Show symmetry in a (x,y) | 3a=f(x,y) set

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The discussion focuses on demonstrating the symmetry of the set S = { (x,y) in Z | 5x+7y is divisible by 3 }. The initial approach involved manipulating the equation 3a = 5x + 7y to derive a corresponding expression for 5y + 7x. The solution was found by showing that 5y + 7x can be expressed in terms of 5x + 7y plus an additional term, 2x - 2y. This term was confirmed to be a multiple of 3, proving the symmetry of the set S. The conclusion is that S exhibits symmetry as required.
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Homework Statement



Let S = { (x,y) in Z | 5x+7y is divisible by 3 }

Show that S is symmetrical.

Homework Equations



None apart from basic algebraic knowledge.

The Attempt at a Solution


[/B]
The only thing I can think of is starting with 3a = 5x+7y and putting x (or y) into the corresponding number 5y+7x and show that that turns out to be a multiple of 3, to show that it is symmetrical. But that leads me nowhere as far as I can tell:

5y+7x = 5y + 7 * (3a-7y) / 5
 
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Nevermind, I got it:

5y+7x = 5x+7y + (2x-2y) and you isolate 2x-2y from the given 3a equation. Now this is a multiple of 3.
 
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