<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] Let $\,x=a+b+c+d\,$ then:
$$ab+c+d=ab\color{red}{-a-b+1-1+a+b}+c+d=(a-1)(b-1)-1+x\,$$
The system can then be written as:
$$
\begin{cases}
   (a-1)(b-1)=16-x
\\ (b-1)(c-1)=25-x
\\ (c-1)(d-1)=43-x
\\ (d-1)(a-1)=14-x
\end{cases}
$$
It follows that:
$$
(a-1)(b-1)(c-1)(d-1) \;=\; (16-x)(43-x) \;=\; (25-x)(14-x)
$$
The latter equality gives $\,2\left(169 - 10 x\right) = 0\,$.
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