Intermediate Algebra, Chapter 4, 4.2, Section 4.2, Problem 32

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

-2x + 5y + z =& -3 \\
5x + 14y - z =& -11 \\
7x + 9y - 2z =& -5

\end{aligned}
\end{equation}
$. If the system is inconsistent or has dependent equations, say so.


$
\begin{equation}
\begin{aligned}

-2x + 5y + z =& -3
&& \text{Equation 1}
\\
5x + 14y - z =& -11
&& \text{Equation 2}
\\
\hline

\end{aligned}
\end{equation}
$




$
\begin{equation}
\begin{aligned}

3x + 19y \phantom{-z} =& -14
&& \text{Add}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

-4x + 10y + 2z =& -6
&& 2 \times \text{ Equation 1}
\\
7x + 9y - 2z =& -5
&& \text{Equation 3}
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

3x + 19y \phantom{-2z} =& -11
&& \text{Add}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

3x + 19y =& -14
&& \text{New Equation 2}
\\
3x + 19y =& -11
&& \text{New Equation 3}

\end{aligned}
\end{equation}
$


We write the equations in two variable as a system.


$
\begin{equation}
\begin{aligned}

3x + 19y =& -14
&&
\\
-3x - 19y =& 11
&& -1 \times \text{ Equation 5}
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

\phantom{3x - 19} 0 =& -3
&& \text{Add; False}
\end{aligned}
\end{equation}
$


Adding the two new equations give false statement, $0 = -3$. It shows that it has no solution. Thus, the system is inconsistent. The solution set is $\cancel{0}$.