SOLUTION: find out the three nos such that the product of the first and second is 24 second and the third is 48 and that of the first and and third is 32

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Question 969390: find out the three nos such that the product of the first and second is 24 second and the third is 48 and that of the first and and third is 32
Found 2 solutions by lwsshak3, ikleyn:
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
find out the three nos such that the product of the first and second is 24 second and the third is 48 and that of the first and and third is 32
***
let x=1st no.
let y= 2nd no.
let z=3rd no.
***
xy=24
yz=48
xz=32
...
y=24/x
y=48/z
24/x=48/z
..
z=32/x
24/x=48/(32/x)
24/x=48x/32
48x^2=24*32=768
x^2=16
x=√16=4
y=24/x=6
z=32/x=8
..
1st no.=4
2nd no.=6
3rd no.=8

Answer by ikleyn(53909) About Me  (Show Source):
You can put this solution on YOUR website!
.
find out the three nos such that the product of the first and second is 24 second and the third is 48
and that of the first and and third is 32
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        I am looking at the solution by @lwsshar3, and I see that the other possible solution
        (x,y,z) = (-4,-6,-8) is missed.

        So, his solution is not as accurate as a Math solution should be.

        It is also long, boring and non-interesting: there is a lot of unnecessary calculations.

        There is more short, more effective, more impressive and more instructive solution. 


Let x be the 1st number, let y be the 2nd number, and let z be the 3rd number..


From the problem's formulation, we have these equations

    xy = 24,    (1)
    yz = 48,    (2)
    xz = 32.    (3)


Multiply these equations.  You will get

    x%5E2%2Ay%5E2%2Az%5E2 = 24*48*32 = %283%2A8%29%2A%283%2A2%2A8%29%2A%282%5E2%2A8%29,

or

    %28x%2Ay%2Az%29%5E2 = 2%5E12%2A3%5E2.


Hence,  x*y*z = sqrt%282%5E12%2A3%5E2%29 = +/- 2%5E6%2A3 = +/- 64*3,

or       x*y*z = +/- 192.


Now consider two cases.


(a)  xyz = 192.    (4)


     To get x, divide equation (4) by equation (2).  You will get x = 192/48 = 4.

     To get y, divide equation (4) by equation (3).  You will get y = 192/32 = 6.

     To get z, divide equation (4) by equation (1).  You will get z = 192/24 = 8.


     So, in case (a), the solution is  (x,y,z) = (4,6,8).



(b)  xyz = -192.   (5)


     To get x, divide equation (5) by equation (2).  You will get x = -192/48 = -4.

     To get y, divide equation (5) by equation (3).  You will get y = -192/32 = -6.

     To get z, divide equation (5) by equation (1).  You will get z = -192/24 = -8.


     So, in case (b), the solution is  (x,y,z) = (-4,-6,-8).


Thus, the problem has two solutions.

One solution is  (x,y,z) = (4,6,8).  Another solution is  (x,y,z,) = (-4,-6, -8).

Solved correctly and completely.

This method of solution is a standard and is the expected to this given problem.