Linear Diophantine Equations in Three Variables $$
3x+6y+5z=7
$$
The general solution to this linear Diophantine equation is as described 
here (Page 7-8) is:
$$
x = 5k+2l+14
$$
$$
y = -l
$$
$$
z = -7-k
$$
$$
k,l \in \mathbb{Z}
$$
If I plug the original equation into Wolframalpha the solution is:
$$
y = 5n+2x+2 
$$
$$
z =-6n-3x-1
$$
$$
n \in \mathbb{Z}
$$
I can rewrite this as:
$$
x = l
$$
$$
y = 5k+2l+2
$$
$$
z = -6k-3l-1
$$
$$
k,l \in \mathbb{Z}
$$
However now two equations depend on two variables ($k,l$) and one on one variable $l$.
In the first solution one equation depends on two variables and two on one variable.
Questions:
How can I come from a representation like the one from wolfram alpha for the general solution to one where all equations depend on one distinct variable except one equation.
Is there always such a representation?
 A: 1142388    
$3x+6y+5z=7$
$3x=7-6y-5z$
$x=\frac{7-6y-5z}3=2-2y-2z+\frac{1+z}3$
New variable $a=\frac{1+z}3$
$y$ had no fractional residue,
so set $y=b$, another new variable.
$z=3a-1$
$x=\frac{7-6b-5(3a-1)}3=4-2b-5a$
Is ${3(4-2b-5a)+6b+5(3a-1)}=7$
true?   
A: A general method, consists in put the matrix $\begin{bmatrix}3&6&5\end{bmatrix}$ in Smith normal form:
$$\begin{bmatrix}3&6&5\end{bmatrix}=\begin{bmatrix}1&0&0\end{bmatrix}\begin{bmatrix}3&6&5\\1&2&2\\0&1&0\end{bmatrix}.$$
A solution of the equation $3x+6y+5z=7$ is clearly $(x,y,z)=(0,7,-7)$.
Clearly, a triple $(x,y,z)$ satisfies $3x+6y+5z=0$ if and only if:
$$\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3&6&5\\1&2&2\\0&1&0\end{bmatrix}^{-1}\begin{bmatrix}0\\k\\l\end{bmatrix}=\begin{bmatrix}2&-5&-2\\0&0&1\\-1&3&0\end{bmatrix}\begin{bmatrix}0\\k\\l\end{bmatrix}=\begin{bmatrix}-5k-2l\\l\\3k\end{bmatrix},$$
Thus, the set of all solutions of $3x+6y+5z=7$ are the triples of the form:
$$(x,y,z)=(-5k-2l,7+l,-7+3k),$$
that's:
$$\left\{\begin{matrix}x=-5k-2l\\y=7+l\\z=-7+3k\end{matrix}\right.$$
