# How do you solve the system x-y-z=6, -x+3y+2z=-11, and 3x+2y+z=1?

Dec 4, 2016

The solution is $\left(2 , - 1 , - 3\right)$

#### Explanation:

(A) $x - y - z = 6$
(B) $- x + 3 y + 2 z = - 11$
(C) $3 x + 2 y + z = 1$

Add equations A and B together to eliminate x.

$\textcolor{w h i t e}{{a}^{2}} x - \textcolor{w h i t e}{a} y - \textcolor{w h i t e}{a} z = \textcolor{w h i t e}{a {a}^{2}} 6$
$\underline{- x + 3 y + 2 z = - 11}$
$\textcolor{w h i t e}{a a a a a} 2 y + \textcolor{w h i t e}{a} z = - 5 \textcolor{w h i t e}{a a a}$Equation D

Multiply equation B by 3 and add it to equation C to also eliminate z.

$3 \cdot \left(- x + 3 y + 2 z = - 11\right)$

$- 3 x + 9 y + 6 z = - 33$
$\underline{\textcolor{w h i t e}{{a}^{2}} 3 x + 2 y + \textcolor{w h i t e}{a} z = = 1}$
$\textcolor{w h i t e}{a a a a a} 11 y + \textcolor{w h i t e}{a} 7 z = - 32 \textcolor{w h i t e}{a a a}$Equation E

Multiply equation D by -7 and add it to equation E to eliminate z.

$- 7 \cdot \left(2 y + \textcolor{w h i t e}{a} z = - 5\right)$

$- 14 y - 7 z = \textcolor{w h i t e}{a {a}^{2}} 35$
$\textcolor{w h i t e}{{a}^{2}} \underline{11 y + 7 z = - 32}$
$\textcolor{w h i t e}{l} - 3 y \textcolor{w h i t e}{a a a a} = \textcolor{w h i t e}{a} 3$

$\frac{- 3 y}{-} 3 = \frac{3}{-} 3$
$y = - 1$

Substitute y

=-1 into either equation D or E to solve for z. I choose equation D.

$2 \left(- 1\right) + z = - 5$

$- 2 + z = - 5$
$+ 2 \textcolor{w h i t e}{a a a a a a} + 2$
$z = - 3$

Substitute $y = - 1$ and $z = - 3$ into equation A, B or C to find x.
I choose equation A.

$x - \left(- 1\right) - \left(- 3\right) = 6$
$x + 1 + 3 = 6$
$x + 4 \textcolor{w h i t e}{a a {a}^{2}} = 6$
$\textcolor{w h i t e}{a} - 4 \textcolor{w h i t e}{a a a a} - 4$
$x = 2$

The solution is $\left(2 , - 1 , - 3\right)$