Let A = the present age of the princess.
Let B = the present age of the prince.

The problem talks about 
(1) a certain number of year in the future,
and 
(2) a certain number of years in the past.

Let x = the number of years in the future that is spoken of.
Let y = the number of years in the past that is spoken of.

A princess is (now) as old as the prince will be
(in x years).

Therefore

A = B + x

In x years, the princess will be twice the age that 
the prince was y years ago.

Therefore,

A+x = 2(B-y)

And y years ago, the princess was half the sum of their 
present ages:



We have 4 unknowns but only 3 equations.  So we
suspect that there may be more than one solution.



Simplifying those

A -  B - x      = 0
A - 2B + x + 2y = 0
A -  B     - 2y = 0

That give the matrix:

[1 -1 -1  0 | 0]
[1 -2  1  2 | 0]
[1 -1  0 -2 | 0]

which simplifies to rref form:

[1  0  0 -8 | 0]
[0  1  0 -6 | 0]
[0  0  1 -2 | 0]

A - 8y = 0
B - 6y = 0
x - 2y = 0

A = 8y   
B = 6y
x = 2y

(A, B, x) = (8y, 6y, 2y)

The first solution that makes sense is when y=1
Then A=8, B=6, x=2, and y=1

The princess' present age is 8
The prince's present age is 6
The number of years in the future talked about is 2
The number of years in the past talked about is 1.

Checking:
The sum of their present ages is 14.
Then half the sum of their present ages is 7.
And indeed, the princess was 7 1 year ago.
In x=2 years the Princess will be 10, and 
that is twice the age the Prince was y=1 
year ago, for he was 5 then.

The next answer is when y=2

(A, B, x) = (8y, 6y, 2y)

The next solution that is when y=2
Then A=16, B=12, x=4, and y=2

The princess' present age is 16
The prince's present age is 12
The number of years in the future talked about is 4
The number of years in the past talked about is 2.

Checking:
The sum of their present ages is 28.
Then half the sum of their present ages is 14.
And indeed, the princess was 14 2 years ago.
In x=4 years the Princess will be 20, and 
that is twice the age the Prince was y=2 
years ago, for he was 10 then.
 
There are many solutions, all found by substituting
arbitrary values for y. 

Edwin