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 $(2,-1)$.