Intermediate Algebra, Chapter 4, 4.2, Section 4.2, Problem 30

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

5x - z =& 38 \\
\\
\frac{2}{3}y + \frac{1}{4}z =& -17 \\
\\
\frac{1}{5}y + \frac{5}{6}z =& 4

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


$
\begin{equation}
\begin{aligned}

- \frac{1}{5} y + \frac{3}{40}z =& \frac{51}{10}
&& - \frac{3}{10} \times \text{ Equation 2}
\\
\\
\frac{1}{5}y + \frac{5}{6}z =& 4
&& \text{Equation 3}
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

\frac{91}{120} z =& \frac{91}{10}
&& \text{Add}
\\
\\
z =& 12
&& \text{Multiply each side by } \frac{120}{91}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

\frac{2}{3}y + \frac{1}{4}(12) =& -17
&& \text{Substitute } z = 12 \text{ in Equation 2}
\\
\\
\frac{2}{3}y + 3 =& -17
&& \text{Multiply}
\\
\\
\frac{2}{3}y =& -20
&& \text{Subtract each side by $3$}
\\
\\
y =& -30
&& \text{Multiply each side by } \frac{3}{2}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

5x - 12 =& 38
&& \text{Substitute } z = 12 \text{ in Equation 1}
\\
5x =& 50
&& \text{Add each side by $12$}
\\
x =& 10
&& \text{Divide each side by $5$}

\end{aligned}
\end{equation}
$



The ordered triple is $\displaystyle \left( 10,-30,12 \right)$.