Elements of the set {6a + 2b} where a and b....

[PeroK]

I just want to say I would do this entirely differently. First, $6a +2b$ is an even integer. So, the set is a subset of the even integers. To show that it is all even integers, we take any even integer $2k$ and this is in our set with $a = 0$ and $b = k$.

Is that not the obvious approach here?

 

[fresh_42]

Is that not the obvious approach here?

This depends on whether you want to emphasize on:
$$a\mathbb{Z}+b\mathbb{Z} \subseteq d\mathbb{Z} \subseteq a\mathbb{Z}+b\mathbb{Z} \Longrightarrow a\mathbb{Z}+b\mathbb{Z} =d\mathbb{Z}$$
or on the fact that $\mathbb{Z}$ is a principle ideal domain with
$$
a\mathbb{Z}+b\mathbb{Z} =\operatorname{gcd}(a,b)\mathbb{Z} \quad\text{ and }\quad a\mathbb{Z}\cap b\mathbb{Z} =\operatorname{lcm}(a,b)\mathbb{Z}
$$

However, as always in real life, the truth originates in a completely different post. It was an example of how $\{2x\in \mathbb{Z}\,|\,|x|\leq 5\}$ could be written better than it was in an OP. @Math100 took this as a template to write equations of sets. Your approach needs a different template to make it written properly because it partly specifies the coefficients which are arbitrary at the beginning. I tried to avoid confusion and pointed out the gcd, not the equality of sets.