Understanding why single-variable expansion of modular arithmetic is valid. I was reading: https://arxiv.org/pdf/1206.5114.pdf page 10 and came across an odd theorem. 
It states, that the system:
$$ \begin{matrix} 2x_1+ 3x_2  \equiv 1 \mod 5  \\ 3x_1 + 5x_2 \equiv 1 \mod 7 \end{matrix} $$ 
Can be "re-written" as: 
$$ \begin{matrix} 2x_1+ 3x_2 +5x_3 =  1   \\ 3x_1 + 5x_2 + 7x_3 = 1  \end{matrix} $$ 
Now what exactly is meant by "re-written here?" is it that for every solution in the first system there exists a solution in the second system? Or that for every solution in the second system there exists a solution in the first system? Or both? 
I find at least the notion of "BOTH" to be impossible. For the simple reason that if we let $X = 2x_1 + 3x_2$ and let $Y = 3x_1 + 5x_2$ 
Then we are claiming that if 
$$ X \equiv 1 \mod 5$$
$$ Y \equiv 1 \mod 7$$
That there exists an $x_3$ such that 
$$ X + 5x_3 = 1 $$
$$ Y + 7x_3 = 1$$ 
Now to me this seems very odd, since if we analyze 
$$ X \equiv 1 \mod 5$$ 
To me all that says is there is SOME integer $s$ such that $X = 5s + 1$ and similarly there is SOME integer $r$ such that $Y = 7r + 1$. The idea that $s = r$ is necessary seems blatantly false (let X = 11, Y = 8 as a counter example). With BOTH ruled out, (And our counter-example ruling out that every solution the first system implies a solution to the second system) it's pretty clear that at best every solution to the second system is a solution to first.
Now certainly I would call that a severe loss of information and not a simple "re-write", so does this paper have a mistake and they meant to communicate something else? or am I missing a major detail? or is the analysis up to here correct and the term "re-written" has a hidden loss of information not made clear by the paper. 

The solutions to $$2x_1+3x_2+5x_3=1, 3x_1+5x_2+7x_3=1$$ satisfy $$2x_1+3x_2≡1, \mod5 $$
$$3x_1+5x_2≡1,  \mod7$$
But the solutions to the second system are not necessarily solutions to the first system because you may have  $$2x_1+3x_2+5x_3=1, 3x_1+5x_2+7x_4=1$$ where $x_3\ne x_4$
