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.
 A: One redefines the variables. For instance, put
$$ x' := 2x,\quad y' := y+x\quad\mbox{and}\quad z' := 2y. $$
This results in
$$(x,x,y,y)+(x,x,x,y) = (x',x',y',z'). $$
Since the $x,y,z$ are arbitrary, the $x',y',z'$ are also arbitrary.

More formally, verify that $U+W$ is of the required form. On the one hand
$$(x,x,y,z) = (x,x,y,y) + (0,0,0,z-y)\in U+W.$$
Conversely,
$$(x,x,y,y)+(x,x,x,y) = (2x,2x,y+x,2y) =: (x',x',y',z').$$
NB! The symbol $2x$ in this context means $x+x$. This is not necessarily arithmetic with real numbers or what have you.
A: 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:
$$\boxed{U + W = \{(x, x, y, z) : x, y, z \in F\}} .$$
