Prove $x + y$ is divisible by $11$. Is my solution correct? If $x$ & $y$ are natural numbers, and $56 x = 65 y$, prove that $x + y$ is divisible by $11$.
Solution)
$56$ and $65$ are relatively prime
So, $65∣x$ and $56∣y$
Let  $x = 65m$ and $y = 56n$
Then,
$56x = 65y$
$56.65m = 65.56n$,
$m = n$ 
Thus, the solutions are of the form $x = 65k$,$y = 56k$ for integers $k$, and
$x+y = (65+56)k = 121k = 11(11k)$.
Thus, $x+y$ is even divisible by $11$

Your solution is correct. Another way to prove is as follows;
We have
$$56 \equiv1\pmod{11}$$
Hence,
\begin{align}
(x+y) & \equiv 56(x+y)\pmod{11}(x+y) \equiv 56x+56y\pmod{11}\\
& \equiv 65y+56y\pmod{11} \equiv 121y \pmod{11} \equiv 0 \pmod{11}
\end{align}
