Intermediate Algebra, Chapter 4, Review Exercises, Section Review Exercises, Problem 16

Solve the system $\begin{equation} \begin{aligned} 4x - y =& 2 \\ 3y + z =& 9 \\ x + 2z =& 7 \end{aligned} \end{equation}$. If a system is inconsistent or has dependent equations, say so.


$\begin{equation} \begin{aligned} 12x - 3y \phantom{+z} =& 6 && 3 \times \text{ Equation 1} \\\ 3y + z =& 9 && \text{Equation 2} \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} \phantom{12x - 3y + z = 15} \\ \hline \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} 12x \phantom{-3y} + z =& 15 && \text{Add; New Equation 2} \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} 12x + z =& 15 && \text{Equation 2} \\ x + 2z =& 7 && \text{Equation 3} \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} -24x - 2z =& -30 && -2 \times \text{ Equation 2} \\ x + 2z =& 7 && \text{Equation 3} \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} \phantom{-23x + 2z = -23} \\ \hline \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} -23x + \phantom{+ 2z} =& -23 && \text{Add} \\ x =& 1 && \text{Divide each side by $-23$} \\ 1 + 2z =& 7 && \text{Substitute $x = 1$ in Equation 3} \\ 2z =& 6 && \text{Subtract each side by $1$} \\ z =& 3 && \text{Divide each side by $2$} \\ \\ 4(1) - y =& 2 && \text{Substitute $x = 1$ in Equation 1} \\ 4 - y =& 2 && \text{Multiply} \\ -y =& -2 && \text{Subtract each side by $4$} \\ y =& 2 && \text{Divide each side by $-1$} \end{aligned} \end{equation}$


The ordered triple is $(1,2,3)$.