Intermediate Algebra, Chapter 4, 4.2, Section 4.2, Problem 26

Solve the system of equations $
\begin{equation}
\begin{aligned}

-4x + 3y - z =& 4 \\
-5x - 3y + z =& -4 \\
-2x - 3z =& 12

\end{aligned}
\end{equation}
$.


$
\begin{equation}
\begin{aligned}

-4x + 3y - z =& 4
&& \text{Equation 1}
\\
-5x - 3y + z =& 14
&& \text{Equation 2}
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

-9z \phantom{-3y + z} =& 0
&& \text{Add}
\\
x =& 0
&& \text{Divide each side by $-9$}
\end{aligned}
\end{equation}
$




$
\begin{equation}
\begin{aligned}

-2(0) - 3z =& 12
&& \text{Substitute } x = 0 \text{ in Equation 3}
\\
0 -3z =& 12
&& \text{Multiply}
\\
\\
z =& -4
&& \text{Divide each side by $-3$}

\end{aligned}
\end{equation}
$




$
\begin{equation}
\begin{aligned}

-4(0) + 3y - (-4) =& 4
&& \text{Substitute } x = 0 \text{ and } z = -4 \text{ in Equation 1}
\\
0 + 3y + 4 =& 4
&& \text{Multiply}
\\
\\
3y =& 0
&& \text{Subtract each side by $4$}
\\
y =& 0
&& \text{Divide each side by $3$}

\end{aligned}
\end{equation}
$


The ordered triple is $\displaystyle \left( 0,0,-4 \right)$.