Find a basis for the subspace sum and then calculate its dimension. 
By definition, $U + V = \{\mathbf{u} + \mathbf{v} : \mathbf{u} \in U\ \; \& \; \mathbf{v} \in V\}$.
  Let $U = \{ \; u_1 = (1, 1, 0, \color{green}0), u_2 = (-3, 7, 2, \color{green}1) \;\}, V = \{ \; \mathbf{v} = (x_1,x_2,x_3,0) : x_i \in \mathbb{R} \; \}$.
  Find $dim(U + V)$. (Source)


My work: $U + V = \{\; \mathbf{u_1 + v} \; , \; \mathbf{u_2 + v}  \; \}
= \{\; (1 + x_1, 1 + x_2, x_3, \color{green}{0}) \; , \; (-3 + x_1,7 + x_2, 2 + x_3, \color{green}{1}) \; \}$.
Since the last entry of $\mathbf{u_1 + v} $ is $ \color{green}{0}$ and that of $\mathbf{u_2 + v}  $ is $ \color{green}{1}$, thus $U + V$ is lin-indept.
Then how and why would you divine/previse $U + V = \mathbb{R^4}$? Then what's being done?
I also tried to determine the $\operatorname{span}(U + V)$  by solving for $a_1$ and $a_2$, in terms of the $b_i$s, in: $a_1(\mathbf{u_1 + v}) + a_2(\mathbf{u_2 + v}) = (b_1, b_2, b_3, b_4) \implies$
$\left[ \begin{array}{cc|c}
1 + x_1 & -3 + x_1 & b_1\\
1 + x_2 & 7 + x_2 & b_2 \\
x_3 & 2 + x_3 & b_3 \\
0 & 1 & b_4 \end{array} \right] $. Too involved?
This question precedes dimensions/theorems of the 4 subspaces, Orthogonality, Determinants, eigenvalues and eigenvectors, and linear transformations. Please omit them.

2nd Supplementary to Pete L. Clark's Answer :
$\Large{{2.1.}}$ Doesn't $U+V = \mathbb{R}^4$ straightaway imply: $\{\mathbf{e_i}\}_{1 \le i \le 4}$ is a basis for $U + V$?
What's the answer doing? Why bother to note that $\mathbf{e_4} = \cdots $ is a linear combination of $ \cdots $?
$\Large{{3.1.}}$ What's wrong with my work intended for finding a basis for $(U + V)$? 
 A: The subspace $V$ has dimension $3$ inside $\mathbb{R}^4$ (thus it is a hyperplane).  Since $V \subseteq V+U \subseteq \mathbb{R}^4$, the first subspace has dimension $3$ and the last subspace has dimension $4$, we must have either that $\operatorname{dim} V+U = 3$ -- and thus $V+U = V$ -- or $\operatorname{dim} V+U = 4$ -- and thus $V+U = \mathbb{R}^4$.  Moreover, $V+U = V$ holds if and only if $U \subset V$.  
So it all boils down to whether $U$ is contained in $V$.  This is immediate to check: does every vector in $U$ have last coordinate equal to $0$?  Clearly not -- look at $u_2$ -- so $U+V = \mathbb{R}^4$.
Added: With regard to the "supplementary": 1) Previously I had a "$W"$ in my answer.  That was a typo: all instances of it have been changed to $U$.  2) Yes, if $U+V = \mathbb{R}^4$ then $\{e_1,e_2,e_3,e_4\}$ is a basis for $U+V$.  You asked how one would know that $U+V = \mathbb{R}^4$, so my answer gave a more detailed and motivated argument for that.  Your work doesn't make much sense to me, since you start by writing that the set $U+V$ has just two elements $s_1$ and $s_2$.  Clearly this is not the case.  Nor is it clear what $s_1$ and $s_2$ mean, since you have unspecified variables $x_1$, $x_2$, $x_3$ in them.  
A: You know a basis of $V$: $\{e_1,e_2,e_3\}$; thus a spanning set for $U+V$ is
$$
\{u_1,u_2,e_1,e_2,e_3\}
$$
Perform Gaussian elimination on $\begin{bmatrix}
1&-3&1&0&0\\
1&7&0&1&0\\
0&2&0&0&1\\
0&1&0&0&0
\end{bmatrix}
$
and you'll find your basis:
$RREF = \begin{bmatrix}
    1 & ~ & ~ & 1  & ~ \\
    ~ & 1 & ~ & ~  & ~ \\
    ~ & ~ & 1 & -1 & ~ \\
    ~ & ~ & ~ & ~  & 1 \\
    \end{bmatrix}$ 
Of course, the specific case could be managed in a different way, since the subspace $U$ is clearly not contained in $V$, so by dimensionality reasons it must be $U+V=\mathbb{R}^4$. The method with Gaussian elimination works in any case: all you need is spanning sets for $U$ and $V$.
