Representing $\{x | Ax = b\}$ as the nullspace plus an offset During a class on linear algebra my prof wrote $$\{x | Ax = b\} = x_p + N(A), x\in \mathbb{R}^n, A  \in \mathbb{R}^{m\times n}$$ (exactly as written)
For the last two hours I have been trying to understand how you can go from left hand side to right hand side in non-hand wavy style as was demonstrated in the lecture. 
The prof is a good mathematician so I don't think he made a mistake, so the problem is me. I hope someone can help me out. 
Here's my confusion:

Let $Ax_p = b$, and $Ax_n = 0$,
then $\{x \in \mathbb{R}^n  | Ax = b\} = \{x_p \in \mathbb{R}^n | Ax_p = b\}\cup \{x_n \in \mathbb{R}^n | Ax_n = 0\} $ 
At the final step I just do not feel comfortable writing $\{x_p \in \mathbb{R}^n | Ax_p = b\}\cup \{x_n \in \mathbb{R}^n | Ax_n = 0\} = x_p + N(A)$
First:  $\{x_p \in \mathbb{R}^n | Ax_p = b\} \neq x_p$
Second: $\{x_n \in \mathbb{R}^n | Ax_n = 0\} \neq N(A) = \{x \in \mathbb{R}^n | Ax = 0\}$
Third: $\{x_p \in \mathbb{R}^n | Ax_p = b\}\cup \{x_n \in \mathbb{R}^n | Ax_n = 0\} \neq \{x_p \in \mathbb{R}^n | Ax_p = b\} + \{x_n \in \mathbb{R}^n | Ax_n = 0\}$
Can someone please point me to the right direction and show me how this relation can be made true?
 A: The expression $x_p + N(A)$ does not stand for a union of sets; it is the set of all the points that result from adding $x_p$ to some vector $x_n$ in the nullspace. Here we are speaking of one specific vector $x_p$ which solves the equation $Ax=b$.
(That set can be called the translation of the nullspace of $A$ by $x_p$.)
By contrast, the union you wrote stands for the set of all points which are either a(any) solution to the equation $Ax=b$ or belong to the nullspace of A.
Hint:
You want to show the following equality of sets, for a given $x_p$ which satisfies $Ax_p=b$:
$$\{x\in\mathbb R^n | Ax=b\} = \{x \in \mathbb R^n| x = x_p + x_n \text{, with } Ax_n=0  \}.$$
The old-fashioned way of tackling this problem is to show the two set inclusions $\subseteq, \supseteq$ that imply the equality of the sets.


*

*For the first inclusion $(\subseteq)$: if $x$ is such that $Ax=b$ then show that it can be written as a sum of the known solution $x_p$ and some other vector in the nullspace. Think about it for a while. (Further hint: $x = x + x_p - x_p $)

*The other inclusion $(\supseteq)$ is much easier: if $y = x_p + x_n$, where $x_p$ is the known solution and $x_n$ belongs to the nullspace, show that it satisfies $Ay=b$.
