2011/025) solve x = a/(b+c) = b/(c+a) = c/(a+b)

x cannot be zero as if x= 0 then a = b =c so a/(b+c) = b/(c+a) = c/(a+b) are indeterminate form

so we take reciprocals

1/x = (b+c)/ a= (c+a)/b = (a+b)/c
add 1 on all 2 sides

1/x +1 = (a+b+c)/a = (a+b+c)/b = (a+b+c)/ c

so a+b+c = 0 = 1/x + 1 or a= b = c

a+b+c = 0 = 1/x + 1 => x= -1 => b+ c = - a,if a = 1 then b= k and c = -1 - k

this satisfies all 1/x = (b+c)/ a= (c+a)/b = (a+b)/c
add 1 on all 2 sides


if a=b= c then we get x = 1/2 so solution (x=1/2, a= k. b= k,c =k ) for any k

if x = -1 then (x= -1, a = 1 , b= k, c = - 1 -k) or any permutation or any multiple of the same.