Solve the following system of linear equations in three variables

`1/x - 2/y + 4 = 0; 1/y - 1/z + 1 = 0; 2/z + 3/x = 14`

#### Solution

Let `1/x` = p, `1/y` = q and `1/z` = r

p – 2q + 4 = 0

p – 2q = – 4 ...(1)

q – r + 1 = 0

q – r = – 1 ...(2)

3p + 2r = 14 ...(3)

(1) × 1 ⇒ p – 2q + 0 = – 4 ...(1)

(2) × 2 ⇒ 0 + 2q – 2r = – 2 ...(2)

(1) + (2) ⇒ p + 0 – 2r = – 6

p – 2r = – 6 ...(4)

(3) ⇒ 3p + 2r = 14 ...(3)

(3) + (4) ⇒ 4p + 0 = 8

4p = 8

p = `8/4` = 2

Substituting the value of p = 2 in (1)

2 – 2q = – 4

– 2q = – 4 – 2

– 2q = – 6

q = `6/2` = 3

Substituting the value of q = 3 in (2)

3 – r = 1

– r = – 1 – 3

r = 4

But `1/x` = p `1/x` = 2 2x = 1 x = `1/2` |
`1/y` = q `1/y` = 3 3y = 1 y = `1/3` |
`1/z` = r `1/z` = 4 4y = 1 z = `1/4` |

The value of x = `1/2`, y = `1/3` and z = `1/4`