Bunuel wrote:
If\(\frac{ a}{(b +c)} =\frac{ b}{(c +a)} =\frac{ c}{(a +b)} =r\) then r cannot take any value except:
(A) 1/2
(B) –1
(C) 1/2 or –1
(D) –1/2 or –1
(E) -3/2 or-1
Stonely I got it wrong but tried to come up with a decent solution.
First I did the same as you. a = b = c = 1, gives us r = 1/2.
Then to the second part.
Consider r is equal to -1. This means that
\(\frac{ a}{(b +c)} =-1\)
thus
\( a =-1(b+c)\)
and finally
\( a + b + c =0\)
(The equations II. and III. are identical).
Any value for a, b and c which satisfies
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