Eliminate $x, y, z, u$ from the equations $x = b y + c z + d u, y = c z + d u + a x, z = d u + a x + b y, u = a x + b y + c z$ .

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Solution

Just transform the equations, to eliminate $x, y, z$
$x = b y + c z + d u$
$x + a x = b y + c z + d u + a x$ ...... $(i)$
$y = c z + d u + a x$
$y + b y = c z + d u + a x + b y$ ...... $(i i)$
$z = d u + a x + b y$
$z + c z = d u + a x + b y + c z$ ...... $(i i i)$
$u = a x + b y + c z$
$u + d u = a x + b y + c z + d u$ ...... $(i v)$

Let $k = a x + b y + c z + d u$

$x (1 + a) = k$ ........ [From $(i)$ ]

$a x = a \frac{k}{1 + a}$

Similarly, we get
$b y = b \frac{k}{1 + b}$ ........ [From $(i i)$ ]

$c z = c \frac{k}{1 + c}$ ........ [From $(i i i)$ ]

$d u = d \frac{k}{1 + d}$ ........ [From $(i v)$ ]

Add these equations, we get

$a x + b y + c z + d u = k (\frac{a}{1 + a} + \frac{b}{1 + b} + \frac{c}{1 + c} + \frac{d}{1 + d})$

$\Rightarrow k = k (\frac{a}{1 + a} + \frac{b}{1 + b} + \frac{c}{1 + c} + \frac{d}{1 + d})$

Therefore, $\frac{a}{1 + a} + \frac{b}{1 + b} + \frac{c}{1 + c} + \frac{d}{1 + d} = 1$

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x = b y + c z + d u, y = c z + a x + d u, z = a x + b y + d u, u = a x + b y + c z .

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