Question 1121449
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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 = +/- {{{sqrt(6625476)}}} = +/- 2574.       (4)


First, consider the case  xyz = 2574      (5)


Divide equation (4) by equation (1).  You will get  z = {{{2574/117}}} = 22.


Next divide equation (4) by equation (2).  You will get  x = {{{2574/286}}} = 9.


Finally, divide equation (4) by equation (3).  You will get  y = {{{2574/198}}} = 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).


<U>Answer</U>.  The problem has two solutions  (x,y,z) = (9,13,22)  and  (x,y,z) = (-9,-13,-22).
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It is a classic problem and its standard/canonical solution.


To see other similar problems solved in this way, &nbsp;look into the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/geometry/Find-the-volume-of-a-rectangular-box-if-the-areas-of-its-faces-are-given.lesson>Find the volume of the rectangular box if the areas of its faces are given</A>

in this site.