I'm sorry Smeagol but you can't have the ring no matter how hard the riddle

20 pounds I believe.

I say 20 too
x is the weight of the picture right?
So:
x=10+[x÷2]
Then using algebra magic you multiply the 2 out so:
2x=20+x
Then move the x to the other side and change the sign to negative:
2x-x=20
Therefore:
x=20
Boom, picture weighs 20 pounds

Which makes it worth 50 words per pound.

I mean, grammatically speaking it weighs 10 pounds.

The obvious answer is 20lbs
10+x=y.
2x=Y
10+x=2x
2x-x=10
x=10
10+10=20.
y=20
But math isn’t so simple- nor is language. Math is contextual to the problem. That is why engineering is its own discipline beyond mathematics. Knowing which math to apply when is it’s own set of rules. Likewise- higher math doesn’t deal in physical properties but in concepts- to the point that we don’t need an analog in reality- it goes past “if you have one Apple... and then someone gives you another apple... how many apples do you have?”

So- this exercise exists in thought. We aren’t discussing a real picture. If we were- we could weigh it, or we could have known parameters such as it’s composition and relative dimensions to extrapolate from- much that we can say a distant star has a given composition and density and size- so the mass of it must likely be within a certain range. Since this picture exists as an abstract- we define the parameters of the picture.

So we have to make some assumptions like the weight of the picture is fixed... blah blah. But I mean... there are several possible answers and it really depends on how YOU want to define the math that applies and what constraints are on this picture.

It weighs nothing.

I use "it" very loosely.

from a computer logic standpoint it would be something like
int x=10
x=x+(x/2);
the computer does the math before setting the variable to a new number and inserts the variables before it does the math so it looks at that and goes
x is 10
x=x+(x/2)
10+(10/2)
x=10+(10/2)
x=15

@fly_to_mordor so are you going to chime in with the answer?