Intermediate Algebra, Chapter 4, Review Exercises, Section Review Exercises, Problem 16

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

4x - y =& 2 \\
3y + z =& 9 \\
x + 2z =& 7

\end{aligned}
\end{equation}
$. If a system is inconsistent or has dependent equations, say so.


$
\begin{equation}
\begin{aligned}

12x - 3y \phantom{+z} =& 6
&& 3 \times \text{ Equation 1}
\\\
3y + z =& 9
&& \text{Equation 2}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

\phantom{12x - 3y + z = 15}
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

12x \phantom{-3y} + z =& 15
&& \text{Add; New Equation 2}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

12x + z =& 15
&& \text{Equation 2}
\\
x + 2z =& 7
&& \text{Equation 3}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

-24x - 2z =& -30
&& -2 \times \text{ Equation 2}
\\
x + 2z =& 7
&& \text{Equation 3}


\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

\phantom{-23x + 2z = -23}
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

-23x + \phantom{+ 2z} =& -23
&& \text{Add}
\\
x =& 1
&& \text{Divide each side by $-23$}
\\
1 + 2z =& 7
&& \text{Substitute $x = 1$ in Equation 3}
\\
2z =& 6
&& \text{Subtract each side by $1$}
\\
z =& 3
&& \text{Divide each side by $2$}
\\
\\
4(1) - y =& 2
&& \text{Substitute $x = 1$ in Equation 1}
\\
4 - y =& 2
&& \text{Multiply}
\\
-y =& -2
&& \text{Subtract each side by $4$}
\\
y =& 2
&& \text{Divide each side by $-1$}


\end{aligned}
\end{equation}
$


The ordered triple is $(1,2,3)$.