Intermediate Algebra, Chapter 4, 4.2, Section 4.2, Problem 12

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

2x + y + 2z =& 1 \\
x + 2y + z =& 2 \\
x - y - z =& 0

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


$
\begin{equation}
\begin{aligned}

2x + y + 2z =& 1
&& \text{Equation 1}
\\
2x - 2y - 2z =& 0
&& 2 \times \text{Equation 3}
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

4x - y \phantom{-2z} =& 1
&& \text{Add}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

x + 2y + z =& 2
&& \text{Equation 2}
\\
x - y - z =& 0
&& \text{Equation 3}
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

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

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

4x - y =& 1
&& \text{Equation 4}
\\
2x + y =& 2
&& \text{Equation 5}

\end{aligned}
\end{equation}
$


We write the equations in two variables as a system


$
\begin{equation}
\begin{aligned}

4x - y =& 1
&&
\\
2x + y =& 2
&&
\\
\hline

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

6x \phantom{+y} =& 3
&& \text{Add}
\\
x =& \frac{1}{2}
&& \text{Divide each side by $6$}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

2 \left( \frac{1}{2} \right) + y =& 2
&& \text{Substitute } x = \frac{1}{2} \text{ in Equation 5}
\\
1 + y =& 2
&& \text{Multiply}
\\
y =& 1
&& \text{Subtract each side by $1$}

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

2 \left( \frac{1}{2} \right) + 1 + 2z =& 1
&& \text{Substitute } x = \frac{1}{2} \text{ and } y = 1 \text{ in Equation 1}
\\
1 + 1 + 2z =& 1
&& \text{Multiply}
\\
2 + 2z =& 1
&& \text{Combine like terms}
\\
2z =& -1
&& \text{Subtract each side by $2$}
\\
z =& - \frac{1}{2}
&& \text{Divide each side by } 2

\end{aligned}
\end{equation}
$


The ordered triple is $\displaystyle \left( \frac{1}{2}, 1, - \frac{1}{2} \right)$.

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