British Maths Olympiad (BMO) 2004 Round 1 Question 1 alternative approaches? The questions states: 

Solve the simultaneous equations (which I respectively label as $
> \ref{1}, \ref{2}, \ref{3}, \ref{4}$)
$$\begin{align} ab + c + d &= 3 \tag{1} \label{1} \\ bc + d + a &= 5
 \tag{2} \label{2} \\  cd + a + b &= 2 \tag{3} \label{3} \\  da + b + c
 &= 6 \tag{4} \label{4} \end{align}$$

where $a,b,c,d$ are real numbers.
I solved this system after quite a while by taking  
$eqns$ 1 - 3 = $eqns$ 4 - 2 
which yields $a + c = 2$ 
You can then substitute that in and find the other variables
I also noticed that $(a+1)(b+1) + (a+1)(d+1) + (c+1)(b+1) + (c+1)(d+1) = 20$ but that line didnt really help me.
I'm interested in seeing the other approaches people can take with this system.
Additionally, is there a sufficient enough hint to take another route? Did I miss an easy solution?
 A: Step 1: Obtain $a+c=2$.
Step 2:
Note that
$$ab+bc+cd+ad=(a+c)(b+d)$$
Adding four equations gives
$$(a+c)(b+d)+2(a+c)+2(b+d)=16$$
$$(a+c)(b+d+2)+2(b+d+2)=20$$
$$(a+c+2)(b+d+2)=20$$
With $a+c=2$, we have $b+d=3$.
Step 3:
Further manipulating, 1-2+3-4 gives
$$(a-c)(b+d)=6$$
Thus, $a-c=2$.
Therefore, we have $a=2, c=0$.
Step 4: Put $a,c$ into 1,
$$2b+d=3$$
With $b+d=3$, $b=0, d=3$.
A: Here's a way after you get $a+c=2$. 
Take equation $2$ subtract equation $1$ to get $(b-1)(c-a)=2$.
Likewise, take equation $3$ subtract equation $4$ to get $(d-1)(c-a) = -4$.
Finally, we see that
\begin{align}
\frac{b-1}{d-1}=\frac{(b-1)(c-a)}{(d-1)(c-a)}= -\frac{1}{2} \ \ \implies \ \ 2b+d =3. 
\end{align}
Next, take equation $1$ plus equation $2$ to get $b(a+c)+(a+c)+2d=8$ which implies $b+d = 3$ since $a+c=2$. 
Now, we see that $b=0$ and $d=3$. Using equation $2$, we have that $a=2$ and $c=0$. 
A: My approach was to set $A=a-1,B=b-1,C=c-1,D=d-1$ so that we get
\begin{align}
  AB + A + B + C + D &= 0 \\
  BC + A + B + C + D &= 2 \\
  CD + A + B + C + D &= -1 \\
  AD + A + B + C + D &= 3.
\end{align}
Letting $S = A+B+C+D$, we have $ABCD = (-S)(-1-S) = (2-S)(3-S)$, so $S=1$, and therefore 
\begin{align}
   AB &= -1 \\
   BC &= 1 \\
   CD &= -2 \\
   AD &= 2.
\end{align}
This gives us $A = -\frac1B$, $C = \frac1B$, and $D = -\frac2C = -2B$.
From $A+B+C+D=1$, we have $-\frac1B + B + \frac1B - 2B = 1$, or $B = -1$. Then we can solve for $A,C,D$ and finally get $a,b,c,d$.
