Intermediate Algebra, Chapter 4, 4.2, Section 4.2, Problem 36

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

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

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


$
\begin{equation}
\begin{aligned}

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

\end{aligned}
\end{equation}
$




$
\begin{equation}
\begin{aligned}

2x - 5y \phantom{+2z} =& -8
&& \text{Add}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

-3x - 9y - 3z =& -6
&& -3 \times \text{ Equation 1}
\\
5x + 2y + 3z =& -2
&& \text{Equation 3}
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

2x - 7y \phantom{+3z} =& -8
&& \text{Add}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

2x - 5y =& -8
&& \text{New Equation 2}
\\
2x -7y =& -8
&& \text{New Equation 3}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

2x - 5y =& -8
&&
\\
-2x + 7y =& 8
&& -1 \times \text{ New Equation 3}
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

\phantom{2x + } 2y =& 0
&& \text{Add}
\\
y =& 0
&& \text{Divide each side by $2$}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

2x - 5(0) =& -8
&& \text{Substitute } y = 0 \text{ in New Equation 2}
\\
2x =& -8
&& \text{Multiply}
\\
\\
x =& -4
&& \text{Divide each side by $2$}

\end{aligned}
\end{equation}
$




$
\begin{equation}
\begin{aligned}

-4 + 3(0) + z =& 2
&& \text{Substitute } x = -4 \text{ and } y = 0 \text{ in Equation 1}
\\
-4 + 0 + z =& 2
&& \text{Multiply}
\\
z =& 6
&& \text{Add each side by $4$}

\end{aligned}
\end{equation}
$


The ordered triple is $\displaystyle \left( -4,0,6 \right)$.