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.


$\begin{equation} \begin{aligned} y =& -7(-1) - 4 && \text{Substitute } x = -1 \\ y =& 7 - 4 && \text{Multiply} \\ y =& 3 && \text{Subtract} \end{aligned} \end{equation}$