Beginning Algebra With Applications, Chapter 6, 6.2, Section 6.2, Problem 62

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

-0.1x+0.3y =& 1.1 \\
0.4x - 0.1y =& -2.2

\end{aligned}
\end{equation}
$


$
\begin{equation}
\begin{aligned}

-0.1x+0.3y =& 1.1
\qquad \text{Solve equation 1 for } y
\\
\\
0.3y =& 0.1x + 1.1
\\
\\
y =& \frac{0.1x + 1.1}{0.3}
\\
\\
y =& \frac{1}{3}x + \frac{11}{3}

\end{aligned}
\end{equation}
$




$
\begin{equation}
\begin{aligned}

0.4x- 0.1y =& -2.2
\qquad \text{Substitute } \frac{1}{3}x + \frac{11}{3} \text{ for $y$ in equation 2}
\\
\\
0.4x - 0.1 \left( \frac{1}{3}x + \frac{11}{3} \right) =& -2.2
\\
\\
0.4x - \frac{1}{30} x - \frac{11}{30} =& -2.2
\\
\\
\frac{11}{30}x =& -2.2 + \frac{11}{30}
\\
\\
\frac{11}{30}x =& \frac{-11}{6}
\\
\\
x =& -5

\end{aligned}
\end{equation}
$


Substitute the value of $x$ in equation 1


$
\begin{equation}
\begin{aligned}

y =& \frac{1}{3} (-5) + \frac{11}{3}
\\
\\
y =& \frac{-5}{3} + \frac{11}{3}
\\
\\
y =& \frac{6}{3}
\\
\\
y =& 2

\end{aligned}
\end{equation}
$


The solution is $(-5,2)$.