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.
Found 4 solutions by ankor@dixie-net.com, MathTherapy, josgarithmetic, ikleyn:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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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

Answer by MathTherapy(10898) About Me  (Show Source):
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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.
************
Let 1st, 2nd, and 3rd numbers, be F, S, and T, respectively
Since the "product of the first and the second is 28.....", then, matrix%281%2C4%2C+FS+=+28___S+=+28%2FF%2C+%22-----%22%2C+%22eq%22%2C+%22%28i%29%22%29
Since "product of the second and third is 84.....", then, matrix%281%2C4%2C+ST+=+84___T+=+84%2FS%2C+%22-----------%22%2C+%22eq%22%2C+%22%28ii%29%22%29 
And, since "the product of the third and the first is 48.....", then, matrix%281%2C4%2C+TF+=+48___T+=+48%2FF%2C+%22---%22%2C+%22eq%22%2C+%22%28iii%29%22%29
  As T+=+84%2FS and T+=+48%2FF, 84%2FS+=+48%2FF
                                     84%2F%2828%2FF%29+=+48%2FF ----- Substituting 28%2FF for S
                               84%28F%2F28%29+=+48%2FF
                          3cross%2884%29%28F%2Fcross%2828%29%29+=+48%2FF
                                       3F+=+48%2FF
                                    3F%5E2+=+48 ----- Cross-multiplying
                                      F%5E2+=+48%2F3+=+16
      First number, or F+=+0+%2B-+sqrt%2816%29+=+highlight%280+%2B-+4%29%29 
Second number, or matrix%281%2C3%2C+S+=+28%2FF+=+28%2F4%2C+or%2C+28%2F%28-+4%29+=+highlight%280+%2B-+7%29%29
    Third number, or 

Numbers: First, Second, Third:  4,   7,   12, or
                                                   - 4, - 7, - 12

Answer by josgarithmetic(39851) About Me  (Show Source):
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x, y, z
system%28xy=28%2Cyz=84%2Cxz=48%29

system%28x=28%2Fy%2Cy=84%2Fz%29

Substitute.
%2828%2Fy%29z=48
%2828%2F%2884%2Fz%29%29z=48
%2828%2F84%29%28z%5E2%29=48
%284%2F12%29z%5E2=48
%281%2F3%29z%5E2=48
z%5E2=48%2A3=24%2A2%2A3=2%2A3%2A4%2A2%2A3
z%5E2=2%5E4%2A3%5E2
z=3%2A4
highlight%28z=12%29----------------Use this to find x and y.

Answer by ikleyn(54062) About Me  (Show Source):
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.
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.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


        There is an elegant way to solve this problem. 


Let x be the first number, y be the second number and z be the third number.

We are given

    xy = 28,    (1)

    yz = 84,    (2)

    xz = 48.    (3)


Multiply equations (1), (2) and (3).  You will get

    x%5E2%2Ay%5E2%2Az%5E2 = 28*84*48,  

    %28x%2Ay%2Az%29%5E2 = %282%5E2%2A7%29%2A%282%5E2%2A3%2A7%29%2A%282%5E4%2A3%29 = 2%5E8%2A3%5E2%2A7%5E2 = %282%5E4%2A3%2A7%29%5E2,

    x%2Ay%2Az = +/- %282%5E4%2A3%2A7%29 = +/- 336.    (4)


Now, to find x, divide equation (4) by equation (2).  You will get  x = +/- 336/84 = +/- 4.

     To find y, divide equation (4) by equation (3).  You will get  y = +/- 336/48 = +/- 7.

     To find z, divide equation (4) by equation (1).  You will get  z = +/- 336/28 = +/- 12.


Thus, the problem has two solutions for (x,y,z).


One solution is  (x,y,z) = (4,7,12).

Another solution is (x,y,z) = (-4,-7,-12).

Solved. This elegant solution is what is expected for this problem.

Usually, students learn such tricks in Math circles, Math schools, or from math-circle literature.