College Algebra, Chapter 7, 7.4, Section 7.4, Problem 36

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

\frac{1}{2} x + \frac{1}{3} y =& 1
\\
\\
\frac{1}{4} x - \frac{1}{6}y =& \frac{-3}{2}

\end{aligned}
\end{equation} \right.$ using Cramer's Rule.

For this system we have


$
\begin{equation}
\begin{aligned}

|D| =& \left| \begin{array}{cc}
\displaystyle \frac{1}{2} & \displaystyle \frac{1}{3} \\
\displaystyle \frac{1}{4} & \displaystyle \frac{-1}{6}
\end{array} \right| = \frac{1}{2} \cdot \left( \frac{-1}{6} \right) - \frac{1}{3} \cdot \frac{1}{4} = \frac{-1}{6}
\\
\\
|D_{x}| =& \left| \begin{array}{cc}
1 & \displaystyle \frac{1}{3} \\
\displaystyle \frac{-3}{2} & \displaystyle \frac{-1}{6}
\end{array} \right| = 1 \cdot \left( \frac{-1}{6} \right) - \frac{1}{3} \cdot \left( \frac{-3}{2}\right) = \frac{1}{3}
\\
\\
|D_{y}| =& \left| \begin{array}{cc}
\displaystyle \frac{1}{2} & 1 \\
\displaystyle \frac{1}{4} & \displaystyle \frac{-3}{2}
\end{array} \right| = \frac{1}{2} \cdot \left( \frac{-3}{2} \right) - 1 \cdot \frac{1}{4} = -1

\end{aligned}
\end{equation}
$


The solution is


$
\begin{equation}
\begin{aligned}

x =& \frac{|D_x|}{|D|} = \frac{\displaystyle \frac{1}{3}}{\displaystyle \frac{-1}{6}} = 2
\\
\\
y =& \frac{|D_y|}{|D|} = \frac{-1}{\displaystyle \frac{-1}{6}} = 6

\end{aligned}
\end{equation}
$