Question

How do you solve (1) `x + 2y + z = 1`, (2) `2x - 3y - z = 6`, (3) `3x + 5y + 4z = 5`, (4) `4x + y + z = 8`?

Answer 1

`x=2`, `y = -1` and `z = 1`

Explanation:

Given:

(1): `color(white)(X)x+2y+z=1`

(2): `color(white)(X)2x-3y-z=6`

(3): `color(white)(X)3x+5y+4z=5`

(4): `color(white)(X)4x+y+z=8`

First note that we have `4` equations in `3` unknowns. So either there is some redundancy, or there is no solution.

Note that: (1) + (1) + (2) is `4x+y+z=8`, identical to (4). So there's our expected redundancy. We can ignore (4) and work with the first three equations.

Subtract `2 xx` (1) from (2) to get:

(5): `color(white)(X)-7y-3z=4`

Subtract `3 xx` (1) from (3) to get:

(6): `color(white)(X)-y+z=2`

Add `3xx` (6) to (5) to get:

`-10y = 10`

So `y = -1`

Hence from (6) we have `z = 2+y = 2-1=1`

Then from (1) we have `x = 1-2y-z = 1+2-1 = 2`

So: `x=2`, `y = -1` and `z = 1`