Intermediate Algebra, Chapter 4, 4.2, Section 4.2, Problem 20

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

2x - y + 3z =& 6 \\
x + 2y - z =& 8 \\
2y + z =& 1

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


$
\begin{equation}
\begin{aligned}

2x - y + 3z =& 6
&& \text{Equation 1}
\\
-2x - 4y + 2z =& -16
&& -2 \times \text{ Equation 2}
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

\phantom{-2x} - 5y + 5z =& -10
&& \text{Add}
\\
-y + z =& -2
&& \text{Reduce to lowest terms}

\end{aligned}
\end{equation}
$


We write the equations in two variables as a system


$
\begin{equation}
\begin{aligned}

-y + z =& -2
&& \text{Equation 4}
\\
2y + z =& 1
&& \text{Equation 3}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

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

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

\phantom{2y + } 3z =& -3
&& \text{Add}
\\
z =& -1
&& \text{Divide each side by $3$}

\end{aligned}
\end{equation}
$





$
\begin{equation}
\begin{aligned}

2y + (-1) =& 1
&& \text{Substitute } z = -1 \text{ in Equation 3}
\\
2y =& 2
&& \text{Add each side by $1$}
\\
y =& 1
&& \text{Divide each side by $2$}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

2x - 1 + 3(-1) =& 6
&& \text{Substitute } y = 1 \text{ and } z = -1 \text{ in Equation 1}
\\
2x - 1 - 3 =& 6
&& \text{Multiply}
\\
2x - 4 =& 6
&& \text{Combine like terms}
\\
2x =& 10
&& \text{Add each side by $4$}
\\
x =& 5
&& \text{Divide each side by $2$}

\end{aligned}
\end{equation}
$



The ordered triple is $\displaystyle \left( 5,1,-1 \right)$.