Number Theory Proof Divisibility

Question

if $m\mid(35n+26)$, $m\mid(7n+3)$, and $m>1$, prove that $m=11$.

This is what I have so far but I don't know if my steps would lead me to conclusion I want. I don't know how to proceed.

Answer 1

writing $$5n+26=k_1m$$ and $$7n+3=k_2m$$ then we get ( by multiplication of the second equation by $-5$ and adding both $$11=m(k_1-5k_2)$$ Can you proceed?

Answer 2

We have $m \mid ((35n+26)-5*(7n+3))=11$ so $m=11$.