Intermediate Algebra, Chapter 4, 4.1, Section 4.1, Problem 32

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

& \frac{1}{4}x - \frac{1}{5}y = 9
\\
\\
& 5x - y = 0


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

We solve $y$ in equation 2


$
\begin{equation}
\begin{aligned}

5x - y =& 0
&& \text{Given equation}
\\
-y =& -5x
&& \text{Subtract each side by $5x$}
\\
y =& 5x
&& \text{Multiply each side by $-1$}

\end{aligned}
\end{equation}
$


Since equation 2 is solved for $y$, we substitute $5x$ for $y$ in equation 1.


$
\begin{equation}
\begin{aligned}

\frac{1}{4}x - \frac{1}{5} (5x) =& 9
&& \text{Substitute } y = 5x
\\
\\
\frac{1}{4}x - x =& 9
&& \text{Multiply}
\\
\\
x - 4x =& 36
&& \text{Multiply each side by $4$}
\\
\\
-3x =& 36
&& \text{Combine like terms}
\\
x =& -12
&& \text{Divide each side by $-3$}

\end{aligned}
\end{equation}
$


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


$
\begin{equation}
\begin{aligned}

y =& 5(-12)
&& \text{Substitute } x = -12
\\
y =& -60
&& \text{Multiply}


\end{aligned}
\end{equation}
$