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Question
Discuss the nature of solutions of the following system of equations
`(y + z)/4 = (z + x)/3 = (x + y)/2`; x + y + z = 27
Solution
`(y + z)/4 = (z + x)/3`
3y + 3z = 4z + 4x
– 4x + 3y + 3z – 4z = 0
– 4x + 3y – z = 0
4x – 3y + z = 0 ...(1)
`(z + x)/3 = (x + y)/2`
2z + 2x = 3x + 3y
– 3x + 2x – 3y + 2z = 0
– x – 3y + 2z = 0
x + 3y – 2z = 0 ...(2)
x + y + z = 27 ....(3)
4x – 3y + z = 0 ....(1)
x + 3y – 2z = 0 ....(2)
(1) + 2 ⇒ 5x + 0 – z = 0 ....(4)
(3) × 3 ⇒ 3x + 3y + 3z = 81 ....(3)
(2) × 1 ⇒ x + 3y – 2z = 0 ....(2)
(–) (–) (+)
(3) – 2 ⇒ 2x + 5z = 81 ....(5)
(4) × (5) ⇒ 25x – 5z = 0 ....(4)
(5) × 1 ⇒ 2x + 5z = 81 ....(5)
(4) + 5 ⇒ 27x = 81
x = `81/27` = 3
Substituting the value of x in (5)
6 + 5z = 81
5z = 81 – 6
5z = 75
z = `75/5` = 15
Substituting the value of x = 3
and z = 15 in (3)
3 + y + 15 = 27
y + 18 = 27
y = 27 – 18
= 9
The value of x = 3, y = 9 and z = 15
This system of equations have unique solution.
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