$17 | 29x + 18y \implies 17 | 23x + 9y$ So, basically the title.
Prove that $17 | 29x + 18y \implies  17 | 23x + 9y$.
Plus, does then $17 | 11x + 8y$ ?
Can you please explain what's the idea for doing these type of problems .. ?
 A: For the first part :
$17 \mid 29x+18y \Rightarrow 17 \mid 17x+29x+18y \Rightarrow 17 \mid 46x+18y \Rightarrow 17 \mid 2\cdot (23x+9y) \Rightarrow 17 \mid 23x+9y$
For the second part :
$17 \mid 29x+18y \Rightarrow 17 \mid 17x+17y+12x+y \Rightarrow 17 \mid 12x+y$
On the other hand :
$17 \mid 23x+9y \Rightarrow 17 \mid 12x+y+11x+8y \Rightarrow 17 \mid 11x+8y$
A: For the first part:
Let us first reduce it mod 17. Hence 
$$29x+18y \equiv 12x+y \pmod{17}$$
$$23x+9y \equiv 6x+9y \pmod{17}$$
since $17$ is relative ly prime to $2$, just multiply the second equation by $2$. What can you conclude?
A: $\left.\begin{align} 
\!\!\bmod 17\!:\ \        0\, \equiv &\, \ 18y + 29x\\
\equiv &\,\ \ \ \ \ y+12x \\
\iff\ \color{#c00}y\,\equiv&\ {-}12x\equiv\color{#c00}{5x}
\end{align}\right\}\ $ If so then  $\ \left\{ \begin{align} 
       &\,\ 23x + 9\,\color{#c00}y\\ 
\equiv &\ \ \,\ 6x + 9(\color{#c00}{5x})\\
\equiv &\,\ 51x\,\equiv\, 0
\end{align}\right\}\ $ and $\ \left\{ \begin{align} 
       &\,\ 11x + 8\,\color{#c00}y\\ 
\equiv &\, \ 11x + 8(\color{#c00}{5x})\\
\equiv &\,\ 51x\,\equiv\, 0
\end{align}\right\}$
A: Here is the ansatz:

If $a(29x + 18y)\equiv 23x + 9y \bmod 17$ for some $a \in \mathbb Z$, then the result follows easily.

This has to work for all $x,y$ and so $29a \equiv 23 \bmod 17$, which gives $a \equiv 9 \bmod 17$. This $a$ also works for $18a\equiv 9 \bmod 17$.
This argument actually shows

$17 \mid 29x + 18y \iff 17 \mid 23x + 9y$

Here is an explanation.
Consider the matrix $\pmatrix{ 29 & 18 \\ 23 & 9}$.
Its determinant is $-153$, which is $0$ mod $17$.
Therefore, its two rows are linearly dependent mod $17$.
In particular, the second row is a nonzero multiple of the first row, mod $17$, which implies the result, even without computing an explicit multiplier. (Here it is important that $17$ is prime.)
For the second part, consider the matrix $\pmatrix{ 29 & 18 \\ 11 & 8}$.
