1/(a+1 /(b+1/(c+1/2)))=16/23 find value of a +b +c

Answer 1

c+1/2 = (2c +1 )/2

THe reciprocal of that is 2 / (2c+1)

b + 2/(2c+1) = [b(2c+1) + 2] / (2c+1)

= [2bc + b + 2]/(2c+1)

the reciprocal of that is (2c+1)/ [ 2bc+b+2]

a + (2c+1)/ [2bc + b + 2] =

{a[2bc+b+2]+ (2c+1)} / [ 2bc +b +2] =

{2abc + ab + 2a + 2c + 1 } / [ 2bc + b+ 2] =

The reciprocal of that is [ 2bc + b+ 2]/ {2abc + ab + 2a + 2c + 1 } =

[ 2bc + b+ 2]/ {a(2bc + b + 2) + 2c + 1 }

So the numerator is 2bc + b + 2 = 16.

Plugging this into the denominator:

a(16) + 2c + 1 = 23

16a + 2c = 22

8a + c = 11

c = 11 - 8a

Plugging this back in...

2b(11 - 8a) + b + 2 = 16

22b - 16ab + b = 14

23b - 16ab = 14

b (23 - 16a) = 14

b = 14 / (23 - 16a)

So

a + b + c = a + 14/(23-16a) + 11 - 8a =

[a(23-16a) + 14 + (11 - 8a)(23 - 16a) ]/ (23 - 16a)

[ 23a - 16a^2 + 14 + 253 - 176a - 184a + 128 a^2 ] / (23 - 16a)

[ 112a^2 - 383a +267]/(23 - 16a)