Intermediate Algebra, Chapter 4, 4.2, Section 4.2, Problem 28

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

5x - 2z =& 8 \\
4y + 3z =& -9 \\
\frac{1}{2}x + \frac{2}{3}y =& -1

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


$
\begin{equation}
\begin{aligned}

15x \phantom{+8y} - 6z =& 24
&& 3 \times \text{ Equation 1}
\\
\phantom{15x + } 8y + 6z =& -18
&& 2 \times \text{ Equation 2}
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

15x + 8y \phantom{+ 6z} =& 6
&& \text{Add}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

15x + 8y =& 6
&& \text{Equation 4}
\\
\frac{1}{2}x + \frac{2}{3}y =& -1
&& \text{Equation 3}


\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

15x + 8y =& 6
&& \text{Equation 4}
\\
-6x - 8y =& 12
&& 12 \times \text{ Equation 3}
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

9x \phantom{-8y} =& 18
&& \text{Add}
\\
x =& 2
&& \text{Divide each side by $9$}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

5(2) - 2z =& 8
&& \text{Substitute } x = 2 \text{ in Equation 1}
\\
10 - 2z =& 8
&& \text{Multiply}
\\
-2z =& -2
&& \text{Subtract each side by $10$}
\\
z =& 1
&& \text{Divide each side by $-2$}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

4y + 3(1) =& -9
&& \text{Substitute } z = 1 \text{ in Equation 2}
\\
4y + 3 =& -9
&& \text{Multiply}
\\
4y =& -12
&& \text{Subtract each side by $3$}
\\
y =& -3
&& \text{Divide each side by $-3$}

\end{aligned}
\end{equation}
$



The ordered triple is $\displaystyle \left( 2, -3, 1 \right)$.