Beginning Algebra With Applications, Chapter 6, 6.2, Section 6.2, Problem 36

Solve the system
$\begin{equation} \begin{aligned} 3x+y =& 4 \\ 4x-3y =& 1 \end{aligned} \end{equation}$

by substitution.



$\begin{equation} \begin{aligned} 3x+y =& 4 && \text{Solve equation 1 for $y$} \\ y =& 4-3x && \\ 4x-3y =& 1 && \text{Substitute $4-3x$ for $y$ in equation 2} \\ 4x-3(4-3x) =& 1 && \\ 4x-12+9x =& 1 && \\ 13x =& 1+12 && \\ 13x =& 13 && \\ x =& 1 && \end{aligned} \end{equation}$


Substitute value of $x$ in equation 1


$\begin{equation} \begin{aligned} y =& 4-3(1) \\ y =& 4-3 \\ y =& 1 \end{aligned} \end{equation}$


The solution is $(1,1)$.