If $\gcd(x;y)=1$, why is $\gcd[(ax+by);(cx+dy)]=1$?

Question

I found this in a book:

$\gcd(x;y)=1$

prove that $\gcd[(x+4y);(2x+7y)]=1$

How can I prove this?

My reasoning is $\gcd(x;y)=1$ so with Bezout's we find $ax+by=1$ but I don't know how to continue?

Thanks

Answer 1

Let $d = \gcd(x+4y,2x+7y)$. You have $$2(x+4y)-(2x+7y)=y$$

so $d\mid y$. But $d\mid (x+4y)$, so $d\mid x$. So $d\mid\gcd(x,y)$, so $d=1$.