Find all functions $ f: \mathbb{R} \to \mathbb{R}$ satisfying $f(x + y) = x + f(y)$ I'm engaging in the quest for understanding functional equations and I am trying to solve the problem:

Find all functions $ f: \mathbb{R} \to \mathbb{R}$ satisfying $f(x + y) = x + f(y)$

This is what I have done so far:
Let $y = 0$, then:
$f(x + y) = x + f(y) \implies f(x) = x + f(0)$
Let $y = -x$, then:
$f(x + y) = x + f(y) \implies f(0) = x + f(-x)$
Then
$f(x) = x + x + f(-x) = 2x + f(-x) \implies f(x) = 2x + f(-x)$
I think that what I obtained is not the final answer since $f(-x)$ is a function itself, but I'm stuck. I have been trying to obtain a fixed value for $f(0)$ so I can substitute it in my first equality, but I don't think it is possible. 
How can I proceed to obtain a general form of the equation? Is it even possible to obtain a general form?
 A: The solutions to this functional equation are precisely those functions $f:\mathbb R\to\mathbb R$ that are of the form $f(x)=x+c$ for all $x\in\mathbb R$, where $c\in\mathbb R$ is some constant.
More precisely, if $f:\mathbb R\to\mathbb R$, then the following are equivalent:
(1) $f(x+y)=x+f(y)$ for all $x,y\in\mathbb R$;
(2) there exists some $c\in\mathbb R$ such that $f(x)=x+c$ for all $x\in\mathbb R$.
Proof:
(1) $\Rightarrow$ (2) If (1) holds, then, for any $x\in\mathbb R$, $f(x)=f(x+0)=x+f(0)$. Define $c\equiv f(0)$.
(2) $\Rightarrow$ (1) If (2) holds, then there exists some $c\in\mathbb R$ such that $f(x)=x+c$ for all $x\in\mathbb R$. Therefore, for any $x,y\in\mathbb R$: $$f(x+y)=(x+y)+c=x+(y+c)=x+f(y).$$
A: We shall first show that $f(x)$ is differentiable, and further that the derivative is $1$ at all $x$. This is easily seen as
$$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{f(h+x)-f(x)}{h}=\lim_{h\to 0}\frac{h+f(x)-f(x)}{h}=\lim_{h\to 0}\frac{h}{h}=\lim_{h\to 0}1=1.$$
Thus $f(x)=x+c$ for some constant $c$. Can we improve this upon this? Unfortunately, this is as much as we can learn about $f(x)$. This is because for any constant $c$,
$$f(x+y)=x+y+c=x+f(y)$$
and thus $f(x)$ satisfies the functional equation.
A: I am assuming that
$$f(x + y) = x + f(y)$$
for all real numbers $x$ and $y$.
Set $y = 0$.  Then
$$f(x) = x + f(0).$$
Let $c = f(0)$, so $f(x) = x + c$.
Plugging back into the functional equation, we are given, we get
$$x + y + c = x + y + c.$$
So the solution is all functions of the form $f(x) = x + c$, where $c$ is a constant.
