Question

Solve the following system of linear equations in three variables

(i) x + y + z = 5; 2x – y + z = 9; x – 2y + 3z = 16 

(ii) 1/x – 2/y + 4 = 0; 1/y – 1/z + 1 = 0; 2/z + 3/x = 14

(iii) x + 20 = 3y/2 + 10 = 2z + 5 = 110 – (y + z)

Answer 1

(i) x + y + z = 5 … (1) 

2x – y + z = 9 … (2) 

x – 2y + 3z = 16 … (3)

Substitute z = 4 in (4) 

3x + 2(4) = 14

3x + 8 = 14 

3x = 6 

x = 2 

Substitute x = 2, z = 4 in (1) 

2 + y + 4 = 5 ⇒ y = -1 

x = 2, y = -1, z = 4

(ii) 1/x – 2/y + 4 = 0; 1/y – 1/z + 1 = 0; 2/z + 3/x = 14

1/y = b 

1/z = c in (1), (2) & (3) 

a – 2b + 4 = 0 ⇒ a – 2b = -4 … (1) 

b – c + 1 = 0 ⇒ b – c = -1 … (2)

2c + 3a = 14 ⇒ 2c + 3a = 14 … (3) 

(iii) x + 20 = 3y/2 + 10 = 2z + 5 = 110 – (y + z)

x = 3y/2 – 10 … (1) 

2z + 5 = 110 – (y + z) 

2z = 105 – y – z 

y = 105 – 3z … (2) 

Substitute (2) in (1), x = 315/2 – 9z/2 – 10 

= 2z + 5 – 20 

∴ 315 – 9z – 20 = 4z – 30 

13 z = 315 – 20 + 30 

= 325 

z = 325/13 = 25 

x + 20 = 2z + 5 

x + 20 = 50 + 5 

x = 35 

Substitute z = 25 in (2) 

y = 105 – 3z = 105 – 75 = 30 

y = 30

x = 35, y = 30, z = 25 

The system has unique solutions