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}$