# How do you solve 3x + 6z = 0, -2x + y = 5, y + 2z = 3?

Dec 31, 2016

$\left(x , y , z\right) = \left(- 2 , 1 , 1\right)$

#### Explanation:

Given
[1]$\textcolor{w h i t e}{\text{XXX}} \textcolor{w h i t e}{+} 3 x \textcolor{w h i t e}{+ 0 y} + 6 z = 0$
[2]$\textcolor{w h i t e}{\text{XXX}} - 2 x + \textcolor{w h i t e}{1} y \textcolor{w h i t e}{+ 0 z} = 5$
[3]$\textcolor{w h i t e}{\text{XXX}} \textcolor{w h i t e}{+ 0 x + 1} y + 2 z = 3$

Statement [2]$\rightarrow$
[4]$\textcolor{w h i t e}{\text{XXX}} y = 2 x + 5$

Substituting from [4] into [3]
[5]$\textcolor{w h i t e}{\text{XXX}} 2 x + 5 + 2 z = 3$
Simplifying
[6]$\textcolor{w h i t e}{\text{XXX}} 2 x + 2 z = - 2$
or
[7]$\textcolor{w h i t e}{\text{XXX}} x + z = - 1$
or
[8]$\textcolor{w h i t e}{\text{XXX}} z = - x - 1$

Substituting from [8] into [1]
[9]$\textcolor{w h i t e}{\text{XXX}} 3 x + 6 \left(- x - 1\right) = 0$
Simplifying
[10]$\textcolor{w h i t e}{\text{XXX}} - 3 x = 6$
or
[11]$\textcolor{w h i t e}{\text{XXX}} x = - 2$

Substituting from [11] into [8]
[12]$\textcolor{w h i t e}{\text{XXX}} z = - \left(- 2\right) - 1 = + 2 - 1 = 1$

Substituting from [11] into [4]
[13]$\textcolor{w h i t e}{\text{XXX}} y = 2 \left(- 2\right) + 5 = - 4 + 5 = 1$