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