Solving two variables system equations with parameter above $\mathbb{Z}_7$

Question

Let:
$$(1+5a)x +y = 1$$ $$a^2x + y = 2$$

Eliminating the $y$ variable we have:
$$(-a^2 +5a +1)x = 6$$

Now, I should have find $y$ such that $(-a^2 +5a +1)y = 1$, but obvoiusly I can't do that because the parameter $a$.

What should I do instead?

Answer 1

First off, you need to solve $$(-a^2+5a+1)x=6\tag{$\star$}$$ instead of $(-a^2+5a+1)y=1.$ I'm not sure where that equation could have come from.

You'll need to proceed casewise, as the value of $a$ will make a difference. For example, one readily sees that if $a=0,$ then $(\star)$ becomes $x=6,$ from which it follows that $y=2.$ On the other hand, note that you can equivalently rewrite $(\star)$ as $$(-a^2+5a-6)x=-1\\(a^2-5a+6)x=1\\(a-2)(a-3)x=1$$ Clearly, then, if $a=2$ or $a=3,$ then there is no solution. There are $4$ cases left to consider. I leave them to you.

Nota Bene: If you are supposed to give a solution that is independent of $a,$ then something probably went wrong in the process of obtaining this system of equations. If you let me know how the problem was presented, and what you've done so far, I may be able to help you isolate any errors you've made.