A fun arithmetic problem?

Question

Mathematics

There are two similar tasks:
The older brother said to the younger one: "Give me 8 kopecks, then I will have twice as much money as you." And the younger objected: “You better give me 8 kopecks, then we will have equal money.” How much money does each brother have?
Answer: 56 and 40. The
matches are in two piles. If you transfer 2 matches from the first pile to the second, then there will be 5 times more matches in the second than in the first. If, however, 5 matches are transferred from the second pile to the first, then in the first there will be three times more than in the second. How many matches are in each pile?

Answer: 4 and 8.
In the explanation for the first it is written:
The younger one asks the older one for 8 kopecks, claiming that they will have the same amount of money. Therefore, the younger one has 2*8=16 kopecks less than the older one. If the younger gives 8 kopecks to the elder, then the latter will have twice as much, i.e. no longer by 16 kopecks, but by 32 kopecks. Therefore, one half of the elder's money is known to us. So the older one had 56 kopecks, and the younger one had 40 kopecks.
I solved both problems, but with a system of equations. I understood the logic of the solution from the explanation for the first problem, but by analogy I can’t solve the second one (therefore I doubt that I understood the first one).
Can someone explain to me the solution by analogy to the second problem?

Answer 1

The solution of the first problem is based on the fact that in one of the cases two quantities become equal, and therefore the initial difference (16) is known above, from which you can dance.
In the second problem, there is no such case, and therefore it is impossible to obtain a solution similar in beauty. Is it possible to "virtually reduce" one of the heaps by three or five times, then the shifting operation will become more complicated, but the difference will be known again. But such a solution will be much more difficult to understand than a system of equations.

Answer 2

I got great answers -8 and -4 pieces)))
m.wolframalpha.com/input/?i=x-2%3D5%28y%2B2%29+%3B+3%28x%2B5%29%3Dy- 5&x=0&y=0

Answer 3

When viewed from the point of view of describing the solution, the tasks are completely different.
If, in fact, they are both solved by compiling a system of equations, it’s just that in the first case everything turns out beautifully, because is a common factor. Look closely, the description of the first problem clearly describes the algorithm for solving the system of equations.
Most likely, the author did not even bother about the description of the solution, but simply took the numbers from the ceiling and solved the system of equations with the help.

Answer 4

(y+2)/(x-2) = 5
y+2 = 5x - 10
y = 5x - 12
(x+5)/(y-5) = 3
x+5 = 3y - 15
x = 3y - 20
x = 15x - 36 - 20
14x = 56
x = 4
y = 20 - 12 = 8