Let x and y be the unknown numbers.
We know that one number is 10 more than the other, this can be express as:

We also know that twice the sum of both numbers is equal to 8, the sum of the two numbers is:

Twice this sum is:

And this have to be equal to 8, then we have the equation:

Hence we have the system of equations:

To solve this system we take the expression for y from the first equation and plug it in the second, then we solve for x:

Now that we know the value of x we plug it in the expression for y:

Therefore the numbers we are looking for are -3 and 7