Help to solve a system equation: $x-y+xy=-4$; $xy(x-y)=-21$. I need to solve a system equation. Here's how it looks:
$x-y+xy=-4$
$xy(x-y)=-21$
I tried to substitute $x-y$ with $w$ and $xy$ with $t$ to simplify everything. After that I got this system equation:
$w+t=-4$
$tw=-21$
I solved this new system equation and got these results: 
$t_1=-7, w_1=3$
$t_2=3, w_2=-7$
After all these steps I ended up with 2 new system equations:
1.
$xy=-7 (\leftarrow t)$
$x-y=3 (\leftarrow w)$
2.
$xy=3 (\leftarrow t)$
$x-y=-7 (\leftarrow w)$
Looks like the first one doesn't have any solutions. And I can't solve the second one. Am I doing wrong steps? Please help me to solve this system equation. Thanks.
 A: The first equation is linear in $y$; we have
$$
y(x-1)+x+4=0.
$$
Since $x=1$ is imposssible here, we can write $y=\frac{x+4}{1-x}$ and substitute this into the second equation. Then one obtains
$$
(x^2 + 7x - 3)(x^2 - 3x + 7)=0.
$$
A: HINTS: 
We have the equations $$x-y+xy=-4\implies x-y=-4-xy\tag{1}$$ and $$xy(x-y)=-21\tag{2}$$ so substituting $(1)$ into $(2)$, we get $$xy(-4-xy)=-21\implies(xy)^2+4xy-21=0$$ Let $u=xy$. Then $$u^2+4u-21=0$$ and so...
A: \begin{align} 
x-y+xy&=-4
,\\
xy(x-y)&=-21
.
\end{align}  
We can consider $u,v$
\begin{align} 
u&=x-y
,\\
v&=xy
\end{align}
as two roots of the quadratic equation,
\begin{align}
z^2+az+b&=0
\tag{1}\label{1}
,\\
a&=-(u+v)=-(x-y+xy)=4
,\\
b&=uv=xy(x-y)=-21
,
\end{align}
so \eqref{1} is
\begin{align}
z^2+4z-21&=0
\end{align}
which has two real solutions
\begin{align}
z_{1,2}&=3,-7
.
\end{align}
Since we can not distinguish 
which of the two is $u$ and which is $v$,
we need to consider both cases:
\begin{align}
\text{Case 1.}\quad
&
\begin{cases}
u&=x-y=3
,\\
v&=xy=-7
.
\end{cases}
\end{align}
And 
\begin{align}
\text{Case 2.}\quad
&
\begin{cases}
u&=x-y=-7
,\\
v&=xy=3
.
\end{cases}
\end{align}
We can exploit the quadratic equation again (twice),
considering two roots, $x$ and $-y$,
since we can get the values of the sum and product of named roots
as
\begin{align}
\text{case 1.}\quad
&
\begin{cases}
x+(-y)=3
,\\
x(-y)=-xy=7
,
\end{cases}
\end{align}
and
\begin{align}
\text{case 2.}\quad
&
\begin{cases}
x+(-y)=-7
,\\
x(-y)=-xy=-3
.
\end{cases}
\end{align}
The two quadratic equations are then
\begin{align}
\text{case 1:}\quad
t^2-3t+7&=0
,\\
\text{case 2:}\quad
t^2+7t-3&=0
.
\end{align}
The first case does not have real roots,
but the second one provides two real roots,
\begin{align}
t_{1,2}&=
-\tfrac72\pm\tfrac12\,\sqrt{61}
.
\end{align}
And as before, we have to consider two options,
\begin{align}
\begin{cases}
x&=-\tfrac72+\tfrac12\,\sqrt{61}
,\\
y&=-(-\tfrac72-\tfrac12\,\sqrt{61})
.
\end{cases}
\end{align}
and 
\begin{align}
\begin{cases}
x&=-\tfrac72-\tfrac12\,\sqrt{61}
,\\
y&=-(-\tfrac72+\tfrac12\,\sqrt{61})
.
\end{cases}
\end{align}
The final substitution into original pair of equations
confirms that the two solutions are indeed
\begin{align}
 \begin{cases}
  x&=-\tfrac72+\tfrac12\,\sqrt{61}
  \\
  y&=\phantom{-}\tfrac72+\tfrac12\,\sqrt{61}
 \end{cases}
\end{align}
and
\begin{align}
 \begin{cases}
  x&=-\tfrac72-\tfrac12\,\sqrt{61}
  ,\\
  y&=\phantom{-}\tfrac72-\tfrac12\,\sqrt{61}
  .
 \end{cases}
\end{align}
