$ such that $w$ and $x$ are even numbers such that $4\leq x<w\leq 16$, and then for each choice of $w$ and $x$, find the number of solutions to $a+b=w$ and the number of solutions to $c+d=x$ such that $a,b,c,d\in\{2,4,6,8\}$.
