Determining $x + y + z$ $x$, $y$ and $z$ are different prime numbers;
$$x(z-y) = 18$$
$$y(z-x) = 40$$
Determine $x+y+z$. 
I conceive of a way to solve this problem, which is to analyze the cases. For instance, Let's recall $x = 2$, then $z-y$ should be $9$. It will also give us $y = 5$. However, it doesn't satisfy the condition due to $z = 14$, which is not a prime number. How would you solve this question? 
Regards!

Given that the are different integers and prime the $x(z-y) = 18$ means $x = 2$ or $x=3$. $z > y$ and $z -y = 9$ or $z-y= 6$.
$y(z-x) =40$ mean $y = 2$ or $5$. $z> x$ and $z-x = 20$ or $z-x =8$.
There are only  so many cases to check.
1: $x =2; y=5$; So $z-5=18;z-2 = 8$ so $z = 23=10$. This is impossible.
2: $x=3; y = 2$; So $z-2 = 6; z-3= 20$ so $z = 8 = 23$.  This is impossible.
3: $x=3; y = 5$; So $x-6 =6; z-3=8$ so $z = 11$.  This is possible.  
As those are the only three options  $x = 3; y = 5; z = 11$.
