<s>[INST] A Junior Olympiad like system of linear equations Given a system of linear equations
$$\begin{align}\frac{x}{3}+\frac{y}{5}+\frac{z}{9}+\frac{w}{17} &=1 \\
\frac{x}{4}+\frac{y}{6}+\frac{z}{10}+\frac{w}{18} &=\frac{1}{2} \\
\frac{x}{5}+\frac{y}{7}+\frac{z}{11}+\frac{w}{19} &=\frac{1}{3} \\
\frac{x}{6}+\frac{y}{8}+\frac{z}{12}+\frac{w}{20} &=\frac{1}{4} \\
\end{align}$$
Determine $$ \frac{x}{10}+\frac{y}{12}+\frac{z}{16}+\frac{w}{24}$$
This is a problem I was asked to solve a bit long before. Since I didn’t come up with any other ideas than using Gauss Jordan elimination, I did so. The answer is $\frac{1}{36}$. Could someone provide me an elegant solution to this problem?
 [/INST] Let $G(s) = \frac{x}{s+2} + \frac{y}{s+4} + \frac{z}{s+8} + \frac{w}{s+16} - \frac 1s$, and $F(s) = s(s+2)(s+4)(s+8)(s+16)G(s)$.
Clearly $F(s)$ is a degree 4 polynomial with roots being 1, 2, 3, and 4. Also, $F(0) = - 2\times4\times8\times16 = -2^{10}$. Hence
$$
F(s) = -\frac{2^{10}}{24}(s-1)(s-2)(s-3)(s-4) = -\frac{2^7}{3}(s-1)(s-2)(s-3)(s-4).
$$
We are interested in $F(8) = -\frac{2^7}{3}\times7\times6\times5\times4 = -2^{10}\times5\times7$.
Hence
$$
\frac{x}{10}+\frac{y}{12}+\frac{z}{16}+\frac{w}{24} - \frac{1}{8} = \frac{F(8)}{8\times10\times12\times16\times24} = \frac{-2^{10}\times 5\times 7}{2^{13}\times3^2\times5} = -\frac{7}{72},
$$
and
$$
\frac{x}{10}+\frac{y}{12}+\frac{z}{16}+\frac{w}{24} = \frac18 - \frac{7}{72} = \frac{1}{36}.
$$
</s>