Let x, and y be the two numbers the problem is asking for, then we can set the following system of equations:

Solving the second equation for x, we get:

Now, substituting x=2y in the first equation we get:

Solving for y we get:

Finally, we substitute y=10 in the first equation on the board and get:

Solving for x we get:

Answer: the numbers we are looking for are 20 and 10.