The question you opened
Four times as old, then twice as old
A father is 4 times as old as his son. In 20 years he will be twice as old.
How old is the son now?
10 years old
The son is 10 and the father is 40. In 20 years they are 30 and 60, and 60 is twice 30, so both conditions hold.
A father is four times as old as his son, and in twenty years he will be twice as old. The son is 10 and the father is 40. Set it up with one letter. Let the son be s, so the father is 4s. In twenty years the son is s plus 20 and the father is 4s plus 20, and at that moment the father is twice the son. So 4s plus 20 equals 2 times s plus 20, which is 2s plus 40.
Subtract 2s from both sides to get 2s plus 20 equals 40, so 2s equals 20 and s equals 10. The father is 40. Check it: in twenty years they are 30 and 60, and 60 is twice 30. There is a faster route that uses the one thing about ages that never moves. The gap between two people is fixed for life. Here the father is three times the son older than him, since 4s minus s is 3s.
In twenty years he is twice as old, which means the gap equals the son's age at that point. So s plus 20 equals 3s, giving 2s equals 20 and s equals 10, in one line. That fixed gap is the whole idea worth taking away. Ratios of ages shrink steadily as people get older, because both ages grow by the same amount while only one of them is being divided.
A father four times as old becomes three times as old, then twice, then eventually barely more. He is never twice as old twice. The same structure appears outside ages. Two bank balances growing by the same fixed amount each month. Two queues shortening at the same rate. Whenever both quantities change by an equal amount, the difference is constant and the ratio is the thing that moves. Solving for the difference first is usually the shorter road.
Technique: The age gap never changes