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

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

x + 3y - 6z =& 7 \\
2x - y + z =& 1 \\
x + 2y + 2z =& -1

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


$
\begin{equation}
\begin{aligned}

-2x - 6y + 12z =& -14
&& -2 \times \text{ Equation 1}
\\
2x - y + z =& 1
&& \text{ Equation 2}
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

\phantom{2x} -7z + 13z =& -13
&& \text{Add}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

-x - 3y + 6z =& -7
&& -1 \times \text{ Equation 1}
\\
x + 2y + 2z =& -1
&& \text{Equation 3}
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

\phantom{x} -y + 8x =& -8
&& \text{Add}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

-7y + 13z =& -13
&& \text{Equation 4}
\\
-y + 8z =& -8
&& \text{Equation 5}

\end{aligned}
\end{equation}
$


We write the equations in two variables as a system


$
\begin{equation}
\begin{aligned}

-7y + 13z =& -13
&&
\\
7y - 56z =& 56
&& \text{Multiply each side by $-7$}
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

\phantom{7y} - 43z =& 43
&& \text{Add}
\\
z =& -1
&& \text{Divide each side by $-43$}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

-y + 8(-1) =& -8
&& \text{Substitute $z = -1$}
\\
-y - 8 =& -8
&& \text{Multiply}
\\
-y =& 0
&& \text{Add each side by $8$}
\\
y =& 0
&& \text{Divide each side by $-1$}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

x + 3(0) - 6(-1) =& 7
&& \text{Substitute $y = 0$ and $z = -1$ in Equation 1}
\\
x + 6 =& 7
&& \text{Multiply}
\\
x =& 1
&& \text{Subtract each side by $6$}

\end{aligned}
\end{equation}
$


The ordered triple is $(1,0,-1)$.