<s>[INST] Finding $a+b+c+d$, where $ab+c+d=15$, $bc+d+a=24$, $cd+a+b=42$, $da+b+c=13$ Let $a,b,c,d \in \mathbb{R}$. Consider the following constraints:
\begin{cases} ab+c+d=15 \\ bc+d+a=24 \\ cd+a+b=42 \\da+b+c=13 \end{cases}
Calculate the value of $a+b+c+d$.
It is easy to use the Gröbner basis to get the value:
\begin{cases}
10849-4501 d+380d^2,-39409+2320c+3420d,-20+29b-9d,1801+2320 a-380 d\}
\end{cases}
so the value of $a+b+c+d$ is $\frac{169}{10}$.
What I am curious about is how to use high schools mathematics to get an answer without too much complicated mathematical calculations ?
 [/INST] Playing around with the four LHSs, I tried to obtain polynomials with some symmetry. I first noted that
$$\left((ab+c+d)+(cd+a+b)\right)-\left((bc+d+a)+(da+b+c)\right)=(b-d)(a-c)$$
then after replacing the $+$ sign with $\cdot$, we get
$$(ab+c+d)\cdot(cd+a+b)-(bc+d+a)\cdot(da+b+c)=(b-d)(a-c)(a+b+d+c-1).$$
Putting all together,
$$\begin{align}
&\frac{(ab+c+d)(cd+a+b)-(bc+d+a)(da+b+c)}{(ab+c+d)+(cd+a+b)-(bc+d+a)-(da+b+c)}\\&\qquad=
\frac{(b-d)(a-c)(a+b+d+c-1)}{(b-d)(a-c)}=a+b+c+d-1.
\end{align}$$
Hence with the given values we find
$$a+b+c+d=1+\frac{15\cdot 42-24\cdot 13}{15+42-24-13}=1+\frac{318}{20}=\frac{169}{10}.$$
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