How do I solve this system of residue class equations?

Question

$$\overline{3}x+\overline{2}y=\overline1$$ $$\overline{5}x+y=\overline4$$ both in $\Bbb{Z}_7$

So I multiple the second by 2 to get $\overline{10}x+\overline2y=\overline8$. Then I subtract the first from the second to get $\overline7x=\overline7$

Does $x = \overline1?$ Or do I need to try to solve this by finding the inverse of $\overline7$ in $\Bbb{Z}_7$? Someone told me there are multiple solutions to this system but I'm not sure why that might be the case.

Answer 1

If you multiply the first equation with $\overline{4}$; you obtain that it is equal to the second equation since:

$\overline{3}.\overline{4}=\overline{12}=\overline{5}$;

$\overline{2}.\overline{4}=\overline{8}=\overline{1}$; and

$\overline{1}.\overline{4}=\overline{4}$.

So the two equations are equivalent, and you have more than one solution. In particular, since you work in a field, for every fixed $\overline{x}$ you have a unique $\overline{y}$ solving the equation.

The solution are: $(\overline{1},\overline{6})$, $(\overline{2},\overline{1})$, $(\overline{3},\overline{3})$, $(\overline{4},\overline{5})$, $(\overline{5},\overline{0})$, $(\overline{6},\overline{2})$, $(\overline{0},\overline{4})$.