Intermediate Algebra, Chapter 4, 4.2, Section 4.2, Problem 10

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

x - 2y + 5z =& -7 \\
-2x - 3y + 4z =& -14 \\
-3x + 5y - z =& -7

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


$
\begin{equation}
\begin{aligned}

2x - 4y + 10z =& -14
&& 2 \times \text{ Equation 1}
\\
-2x - 3y + 4z =& -14
&& \text{Equation 2}
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

\phantom{2x} - 7y+ 14z =& -28
&& \text{Add}
\\
-y + 2z =& -4
&& \text{Reduce to lowest terms}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

3x - 6y + 15z =& -21
&& 3 \times \text{ Equation 1}
\\
-3x + 5y - z =& -7
&& \text{Equation 3}
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

\phantom{3x} -y + 14z =& -28
&& \text{Add}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

-y + 2z =& -4
&& \text{Equation 4}
\\
-y + 14z =& -28
&& \text{Equation 5}

\end{aligned}
\end{equation}
$


We write the equations in two variables as a system


$
\begin{equation}
\begin{aligned}

-y + 2z =& -4
&&
\\
y - 14z =& 28
&& -1 \times \text{ Equation 5}
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

\phantom{y} -12z =& 24
&& \text{Add}
\\
z =& -2
&& \text{Divide each side by $-12$}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

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

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

x - 2(0) + 5(-2) =& -7
&& \text{Substitute $y = 0$ and $z = -2$ in Equation 1}
\\
x -0 -10 =& -7
&& \text{Multiply}
\\
x =& 3
&& \text{Add each side by $10$}

\end{aligned}
\end{equation}
$


The ordered triple is $(3,0,-2)$.