Describe the set $A = \{ 7a +3b: a,b \in \mathbb{Z}\}$ So in Hammack's "Book of Proof" there is the following example when he introduces the concept of set-builder notation in chapter 1.1.

Describe the set $A = \{ 7a +3b: a,b \in \mathbb{Z}\}$

In which he then proceeds to state the following:

If n is any integer, then $n = 7n +3(−2n)$, so
$n = 7a+3b$ where $a = n$ and $b = −2n$. Therefore $n \in A$. We’ve now shown that
$A$ contains only integers, and also that every integer is an element of $A$.
Consequently $A = \mathbb{Z}$.

My question is: are the values $a=n$ and $b=-2n$ arbitrary? If not, where did they come from?
 A: Where did $a=n$ and $b=-2n$ come from?
Well, since $7$ and $3$ are relatively prime,
one can use the extended Euclidean algorithm to find a Bezout relation;
e.g., $7(1)+3(-2)=1$.  Thus, any integer $n=7(n)+3(-2n)$.
A: $n$ is arbitrary.
ANd we use the arbtirary $n$ to define $a,b$.  $a, b$ are not arbitrary as they are determined by and are dependent upon $n$.
But the $n$ which determines them is arbitrary.
That is standard and acceptable.
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Take 2:  I'm not sure I understand the question but...
The author suspected (s/he actually knew) that $A = \mathbb Z$.  And to prove it s/he takes $n \in \mathbb Z$ is to be an arbitrary integer and s/he wants to prove that $n \in A$.
So she needs to solve $n = 7a + 3b$.  that's one equation and two unknowns.  That should be doable.  In fact there should be infinitely many solutions.
We need $3b = n - 7a$ so $b = \frac {n-7a}3$ so we need to choose $a$ so that $\frac {n-7a}3$ is an integer.  But $7a= 3(2a) + a$ that means $\frac {n-7a}3 = \frac {n-a- 6a}3 = \frac {n-a}3 -2a$ so we just need for $\frac {n-a}3$ to be an integer.
The easiest way to do that is to let $a=n$ then $b = \frac {n-n}3 - 2n=-2n$.
That that's one solution.  If $a=n$ and $b =-2n$ then $7a + 3b = 7n +3(-2n) = n$.
But that's only one solution. We could have done let $a=4n$ and then $b=\frac {n-4n}3 - 2*4n = -9n$ and so $7a + 3b = 7*4n +3(-9n) =n$ are we could have let $a = -2n$ so $b = \frac {n-(-2)n}3 -2*(-2n)=n+4n=5n$ and $7a + 3b = 7(-2n)+ 3(5n) = n$. etc.
Is that what you were asking?
So anyway.... so for any $n$ we can find $a,b$ where $a=n; b=-2n$ and $7a+3b= 7(n) + 3(-2n) = n\in A$.  So $\mathbb Z \subset A$.
And for any arbitrary $w = 7a + 3b$ for any possible $a$ and $b$ we have $a,b,7,3\in \mathbb Z$ so $w=7a + 3b \in \mathbb Z$ so $A\subset \mathbb Z$.
So $A = \mathbb Z$.
