SOLUTION: Find abc, If: a + 1/b = 7/3 b + 1/c = 4 c + 1/a = 1

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Question 1206800: Find abc,
If:
a + 1/b = 7/3
b + 1/c = 4
c + 1/a = 1
Found 3 solutions by ikleyn, Edwin McCravy, math_tutor2020:
Answer by ikleyn(52747) About Me  (Show Source):
You can put this solution on YOUR website!
.

See the solution under this link

https://www.quora.com/If-a-b-and-c-are-real-numbers-and-a-1-b-7-3-b-1-c-4-c-1-a-1-then-what-is-the-value-of-ABC



Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
system%28%0D%0A%0D%0Aa%2B1%2Fb=7%2F3%2Cb%2B1%2Fc=4%2Cc%2B1%2Fa=1%29

b=4-1%2Fc
c=1-1%2Fa

a%22%22%2B%22%221%2Fb%22%22=%22%227%2F3

a%22%22%2B%22%221%2F%28matrix%281%2C3%2C4%2C%22%22-%22%22%2C1%2Fc%29%29%22%22=%22%227%2F3

a%22%22%2B%22%22%22%22=%22%227%2F3

a%22%22%2B%22%22%22%22=%22%227%2F3

a%22%22%2B%22%22%22%22=%22%227%2F3

a%22%22%2B%22%22%22%22=%22%227%2F3

3a%283a-4%29%2B3%28a-1%29%22%22=%22%227%283a-4%29

9a%5E2-12a%2B3a-3%22%22=%22%2221a-28%29

9a%5E2-9a-3%22%22=%22%2221a-28%29

9a%5E2-30a%2B25%22%22=%22%220

(3a - 5)(3a - 5) = 0

3a - 5 = 0
    3a = 5
     a = 5/3
---------------------
c=1-1%2Fa
c=1-1%5E%22%22%2F%285%2F3%29
c=1-3%2F5
c=2%2F5
----------------------
b=4-1%2Fc
b=4-1%5E%22%22%2F%282%2F5%29
b=4-5%2F2
b=8%2F2-5%2F2
b=3%2F2

matrix%281%2C7%2C++++++a=5%2F3%2C%22%22%2C%22%22%2Cb=3%2F2%2C%22%22%2C%22%22%2Cc=2%2F5%29

Edwin

Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

Answer: abc = 1


Work Shown

Let w = abc

The given equations are
a + 1/b = 7/3
b + 1/c = 4
c + 1/a = 1

Multiply out those equations by multiplying the left hand sides together, and the right hand sides together. Then we'll do a bit of algebra to isolate w = abc.
The basic theme is to somehow pull out the original left hand sides so we can apply substitutions.

(a+1/b)*(b+1/c)*(c+1/a) = (7/3)*4*1
(ab + a/c + b/b + 1/bc)*(c+1/a) = 28/3
(ab + a/c + 1 + 1/bc)*(c+1/a) = 28/3
c(ab + a/c + 1 + 1/bc) + (1/a)*(ab + a/c + 1 + 1/bc) = 28/3
(abc + a + c + 1/b) + (b + 1/c + 1/a + 1/abc) = 28/3
(abc + 1/abc) + (a + 1/b) + (b + 1/c) + (c + 1/a) = 28/3
(w + 1/w) + (a + 1/b) + (b + 1/c) + (c + 1/a) = 28/3
(w^2 + 1)/w + (7/3) + (4) + (1) = 28/3
(w^2 + 1)/w + 22/3 = 28/3
(w^2 + 1)/w = 28/3-22/3
(w^2 + 1)/w = 2
w^2 + 1 = 2w
w^2 - 2w + 1 = 0
(w - 1)^2 = 0
w-1 = 0
w = 1
abc = 1

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Alternative Route

Isolate b in the 1st equation.
a + 1/b = 7/3
1/b = 7/3 - a
1/b = (7 - 3a)/3
b = 3/(7 - 3a)

Isolate 1/c in the 3rd equation.
c + 1/a = 1
c = 1 - 1/a
c = (a - 1)/a
1/c = a/(a - 1)

It might seem strange to isolate 1/c instead of just c.
However it should be fairly obvious why I'm doing this when it comes to the 2nd equation.

b + 1/c = 4
3/(7 - 3a) + a/(a - 1) = 4
I'll skip steps since this post is already quite lengthy.
Solving that equation should get you a = 5/3
Using that value in the 1st and 3rd equations should lead to b = 3/2 and c = 2/5 respectively.

Finally, a*b*c = (5/3)*(3/2)*(2/5) = 1

Other pathways are possible.