Beginning Algebra With Applications, Chapter 6, 6.2, Section 6.2, Problem 38

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

3x-y =& 6 \\
x+3y =& 2

\end{aligned}
\end{equation}
$

by substitution.



$
\begin{equation}
\begin{aligned}

x+3y =& 2
&& \text{Solve equation 2 for $x$}
\\
x =& 2-3y
&&
\\
3x-y =& 6
&& \text{Substitute $2-3y$ for $x$ in equation 1}
\\
3(2-3y)-y =& 6
&&
\\
6-9y-y =& 6
&&
\\
-10y =& 6-6
&&
\\
-10y =& 0
&&
\\
y =& 0
&&

\end{aligned}
\end{equation}
$


Substitute value of $y$ in equation 2


$
\begin{equation}
\begin{aligned}

x =& 2-3(0)
\\
x =& 2

\end{aligned}
\end{equation}
$


The solution is $(2,0)$.