Solve the following system of linear equations in three variables
x + 20 = `(3y)/2 + 10` = 2z + 5 = 110 – (y + z)
Solution
x + 20 = `(3y)/2 + 10`
Multiply by 2
2x + 40 = 3y + 20
2x – 3y = – 40 + 20
2x – 3y = – 20 ...(1)
`(3y)/2 + 10` = 2z + 5
Multiply by 2
3y + 20 = 4z + 10
3y – 4z = 10 – 20
3y – 4z = – 10 ...(2)
2z + 5 = 110 – (y + z)
2z + 5 = 110 – y – z
y + 3z = 110 – 5
y + 3z = 105 ...(3)
(3) × (3) ⇒ 3y + 9z = 315 ...(3)
(2) × (1) ⇒ 3y – 4z = – 10 ...(2)
(–) (+) (+)
(3) – (2) ⇒ 13z = 325
z = `325/13` = 25
Substitute the value of z = 25 in (2)
3y – 4(25) = – 10
3y – 100 = – 10
3y = – 10 + 100
3y = 90
y = `90/3` = 30
∴ The value of x = 35, y = 30 and z = 25
Substitute the value of y = 30 in (1)
2x – 3(30) = – 20
2x – 90 = – 20
2x = – 20 + 90
2x = 70
x = `70/2` = 35
