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?
Best Answer
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 $b+d=3$, $b=0, d=3$.