Help to solve a system equation: $x-y+xy=-4$; $xy(x-y)=-21$.

Question

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.

Answer 1

\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}

Answer 2

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 $$ (x^2 + 7x - 3)(x^2 - 3x + 7)=0. $$

Answer 3

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...