SOLUTION: find the numbers such that the product of the first and the second is 28, product of the second and third is 84 and the product of the third and the first is 48.

Algebra ->  Square-cubic-other-roots -> SOLUTION: find the numbers such that the product of the first and the second is 28, product of the second and third is 84 and the product of the third and the first is 48.      Log On


   



Question 940926: find the numbers such that the product of the first and the second is 28, product of the second and third is 84 and the product of the third and the first is 48.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
find the numbers
a, b, c
such that the product of the first and the second is 28,
ab = 28
a = 28/b
product of the second and third is 84
bc = 84
c = 84/b
and the product of the third and the first is 48.
ac = 48
Replace a with (28/b) and c with (84/b)
28%2Fb * 84%2Fb = 48
2352%2Fb%5E2 = 48
which is
48b^2 = 2352
b^2 = 2352%2F48
b^2 = 49
b = sqrt%2849%29
b = 7
then
a = 28/7 = 4
and
c = 84/7 = 12
:
The numbers: 4, 7, and 12