Question

Find the number of ordered triples of real numbers (x, y, z) that satisfy x + yz = 6, y + xz = 6 and

z + xy = 6.​

Answer 1

Given :   x + yz = 6, y + xz = 6 and z + xy = 6.

To find : number of ordered triples of real numbers (x, y, z) that satisfy

Solution:

x + yz = 6    Eq1

y + xz = 6     Eq2

z + xy = 6.    Eq3

Eq1 - Eq2

=> (x - y)  -z(x - y) = 0

=> (x - y)(1 - z) = 0

=> x = y   or   z  = 1

case 1 :

x = y  

=> x + xz = 6

&   z  + x²  = 6

=> (x - z)  - x(x - z) = 0

=> (x - z)(1 - x) = 0

=>  x  = z    or  x  =  1

x = 1  , y  = 1  

=> 1 + 1(z) = 6

=> z = 5

( 1 ,  1  , 5)  

x = y = z  

=> x + x² = 6

=> x² + x - 6 = 0

=> (x + 3)(x - 2) = 0

=> x = - 3  , x  = 2

( 2  ,  2  , 2)  , (  -3 , - 3 ,  - 3)

Case  2   z  = 1

=> x  + y = 6

    y + x  = 6

   1   +  xy  = 6

=> xy  = 5    x + y  = 6

=> x & y are roots of Equation  

a² - 6a  +  5  = 0

=> a² - a - 5a + 5 = 0

=> (a - 1)(a - 5) = 0

=> a =  1  , 5  

(x , y) = ( 1, 5)  or (5 , 1)

Hence  triplet ( 1 , 5 , 1) or ( 5 , 1 , 1)

Triplet : (x , y , z)

( 1 ,  1,  5)  , ( 1, 5  , 1)  , ( 5 , 1 , 1)

( 2  ,  2  , 2)  , (  -3 , - 3 ,  - 3)

5 possible Triplets

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