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Question
Solve the following system of linear equations in three variables
x + y + z = 5; 2x – y + z = 9; x – 2y + 3z = 16
Solution
x + y + z = 5 ...(1)
2x – y + z = 9 ...(2)
x – 2y + 3z = 16 ...(3)
by adding (1) and (2)
x + y + z = 5 ...(1)
2x – y + z = 9 ..(2)
3x + 0 + 2z = 14
3x + 2z = 14 ...(4)
(3) × (1) ⇒ x – 2y + 3z = 16 ...(3)
(1) × (2) ⇒ 2x + 2y + 2z = 10 ...(1)
(3) × (1) ⇒ 3x + 0 + 5z = 26
3x + 5z = 26 ...(5)
3x + 2z = 14 ...(4)
3x + 5z = 26 ...(5)
(–) (–) (–)
(4) – (5) ⇒ 0 – 3z = – 12
3z = 12 ⇒ z = `12/3` = 4
Substituting z = 4 (4)
3x + 2(– 4) = 14
3x – 8 = 14
3x = 14 – 8
3x = 6
x = `6/3` = 2
Substituting x = 2 and z = 4 in (1)
2 + y + 4 = 5
y + 6 = 5
y = 5 – 6
= – 1
∴ The value of x = 2, y = – 1 and z = 4
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