Intermediate Algebra, Chapter 4, 4.1, Section 4.1, Problem 26

Solve the system $\begin{equation}
\begin{aligned}

& 6x - y = -9 \\
& 4 + 7x = -y

\end{aligned}
\end{equation}
$ by substitution. If the system is inconsistent or has dependent equations.

We solve for $y$ in equation 2


$
\begin{equation}
\begin{aligned}

& 4 + 7x = -y
&& \text{Given equation}
\\
& y = -7x - 4
&& \text{Multiply each side by $-1$}

\end{aligned}
\end{equation}
$


Since equation 2 is solved for $y$, we substitute $-7x - 4$ for $y$ in equation 1.


$
\begin{equation}
\begin{aligned}

6x - (-7x - 4) =& -9
&& \text{Substitute $y = -7x - 4$}
\\
6x + 7x + 4 =& -9
&& \text{Distributive Property}
\\
13x + 4 =& -9
&& \text{Combine like terms}
\\
13x =& -13
&& \text{Subtract each side by $4$}
\\
x =& -1
&& \text{Divide each side by $13$}

\end{aligned}
\end{equation}
$


We found $x$. Now we solve for $y$ in equation 2.


$
\begin{equation}
\begin{aligned}

y =& -7(-1) - 4
&& \text{Substitute } x = -1
\\
y =& 7 - 4
&& \text{Multiply}
\\
y =& 3
&& \text{Subtract}

\end{aligned}
\end{equation}
$