Sum of subsets of a Vector Space 
Image is from "Linear Algebra Done Right" (Sheldon Jay Axler) page 20.
I"m having trouble with example 1.38.
Example 1.37 seems straightforward, I think $(x, 0, 0) + (0,y,0) = (x+0, 0+y, 0+0) = (x,y,0)$ which is what we define $U + W$ to be.
By the same logic, in 1.38, $(x,x,y,y) + (x,x,x,y) = (2x, 2x, y+x, 2y) = (x, x, \frac{y+x}{2}, y)$.
I can't see how we arrive at the "$(x,x,y,z)$" part in the $U+W$ in the solution.

The key point here is that the variable names $x, y, z$ are just placeholders in the definitions of $U$ and $W$: So, for example the $x$ in the definition of $U$ has nothing in particular to do with the $x$ in the definition of $W$. To avoid the confusion that can create, let's write rewrite definitions for $U$ and $W$ using different placeholder variable names for the two sets:
$$U := \{(a, a, b, b) : a, b \in F\}, \qquad W := \{(c, c, c, d) : c, d \in F\} .$$
Now, a generic element of $U + W$ looks like
$$(a, a, b, b) + (c, c, c, d) = (a + c, a + c, b + c, b + d) ,$$
so we could write
$$U + W = \{(a + c, a + c, b + c, b + d) : a, b, c, d \in F \} .$$
Renaming $x := a + c$, $y := b + c$, $z := b + d$ gives the simpler desired description of the set:
$$\boxed{U + W = \{(x, x, y, z) : x, y, z \in F\}} .$$
