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 $\displaystyle \left( 10,-30,12 \right)$.