SOLUTION: a) Find the value of xyz, given that xy = 117, yz = 286 and xz = 198. b) Hence, find the values of x, y and z.

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Question 1121449: a) Find the value of xyz, given that xy = 117, yz = 286 and xz = 198.
b) Hence, find the values of x, y and z.
Found 4 solutions by rothauserc, Alan3354, ikleyn, greenestamps:
Answer by rothauserc(4718)   (Show Source): You can put this solution on YOUR website!
y = 117/x, y = 286/z
:
117/x = 286/z
:
117z = 286x
:
z = 286x/117
:
note that xz = 198, therefore
:
z = 198/x, and
:
198/x = 286x/117
:
286x^2 = 198 * 117
:
x^2 = 198 * 117 /286 = 81
:
x = 9 or -9
:
using given relationships and x = 9
:
9y = 117
:
y = 13
:
9z = 198
:
z = 22
:
******************************************************
there are two solutions
:
x = 9, y = 13, z = 22, xyz = 9 * 13 * 22 = 2574
:
x = -9, y = -13, z = -22, xyz = -9 * -13 * -22 = -2574
*******************************************************
:

Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
a) Find the value of xyz, given that xy = 117, yz = 286 and xz = 198.
117 and 286 both have a factor of 13, and both xy & yz have a factor of y.
--> y = 13
x = 9
z = 22
xyz = 9*13*22
========
b) Hence, find the values of x, y and z.
9, 13, 22
Answer by ikleyn(52855)   (Show Source): You can put this solution on YOUR website!
.
You are given

    xy = 117     (1)

    yz = 286     (2)

    xz = 198     (3)


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


    (xyz)^2 = 117*286*198 = 6625476;   hence,

    xyz = +/-  = +/- 2574.       (4)


First, consider the case  xyz = 2574      (5)


Divide equation (4) by equation (1).  You will get  z =  = 22.


Next divide equation (4) by equation (2).  You will get  x =  = 9.


Finally, divide equation (4) by equation (3).  You will get  y =  = 13.


Thus, in this case the solution is  (x,y,z) = (9,13,22).



Next consider the case   xyz = -2574.

Do the same operations, and you will get the second solution  (x,y,z) = (-9,-13,-22).


Answer.  The problem has two solutions  (x,y,z) = (9,13,22)  and  (x,y,z) = (-9,-13,-22).

------------

It is a classic problem and its standard/canonical solution.

To see other similar problems solved in this way,  look into the lesson
    - Find the volume of the rectangular box if the areas of its faces are given
in this site.


Answer by greenestamps(13203)   (Show Source): You can put this solution on YOUR website!


Without context, the problem PROBABLY is looking for a solution in positive integers. However, if (a,b,c) is a solution, then (-a,-b,-c) is also a solution.

The problem seems to suggest finding the product xyz first and then using that to find x, y, and z separately.

That can be done; but the arithmetic gets ugly. It is far easier to find the values of x, y, and z and then find the value of the product xyz.

The values of the separate variables can be found easily by trial and error, by looking at the possible factorizations of the three given products.

For me, the first thing I see is 117 = 9*13.
Then, having one number that has a factor of 9, I look for another; 198 = 9*22.
And trying 13*22 for the third product gives me the right result, 286.

So the solution in positive integers is 9, 13, and 22.

Finding the product of all three numbers is then simple arithmetic.
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