Intermediate Algebra, Chapter 4, 4.1, Section 4.1, Problem 22

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

& -3x - 5y = -17 \\
& y = 4x + 8

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

Since equation 2 is solved for $y$, we substitute $4x + 8$ for $y$ in equation 1.



$
\begin{equation}
\begin{aligned}

-3x - 5(4x + 8) =& -17
&& \text{Substitute $y = 4x + 8$}
\\
-3x - 20x - 40 =& -17
&& \text{Distributive Property}
\\
-23x - 40 =& -17
&& \text{Combine like terms}
\\
-23x =& 23
&& \text{Add each side by $40$}
\\
x =& -1
&& \text{Divide each side by $-23$}

\end{aligned}
\end{equation}
$


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


$
\begin{equation}
\begin{aligned}

y =& 4(-1) + 8
&& \text{Substitute $x = -1$}
\\
y =& -4 + 8
&& \text{Multiply}
\\
y =& 4
&& \text{Add}

\end{aligned}
\end{equation}
$