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