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
Answer:
x + yz = 6 ...(i)
y + xz = 6 ...(ii)
z + xy = 6 ...(iii)
(i) – (ii) ⇒ x – y – z(x – y) = 0
⇒ (x – y)(1 – z) = 0
⇒ z = 1 or x = y
If z = 1 we have x + y = 6 and 1+ xy = 6
⇒ (x, y) = (5, 1) or (1, 5)
∴ (x, y, z) = (5, 1, 1) or (1, 5, 1).
if x = y, we have x + xz = 6 and z + x ^2 = 6
⇒ x ^2 – x – xz + z = 0 ⇒ (x – z)(x –1) = 0
⇒ x = 1 or x = z
If x = 1, we have y = 1 and z = 5
⇒ (x, y, z) = (1, 1, 5)
If x = z, we have x = y = z
⇒ x + x ^2 = 6
⇒ x = 2 or –3
⇒ (x, y, z) = (2, 2, 2), or (–3, –3, –3)
∴ (x, y, z) = (5, 1, 1), (1, 5 1), (1, 1, 5) (2, 2, 2),
(–3, –3, –3).
Step-by-step explanation:
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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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