Intermediate Algebra, Chapter 4, 4.2, Section 4.2, Problem 22

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

2x + 3y - 4z =& 4 \\
x - 6y + z =& -16 \\
-x + 3z =& 8

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


$
\begin{equation}
\begin{aligned}

4x + 6y - 8z =& 8
&& 2 \times \text{Equation 1}
\\
x - 6y + z =& -16
&& \text{Equation 2}
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

5x \phantom{-6y} -7z =& -8
&& \text{Add}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

5x - 7z =& -8
&& \text{Equation 4}
\\
-x + 3z =& 8
&& \text{Equation 3}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

5x - 7z =& -8
&&
\\
-5x + 15z =& 40
&& 5 \times \text{ Equation 3}
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

\phantom{-5x + } 8z =& 32
&& \text{Add}
\\
z =& 4
&& \text{Divide each side by $8$}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

-x + 3(4) =& 8
&& \text{Substitute } z = 4 \text{ in Equation 3}
\\
-x + 12 =& 8
&& \text{Multiply}
\\
-x =& -4
&& \text{Subtract each side by $12$}
\\
x =& 4
&& \text{Divide each side by $-1$}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

2(4) + 3y - 4(4) =& 4
&& \text{Substitute } x = 4 \text{ and } z = 4 \text{ in Equation 1}
\\
8 + 3y - 16 =& 4
&& \text{Multiply}
\\
3y - 8 =& 4
&& \text{Combine like terms}
\\
3y =& 12
&& \text{Add each side by $8$}
\\
y =& 4
&& \text{Divide each side by $3$}

\end{aligned}
\end{equation}
$



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