Prove that for all integers $x, y$ and $z$, if $z | (5x+11y)$ and $z | (6x+13y)$, then $z | y$.
All I can think now is that because $z | (5x+11y)$ and $z | (6x+13y)$, there exists integers $k$ and $m$ such that $zk = 5x+11y$ and $zm = 6x+13y$. Could anyone give an idea of how to proceed from there?
The question as currently stated doesn't make sense, because $c$ and $b$ are not defined,
but I can prove that $z|y$:
$z|5x+11y\implies z|6(5x+11y)=30x+66y$
$z|6x+13y\implies z|5(6x+13y)=30x+65y$
and both of these imply
$z|(30x+66y)-(30x+65y)=y$.
you can start frm here $$z|5x+11y \to z|(5x+11y) \times 6\\z|6x+13y \to z|(6x+13y)\times(-5)\\$$so $$z|30x+66y\\z|-30x-65y\\\to z|(30x+66y)+(-30x-65y)\\\to z|y$$
If $z|5x + 11y$ and $z|6x + 13y$ then $z|m(5x+11y)+k(6x+13y)$ for any integers $m,k$. So if we can find an $m,k$ so that $m(5x +11y) +k(6x + 13y) = x$ we will be good.
If $m(5x+11y) + k(6x +13y) = (5m + 6k)x + (11m+13k)y= x$ for every integer $x,y$ then $5m+6k =1$ and $11m + 13k = 0$.
So solve: $11(5m+6k) = 55m +66k =11$ and $5(11m+13k) = 55m + 6k = 0$ so $(55m + 66k)-(55m +65k) = k = 11 - 0 =11$. so $k =11$ and $5m + 55 =1$ and $11m + 143=0$ so $m =-13$.
So $z|11(6x+13y) - 13(5x+11y)=x$.
