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

Aug 6, 2016

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

#### Explanation:

Given:

(1): $\textcolor{w h i t e}{X} x + 2 y + z = 1$

(2): $\textcolor{w h i t e}{X} 2 x - 3 y - z = 6$

(3): $\textcolor{w h i t e}{X} 3 x + 5 y + 4 z = 5$

(4): $\textcolor{w h i t e}{X} 4 x + y + z = 8$

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 $4 x + y + z = 8$, identical to (4). So there's our expected redundancy. We can ignore (4) and work with the first three equations.

Subtract $2 \times$ (1) from (2) to get:

(5): $\textcolor{w h i t e}{X} - 7 y - 3 z = 4$

Subtract $3 \times$ (1) from (3) to get:

(6): $\textcolor{w h i t e}{X} - y + z = 2$

Add $3 \times$ (6) to (5) to get:

$- 10 y = 10$

So $y = - 1$

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

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

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