Beginning Algebra With Applications, Chapter 6, Test, Section Test, Problem 8

Solve by substitution: $\begin{equation} \begin{aligned} 3x + 5y =& 1 \\ 2x-y =& 5 \end{aligned} \end{equation}$


$\begin{equation} \begin{aligned} 2x-y =& 5 && \text{Solve equation 2 for $y$} \\ -y =& 5-2x && \\ y =& 2x-5 && \\ 3x + 5y =& 13 && \text{Substitute $2x-5$ for $y$ in equation 1} \\ 3x + 5 (2x-5) =& 1 && \text{Solve for } x \\ 3x+10x - 25 =& 1 && \\ 13x =& 26 && \\ x =& 2 && \end{aligned} \end{equation}$


Substitute the value of $x$ in equation 2


$\begin{equation} \begin{aligned} y =& 2(2)-5 \qquad \text{Solve for } y \\ y =& 4-5 \\ y =& -1 \end{aligned} \end{equation}$



The solution is $(2,-1)$.