Number Of Solutions Of Homogeneous And Non-Homogeneous System let there be a matrix $A^{n*m}$ that $Ax=b$ 
the solution set of the homogeneous system  $H=(h\in F^m; Ah=0)$ 

the solution set of the non-homogeneous system $L=(l \in F^m; Al=b)$
How do |L| and |H| correlate?
Because $L=l_1+H \rightarrow |L|<|H|$  
How can I prove that if $L \neq \emptyset \rightarrow |L|=|H|$?
(Update: Can you please answer without affine space)
 A: Suppose $L \neq \emptyset$, and fix any reference element $\ell_0 \in L$. Then, the translation map $L \to H$ defined by $\ell \mapsto \ell - \ell_0$ is a bijection.
Remark Of course, you'll get different maps for different choices of $\ell_0$. This sort of thinking motivates the slogan that "an affine space is a vector space that has forgotten its origin"---any choice of element $\ell_0$ in an affine space $L$ can be thought of as making a choice of origin and identify $L$ with a vector space $H$.
A: $H$ is an vectoriel space
$L$ is an afine space which direction is $H$
If $L \neq \emptyset$ there is a solution $x$ to $Al=b$ that means that $L=(l \text{ such as }l=x+h \text{ for } h\in H)$
There you can see that $|H|=|L|$
Without affine space :
We suppose that $L \neq \emptyset$ so there is a solution $x$ to $Al=b$
For each $h \in H$, $A(x+h)=Ax + Ah = b + 0 = b$ so $x+h\in L$. Thus $|H| \le |L|$
For each $l \in L$, $b=Al=A(l-x+x)=A(l-x)+Ax=A(l+x)+b$ so $A(l+x)=0$ so $l+x\in H$. Thus $|L| \le |H|$
In conclusion $|L|=|H|$
