Find the number of ordered triples (x, y, z) of real numbers that satisfy the system of
equation:
x+y+z = 7; 2? + y2 + 22 = 27; xyz = 5.
Answers
Given : x + y + z = 7
x² + y² + z² = 27
xyz = 5
To Find : number of ordered triples (x, y, z)
Solution:
x + y + z = 7
=> (y + z) = 7 - x
squaring both sides
y² + z² + 2yz = 49 + x² - 14x
x² + y² + z² = 27 => y² + z² = 27 - x²
xyz = 5 => yz = 5/x
27 - x² + 2(5/x) = 49 + x² - 14x
=> 27x - x³ + 10 = 49x + x³ - 14x²
=> 2x³ - 14x² + 22x - 10 = 0
=> x³ - 7x² + 11x - 5 = 0
=> (x - 5)(x - 1)² = 0
x = 5 , x = 1 , 1
5 + y + z = 7 => y + z = 2
5² + y² + z² = 27 => y² + z² = 2
y² + (2 - y)² = 2
=> y² + 4 + y² - 4y = 2
=> 2y² - 4y + 2 = 0
=> y² - 2y + 1 = 0
=> (y - 1)² = 0
=> y = 1 Hence z = 1
x = 1
1 + y + z = 7 => y + z = 6
1² + y² + z² = 27 => y² + z² = 26
y² + (6 - y)² = 26
=> y² + 36 + y² - 12y = 26
=> 2y² -12y + 10 = 0
=> y² - 6y + 5 = 0
=> (y - 1)(y - 5) = 0
=> y = 1 Hence z = 5
y = 5 hence z = 1
Solution are
x y z
5 1 1
1 5 1
1 1 5
the number of ordered triples = 3
( 5 , 1 , 1) , ( 1 , 5 , 1) , ( 1 , 1 , 5)
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