Intermediate Algebra, Chapter 4, 4.2, Section 4.2, Problem 8

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

5x - 2y + 3z =& -9 \\
4x + 3y + 5z =& 4 \\
2x + 4y - 2z =& 14

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


$
\begin{equation}
\begin{aligned}

-25x + 10y - 15z =& 45
&& -5 \times \text{ Equation 1}
\\
12x + 9y + 15z =& 12
&& 3 \times \text{ Equation 2}
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

-13x + 19y \phantom{+ 15z} =& 57
&& \text{Add}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

10x - 4y + 6z =& -18
&& 2 \times \text{ Equation 1}
\\
6x + 12y - 6z =& 42
&& 3 \times \text{Equation 3}
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

16x + 8y \phantom{-6z} =& 24
&& \text{Add}
\\
2x + y =& 3
&& \text{Reduce to lowest terms}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

-13x + 19y =& 57
&& \text{Equation 4}
\\
2x + y =& 3
&& \text{Equation 5}

\end{aligned}
\end{equation}
$


We write the equations in two variables as a system


$
\begin{equation}
\begin{aligned}

-13x + 19y =& 57
&&
\\
-38x - 19y =& -57
&& -19 \times \text{ Equation 5}
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

-51x \phantom{-19y} =& 0
&& \text{Add}
\\
x =& 0
&& \text{Divide each side by $-51$}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

2(0) + y =& 3
&& \text{Substitute $x = 0$ in Equation 5}
\\
y =& 3
&&

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

5(0) -2(3) + 3z =& -9
&& \text{Substitute $x = 0$ and $y = 3$ in Equation 1}
\\
-6 + 3z =& -9
&& \text{Multiply}
\\
3z =& -3
&& \text{Add each side by $6$}
\\
z =& -1
&& \text{Divide each side by $3$}

\end{aligned}
\end{equation}
$


The ordered triple is $(0,3,-1)$.