Beginning Algebra With Applications, Chapter 6, 6.2, Section 6.2, Problem 60

Solve the system of equations: $
\begin{equation}
\begin{aligned}

0.8x-0.1y =& 0.3 \\
0.5x - 0.2y =& -0.5

\end{aligned}
\end{equation}
$


$
\begin{equation}
\begin{aligned}

0.5x -0.2y =& -0.5
\qquad \text{Solve equation 2 for } x
\\
\\
0.5x =& 0.2y - 0.5
\\
\\
x =& \frac{0.2y - 0.5}{0.5}
\\
\\
x =& \frac{0.2}{0.5} y - 1

\end{aligned}
\end{equation}
$




$
\begin{equation}
\begin{aligned}

0.8x - 0.1y =& 0.3
\qquad \text{Substitute } \frac{0.2}{0.5}y-1 \text{ for $x$ in equation 1}
\\
\\
0.8 \left( \frac{0.2}{0.5} y - 1 \right) - 0.1y =& 0.3
\\
\\
\frac{0.16}{0.5} y - 0.8 - 0.1y =& 0.3
\\
\\
0.32y - 0.1y =& 0.3+0.8
\\
\\
0.22y =& 1.1
\\
\\
y =& 5

\end{aligned}
\end{equation}
$


Substitute the value of $y$ in equation 2


$
\begin{equation}
\begin{aligned}

x =& \frac{0.2}{0.5} y - 1
\\
\\
x =& \frac{0.2}{0.5} (5) - 1
\\
\\
x =& 2-1
\\
\\
x =& 1

\end{aligned}
\end{equation}
$


The solution is $(1,5)$.