Let $x, y_1, y_2, z_1, z_2$ be all positive integers and $y_1, y_2$ are relatively prime. Show that $x$ divides $z_1, z_2$.
I have checked through numerical example, and feel it is true. However I am not getting any proper way to complete it. What I saw is that, as $y_1, y_2$ are relatively prime integers so there exist two integers $u_1, u_2$ such that $y_1u_1+y_2u_2=1$. Let us now assume, at least one of $z_1, z_2$ be not divisible by $x$. Without loss of generality, let $z_1$ be the one that is not divisible by $x$. What can be done from here, or i was wondering if any other way is there.
It is false, let $x=2$, $y_1=2$, $z_1 = 3$, $y_2=3$, and $z_2 = 2$.
Then $x=2$ divides $y_1z_1=6$ and $y_2z_2 = 6$, but it does not divide $z_1=3$.