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 $(-5,2)$.