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 $(1,1)$.