421. Solve the equations with two unknowns:
a) $x^{2}+y^{2}-4 x+6 y+13=0$
b) $x y-1=x-y$.
$\triangle$ a) Transform the equation:

$$
\left(x^{2}-4 x+4\right)+\left(y^{2}+6 y+9\right)=0, \quad(x-2)^{2}+(y+3)^{2}=0
$$

Then

$$
x-2=0, \quad y+3=0 \Rightarrow x=2, \quad y=-3
$$

b) Gather all terms on the left side and factor the left side of the new equation:

$$
x y-1-x+y=0, \quad(x+1)(y-1)=0
$$

In addition to the obvious solution $(-1 ; 1)$ of the last equation, there are others: if the first factor in the left part of the equation is zero, then the second factor can be any value. We get the set of solutions of the equation $(-1 ; y)$, where $y$ is any number. Similarly, we find another set of solutions of the equation $(x ; 1)$, where $x$ is any number not equal to -1. (The last restriction is due to the fact that the solution of the equation $(-1 ; 1)$, which is already included in the first set of solutions, should not be included in the second set.)

Answer: a) $(2 ;-3) ;$ b) $(-1 ; y)$, where $y$ is any number; $(x ; 1)$, where $x$ is any number not equal to -1.