Suppose $v, w \in V$ where $V$ is a vector space. Explain why there's unique $x \in V$ such that $v + 3x = w$ Let $x = \frac13(w + (-v))$. Then $v + 3(\frac13w + \frac 13(-v)) = v + w + (-v) = (v + (-v)) + w = 0 + w = w$, so that $x$ exists.
Since $x - \frac13(w + (-v)) = 0$, we have that $x$ is inverse of $- \frac13(w + (-v))$. Since scalar multiplication and vector addition are defined in $V$, then $- \frac13(w + (-v)) \in V$. Since every vector in $V$ has a unique inverse, $x$ must be unique.
My explanation is wordy and awkward, but is it convincing?

Your reasoning is fine, but your wording is a bit ugly, as you observe. A better approach might be to suppose that there are two such values, and then show that they must be the same: 
Suppose that 
\begin{align}
v + 3x &= w \text{ and} \\
v + 3x' & = w. 
\end{align}
Subtracting the second equation from the first, we get
\begin{align}
3x - 3x' &= 0 \\
3(x - x') &= 0  \text{ (distributive law) }\\
\frac{1}{3}(3(x - x') &= \frac{1}{3} 0  \text{ (multiplication by a constant preserves equality) }\\
(\frac{1}{3}3)(x - x') &= 0  \text{ (associativity of scalar mult'n) }\\
1 (x - x') &= 0  \text{ (arithmetic) }\\
x - x' &= 0  \text{ (identity rule for scalar mult'n) }\\
x &= x'  \text{ (add $s'$ to both sides; use additive inverses to cancel.) }\\
\end{align}
By the way, once you've done a few of these, everyone agrees you know how to do them, and you never have to write one again (unless you're answering an MSE question or teaching a linear algebra class). 
