My method to solving the Arithmagons is to add all of the numbers in the red-bordered boxes and then halve them. You divide them by two because each number in the purple-bordered boxes represent two numbers.
You then pick one of the numbers in the red-bordered boxes and take it away from the number you got when you did the previous operation. You then get the number that you have in the only purple-bordered box that is not connected to the red-bordered box the you subtracted. This is because the other two numbers are left over, and you can safely say that the number that is left over goes in that box.
You then take away the number that you just got from any of the connecting red-bordered boxes, and you get the other number in the purple-bordered box for that total. You then do the same for the only purple-bordered box that's left over, and you have solved the problem.