Find the number of ordered triples of real numbers (x, y, z) that satisfy x + yz = 6, y + xz = 6 and z + xy = 6.
Answers
Given : x + yz = 6, y + xz = 6 and z + xy = 6.
To find : Number of order triplets which 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
Learn more:
If the ordered triplets of real numbers (x, y, z) satisfy √x-y+z=√x ...
https://brainly.in/question/18708970
The number of triples (x, y, z) such that any one of these numbers is ...
https://brainly.in/question/13612444