Showing that a function is surjective (onto)? For example : $F:\Bbb R\rightarrow\Bbb R$ defined by $F(x) = \frac{2x+1}{3}$ 
I let $F(x)=Y$ which gives $Y=\frac{2x+1}{3}$ then simplify and solve for $x$ , what I have at the end is $x=\frac{3Y-1}{2}$ , now I don't get how does this prove that the function is onto ?
 A: A function $f$ from $A$ to $B$ is called onto, if and only if for every element b $\in B$, there is an element $a \in A$ with $f(a) = b$.
In other words, proving a function is onto comes down to showing that for every element in the codomain there exists an element in the domain which maps to it. Thus, it is an existence proof.
By picking an arbitrary $y$, you are finding a possible candidate for a domain element. Once you substitute it in the function and obtain $y$, an element in the codomain, you have demonstrated that it is a suitable domain element. 
A: I think this question is nicely answered already, but just to restate the facts and maybe give another point of view.
Let $A,B$ be a set (e.g. $A=B= \mathbb{R}$). Then cartesian product $\times$ of $A,B$ is defined as
$$
A\times B = \{(a,b): a \in A, b \in B\}
$$
Then a relation $R$ is defined as $R\subseteq A\times B$. So, basically a set of ordered pairs. A function (I think the most common definition) is as follows 
$f:A\to B$ is function if $f$ is relation ($f\subseteq A\times B$) such that for every $a \in A$ there is exactly one $b\in B$ such that $(a,b) \in f$.
The last expression $(a,b) \in f$ can be rewritten in many different ways, e.g. $(a,b) \in f \leftrightarrow a f b$ or most commonly $f(a) = b$.
There is also geometrical visualization (of relation in general):
Ex.1
Imagine $f \subseteq \mathbb{N}\times \mathbb{N} = \{(n, n): n \in \mathbb{N}\}$, then you can visualize as follows
\begin{align}
&f\\
1 &\mapsto 1\\
2 &\mapsto 2\\
&\vdots\\
n &\mapsto n\\
&\vdots
\end{align}
The question, whether $f$ is surjective (you can read the definition in other answers/comments) is now following:
Is there an arrow pointing to each number on the right?
The answer to Ex.1 is yes, there is.
Ex.2 $f:\mathbb{N} \to \mathbb{N}, f(n) = 2n$ and let us draw it
\begin{align}
&f\\
  &\hphantom{1 \mapsto} 1\\
1 &\mapsto 2\\
  &\phantom{3 \mapsto} 3\\
2 &\mapsto 4\\
&\vdots\\
n &\mapsto 2n\\
&\vdots
\end{align}
As you can see, there are no arrows pointing to odd numbers on the right side. So $f$ from Ex.2 is not surjective.
Of course this is nice only for discrete examples (or even finite one). But this is not advised as proof. This is merely a try to help you imagine, what onto means.
