Beginning Algebra With Applications, Chapter 6, Test, Section Test, Problem 10

Solve by substitution: $
\begin{equation}
\begin{aligned}

3x - 5y =& 13 \\
x+3y =& 1

\end{aligned}
\end{equation}
$


$
\begin{equation}
\begin{aligned}

x+3y =& 1
&& \text{Solve equation 2 for $x$}
\\
x =& 1-3y
&&
\\
3x-5y =& 13
&& \text{Substitute $1-3y$ for $x$ in equation 1}
\\
3(1-3y)-5y =& 13
&& \text{Solve for } y
\\
3-9y - 5y =& 13
&&
\\
-14y =& 10
&&
\\
y =& \frac{-10}{14}
&&
\\
\
y =& \frac{-5}{7}
&&

\end{aligned}
\end{equation}
$


Substitute the value of $y$ in equation 2


$
\begin{equation}
\begin{aligned}

x =& 1-3 \left( \frac{-5}{7} \right)
\\
\\
x =& 1 + \frac{15}{7}
\\
\\
x =& \frac{22}{7}


\end{aligned}
\end{equation}
$



The solution is $\displaystyle \left( \frac{22}{7}, \frac{-5}{7} \right)$.