If 3xy = 9xz = 12yz = 1, what is the value of x, y, and z?

Question

This is just a set of 3 equations:

3xy=1
9xz=1
12yz=1

⇒(18xyz)^2=1
⇒ xyz = ± 1/18
⇒ (x,y,z) = ±(2/3,1/2,1/6)

PLEASE TELL ME HOW DID THE SOLUTION CAME UP WITH xyz = ± 1/18 and (x,y,z) = ±(2/3,1/2,1/6. THANKS!

Answer 1

3 x y = 9 x z = 12 y z = 1 Divide by 3

x y = 3 x z = 4 y z = 1 / 3

x y = 1 / 3 Divide both sides by x

y = 1 / ( 3 x )

3 x z = 1 / 3 Divide both sides by 3 x

z = 1 / ( 9 x )

4 y z = 1 / 3

4 * 1 / ( 3 x ) * 1 / ( 9 x ) = 1 / 3

4 * 1 / ( 27 x ^ 2 ) = 1 / 3

4 / 27 x ^ 2 = 1 / 3 Multiply both sides by 27 x ^ 2

4 = ( 1 / 3 ) * 27 x ^ 2

4 = 27 x ^ 2 / 3

4 = 9 x ^ 2

9 x ^ 2 = 4 Divide both sides by 9

x ^ 2 = 4 / 9

x = + OR - sqrt ( 4 / 9 )

x = + OR - 2 / 3

For x = - 2 / 3

y = 1 / ( 3 x )

y = 1 / [ 3 * ( - 2 ) / 3 ]

y = 1 / - 2

y = - 1 / 2

z = 1 / ( 9 x )

z = 1 / [ 9 * ( - 2 ) / 3 ]

z = 1 / [ ( 9 / 3 ) * - 2 )

z = 1 / [ 3 * ( - 2 ) ]

z = 1 / - 6

z = - 1 / 6

For x = 2 / 3

y = 1 / ( 3 x )

y = 1 / ( 3 * 2 / 3 )

y = 1 / 2

z = 1 / ( 9 x )

z = 1 / ( 9 * 2 / 3 )

z = 1 / [ ( 9 / 3 ) * 2 )

z = 1 / ( 3 * 2 )

z = 1 / 6

You have two set of solutions :

x = - 2 / 3 , y = - 1 / 2 , z = - 1 / 6

and

x = 2 / 3 , y = 1 / 2 , z = 1 / 6