Second system of equations I've solved that system and recieved $y=\frac{1}{4}$, $x = -\frac{4}{5}$ but there are one more pair of $(x, y)$ in book and I don't know what I have to do to find its.
\begin{cases} 2x - 3xy + 4y=0 \\ x + 3xy -3x = 1 \end{cases}
In book there are two answers and the second answer is $(1, -2)$.
I've reached $y=\frac{1}{4}$, $x = -\frac{4}{5}$ through that way:
\begin{array}{lcl}2x - 3xy + 4y = 0\\2x - 3xy = -4y\\x(2 - 3y) = -4y\end{array}
2)
\begin{array}{lcl}x + 3xy - 3x = 1\\-2x + 3xy = 1\\-x(2 - 3y) = 1\end{array}
Get new system:
\begin{cases} x(2 - 3y) = -4y \\ -x(2 - 3y) = 1 \end{cases}
Then I divide first to second and get
\begin{array}{lcl}-1 = -4y\\y = \frac{1}{4}\end{array}
Next solve one of equation and get \begin{array}{lcl} x = -\frac{4}{5}\end{array}
But how to get $x = 1, y = -2$?

We have
$$2x-3xy+4y\quad=x+3xy-3x-1=0\implies 4x-6xy+4y+1=0$$
$$\implies\quad x = \frac{(4 y + 1)}{(6 y - 4)} \land 3 y\ne 2\qquad y = \frac{(4 x + 1)}{(6 x - 4)} \land 3 x\ne2$$
so the solution for one variable in terms of the other yields the same results in that $1)$ either independent variable can be any of an infinity of numbers except $\frac{2}{3}$ and,  $2)$ both variables cannot be integers and the same time.
examples
$$x=\frac{4(1)+1}{6(1)-4}=\frac{5}{2}\rightarrow (\frac{5}{2},1)\land (1,\frac{5}{2})$$
$$y=\frac{4(2)+1}{6(2)-4}=\frac{9}{8}\rightarrow (2,\frac{9}{8})\land (\frac{9}{8},2)$$
