Beginning Algebra With Applications, Chapter 6, Review Exercises, Section Review Exercises, Problem 18

Solve by substitution: $\begin{equation} \begin{aligned} 9x+12y =& -1 \\ x-4y =& -1 \end{aligned} \end{equation}$


$\begin{equation} \begin{aligned} 9x+12y =& -1 && \text{Solve equation 2 for } x \\ \\ x =& 4y - 1 && \\ \\ 9x + 12y =& -1 && \text{Substitute $4y - 1$ for $x$ in equation 1} \\ \\ 9(4y-1) + 12y =& -1 && \text{Solve for } y \\ \\ 36y-9 + 12y =& -1 && \\ \\ 48y =& 8 && \\ \\ y =& \frac{8}{48} && \\ \\ y =& \frac{1}{6} && \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} x =& 4 \left( \frac{1}{6} \right) -1 \qquad \text{Substitute the value of $y$ in equation 2} \\ \\ x =& \frac{2}{3}-1 \\ \\ x =& \frac{-1}{3} \end{aligned} \end{equation}$


The solution is $\displaystyle \left( \frac{-1}{3}, \frac{1}{6} \right)$.