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 $\displaystyle \left( 2, -3, 1 \right)$.