Beginning Algebra With Applications, Chapter 6, 6.2, Section 6.2, Problem 30

Solve the system
$\begin{equation} \begin{aligned} 3x-5y =& 2 \\ 2x-y =& 4 \end{aligned} \end{equation}$

by substitution.



$\begin{equation} \begin{aligned} 2x-y =& 4 && \text{Solve equation 2 for } y \\ -y =& 4-2x && \\ y =& 2x-4 && \\ 3x-5y =& 2 && \text{Substitute $2x-4$ for $y$ in equation 1} \\ 3x-5(2x-4) =& 2 && \\ 3x-10x+20 =& 2 && \\ -7x =& 2-20 && \\ -7x =& -18 && \\ x =& \frac{-18}{-7} && \\ x =& \frac{18}{7} && \end{aligned} \end{equation}$


Substitute value of $x$ in equation 2


$\begin{equation} \begin{aligned} y =& 2 \left( \frac{18}{7} \right) -4 \\ \\ y =& \frac{36}{7}-4 \\ \\ y =& \frac{8}{7} \end{aligned} \end{equation}$


The solution is $\displaystyle \left( \frac{18}{7}, \frac{8}{7} \right)$.