let a,b,c belongs to positive real numbers such that a+1/b=3, b+1/c=4, c+1/a= 9/11 then abc=?
Answers
If a + 1/b = 3 …(1)
b + 1/c = 4 …(2)
c + 1/a = 9/11 …(3)
then what is abc?
From (3), 1/a = (9/11) -c = (9-11c)/c, or
a = c/(9–11c) …(4)
From (1), a + 1/b = 3, or
1/b = 3-a, or
b = 1/(3-a) = 1/[3 - c/(9–11c)]
= 1/[27–33c-c]/(9=11c)
= (9–11c)/(27–34c) …(5)
From (2), b + 1/c = 4 or
b = 4 - (1/c) = (4c-1)/c …(6)
Equate (5) and (6)
(9–11c)/(27–34c) = (4c-1)/c or
c(9–11c) = -(27–34c)(1–4c) or
9c-11c^2 = -[27–34c -108c+136c^2]
9c-11c^2 = -27+34c +108c-136c^2
-27+133c-125c^2 = 0, or
125c^2 - 133c + 27 = 0
c1 = [+133+(133^2–4*125*27)^0.5]/250
= [+133+(17689–13500)^0.5]/250
= [+133+64.7224845]/250
= 197.7224845/250
= 0.790889938
c2 = [+133-(133^2–4*125*27)^0.5]/250
= [+133-(17689–13500)^0.5]/250
= [+133-64.7224845]/250
= 68.2775155/250
= 0.273110062
From (4), a1 = c1/(9–11c1)
= 0.790889938 / [9 - 11*0.790889938]
= 0.790889938/0.300210682
= 2.63444969
Again, from (4), a2= c2/(9–11c2)
= 0.273110062/[9 - 11*0.273110062]
= 0.045550309
From (6), b1 = (4c1-1)/c1
= [4*0.790889938 -1]/0.790889938
= 2.735601565
Again from (6), b2 = (4c2-1)/c2
= [4*0.273110062 -1]/0.273110062
= 0.338472509
So a1b1c1 = 2.63444969 * 2.735601565 * 0.790889938 = 5.699789318
and a2b2c2 = 0.045550309 * 0.338472509 * 0.273110062 = 0.004210681857
Hence abc = 5.699789318 or 0.004210681857.
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