Question

if a b and c are positive numbers satisfying a+1/b=4,b+1/c=1 and c+1/a=7/3 then the value of 6abc.

Answer 1

Answer:

Step-by-step explanation:

We need to form three simultaneous equations:

a + 1/b = 4

a + 1 = 4b

4b - a = 1...............i)

b + 1/c = 1

b + 1 = c

c - b = 1 ....................ii)

c + 1 / a = 7/3

3(c + 1) = 7a

3c + 3 = 7a

3c - 7a = 3..................iii)

From these equations we can solve these equations simultaneously to get a , b and c.

From 1 we can get by substitution.

a = 4b - 1...**

Substitute this in 3 .

3c - 7(4b - 1) = -3

3c - 28b + 7 = -3

3c = -3 - 7 + 28b

3c = -10 + 28b

c = -10/3 + 28b/3

We substitute this in 2 to get :

-10/3 + 28b/3  - b = 1

Multiply through by 3:

-10 + 28b - 3b = 3

25b = 13

b = 13/25

We can get a from ***

a = 4 × 13/25 - 1 = 27/25

From 2:  c = 1 + b

c = 38/25

The value of 6abc = 6 × 38/25 × 13/25 × 27/25 = 5.122