Find $D_{x}$ for the following simultaneous equations: $3 x + 4 y = 8, x - 2 y = 5$

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Solution

In cramer's rule of $a_{1} x + b_{1} y = c_{1}, a_{2} x + b_{2} y = c_{2}$
$x = \frac{D_{x}}{D} = \frac{∣ ∣ ∣ \begin{matrix} c_{1} & b_{1} c_{2} & b_{2} \end{matrix} ∣ ∣ ∣}{∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} a_{2} & b_{2} \end{matrix} ∣ ∣ ∣}$
$y = \frac{D_{y}}{D} = \frac{∣ ∣ ∣ \begin{matrix} a_{1} & c_{1} a_{2} & c_{2} \end{matrix} ∣ ∣ ∣}{∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} a_{2} & b_{2} \end{matrix} ∣ ∣ ∣}$
$D_{x} = ∣ ∣ ∣ \begin{matrix} c_{1} & b_{1} c_{2} & b_{2} \end{matrix} ∣ ∣ ∣$
So here $D_{x} = ∣ ∣ ∣ \begin{matrix} 8 & 4 5 & - 2 \end{matrix} ∣ ∣ ∣ = - 36$

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