Consider the following set of equations $$a_{0}b_{0} \neq 0, \hspace{3mm} a_{0}b_{1}+a_{1}b_{0} = 0, \hspace{3mm} \text{ and } a_{0}b_{2}+a_{1}b_{1}+a_{2}b_{0} = 0 $$ in the variables $\vec{a} = (a_{0}, a_{1}, a_{2}) \in\mathbb{Z}^{3} \text{ and } \vec{b} = (b_{0}, b_{1}, b_{2})\in\mathbb{Z}^{3}$. Let us write the solutions of the above system of equations as $(\vec{a}, \vec{b})\in\mathbb{Z}^{3}\times\mathbb{Z}^{3}$.
To slightly explain my question, I would like to use an example. The density of a subset can be defined as here and is somewhat known. I am only going to use that any arithmetic progression of the form $\pm a\mathbb{Z}+b$ has density $\frac{1}{a}$ and any cofinite subset of $\mathbb{Z}$ has full density.
To satisfy $a_{0}b_{0} \neq 0$, one can pick any non-zero $a_{0}, b_{0}$. So, the set of $a_{0}$ and the set of $b_{0}$ that satisfy the first equation, each have density $1$ respectively.
Now for each fixed choice of $a_{0},b_{0}$ that satisfy the first equation, the second equation is satisfied when $a_{1} = -\frac{a_{0}}{b_{0}} b_{1}$. If we pick any $b_{1} \in b_{0}\mathbb{Z}$ (say, $b_{1} = b_{0}k$), then $a_{1} = -a_{0}k$. But the set $b_{0}\mathbb{Z}$ has density $\frac{1}{b_{0}}$ in $\mathbb{Z}$ and then the set $-a_{0}\mathbb{Z}$ also has density $\frac{1}{a_{0}}$ in $\mathbb{Z}$.
At this point, the set of $a_{0}, b_{0}\in\mathbb{Z}$ that satisfy the first equation, and the set of $a_{0}, b_{0}, a_{1}, b_{1}$ that satisfy the first two equations each have positive density. Or at least that is what I wanted to argue.
We can similarly work out that when first two equations are satisfied, i.e., when $a_{0}, b_{0}$ are some fixed non-zero value, $b_{1} = b_{0}k$ and $a_{1} = -a_{0}k$, then the third equation becomes $$a_{2} = -\frac{a_{0}}{b_{0}}b_{2} -a_{0}b_{0}k^{2},$$ which is satisfied whenever $b_{2}\in b_{0}\mathbb{Z}$ (say, $b_{2} = b_{0}r$) and $a_{2} = -a_{0}r-a_{0}b_{0}k^{2}$.
Even in this case, since $a_{0}, b_{0}, r, k$ are fixed, the set of desired $b_{2}$ which can be anything in $b_{0}\mathbb{Z}$ has density $\frac{1}{b_{0}}$ and the set of desired $a_{2}$ which are anything in $-a_{0}\mathbb{Z}-a_{0}b_{0}k^{2}$ still have density $\frac{1}{a_{0}}$.
Like I explained above, since at every step, we have a positive density of solutions for the variables $a_{0}, b_{0}$, then for $a_{1}, b_{1}$ and then for $a_{2}, b_{2}$, I want to say that the set of solutions $(\vec{a}, \vec{b})\in\mathbb{Z}^{3}\times\mathbb{Z}^{3}$ for the above system of equations has positive density inside $\mathbb{Z}^{3}\times\mathbb{Z}^{3}$.
Does this sort of argument make sense? Is my presentation correct? I started because I thought that such a system of equations should have "plenty" of solutions, which later I learned could make sense in the context of density inside $\mathbb{Z}^{d}$.