Beginning Algebra With Applications, Chapter 6, 6.2, Section 6.2, Problem 40

Solve the system
$
\begin{equation}
\begin{aligned}

3x-y =& 5 \\
2x+5y =& -8

\end{aligned}
\end{equation}
$

by substitution.



$
\begin{equation}
\begin{aligned}

3x-y =& 5
&& \text{Solve equation 1 for $y$}
\\
-y =& 5-3x
&&
\\
y =& 3x-5
\\
2x+5y =& -8
&& \text{Substitute $3x-5$ for $y$ in equation 2}
\\
2x+5(3x-5) =& -8
&&
\\
2x+15x - 25 =& -8
&&
\\
17x =& -8+25
&&
\\
17x =& 17
&&
\\
x =& 1
&&

\end{aligned}
\end{equation}
$


Substitute value of $x$ in equation 1


$
\begin{equation}
\begin{aligned}

y =& 3(1)-5
\\
y =& 3-5
\\
y =& -2

\end{aligned}
\end{equation}
$


The solution is $(1,-2)$.