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