Intermediate Algebra, Chapter 4, 4.2, Section 4.2, Problem 6

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

-3x + y - z =& -10 \\
-4x + 2y + 3z =& -1 \\
2x + 3y - 2z =& -5

\end{aligned}
\end{equation}
$.


$
\begin{equation}
\begin{aligned}

-9x + 3y - 3z =& -30
&& 3 \times \text{ Equation 1}
\\
-4x + 2y + 3z =& -1
&& \text{ Equation 2}
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

-13z + 5y \phantom{+3z} =& -31
&& \text{Add}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

6x - 2y + 2z =& 20
&& -2 \times \text{ Equation 1}
\\
2x + 3y - 2z =& -5
&& \text{Equation 3}
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

8x + y \phantom{-23z} =& 15
&&

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

-13x + 5y =& -31
&& \text{Equation 4}
\\
8x + y =& 15
&& \text{Equation 5}

\end{aligned}
\end{equation}
$


We write the equations in two variables as a system


$
\begin{equation}
\begin{aligned}

-13x + 5y =& -31
&&
\\
-40x - 5y =& -75
&& -5 \times \text{ Equation 5}
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

-53x \phantom{-5y} =& -106
&& \text{Add}
\\
x =& 2
&& \text{Divide each side by $-53$}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

-13(2) + 5y =& -31
&& \text{Substitute $x = 2$ in Equation 4}
\\
-26 + 5y =& -31
&& \text{Multiply}
\\
5y =& -5
&& \text{Add each side by $26$}
\\
y =& -1
&& \text{Divide each side by $5$}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

-3(2) + (-1) - z =& -10
&& \text{Substitute $x = 2$ and $y = -1$ in Equation 1}
\\
-6 - 1 - z =& -10
&& \text{Multiply}
\\
-7 - z =& -10
&& \text{Combine like terms}
\\
-z =& -3
&& \text{Add each side by $7$}
\\
z =& 3
&& \text{Divide each side by $-1$}

\end{aligned}
\end{equation}
$


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