Intermediate Algebra, Chapter 4, 4.2, Section 4.2, Problem 36

Solve the system of equations $\begin{equation} \begin{aligned} x + 3 + z =& 2 \\ 4x + y + 2z =& -4 \\ 5x + 2y + 3z =& -2 \end{aligned} \end{equation}$. If the system is inconsistent or has dependent equations, say so.


$\begin{equation} \begin{aligned} -2x - 6y - 2z =& -4 && -2 \times \text{ Equation 1} \\ 4x + y + 2z =& -4 && \text{Equation 2} \\ \hline \end{aligned} \end{equation}$




$\begin{equation} \begin{aligned} 2x - 5y \phantom{+2z} =& -8 && \text{Add} \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} -3x - 9y - 3z =& -6 && -3 \times \text{ Equation 1} \\ 5x + 2y + 3z =& -2 && \text{Equation 3} \\ \hline \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} 2x - 7y \phantom{+3z} =& -8 && \text{Add} \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} 2x - 5y =& -8 && \text{New Equation 2} \\ 2x -7y =& -8 && \text{New Equation 3} \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} 2x - 5y =& -8 && \\ -2x + 7y =& 8 && -1 \times \text{ New Equation 3} \\ \hline \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} \phantom{2x + } 2y =& 0 && \text{Add} \\ y =& 0 && \text{Divide each side by $2$} \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} 2x - 5(0) =& -8 && \text{Substitute } y = 0 \text{ in New Equation 2} \\ 2x =& -8 && \text{Multiply} \\ \\ x =& -4 && \text{Divide each side by $2$} \end{aligned} \end{equation}$




$\begin{equation} \begin{aligned} -4 + 3(0) + z =& 2 && \text{Substitute } x = -4 \text{ and } y = 0 \text{ in Equation 1} \\ -4 + 0 + z =& 2 && \text{Multiply} \\ z =& 6 && \text{Add each side by $4$} \end{aligned} \end{equation}$


The ordered triple is $\displaystyle \left( -4,0,6 \right)$.