How do I prove the existence of uniqueness of the vector $x$? Exercise 3-1.b in Linear Algebra Done right. Here the exercise:

Let $w, v \in V$. Explain why there exists a unique $x\in V$ such that 
$$w+3x=v$$

At this point, Axler has just outlined the definition of a vector space. In all of these, let $u,v, w\in V$.
Commutativity:
$$u+v=v+u  $$
Associativity:
$$(u+v)+w=(w+u)+v$$
Additive Identity:
There exists a $0\in V$ such that
$$v+0=v$$
Additive inverse:
There exists a $w$ such that 
$$v+w=0$$
Multiplicative Identity:
$$1v=v$$
Distributive Identity:
$$a(u+v)=au+av$$
$$(a+b)v=av+bv$$
For $a, b \in \mathbf{F}$, where $V$ is a vector field over $\mathbf{F}$.

I am not sure how to go about proving the existence and uniqueness of this $x$. Should I break the vector into components somehow? I don't know the dimensions of the field though.
 A: Let's do this using just the items in the definition, explicitly.  These do not include the notion of components.
You definition also should include closure under vector addition and scalar multiplication, that is, when $(a,b) \in V$ then $a+b \in V$ and if $r \in \Bbb R$, $ra \in V$.
$$w \in V \mbox{  (given)}$$ 
$$\exists u \in V: w + u = 0 \mbox{  (additive identity)}$$ 
$$ \exists t \in V: t = v + u \mbox{  (additive closure)}$$
$$ \exists x \in V: x = \frac13 t\mbox{  (scalar multiplication closure)}$$
$x$ satisfies $w+3x = v$:
$$ w + 3x = w + 3 (\frac13 t) = w + (3 (\frac13)) t)= w + t= w + (u+v) \\
=(w+u)+v \mbox{  (additive associativity)}\\
=0+v = v
$$
That shows existence; now we need to show uniqueness.  So assume there is some other solution $y : w+3y=v$. Then:
$$
w+3y = v = w+3x\\
u+ (w+3y) = u+(w+3x)\\
(u+w) + 3y = (u+w) + 3x) \mbox{  (additive associativity)}\\
0+3y = 0 + 3x \\
3y = 3x \\
\frac13(3y) = \frac13 (3x) \\
(\frac13 3) y = (\frac13 3) x\\
y=x
$$
So any other solution is the same $x$ as the one first found, so the solution is unique.
A: Essentially you should "solve" for $x$ using the fact that every vector has a unique additive inverse and that the scalars form a field (say where $3$ is invertible). Then
$$
w+3x=v\iff 3x=v-w\iff x=3^{-1}(v-w)
$$ 
To be more formal you could substitute $x=3^{-1}(v-w)$ to show that it is a solution and then do a standard uniqueness proof to prove that it is the only solution.
A: Assume such a vector $x$ exists.
Then $$w + 3x = v \implies 3x = v - w \implies x =\frac13(v-w)$$
having added $-w$ (the additive inverse of $w$) to both sides of the equality, used commutativity of $+$, and then multiplied both sides of the equality by scalar $\frac13$.
So the only candidate is $x =\frac13(v-w)$.
Plugging it in:
$$w +3x = w + 3\cdot \frac13(v-w) = w + (v - w) = v$$
having used quasi-associativity $\lambda (\mu x) = (\lambda \cdot \mu) u$, and then $1\cdot u = 1$ and finally commutativity of $+$.
Therefore, such a vector $x$ exists and it is equal to $\frac13(v-w)$. 
