Intermediate Algebra, Chapter 4, 4.2, Section 4.2, Problem 24

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

4x -8y =& -7 \\
4y + z =& 7 \\
-8x + z =& -4

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


$
\begin{equation}
\begin{aligned}

4x - 8y \phantom{+ 2z} =& -7
&& \text{Equation 1}
\\
8y + 2z =& 14
&& 2 \times \text{ Equation 2}
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

4x \phantom{8y} +2z =& 7
&& \text{Add}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

4x + 2z =& 7
&& \text{Equation 4}
\\
-8x + z =& -4
&& \text{Equation 3}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

8x + 4z =& 14
&& 2 \times \text{ Equation 4}
\\
-8x + z =& -4
&& \text{Equation 3}
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

\phantom{8x + } 5z =& 10
&& \text{Add}
\\
z =& 2
&& \text{Divide each side by $5$}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

-8x + 2 =& -4
&& \text{Substitute } z = 2 \text{ in Equation 3}
\\
-8x =& -6
&& \text{Subtract each side by $2$}
\\
x =& \frac{-6}{-8}
&& \text{Divide each side by $-8$}
\\
x =& \frac{3}{4}
&& \text{Reduce to lowest terms}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

4 \left( \frac{3}{4} \right) - 8y =& -7
&& \text{Substitute } x = \frac{3}{4} \text{ in Equation 1}
\\
\\
3 - 8y =& -7
&& \text{Multiply}
\\
\\
-8y =& -10
&& \text{Subtract each side by $3$}
\\
\\
y =& \frac{-10}{-8}
&& \text{Divide each side by $-8$}
\\
\\
y =& \frac{5}{4}
&& \text{Reduce to lowest terms}


\end{aligned}
\end{equation}
$



The ordered triple is $\displaystyle \left( \frac{3}{4}, \frac{5}{4}, 2 \right)$.