\(\frac{l}{m+n}=\frac{m}{n+l}=\frac{n}{l+m}=k\) , where k is a real number

Quantity A	Quantity B
k	\(\frac{1}{3}\)

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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16. The given equation is l
m+ n
= m
n + l
= n
l + m
= k . Forming the three equations yields l = (m + n)k, m =
(n + l)k, n = (l + m)k. Summing these three equations yields
l + m + n = (m + n)k + (n + l)k + (l + m)k
= k[(m + n) + (n + l) + (l + m)]
= k(m + n + n + l + l + m)
= k(2m + 2n + 2l)
= 2k(m + n + l)
1 = 2k by canceling m + n + l from each side
1/2 = k
Hence, Column A equals 1/2. Since 1/2 is greater than 1/3, Column A is greater than Column B, and the
answer is (A).

\(\frac{l}{m+n}=\frac{m}{n+l}=\frac{n}{l+m}=k\)....
three equations can be formed :-

\(\frac{l}{m+n}=k........l=(m+n)k\)
\(\frac{m}{l+n}=k........m=(l+n)k\)
\(\frac{n}{m+l}=k........n=(m+l)k\)

add the three equations...
\(l+m+n=(m+n)k+(l+n)k+(m+l)k...........l+m+n=(m+n+l+n+m+l)k..........l+m+n=2(m+n+l)k.........2k=1.....k=\frac{1}{2}=A\) and \(B=\frac{1}{3}\) thus \(A>B\)

A
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Some useful Theory.
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I tried the following approach and got k = -1 , can someone explain where I went wrong.
l(n+l)=m(m+n) { cross multiplying the first 2 terms}
ln+lpow2 = mpow2 +mn
n(l-m)= (m+l)*(m-l)
n= -(m+l)

Substituting this value in the last equation, yields k=-1 , and hence B>A
@chetan2u