<s>[INST] 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?
 [/INST] 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 $c=0$. 
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