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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?

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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 $a,b,c,d$.

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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 $b+d=3$, $b=0, d=3$.

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  • $\begingroup$ This is a nice factorization. What would the next steps be? $\endgroup$ – Abe May 7 '18 at 5:13
  • $\begingroup$ @Abe See my edited answer. $\endgroup$ – Szeto May 7 '18 at 5:25
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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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