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}
$