Intermediate Algebra, Chapter 4, 4.2, Section 4.2, Problem 2

Show that the ordered triple $(-3,1,6)$ is also a solution of equations
$
\begin{equation}
\begin{aligned}

x + 7y - 3z =& -14
&& \text{Equation 2}
\\
2x - 3y + 2z =& 3
&& \text{Equation 3}

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



$
\begin{equation}
\begin{aligned}

x + 7y - 3z =& - 14
&& \text{Equation 2}
\\
-3 + 7(1) - 3(6) =& -14
&& \text{Substitute $x = -3, y = 1$ and $z = 6$}
\\
-3 + 7 - 18 =& -14
&& \text{Multiply}
\\
-14 =& -14
&& \text{True}
\\
\\
\\
2x - 3y + 2z =& 3
&& \text{Equation 3}
\\
2(-3) - 3(1) + 2(6) =& 3
&& \text{Substitute $x = -3, y = 1$ and $z = 6$}
\\
-6-3 + 12 =& 3
&& \text{Multiply}
\\
3 =& 3
&& \text{True}

\end{aligned}
\end{equation}
$


It shows that the ordered triple $(-3,1,6)$ is also a solution to equation 2 and equation 3.