I have the following equations derived from simultaneous equations.
$x=\frac{2}{3-a}$
$y=\frac{3(1-a)}{1-a}$
For $a=3$ we get no solutions, which makes sense to me.
However, for $a=1$ my understanding is we get a family of solutions.
$x=1$
$y=\lambda$
What is the reasoning behind this? Why do we not simply get no solutions? I see that we are essentially dividing $0$ by $0$ but is this not still undefined?
which means you didn't start with $y = \frac {3(1-a)}{1-a}$ but that you arrived at the conclusion, probably from $y(1-a) =3(1-a)$.
Had you started with $y=\frac {3(1-a)}{1-a}$ then you'd be correct. $a =1$ is impossible. But if you start with $y(1-a) = 3(1-a)$ then $a = 1$ means $y*0 = 3*0$ which is perfectly fine.
If you divide both sides by an unknown you must do two things. You must specify that the result you get from the dividing is conditional case that assumes what you are dividing by is not $0$. And you must consider the other conditional case that it is $0$ and we do not divide it.
SO you did NOT correctly derive that $y = \frac {3(1-a)}{1-a}$. You derived that EITHER $a\ne 1$ and $y= \frac{3(1-a)}{1-a}$ OR that $y*0 = 3*0$. The second is not a problem and has "a family of solutions".