Prove if $56x = 65y$ then $x + y$ is divisible by $11$

Why not just $x+y=(56x-55x)+(66y-65y)=11(6y-5x)+(56x-65y)=11(6y-5x)$ (since $56x-65y=0$).

We have $56x\equiv x\pmod{11}$ and $65y\equiv -y \pmod{11}$. If $56x=65y$, it follows that $x\equiv -y\pmod{11}$, which is what we needed to show.

$56$ and $65$ are relatively prime, so if $56x=65y$, then $65\mid x$ and $56\mid y$; say $x=65m$ and $y=56n$. Then

$$56\cdot65m=56x=65y=65\cdot56n\;,$$

so $m=n$. 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 $11^2$.

$56x = 65y \implies x + y = 11(6y - 5x)$

Since $11$ does not divide $56$ and $11$ is prime, $11$ divides $x+y$ if and only if it divides $56(x+y)$. But $56(x+y)=56x+56y=65y+56y=121y$, which in fact is divisible by $11^2$.