Question

pls answer to this question
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Answer 1

Answer:

x=k(y+z)

y=k(z+x)

z=k(x+y)

Now

x+y+z=k(y+z+z+x+x+y)

or, (x+y+z) = 2k(x+y+z)

or, k= 1/2

Answer 2

Process

$\frac{x}{y + z} + \frac{y}{z + x} + \frac{z}{x + y} = k \\now... \\ \frac{x}{y + z} = k = > x = k(y + z) \\ and \\ \frac{y}{z + x} = k = > y = k(z + x) \\ and \\ \frac{z}{x + y} = k = > z = k(x + y) \\ therefore \: now \: (x + y + z) = k(y + z) + k(z + x) + k(x + y) \\ (x + y + z) = 2k(x + y + z) \\ now \: \: if \: (x + y + z) \: is \: not \: equal \: to \: zero \: then \\ 2k = 1 \\ k = \frac{1}{2} \: (proved)$