Showing $5a + 2b \equiv 0 \pmod 7$ is symmetric, pictorially I have a problem regarding the understanding of modulus.
So someone proved $5a + 2b \equiv 0\pmod 7$ is symmetric using the modulo circle,
and my brain cant comprehend how he meant that. I'm talking about circles like this:

I know there is a other way to prove it with $7 \mid 5a + 2b$

The relation is defined on $\Bbb Z$ like this: $a \sim b$ if and only if $5a + 2b \equiv 0 \pmod 7$. On the assumption this holds, we wish to show $b \sim a$, i.e. $5b+2a \equiv 0 \pmod 7$.
Using the definition of modular congruence, then, $\exists k \in \Bbb Z$ such that
$$\frac{5a+2b-0}{7} = k \iff 5a + 2b = 7k$$
Subtract $7a$ and add $7b$ to both sides. Then you have
$$-2a - 5b = 7(k - a + b)$$
Now multiply both sides by $-1$:
$$2a + 5b = 7(-k+a-b)$$
Thus, there an $\ell \in \Bbb Z$ (namely $\ell = -k+a-b$) such that
$$\frac{5b+2a}{7} = \ell$$
ensuring that $5b + 2a \equiv 0 \pmod 7$, i.e. $b \sim a$. Thus, symmetry.

Now, with this in mind, consider the "circle" for equivalence modulo $7$:

If $5a + 2b \equiv 0 \pmod 7$, then the sum $5a+2b$ is in that topmost notch. Moreover, adding multiples of seven to it will make $5a+2b$ cycle back to that same place. So what we do, then, is add $-7a-7b= 7(-a-b) \in 7 \Bbb Z$ to it. This gives us $-2a-5b \equiv 0 \pmod 7$. However, in this notch of the diagram, if $x$ is in it, so is $-x$. (That is, $x \equiv -x \pmod 7$. I leave it to you to justify why.) Thus, $-(-2a-5b)= 2a +5b \equiv 0 \pmod 7$ as desired.

...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.
