Intermediate Algebra, Chapter 4, 4.1, Section 4.1, Problem 30

Solve the system $\begin{equation} \begin{aligned} & x = 6y - 2 \\ & x = \frac{3}{4}y \end{aligned} \end{equation}$ by substitution. If the system is inconsistent or has dependent equations.


Since equation 2 is solved for $x$, we substitute $\displaystyle \frac{3}{4}y$ for $x$ in equation 1.


$\begin{equation} \begin{aligned} \frac{3}{4}y =& 6y - 2 && \text{Substitute } x = \frac{3}{4}y \\ \\ 3y =& 24y - 8 && \text{Multiply each side by $4$} \\ \\ -21y =& -8 && \text{Subtract each side by $24y$} \\ \\ y =& \frac{8}{21} && \text{Divide each side by $-21$} \end{aligned} \end{equation}$


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


$\begin{equation} \begin{aligned} x =& \frac{3}{4} \left( \frac{8}{21} \right) && \text{Substitute } y = \frac{8}{21} \\ \\ x =& \frac{2}{7} && \text{Multiply} \end{aligned} \end{equation}$