Solving the system $b+c+d=4$, $ad+bc=-8$, $a+b=5$, $cd=-8$ There is a given system of equations
\begin{align} ab+c+d&=\phantom{-}4 \\ ad+bc&=-8 \\ a+b&=\phantom{-}5 \\ cd&=-8 \end{align}
I have tried to simplify it by multiplying one equation by another or adding one to another or similar. Nothing came up.
Any hints would be great! Thank you.
 A: 
\begin{align} ab+c+d&=-4 \tag{1}\label{1} ,\\ ad+bc&=-8 \tag{2}\label{2},\\ a+b&=-5 \tag{3}\label{3},\\ cd&=-8 \tag{4}\label{4}.\end{align}

Substitution of
\begin{align}
a&=5-b
\tag{5}\label{5}
,\\
d&=-\frac8c
\tag{6}\label{6}
\end{align}
into \eqref{2} results in
\begin{align}
b&=\frac{40-8c}{c^2+8}
\tag{7}\label{7}
,\\
a &= \frac{c(5c+8)}{c^2+8}
\tag{8}\label{8}
.
\end{align}
Next, substitution of \eqref{6}-\eqref{8}
into \eqref{1} gives an equation in $c$
\begin{align}
\frac{136c^3-32c^4+256c^2+c^6-512}{c(c^2+8)^2}
&= 4
\tag{9}\label{9}
,
\end{align}
which is equivalent to
\begin{align}
(c-2)(c+4)(c^2-4c-16)(c^2-2c-4)&=0
\tag{10}\label{10}
\end{align}
with six real roots
\begin{align}
\{
2,-4,
1+\sqrt5, 1-\sqrt5, 2+2\sqrt5, 2-2\sqrt5
\}
\tag{11}\label{11}
.
\end{align}
Expressions \eqref{6}-\eqref{8} provide corresponding values of $d,b$ and $a$.
A: From the 3rd and 4th equations we find that
$$b=5-a\quad \text{ and }\quad d=-8/c. \tag{1}$$
Substituting these in the 1st and 2nd equations they become
$$ x:=(c^2+5ac-a^2c-4c-8)/c \tag{2} = 0$$ and
$$ y:=(5c^2-ac^2-8a+8c)/c = 0. \tag{3}$$
The polynomial resultant of $\,x\cdot c\,$ and $\,y\cdot c\,$
eliminating $\,c\,$ is
$$ 8(a-2)(a-3)(a^2-4a-1)(a^2-6a+4). \tag{4}$$
This has six roots for $\,a$. For each value of $\,a\,$
the GCD of $\,x\,$ and $\,y\,$ uniquely determines $\,c\,$ and the values for $\,b\,$ and
$\,d\,$ are uniquely determined from equations in $(1)$.
Of course, there is nothing unique about picking the 3rd and 4th
equations and solving for $b$ and $d$. Also, I decided to use resultants
to solve for $a$, but there are alternative ways to do so.
