Intermediate Algebra, Chapter 4, 4.2, Section 4.2, Problem 16

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

\frac{2}{3}x - \frac{1}{4}y + \frac{5}{8}z =& 0 \\
\\
\frac{1}{5}x + \frac{2}{3}y - \frac{1}{4}z =& -7 \\
\\
- \frac{3}{5}x + \frac{4}{3}y - \frac{7}{8}z =& -5

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


$
\begin{equation}
\begin{aligned}

\frac{14}{3}x - \frac{7}{4}y + \frac{35}{8}z =& 0
&& 7 \times \text{ Equation 1}
\\
\\
- \frac{15}{5}x + \frac{20}{3}y - \frac{35}{8}z =& -25
&& 5 \times \text{ Equation 3}
\\
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

\frac{5}{3}x + \frac{59}{12}y \phantom{- \frac{35}{8}} =& -25
&& \text{Add}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

\frac{2}{3}x - \frac{1}{4}y + \frac{5}{8}z =& 0
&&
\\
\\
\frac{1}{2}x + \frac{5}{3}y - \frac{5}{8}z =& - \frac{35}{2}
&& \frac{5}{2} \times \text{ Equation 2}
\\
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

\frac{7}{6}x + \frac{17}{12}y \phantom{ - \frac{5}{8}z} =& - \frac{35}{2}
&& \text{Add}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

\frac{5}{3}x + \frac{59}{12}y =& -25
&& \text{Equation 4}
\\
\\
\frac{7}{6}x + \frac{17}{12}y =& - \frac{35}{2}
&& \text{Equation 5}

\end{aligned}
\end{equation}
$


We write the equations in two variables as a system


$
\begin{equation}
\begin{aligned}

- \frac{85}{3} x - \frac{1003}{12} y =& 425
&& -17 \times \text{ Equation 4}
\\
\\
\frac{413}{6} x + \frac{1003}{12} y =& - \frac{2065}{2}
&& 59 \times \text{ Equation 5}
\\
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

\frac{81}{2}x \phantom{\frac{1003}{12}y} =& - \frac{1215}{2}
&& \text{Add}
\\
\\
x =& -15
&& \text{Multiply each side by } \frac{2}{81}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

\frac{5}{3} (-15) + \frac{59}{12}y =& -25
&& \text{Substitute } x = -15 \text{ in Equation 4}
\\
\\
-25 + \frac{59}{12}y =& -25
&& \text{Multiply}
\\
\\
\frac{59}{12}y =& 0
&& \text{Add each side by $25$}
\\
\\
y =& 0
&& \text{Multiply each side by } \frac{12}{59}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

\frac{2}{3} (-15) - \frac{1}{4} (0) + \frac{5}{8}z =& 0
&& \text{Substitute } x = -15 \text{ and } y = 0 \text{ in Equation 1}
\\
\\
-10 - 0 + \frac{5}{8}z =& 0
&& \text{Multiply}
\\
\\
\frac{5}{8}z =& 10
&& \text{Add each side by $10$}
\\
\\
z =& 16
&& \text{Multiply each side by } \frac{8}{5}

\end{aligned}
\end{equation}
$



The ordered triple is $\displaystyle \left( -15,0,16 \right)$.