John is three times as old as his son, Michael, and in x year time he will be twice as old as him

call Michael's age m
then John's age is 3 m

in x years, Michael will be (m + x) years old

in x years, John's age will be 3 m + x

The problem statement says that
3 m + x = 2 (m + x)
3 m + x = 2 m + 2 x
x = m
for example if today Michael is 10 and John is 30
then x = m = 10
and in 10 years
Michael is 20
and John is 40
and sure enough 40 is twice 20

You are supposed to write down two equations and solve them to obtain rhe age of John (J) and of Michael (M).

Let John's age now be J. Let Michael's age now be M. In x years, John's age will be
J' = J + x
and Michael's age will be
M' = M + x
If J'/M' = 2, then
(J+x)/M+x) = 2
J +x = 2M + 2x
You also know that J = 3M, so
3M + x = 2M + 2x
M = x

No matter what John's age is now, there is an x value that satisfies the two requirements. That is because you have only two equations in three unknowns.

1--J = 3S
2-- J + X = 2(S + X)
3--3S + x = 2S + 2x
4--S = x
5--J = 3x
6--Select an x and solve for J and S
7--x........10...15...20...25...etc.
...S........10...15...20...25...etc.
...J........30...45...60...75...etc.
.S+x........20...30...40...50...etc.
.J+x........40...60...80..100...etc.
(J+x)/(S+x)..2....2....2....2