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