College Algebra, Chapter 7, 7.4, Section 7.4, Problem 34

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

6x + 12y =& 33
\\
4x + 7y =& 20

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

For this system we have


$
\begin{equation}
\begin{aligned}

|D| =& \left| \begin{array}{cc}
6 & 12 \\
4 & 7
\end{array} \right| = 6 \cdot 7 - 12 \cdot 4 = -6
\\
\\
|D_{x}| =& \left| \begin{array}{cc}
33 & 12 \\
20 & 7
\end{array} \right| = 33 \cdot 7 - 12 \cdot 20 = -9
\\
\\
|D_{y}| =& \left| \begin{array}{cc}
6 & 33 \\
4 & 20
\end{array} \right| = 6 \cdot 20 - 33 \cdot 4 = -12

\end{aligned}
\end{equation}
$


The solution is


$
\begin{equation}
\begin{aligned}

x =& \frac{|D_x|}{|D|} = \frac{-9}{-6} = \frac{3}{2}
\\
\\
y =& \frac{|D_y|}{|D|} = \frac{-12}{-6} = 2

\end{aligned}
\end{equation}
$

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